Integrand size = 35, antiderivative size = 370 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 a^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (5 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{24 a^2 d \sqrt {a+b \cos (c+d x)}}-\frac {b \left (5 A b^2+4 a^2 (A+2 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{8 a^3 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 a^3 d}-\frac {5 A b \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{12 a^2 d}+\frac {A \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{3 a d} \] Output:
-1/24*(15*A*b^2+8*a^2*(2*A+3*C))*(a+b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2* d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))/a^3/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+1 /24*(5*A*b^2+8*a^2*(2*A+3*C))*((a+b*cos(d*x+c))/(a+b))^(1/2)*InverseJacobi AM(1/2*d*x+1/2*c,2^(1/2)*(b/(a+b))^(1/2))/a^2/d/(a+b*cos(d*x+c))^(1/2)-1/8 *b*(5*A*b^2+4*a^2*(A+2*C))*((a+b*cos(d*x+c))/(a+b))^(1/2)*EllipticPi(sin(1 /2*d*x+1/2*c),2,2^(1/2)*(b/(a+b))^(1/2))/a^3/d/(a+b*cos(d*x+c))^(1/2)+1/24 *(15*A*b^2+8*a^2*(2*A+3*C))*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/a^3/d-5/12*A *b*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)*tan(d*x+c)/a^2/d+1/3*A*(a+b*cos(d*x+c ))^(1/2)*sec(d*x+c)^2*tan(d*x+c)/a/d
Result contains complex when optimal does not.
Time = 7.32 (sec) , antiderivative size = 604, normalized size of antiderivative = 1.63 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {b \left (\frac {40 a A b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (40 a^2 A+45 A b^2+72 a^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i \left (16 a^2 A+15 A b^2+24 a^2 C\right ) \sqrt {\frac {b-b \cos (c+d x)}{a+b}} \sqrt {-\frac {b+b \cos (c+d x)}{a-b}} \cos (2 (c+d x)) \left (2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )-b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right ) \sin (c+d x)}{a \sqrt {-\frac {1}{a+b}} \sqrt {1-\cos ^2(c+d x)} \sqrt {-\frac {a^2-b^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2}{b^2}} \left (2 a^2-b^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2\right )}\right )}{96 a^3 d}+\frac {\sqrt {a+b \cos (c+d x)} \left (\frac {\sec (c+d x) \left (16 a^2 A \sin (c+d x)+15 A b^2 \sin (c+d x)+24 a^2 C \sin (c+d x)\right )}{24 a^3}-\frac {5 A b \sec (c+d x) \tan (c+d x)}{12 a^2}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 a}\right )}{d} \] Input:
Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[a + b*Cos[c + d*x]] ,x]
Output:
-1/96*(b*((40*a*A*b*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x) /2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + (2*(40*a^2*A + 45*A*b^2 + 7 2*a^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2* b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] - ((2*I)*(16*a^2*A + 15*A*b^2 + 24*a ^2*C)*Sqrt[(b - b*Cos[c + d*x])/(a + b)]*Sqrt[-((b + b*Cos[c + d*x])/(a - b))]*Cos[2*(c + d*x)]*(2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[-(a + b)^(-1)] *Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(2*a*EllipticF[I*ArcSinh[ Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] - b*Ellipt icPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], ( a + b)/(a - b)]))*Sin[c + d*x])/(a*Sqrt[-(a + b)^(-1)]*Sqrt[1 - Cos[c + d* x]^2]*Sqrt[-((a^2 - b^2 - 2*a*(a + b*Cos[c + d*x]) + (a + b*Cos[c + d*x])^ 2)/b^2)]*(2*a^2 - b^2 - 4*a*(a + b*Cos[c + d*x]) + 2*(a + b*Cos[c + d*x])^ 2))))/(a^3*d) + (Sqrt[a + b*Cos[c + d*x]]*((Sec[c + d*x]*(16*a^2*A*Sin[c + d*x] + 15*A*b^2*Sin[c + d*x] + 24*a^2*C*Sin[c + d*x]))/(24*a^3) - (5*A*b* Sec[c + d*x]*Tan[c + d*x])/(12*a^2) + (A*Sec[c + d*x]^2*Tan[c + d*x])/(3*a )))/d
Time = 3.25 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.04, number of steps used = 24, number of rules used = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.686, Rules used = {3042, 3535, 27, 3042, 3534, 27, 3042, 3534, 27, 3042, 3538, 25, 3042, 3134, 3042, 3132, 3481, 3042, 3142, 3042, 3140, 3286, 3042, 3284}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\int -\frac {\left (-3 A b \cos ^2(c+d x)-2 a (2 A+3 C) \cos (c+d x)+5 A b\right ) \sec ^3(c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\int \frac {\left (-3 A b \cos ^2(c+d x)-2 a (2 A+3 C) \cos (c+d x)+5 A b\right ) \sec ^3(c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\int \frac {-3 A b \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a (2 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 A b}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{6 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {\int -\frac {\left (8 (2 A+3 C) a^2+2 A b \cos (c+d x) a+15 A b^2-5 A b^2 \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{2 a}+\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}}{6 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\int \frac {\left (8 (2 A+3 C) a^2+2 A b \cos (c+d x) a+15 A b^2-5 A b^2 \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\int \frac {8 (2 A+3 C) a^2+2 A b \sin \left (c+d x+\frac {\pi }{2}\right ) a+15 A b^2-5 A b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\int -\frac {\left (10 a A \cos (c+d x) b^2+\left (8 (2 A+3 C) a^2+15 A b^2\right ) \cos ^2(c+d x) b+3 \left (4 (A+2 C) a^2+5 A b^2\right ) b\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{a}+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\int \frac {\left (10 a A \cos (c+d x) b^2+\left (8 (2 A+3 C) a^2+15 A b^2\right ) \cos ^2(c+d x) b+3 \left (4 (A+2 C) a^2+5 A b^2\right ) b\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\int \frac {10 a A \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+\left (8 (2 A+3 C) a^2+15 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2 b+3 \left (4 (A+2 C) a^2+5 A b^2\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \int \sqrt {a+b \cos (c+d x)}dx-\frac {\int -\frac {\left (3 b^2 \left (4 (A+2 C) a^2+5 A b^2\right )-a b \left (8 (2 A+3 C) a^2+5 A b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \int \sqrt {a+b \cos (c+d x)}dx+\frac {\int \frac {\left (3 b^2 \left (4 (A+2 C) a^2+5 A b^2\right )-a b \left (8 (2 A+3 C) a^2+5 A b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\int \frac {3 b^2 \left (4 (A+2 C) a^2+5 A b^2\right )-a b \left (8 (2 A+3 C) a^2+5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {3 b^2 \left (4 (A+2 C) a^2+5 