Integrand size = 42, antiderivative size = 296 \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=-\frac {\left (3 a^2 B-6 b^2 B-14 a b C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (3 a^3 B+12 a b^2 B+4 a^2 b C+2 b^3 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \cos (c+d x)}}+\frac {a^2 (5 b B+2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {b (3 a B-2 b C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {a B (a+b \cos (c+d x))^{3/2} \tan (c+d x)}{d} \]
-1/3*b*(3*B*a-2*C*b)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d-1/3*(3*B*a^2-6*B* b^2-14*C*a*b)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(si n(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/d/((a+b*c os(d*x+c))/(a+b))^(1/2)+1/3*(3*B*a^3+12*B*a*b^2+4*C*a^2*b+2*C*b^3)*(cos(1/ 2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1 /2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/d/(a+b*cos(d*x+c))^(1/ 2)+a^2*(5*B*b+2*C*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip ticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b ))^(1/2)/d/(a+b*cos(d*x+c))^(1/2)+a*B*(a+b*cos(d*x+c))^(3/2)*tan(d*x+c)/d
Result contains complex when optimal does not.
Time = 4.41 (sec) , antiderivative size = 442, normalized size of antiderivative = 1.49 \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {\frac {8 b \left (9 a b B+9 a^2 C+b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (27 a^2 b B+6 b^3 B+12 a^3 C+14 a b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 i \left (-3 a^2 B+6 b^2 B+14 a b C\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} \csc (c+d x) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}+4 \sqrt {a+b \cos (c+d x)} \left (3 a^2 B+2 b^2 C \cos (c+d x)\right ) \tan (c+d x)}{12 d} \]
((8*b*(9*a*b*B + 9*a^2*C + b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Ellip ticF[(c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + (2*(27*a^2*b* B + 6*b^3*B + 12*a^3*C + 14*a*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*El lipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + ((2*I) *(-3*a^2*B + 6*b^2*B + 14*a*b*C)*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]* Sqrt[(b*(1 + Cos[c + d*x]))/(-a + b)]*Csc[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*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1 )]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)])))/(a*b*Sqrt[-(a + b)^(-1)] ) + 4*Sqrt[a + b*Cos[c + d*x]]*(3*a^2*B + 2*b^2*C*Cos[c + d*x])*Tan[c + d* x])/(12*d)
Time = 2.76 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.05, number of steps used = 23, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.548, Rules used = {3042, 3508, 3042, 3468, 27, 3042, 3528, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2} (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3468 |
| \(\displaystyle \int \frac {1}{2} \sqrt {a+b \cos (c+d x)} \left (-b (3 a B-2 b C) \cos ^2(c+d x)+2 b (b B+2 a C) \cos (c+d x)+a (5 b B+2 a C)\right ) \sec (c+d x)dx+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \sqrt {a+b \cos (c+d x)} \left (-b (3 a B-2 b C) \cos ^2(c+d x)+2 b (b B+2 a C) \cos (c+d x)+a (5 b B+2 a C)\right ) \sec (c+d x)dx+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (-b (3 a B-2 b C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b (b B+2 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+a (5 b B+2 a C)\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {1}{2} \left (\frac {2}{3} \int \frac {\left (3 (5 b B+2 a C) a^2-b \left (3 B a^2-14 b C a-6 b^2 B\right ) \cos ^2(c+d x)+2 b \left (9 C a^2+9 b B a+b^2 C\right ) \cos (c+d x)\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {\left (3 (5 b B+2 a C) a^2-b \left (3 B a^2-14 b C a-6 b^2 B\right ) \cos ^2(c+d x)+2 b \left (9 C a^2+9 b B a+b^2 C\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {3 (5 b B+2 a C) a^2-b \left (3 B a^2-14 b C a-6 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b \left (9 C a^2+9 b B a+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (-\left (\left (3 a^2 B-14 a b C-6 b^2 B\right ) \int \sqrt {a+b \cos (c+d x)}dx\right )-\frac {\int -\frac {\left (3 b (5 b B+2 a C) a^2+b \left (3 B a^3+4 b C a^2+12 b^2 B a+2 b^3 C\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}\right )-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {\left (3 b (5 b B+2 a C) a^2+b \left (3 B a^3+4 b C a^2+12 b^2 B a+2 b^3 C\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\left (3 a^2 B-14 a b C-6 b^2 B\right ) \int \sqrt {a+b \cos (c+d x)}dx\right )-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {3 b (5 b B+2 a C) a^2+b \left (3 B a^3+4 b C a^2+12 b^2 B a+2 b^3 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\left (3 a^2 B-14 a b C-6 b^2 B\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {3 b (5 b B+2 a C) a^2+b \left (3 B a^3+4 b C a^2+12 b^2 B a+2 b^3 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (3 a^2 B-14 a b C-6 b^2 B\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {3 b (5 b B+2 a C) a^2+b \left (3 B a^3+4 b C a^2+12 b^2 B a+2 b^3 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (3 a^2 B-14 a b C-6 b^2 B\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\int \frac {3 b (5 b B+2 a C) a^2+b \left (3 B a^3+4 b C a^2+12 b^2 B a+2 b^3 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (3 a^2 B-14 a b C-6 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {3 a^2 b (2 a C+5 b B) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx+b \left (3 a^3 B+4 a^2 b C+12 a b^2 B+2 b^3 C\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {2 \left (3 a^2 B-14 