Integrand size = 25, antiderivative size = 410 \[ \int \frac {(e \sin (c+d x))^{3/2}}{a+b \cos (c+d x)} \, dx=\frac {\sqrt [4]{-a^2+b^2} e^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{3/2} d}+\frac {\sqrt [4]{-a^2+b^2} e^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{3/2} d}+\frac {2 a e^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^2 d \sqrt {e \sin (c+d x)}}-\frac {a \left (a^2-b^2\right ) e^2 \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}-\frac {a \left (a^2-b^2\right ) e^2 \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{b^2 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d} \] Output:
(-a^2+b^2)^(1/4)*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)/d+(-a^2+b^2)^(1/4)*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)/d+2*a*e^2*InverseJacobiAM(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)+a*( a^2-b^2)*e^2*EllipticPi(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/(a^2-b*(b-(-a^2+b^2)^(1/2)))/d/(e*sin(d*x+c ))^(1/2)+a*(a^2-b^2)*e^2*EllipticPi(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/(a^2-b*(b+(-a^2+b^2)^(1/2)))/d/ (e*sin(d*x+c))^(1/2)-2*e*(e*sin(d*x+c))^(1/2)/b/d
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 7.15 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.06 \[ \int \frac {(e \sin (c+d x))^{3/2}}{a+b \cos (c+d x)} \, dx=-\frac {\left (\frac {1}{20}-\frac {i}{20}\right ) \cos (c+d x) \left (a+b \sqrt {\cos ^2(c+d x)}\right ) (e \sin (c+d x))^{3/2} \left (-5 \left (a^2-b^2\right ) \left (2 \sqrt [4]{-a^2+b^2} \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \sqrt [4]{-a^2+b^2} \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )+\sqrt [4]{-a^2+b^2} \log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )-\sqrt [4]{-a^2+b^2} \log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )+(4+4 i) \sqrt {b} \sqrt {\sin (c+d x)}\right )+(4+4 i) a b^{3/2} \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sin ^{\frac {5}{2}}(c+d x)\right )}{b^{3/2} \left (-a^2+b^2\right ) d \sqrt {\cos ^2(c+d x)} (a+b \cos (c+d x)) \sin ^{\frac {3}{2}}(c+d x)} \] Input:
Integrate[(e*Sin[c + d*x])^(3/2)/(a + b*Cos[c + d*x]),x]
Output:
((-1/20 + I/20)*Cos[c + d*x]*(a + b*Sqrt[Cos[c + d*x]^2])*(e*Sin[c + d*x]) ^(3/2)*(-5*(a^2 - b^2)*(2*(-a^2 + b^2)^(1/4)*ArcTan[1 - ((1 + I)*Sqrt[b]*S qrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*(-a^2 + b^2)^(1/4)*ArcTan[1 + ( (1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] + (-a^2 + b^2)^(1/ 4)*Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]] - (-a^2 + b^2)^(1/4)*Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]] + (4 + 4*I)*Sqrt[b]*Sqrt[Sin[c + d*x]]) + (4 + 4*I)*a*b^(3/2)*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] ^(5/2)))/(b^(3/2)*(-a^2 + b^2)*d*Sqrt[Cos[c + d*x]^2]*(a + b*Cos[c + d*x]) *Sin[c + d*x]^(3/2))
Time = 1.72 (sec) , antiderivative size = 407, normalized size of antiderivative = 0.99, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {3042, 3174, 25, 3042, 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.