A b^2\right )-a b \left (8 (2 A+3 C) a^2+5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {3 b^2 \left (4 (A+2 C) a^2+5 A b^2\right )-a b \left (8 (2 A+3 C) a^2+5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {3 b^2 \left (4 (A+2 C) a^2+5 A b^2\right )-a b \left (8 (2 A+3 C) a^2+5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {3 b^2 \left (4 a^2 (A+2 C)+5 A b^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx-a b \left (8 a^2 (2 A+3 C)+5 A b^2\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}+\frac {2 \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {3 b^2 \left (4 a^2 (A+2 C)+5 A b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a b \left (8 a^2 (2 A+3 C)+5 A b^2\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {3 b^2 \left (4 a^2 (A+2 C)+5 A b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a b \left (8 a^2 (2 A+3 C)+5 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {3 b^2 \left (4 a^2 (A+2 C)+5 A b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a b \left (8 a^2 (2 A+3 C)+5 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {3 b^2 \left (4 a^2 (A+2 C)+5 A b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a b \left (8 a^2 (2 A+3 C)+5 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\frac {3 b^2 \left (4 a^2 (A+2 C)+5 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 a b \left (8 a^2 (2 A+3 C)+5 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\frac {3 b^2 \left (4 a^2 (A+2 C)+5 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 a b \left (8 a^2 (2 A+3 C)+5 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}}{6 a}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d}-\frac {\frac {5 A b \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {2 \left (8 a^2 (2 A+3 C)+15 A b^2\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\frac {6 b^2 \left (4 a^2 (A+2 C)+5 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {2 a b \left (8 a^2 (2 A+3 C)+5 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}}{2 a}}{4 a}}{6 a}\) |
Input:
Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[a + b*Cos[c + d*x]],x]
Output:
(A*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/(3*a*d) - ((5*A*b *Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (-1/2*((2*( 15*A*b^2 + 8*a^2*(2*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x) /2, (2*b)/(a + b)])/(d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((-2*a*b*(5*A *b^2 + 8*a^2*(2*A + 3*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]) + (6*b^2*(5*A*b^2 + 4*a^2*(A + 2*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d* x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]))/b)/a + ((15*A*b^2 + 8* a^2*(2*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/(a*d))/(4*a))/(6*a )
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[Sqrt[a/(a + b) + ( b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 , 0] && !GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]] Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && !GtQ[a + b, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt [c + d*Sin[e + f*x]] Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !GtQ[c + d, 0]
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin [e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d *(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1561\) vs. \(2(354)=708\).
Time = 3.95 (sec) , antiderivative size = 1562, normalized size of antiderivative = 4.22
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1562\) |
| parts | \(\text {Expression too large to display}\) | \(1624\) |
Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),x,method=_RETUR NVERBOSE)
Output:
-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*(-1/3* cos(1/2*d*x+1/2*c)/a*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2 )^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^3+5/12*b/a^2*cos(1/2*d*x+1/2*c)*(-2*b*s in(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c )^2-1)^2-1/24*(16*a^2+15*b^2)/a^3*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2 *c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)+5/48*b^ 2/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^ (1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipti cF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2) *((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+( a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b)) ^(1/2))-1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/( a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*E llipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/3/a*(sin(1/2*d*x+1/2*c)^ 2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/ 2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(- 2*b/(a-b))^(1/2))-5/16*b^2/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2* d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d* x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+5/16/a^ 3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(...
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),x, algori thm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+b*cos(d*x+c))**(1/2),x)
Output:
Timed out
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{4}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),x, algori thm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^4/sqrt(b*cos(d*x + c) + a), x)
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{4}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),x, algori thm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^4/sqrt(b*cos(d*x + c) + a), x)
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^4\,\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \] Input:
int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + b*cos(c + d*x))^(1/2)),x)
Output:
int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + b*cos(c + d*x))^(1/2)), x)
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}}{\cos \left (d x +c \right ) b +a}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{4}}{\cos \left (d x +c \right ) b +a}d x \right ) a \] Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),x)
Output:
int((sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**2*sec(c + d*x)**4)/(cos(c + d* x)*b + a),x)*c + int((sqrt(cos(c + d*x)*b + a)*sec(c + d*x)**4)/(cos(c + d *x)*b + a),x)*a