a b C-6 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {3 a^2 b (2 a C+5 b B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (3 a^3 B+4 a^2 b C+12 a b^2 B+2 b^3 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (3 a^2 B-14 a b C-6 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {3 a^2 b (2 a C+5 b B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b \left (3 a^3 B+4 a^2 b C+12 a b^2 B+2 b^3 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (3 a^2 B-14 a b C-6 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {3 a^2 b (2 a C+5 b B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b \left (3 a^3 B+4 a^2 b C+12 a b^2 B+2 b^3 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (3 a^2 B-14 a b C-6 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {3 a^2 b (2 a C+5 b B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \left (3 a^3 B+4 a^2 b C+12 a b^2 B+2 b^3 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (3 a^2 B-14 a b C-6 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\frac {3 a^2 b (2 a C+5 b B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (3 a^3 B+4 a^2 b C+12 a b^2 B+2 b^3 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (3 a^2 B-14 a b C-6 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\frac {3 a^2 b (2 a C+5 b B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (3 a^3 B+4 a^2 b C+12 a b^2 B+2 b^3 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (3 a^2 B-14 a b C-6 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {\frac {6 a^2 b (2 a C+5 b B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (3 a^3 B+4 a^2 b C+12 a b^2 B+2 b^3 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (3 a^2 B-14 a b C-6 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )-\frac {2 b (3 a B-2 b C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )+\frac {a B \tan (c+d x) (a+b \cos (c+d x))^{3/2}}{d}\) |
(((-2*(3*a^2*B - 6*b^2*B - 14*a*b*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*b* (3*a^3*B + 12*a*b^2*B + 4*a^2*b*C + 2*b^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*a^2*b*(5*b*B + 2*a*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)/3 - (2*b*( 3*a*B - 2*b*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d))/2 + (a*B*(a + b*Cos[c + d*x])^(3/2)*Tan[c + d*x])/d
3.9.31.3.1 Defintions of rubi rules used
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_)])^(n_), x_Symbol] :> Si mp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a *B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2 , 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
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[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 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[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*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] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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. \(1489\) vs. \(2(361)=722\).
Time = 25.42 (sec) , antiderivative size = 1490, normalized size of antiderivative = 5.03
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1490\) |
| default | \(\text {Expression too large to display}\) | \(1563\) |
int((a+cos(d*x+c)*b)^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x,me thod=_RETURNVERBOSE)
-B*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*cos(1/2* d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a^2*b+(-2*a^3-2*a^2*b)*sin(1/2*d*x+1/2*c)^ 2*cos(1/2*d*x+1/2*c)-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d* x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^( 1/2))*a^3+4*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2-Ellipti cE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+EllipticE(cos(1/2*d*x+1/2*c) ,(-2*b/(a-b))^(1/2))*a^2*b+2*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/ 2))*a*b^2-2*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-5*Ellipti cPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*a^2*b)*sin(1/2*d*x+1/2*c)^2+( sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^ (1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+4*a*b^2*(sin(1/ 2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)* EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-(sin(1/2*d*x+1/2*c)^2)^(1 /2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2* d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b )*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2 *b/(a-b))^(1/2))*a^2*b+2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2* d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^ (1/2))*a*b^2-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c) ^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*...
\[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{3} \,d x } \]
integrate((a+b*cos(d*x+c))^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 3,x, algorithm="fricas")
integral((C*b^2*cos(d*x + c)^4 + B*a^2*cos(d*x + c) + (2*C*a*b + B*b^2)*co s(d*x + c)^3 + (C*a^2 + 2*B*a*b)*cos(d*x + c)^2)*sqrt(b*cos(d*x + c) + a)* sec(d*x + c)^3, x)
Timed out. \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Timed out} \]
\[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{3} \,d x } \]
integrate((a+b*cos(d*x+c))^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 3,x, algorithm="maxima")
\[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{3} \,d x } \]
integrate((a+b*cos(d*x+c))^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 3,x, algorithm="giac")
Timed out. \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^3} \,d x \]