Integrand size = 35, antiderivative size = 278 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {3 A b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 a^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {A b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a d \sqrt {a+b \cos (c+d x)}}+\frac {\left (3 A b^2+4 a^2 (A+2 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^2 d \sqrt {a+b \cos (c+d x)}}-\frac {3 A b \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 a^2 d}+\frac {A \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{2 a d} \] Output:
3/4*A*b*(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^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-1/4*A*b*((a+b*cos(d*x+c))/ (a+b))^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2)*(b/(a+b))^(1/2))/a/d/(a +b*cos(d*x+c))^(1/2)+1/4*(3*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^2/d/(a+b* cos(d*x+c))^(1/2)-3/4*A*b*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/a^2/d+1/2*A*(a +b*cos(d*x+c))^(1/2)*sec(d*x+c)*tan(d*x+c)/a/d
Result contains complex when optimal does not.
Time = 8.05 (sec) , antiderivative size = 603, normalized size of antiderivative = 2.17 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {\cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \left (\frac {8 a A b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 a^2 A+9 A b^2+16 a^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {6 i A b^2 \sqrt {\frac {b-b \cos (c+d x)}{a+b}} \sqrt {-\frac {b+b \cos (c+d x)}{a-b}} \cos (2 (c+d x)) \left (2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )-b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right ) \sin (c+d x)}{a \sqrt {-\frac {1}{a+b}} \sqrt {1-\cos ^2(c+d x)} \sqrt {-\frac {a^2-b^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2}{b^2}} \left (2 a^2-b^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2\right )}\right )}{8 a^2 d (2 A+C+C \cos (2 c+2 d x))}+\frac {\cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \left (C+A \sec ^2(c+d x)\right ) \left (-\frac {3 A b \tan (c+d x)}{2 a^2}+\frac {A \sec (c+d x) \tan (c+d x)}{a}\right )}{d (2 A+C+C \cos (2 c+2 d x))} \] Input:
Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/Sqrt[a + b*Cos[c + d*x]] ,x]
Output:
(Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*((8*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*(8*a^2*A + 9*A*b^2 + 16*a^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Ell ipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] - ((6*I)* A*b^2*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))))/(8*a^2*d*(2*A + C + C*Cos[2*c + 2*d*x])) + (Cos[c + d*x]^2*Sqrt[a + b*Cos[c + d*x]]*(C + A*Sec[c + d*x]^2)*((-3*A*b*Tan[c + d*x])/(2*a^2) + (A *Sec[c + d*x]*Tan[c + d*x])/a))/(d*(2*A + C + C*Cos[2*c + 2*d*x]))
Time = 2.41 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.03, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 3535, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\int -\frac {\left (-A b \cos ^2(c+d x)-2 a (A+2 C) \cos (c+d x)+3 A b\right ) \sec ^2(c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\int \frac {\left (-A b \cos ^2(c+d x)-2 a (A+2 C) \cos (c+d x)+3 A b\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\int \frac {-A b \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a (A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 A b}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\int -\frac {\left (4 (A+2 C) a^2+2 A b \cos (c+d x) a+3 A b^2+3 A b^2 \cos ^2(c+d x)\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{a}+\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}}{4 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\int \frac {\left (4 (A+2 C) a^2+2 A b \cos (c+d x) a+3 A b^2+3 A b^2 \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\int \frac {4 (A+2 C) a^2+2 A b \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 A b^2+3 A b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {3 A b \int \sqrt {a+b \cos (c+d x)}dx-\frac {\int -\frac {\left (b \left (4 (A+2 C) a^2+3 A b^2\right )-a A b^2 \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {\left (b \left (4 (A+2 C) a^2+3 A b^2\right )-a A b^2 \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}+3 A b \int \sqrt {a+b \cos (c+d x)}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {b \left (4 (A+2 C) a^2+3 A b^2\right )-a A b^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+3 A b \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {b \left (4 (A+2 C) a^2+3 A b^2\right )-a A b^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {3 A b \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {b \left (4 (A+2 C) a^2+3 A b^2\right )-a A b^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {3 A b \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {b \left (4 (A+2 C) a^2+3 A b^2\right )-a A b^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {6 A b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b \left (4 a^2 (A+2 C)+3 A b^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx-a A b^2 \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}+\frac {6 A b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b \left (4 a^2 (A+2 C)+3 A b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a A b^2 \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {6 A b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b \left (4 a^2 (A+2 C)+3 A b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a A b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {6 A b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b \left (4 a^2 (A+2 C)+3 A b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a A b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {6 A b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b \left (4 a^2 (A+2 C)+3 A b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a A b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {6 A b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\frac {b \left (4 a^2 (A+2 C)+3 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 a A b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {6 A b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\frac {b \left (4 a^2 (A+2 C)+3 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 a A b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {6 A b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {A \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {3 A b \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\frac {2 b \left (4 a^2 (A+2 C)+3 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {2 a A b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {6 A b \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
Input:
Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/Sqrt[a + b*Cos[c + d*x]],x]
Output:
(A*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (-1/2*((6 *A*b*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(d*Sq rt[(a + b*Cos[c + d*x])/(a + b)]) + ((-2*a*A*b^2*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 ]]) + (2*b*(3*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 + (3*A*b*Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/(a*d))/(4*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. \(813\) vs. \(2(266)=532\).
Time = 3.43 (sec) , antiderivative size = 814, normalized size of antiderivative = 2.93
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(814\) |
| parts | \(\text {Expression too large to display}\) | \(878\) |
Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(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*(-1/2* 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)^2+3/4*b/a^2*cos(1/2*d*x+1/2*c)*(-2*b*si n(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)-1/8*b/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) *EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+3/8/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))-3/8*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))-1/2*(s in(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)*EllipticPi(cos( 1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))-3/8/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)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b ))^(1/2))*b^2)-2*C*(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)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)))/sin(1/2*d*x+...
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c))^(1/2),x, algori thm="fricas")
Output:
Timed out
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\sqrt {a + b \cos {\left (c + d x \right )}}}\, dx \] Input:
integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+b*cos(d*x+c))**(1/2),x)
Output:
Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)**3/sqrt(a + b*cos(c + d*x)), x)
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{3}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(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)^3/sqrt(b*cos(d*x + c) + a), x)
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{3}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(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)^3/sqrt(b*cos(d*x + c) + a), x)
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^3\,\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \] Input:
int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b*cos(c + d*x))^(1/2)),x)
Output:
int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b*cos(c + d*x))^(1/2)), x)
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right ) b +a}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right ) b +a}d x \right ) a \] Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(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)**3)/(cos(c + d* x)*b + a),x)*c + int((sqrt(cos(c + d*x)*b + a)*sec(c + d*x)**3)/(cos(c + d *x)*b + a),x)*a