| \(\displaystyle \int \frac {(e \sin (c+d x))^{3/2}}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}{a-b \sin \left (c+d x-\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3174 |
| \(\displaystyle -\frac {e^2 \int -\frac {b+a \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {e^2 \int \frac {b+a \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \int \frac {b-a \sin \left (c+d x-\frac {\pi }{2}\right )}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3346 |
| \(\displaystyle \frac {e^2 \left (\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{b}-\frac {\left (a^2-b^2\right ) \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{b}\right )}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \left (\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{b}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {e^2 \left (\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{b \sqrt {e \sin (c+d x)}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \left (\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{b \sqrt {e \sin (c+d x)}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{b}\right )}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3181 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (-\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b^2 \sin ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2\right )}d(e \sin (c+d x))}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (-\frac {2 b e \int \frac {1}{b^2 e^4 \sin ^4(c+d x)+\left (a^2-b^2\right ) e^2}d\sqrt {e \sin (c+d x)}}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (-\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\int \frac {1}{b e^2 \sin ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (-\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{b d \sqrt {e \sin (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}+\frac {a \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {b^2-a^2} \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {b^2-a^2} \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}}\right )}{b}\right )}{b}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d}\) |
Input:
Int[(e*Sin[c + d*x])^(3/2)/(a + b*Cos[c + d*x]),x]
Output:
(-2*e*Sqrt[e*Sin[c + d*x]])/(b*d) + (e^2*((2*a*EllipticF[(c - Pi/2 + d*x)/ 2, 2]*Sqrt[Sin[c + d*x]])/(b*d*Sqrt[e*Sin[c + d*x]]) - ((a^2 - b^2)*((-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*EllipticP i[(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*Ell ipticPi[(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]])))/ b))/b
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 + p))), x] + Simp[g^2*((p - 1)/(b*(m + p))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; F reeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[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]
Time = 1.57 (sec) , antiderivative size = 655, normalized size of antiderivative = 1.60
| method | result | size |
| default | \(\frac {-2 e b \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{b^{2}}-\frac {e^{2} \left (a^{2}-b^{2}\right ) \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \sin \left (d x +c \right )+\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {e \sin \left (d x +c \right )}\, \sqrt {2}+\sqrt {\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}}}{e \sin \left (d x +c \right )-\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {e \sin \left (d x +c \right )}\, \sqrt {2}+\sqrt {\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8 b^{2} \left (a^{2} e^{2}-b^{2} e^{2}\right )}\right )+\frac {\sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, a \,e^{2} \left (-\frac {\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )}{b^{2} \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}}-\frac {\left (a^{2}-b^{2}\right ) \left (-\frac {\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {1}{1-\frac {\sqrt {-a^{2}+b^{2}}}{b}}, \frac {\sqrt {2}}{2}\right )}{2 b \sqrt {-a^{2}+b^{2}}\, \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (1-\frac {\sqrt {-a^{2}+b^{2}}}{b}\right )}+\frac {\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {1}{1+\frac {\sqrt {-a^{2}+b^{2}}}{b}}, \frac {\sqrt {2}}{2}\right )}{2 b \sqrt {-a^{2}+b^{2}}\, \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (1+\frac {\sqrt {-a^{2}+b^{2}}}{b}\right )}\right )}{b^{2}}\right )}{\cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(655\) |
Input:
int((e*sin(d*x+c))^(3/2)/(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)
Output:
(-2*e*b*(1/b^2*(e*sin(d*x+c))^(1/2)-1/8*e^2*(a^2-b^2)/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)*a*e^2*(-1/b^2*(1-sin(d*x+c))^(1/2)*(2+2*s in(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*Ellipt icF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-(a^2-b^2)/b^2*(-1/2/b/(-a^2+b^2)^(1/ 2)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(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))+1/2/b/(-a^2+b^2)^(1/2)*(1-sin(d *x+c))^(1/2)*(2+2*sin(d*x+c))^(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))))/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d
Timed out. \[ \int \frac {(e \sin (c+d x))^{3/2}}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:
integrate((e*sin(d*x+c))^(3/2)/(a+b*cos(d*x+c)),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {(e \sin (c+d x))^{3/2}}{a+b \cos (c+d x)} \, dx=\int \frac {\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}{a + b \cos {\left (c + d x \right )}}\, dx \] Input:
integrate((e*sin(d*x+c))**(3/2)/(a+b*cos(d*x+c)),x)
Output:
Integral((e*sin(c + d*x))**(3/2)/(a + b*cos(c + d*x)), x)
\[ \int \frac {(e \sin (c+d x))^{3/2}}{a+b \cos (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:
integrate((e*sin(d*x+c))^(3/2)/(a+b*cos(d*x+c)),x, algorithm="maxima")
Output:
integrate((e*sin(d*x + c))^(3/2)/(b*cos(d*x + c) + a), x)
\[ \int \frac {(e \sin (c+d x))^{3/2}}{a+b \cos (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:
integrate((e*sin(d*x+c))^(3/2)/(a+b*cos(d*x+c)),x, algorithm="giac")
Output:
integrate((e*sin(d*x + c))^(3/2)/(b*cos(d*x + c) + a), x)
Timed out. \[ \int \frac {(e \sin (c+d x))^{3/2}}{a+b \cos (c+d x)} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}}{a+b\,\cos \left (c+d\,x\right )} \,d x \] Input:
int((e*sin(c + d*x))^(3/2)/(a + b*cos(c + d*x)),x)
Output:
int((e*sin(c + d*x))^(3/2)/(a + b*cos(c + d*x)), x)
\[ \int \frac {(e \sin (c+d x))^{3/2}}{a+b \cos (c+d x)} \, dx=\sqrt {e}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )}\, \sin \left (d x +c \right )}{\cos \left (d x +c \right ) b +a}d x \right ) e \] Input:
int((e*sin(d*x+c))^(3/2)/(a+b*cos(d*x+c)),x)
Output:
sqrt(e)*int((sqrt(sin(c + d*x))*sin(c + d*x))/(cos(c + d*x)*b + a),x)*e