Integrand size = 25, antiderivative size = 452 \[ \int \frac {1}{(a+b \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx=-\frac {a^{3/2} b \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\left (a^2-b^2\right )^{7/4} d e^{5/2}}-\frac {a^{3/2} b \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\left (a^2-b^2\right )^{7/4} d e^{5/2}}+\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}+\frac {2 a \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {a b^2 \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right ) \left (a^2-b^2-a \sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {a b^2 \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right ) \left (a^2-b^2+a \sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}} \] Output:
-a^(3/2)*b*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/(a ^2-b^2)^(7/4)/d/e^(5/2)-a^(3/2)*b*arctanh(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^ 2-b^2)^(1/4)/e^(1/2))/(a^2-b^2)^(7/4)/d/e^(5/2)+2/3*(b-a*cos(d*x+c))/(a^2- b^2)/d/e/(e*sin(d*x+c))^(3/2)+2/3*a*InverseJacobiAM(1/2*c-1/4*Pi+1/2*d*x,2 ^(1/2))*sin(d*x+c)^(1/2)/(a^2-b^2)/d/e^2/(e*sin(d*x+c))^(1/2)-a*b^2*Ellipt icPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*a/(a-(a^2-b^2)^(1/2)),2^(1/2))*sin(d*x+c) ^(1/2)/(a^2-b^2)/(a^2-b^2-a*(a^2-b^2)^(1/2))/d/e^2/(e*sin(d*x+c))^(1/2)-a* b^2*EllipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*a/(a+(a^2-b^2)^(1/2)),2^(1/2))* sin(d*x+c)^(1/2)/(a^2-b^2)/(a^2-b^2+a*(a^2-b^2)^(1/2))/d/e^2/(e*sin(d*x+c) )^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 10.54 (sec) , antiderivative size = 1233, normalized size of antiderivative = 2.73 \[ \int \frac {1}{(a+b \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:
Integrate[1/((a + b*Sec[c + d*x])*(e*Sin[c + d*x])^(5/2)),x]
Output:
-1/3*(a*(b + a*Cos[c + d*x])*Sec[c + d*x]*Sin[c + d*x]^(5/2)*((-2*a*Cos[c + d*x]^2*(b + a*Sqrt[1 - Sin[c + d*x]^2])*((b*(-2*ArcTan[1 - (Sqrt[2]*Sqrt [a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*Sqrt[a ]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - Sqrt[2] *Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + a*Sin[c + d*x]] + Log[Sqr t[-a^2 + b^2] + Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + a* Sin[c + d*x]]))/(4*Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(3/4)) - (5*a*(a^2 - b^2)* AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^ 2)]*Sqrt[Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]^2])/((5*(a^2 - b^2)*AppellF1[ 1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)] + 2*( 2*a^2*AppellF1[5/4, -1/2, 2, 9/4, Sin[c + d*x]^2, (a^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, (a^2*S in[c + d*x]^2)/(a^2 - b^2)])*Sin[c + d*x]^2)*(b^2 + a^2*(-1 + Sin[c + d*x] ^2)))))/((b + a*Cos[c + d*x])*(1 - Sin[c + d*x]^2)) + (4*b*Cos[c + d*x]*(b + a*Sqrt[1 - Sin[c + d*x]^2])*(((-1/8 + I/8)*Sqrt[a]*(2*ArcTan[1 - ((1 + I)*Sqrt[a]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)* Sqrt[a]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] + Log[Sqrt[a^2 - b^2] - (1 + I)*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*a*Sin[c + d*x]] - Lo g[Sqrt[a^2 - b^2] + (1 + I)*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*a*Sin[c + d*x]]))/(a^2 - b^2)^(3/4) + (5*b*(a^2 - b^2)*AppellF1[1/4,...
Time = 1.98 (sec) , antiderivative size = 428, normalized size of antiderivative = 0.95, number of steps used = 25, number of rules used = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.960, Rules used = {3042, 4360, 25, 25, 3042, 25, 3345, 27, 3042, 3346, 3042, 3121, 3042, 3120, 3181, 25, 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.
