Integrand size = 25, antiderivative size = 418 \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2} \, dx=\frac {a e^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{3/2} \left (-a^2+b^2\right )^{3/4} d}+\frac {a e^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{3/2} \left (-a^2+b^2\right )^{3/4} d}-\frac {e^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^2 d \sqrt {e \sin (c+d x)}}+\frac {a^2 e^2 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {a^2 e^2 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{2 b^2 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))} \]
1/2*a*e^(3/2)*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2) )/b^(3/2)/(-a^2+b^2)^(3/4)/d+1/2*a*e^(3/2)*arctanh(b^(1/2)*(e*sin(d*x+c))^ (1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(3/2)/(-a^2+b^2)^(3/4)/d+e^2*(sin(1/2*c+ 1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4 *Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/b^2/d/(e*sin(d*x+c))^(1/2)-1/2*a^2* e^2*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*Elliptic Pi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c)^ (1/2)/b^2/d/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(e*sin(d*x+c))^(1/2)-1/2*a^2*e^2* (sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticPi(c os(1/2*c+1/4*Pi+1/2*d*x),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2 )/b^2/d/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(e*sin(d*x+c))^(1/2)+e*(e*sin(d*x+c)) ^(1/2)/b/d/(a+b*cos(d*x+c))
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 7.49 (sec) , antiderivative size = 557, normalized size of antiderivative = 1.33 \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2} \, dx=\frac {(e \sin (c+d x))^{3/2} \left (\csc (c+d x)-\frac {\left (a+b \sqrt {\cos ^2(c+d x)}\right ) \left (\frac {a \left (-2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )\right )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}+\frac {5 b \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\cos ^2(c+d x)} \sqrt {\sin (c+d x)}}{\left (a^2-b^2+b^2 \sin ^2(c+d x)\right ) \left (-5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )+2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},-\frac {1}{2},2,\frac {9}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )+\left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )\right ) \sin ^2(c+d x)\right )}\right )}{\sin ^{\frac {3}{2}}(c+d x)}\right )}{b d (a+b \cos (c+d x))} \]
((e*Sin[c + d*x])^(3/2)*(Csc[c + d*x] - ((a + b*Sqrt[Cos[c + d*x]^2])*((a* (-2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] + 2 *ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - Log[ Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + b *Sin[c + d*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*S qrt[Sin[c + d*x]] + b*Sin[c + d*x]]))/(4*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4) ) + (5*b*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[ c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]])/((a^2 - b^2 + b^2*Sin[c + d*x]^2)*(-5*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin [c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)] + 2*(2*b^2*AppellF1[5/4, - 1/2, 2, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)] + (a^2 - b ^2)*AppellF1[5/4, 1/2, 1, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)])*Sin[c + d*x]^2))))/Sin[c + d*x]^(3/2)))/(b*d*(a + b*Cos[c + d*x]) )
Time = 1.81 (sec) , antiderivative size = 411, normalized size of antiderivative = 0.98, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.760, Rules used = {3042, 3172, 25, 3042, 25, 3346, 3042, 3121, 3042, 3120, 3181, 266, 756, 218, 221, 3042, 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.
| \(\displaystyle \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}{\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3172 |
| \(\displaystyle \frac {e^2 \int -\frac {\cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}-\frac {e^2 \int \frac {\cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{2 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}-\frac {e^2 \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{2 b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {e^2 \int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\sqrt {e \cos \left (\frac {1}{2} (2 c-\pi )+d x\right )} \left (a-b \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )}dx}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3346 |
| \(\displaystyle \frac {e^2 \left (\frac {a \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{b}-\frac {\int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{b}\right )}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \left (\frac {a \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}-\frac {\int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{b}\right )}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {e^2 \left (\frac {a \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}-\frac {\sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{b \sqrt {e \sin (c+d x)}}\right )}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \left (\frac {a \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}-\frac {\sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{b \sqrt {e \sin (c+d x)}}\right )}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {e^2 \left (\frac {a \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}\right )}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3181 |
| \(\displaystyle \frac {e^2 \left (\frac {a \left (-\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b^2 \sin ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2\right )}d(e \sin (c+d x))}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}\right )}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {e^2 \left (\frac {a \left (-\frac {2 b e \int \frac {1}{b^2 e^4 \sin ^4(c+d x)+\left (a^2-b^2\right ) e^2}d\sqrt {e \sin (c+d x)}}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}\right )}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {e^2 \left (\frac {a \left (-\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\int \frac {1}{b e^2 \sin ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}\right )}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {e^2 \left (\frac {a \left (-\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}\right )}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {e^2 \left (\frac {a \left (-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}\right )}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \left (\frac {a \left (-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}\right )}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle \frac {e^2 \left (\frac {a \left (-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}\right )}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \left (\frac {a \left (-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}\right )}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {e^2 \left (\frac {a \left (-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}+\frac {a \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {b^2-a^2} \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {b^2-a^2} \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}}\right )}{b}-\frac {2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}\right )}{2 b}+\frac {e \sqrt {e \sin (c+d x)}}{b d (a+b \cos (c+d x))}\) |
(e*Sqrt[e*Sin[c + d*x]])/(b*d*(a + b*Cos[c + d*x])) + (e^2*((-2*EllipticF[ (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(b*d*Sqrt[e*Sin[c + d*x]]) + (a *((-2*b*e*(-1/2*ArcTan[(Sqrt[b]*Sqrt[e]*Sin[c + d*x])/(-a^2 + b^2)^(1/4)]/ (Sqrt[b]*(-a^2 + b^2)^(3/4)*e^(3/2)) - ArcTanh[(Sqrt[b]*Sqrt[e]*Sin[c + d* x])/(-a^2 + b^2)^(1/4)]/(2*Sqrt[b]*(-a^2 + b^2)^(3/4)*e^(3/2))))/d + (a*El lipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(Sqrt[-a^2 + b^2]*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) - (a*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[S in[c + d*x]])/(Sqrt[-a^2 + b^2]*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d* x]])))/b))/(2*b)
3.1.72.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + 1))), x] + Simp[g^2*((p - 1)/(b*(m + 1))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; Fre eQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && I ntegersQ[2*m, 2*p]
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)* (x_)])), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Simp[-a/(2*q) Int[1/( Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Simp[b*(g/f) Subst[ Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - S imp[a/(2*q) Int[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x])] / ; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 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[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((c_.) + (d_.)*sin[(e_.) + (f_.)* (x_)]))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d/b Int [(g*Cos[e + f*x])^p, x], x] + Simp[(b*c - a*d)/b Int[(g*Cos[e + f*x])^p/( a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1369\) vs. \(2(449)=898\).
Time = 4.17 (sec) , antiderivative size = 1370, normalized size of antiderivative = 3.28
(-4*e^3*a*b*(-1/4/b^2*(e*sin(d*x+c))^(1/2)/(-b^2*cos(d*x+c)^2*e^2+a^2*e^2) +1/32/b^2*(e^2*(a^2-b^2)/b^2)^(1/4)/(a^2*e^2-b^2*e^2)*2^(1/2)*(ln((e*sin(d *x+c)+(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2-b^2 )/b^2)^(1/2))/(e*sin(d*x+c)-(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2) *2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^ (1/4)*(e*sin(d*x+c))^(1/2)+1)+2*arctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*( e*sin(d*x+c))^(1/2)-1)))+(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*e^2*(1/b^2*(1-s in(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*s in(d*x+c))^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-(-3*a^2+b^2)/ b^2*(-1/2/(-a^2+b^2)^(1/2)/b*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*s in(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(1-(-a^2+b^2)^(1/2)/b)*E llipticPi((1-sin(d*x+c))^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))+1/2/( -a^2+b^2)^(1/2)/b*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^( 1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(1+(-a^2+b^2)^(1/2)/b)*EllipticPi(( 1-sin(d*x+c))^(1/2),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2)))-2*a^2*(a^2-b^2) /b^2*(1/2*b^2/e/a^2/(a^2-b^2)*(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(-b^2*cos( d*x+c)^2+a^2)+1/4/a^2/(a^2-b^2)*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2 )*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*EllipticF((1-sin(d*x+ c))^(1/2),1/2*2^(1/2))-5/8/(a^2-b^2)/(-a^2+b^2)^(1/2)/b*(1-sin(d*x+c))^(1/ 2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^...
Timed out. \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{3/2}}{(a+b \cos (c+d x))^2} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]