\(\displaystyle \int \frac {1}{(e \sin (c+d x))^{5/2} (a+b \sec (c+d x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2} \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -\frac {\cos (c+d x)}{(e \sin (c+d x))^{5/2} (-a \cos (c+d x)-b)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\frac {\cos (c+d x)}{(b+a \cos (c+d x)) (e \sin (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\cos (c+d x)}{(e \sin (c+d x))^{5/2} (a \cos (c+d x)+b)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2} \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\left (e \cos \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )^{5/2} \left (b-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )}dx\) |
\(\Big \downarrow \) 3345 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {2 \int \frac {2 a b-a^2 \cos (c+d x)}{2 (b+a \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {\int \frac {2 a b-a^2 \cos (c+d x)}{(b+a \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {\int \frac {\sin \left (c+d x-\frac {\pi }{2}\right ) a^2+2 b a}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3346 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \int \frac {1}{(b+a \cos (c+d x)) \sqrt {e \sin (c+d x)}}dx-a \int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx-a \int \frac {1}{\sqrt {e \sin (c+d x)}}dx}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{\sqrt {e \sin (c+d x)}}}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{\sqrt {e \sin (c+d x)}}}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx-\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3181 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \left (-\frac {a e \int -\frac {1}{\sqrt {e \sin (c+d x)} \left (\left (a^2-b^2\right ) e^2-a^2 e^2 \sin ^2(c+d x)\right )}d(e \sin (c+d x))}{d}-\frac {b \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 \sqrt {a^2-b^2}}-\frac {b \int \frac {1}{\sqrt {e \sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 \sqrt {a^2-b^2}}\right )-\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \left (\frac {a e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\left (a^2-b^2\right ) e^2-a^2 e^2 \sin ^2(c+d x)\right )}d(e \sin (c+d x))}{d}-\frac {b \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 \sqrt {a^2-b^2}}-\frac {b \int \frac {1}{\sqrt {e \sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 \sqrt {a^2-b^2}}\right )-\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \left (\frac {2 a e \int \frac {1}{\left (a^2-b^2\right ) e^2-a^2 e^4 \sin ^4(c+d x)}d\sqrt {e \sin (c+d x)}}{d}-\frac {b \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 \sqrt {a^2-b^2}}-\frac {b \int \frac {1}{\sqrt {e \sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 \sqrt {a^2-b^2}}\right )-\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \left (\frac {2 a e \left (\frac {\int \frac {1}{\sqrt {a^2-b^2} e-a e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {a^2-b^2}}+\frac {\int \frac {1}{a e^2 \sin ^2(c+d x)+\sqrt {a^2-b^2} e}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {a^2-b^2}}\right )}{d}-\frac {b \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 \sqrt {a^2-b^2}}-\frac {b \int \frac {1}{\sqrt {e \sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 \sqrt {a^2-b^2}}\right )-\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \left (\frac {2 a e \left (\frac {\int \frac {1}{\sqrt {a^2-b^2} e-a e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {a^2-b^2}}+\frac {\arctan \left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 \sqrt {a} e^{3/2} \left (a^2-b^2\right )^{3/4}}\right )}{d}-\frac {b \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 \sqrt {a^2-b^2}}-\frac {b \int \frac {1}{\sqrt {e \sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 \sqrt {a^2-b^2}}\right )-\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \left (-\frac {b \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 \sqrt {a^2-b^2}}-\frac {b \int \frac {1}{\sqrt {e \sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 \sqrt {a^2-b^2}}+\frac {2 a e \left (\frac {\arctan \left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 \sqrt {a} e^{3/2} \left (a^2-b^2\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 \sqrt {a} e^{3/2} \left (a^2-b^2\right )^{3/4}}\right )}{d}\right )-\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \left (-\frac {b \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 \sqrt {a^2-b^2}}-\frac {b \int \frac {1}{\sqrt {e \sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 \sqrt {a^2-b^2}}+\frac {2 a e \left (\frac {\arctan \left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 \sqrt {a} e^{3/2} \left (a^2-b^2\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 \sqrt {a} e^{3/2} \left (a^2-b^2\right )^{3/4}}\right )}{d}\right )-\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \left (-\frac {b \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 \sqrt {a^2-b^2} \sqrt {e \sin (c+d x)}}-\frac {b \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 \sqrt {a^2-b^2} \sqrt {e \sin (c+d x)}}+\frac {2 a e \left (\frac {\arctan \left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 \sqrt {a} e^{3/2} \left (a^2-b^2\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 \sqrt {a} e^{3/2} \left (a^2-b^2\right )^{3/4}}\right )}{d}\right )-\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \left (-\frac {b \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 \sqrt {a^2-b^2} \sqrt {e \sin (c+d x)}}-\frac {b \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 \sqrt {a^2-b^2} \sqrt {e \sin (c+d x)}}+\frac {2 a e \left (\frac {\arctan \left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 \sqrt {a} e^{3/2} \left (a^2-b^2\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 \sqrt {a} e^{3/2} \left (a^2-b^2\right )^{3/4}}\right )}{d}\right )-\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{3 e^2 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}-\frac {3 a b \left (\frac {2 a e \left (\frac {\arctan \left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 \sqrt {a} e^{3/2} \left (a^2-b^2\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 \sqrt {a} e^{3/2} \left (a^2-b^2\right )^{3/4}}\right )}{d}+\frac {b \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {a^2-b^2} \left (a-\sqrt {a^2-b^2}\right ) \sqrt {e \sin (c+d x)}}-\frac {b \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {a^2-b^2} \left (\sqrt {a^2-b^2}+a\right ) \sqrt {e \sin (c+d x)}}\right )-\frac {2 a \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}}{3 e^2 \left (a^2-b^2\right )}\) |
Input:
Int[1/((a + b*Sec[c + d*x])*(e*Sin[c + d*x])^(5/2)),x]
Output:
(2*(b - a*Cos[c + d*x]))/(3*(a^2 - b^2)*d*e*(e*Sin[c + d*x])^(3/2)) - ((-2 *a*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(d*Sqrt[e*Sin[c + d*x]]) + 3*a*b*((2*a*e*(ArcTan[(Sqrt[a]*Sqrt[e]*Sin[c + d*x])/(a^2 - b^2)^ (1/4)]/(2*Sqrt[a]*(a^2 - b^2)^(3/4)*e^(3/2)) + ArcTanh[(Sqrt[a]*Sqrt[e]*Si n[c + d*x])/(a^2 - b^2)^(1/4)]/(2*Sqrt[a]*(a^2 - b^2)^(3/4)*e^(3/2))))/d + (b*EllipticPi[(2*a)/(a - Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Si n[c + d*x]])/(Sqrt[a^2 - b^2]*(a - Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d*x]] ) - (b*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt [Sin[c + d*x]])/(Sqrt[a^2 - b^2]*(a + Sqrt[a^2 - b^2])*d*Sqrt[e*Sin[c + d* x]])))/(3*(a^2 - b^2)*e^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[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_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*Co s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c - b*d)* Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Simp[1/(g^2*(a^2 - b^2)*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && Lt Q[p, -1] && IntegerQ[2*m]
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]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 0.44 (sec) , antiderivative size = 607, normalized size of antiderivative = 1.34
method | result | size |
default | \(\frac {2 b e \left (\frac {a^{2} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sqrt {e \sin \left (d x +c \right )}+\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}{\sqrt {e \sin \left (d x +c \right )}-\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )\right )}{4 e^{2} \left (a +b \right ) \left (a -b \right ) \left (-a^{2} e^{2}+b^{2} e^{2}\right )}+\frac {1}{3 e^{2} \left (a^{2}-b^{2}\right ) \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}\right )+\frac {\sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, a \left (\frac {\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+2 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )}{3 \left (a^{2}-b^{2}\right ) \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (\cos \left (d x +c \right )^{2}-1\right )}-\frac {b^{2} \left (-\frac {\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {1}{1-\frac {\sqrt {a^{2}-b^{2}}}{a}}, \frac {\sqrt {2}}{2}\right )}{2 \sqrt {a^{2}-b^{2}}\, a \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (1-\frac {\sqrt {a^{2}-b^{2}}}{a}\right )}+\frac {\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {1}{1+\frac {\sqrt {a^{2}-b^{2}}}{a}}, \frac {\sqrt {2}}{2}\right )}{2 \sqrt {a^{2}-b^{2}}\, a \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (1+\frac {\sqrt {a^{2}-b^{2}}}{a}\right )}\right )}{\left (a +b \right ) \left (a -b \right )}\right )}{e^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(607\) |
Input:
int(1/(a+b*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
(2*b*e*(1/4/e^2/(a+b)/(a-b)*a^2*(e^2*(a^2-b^2)/a^2)^(1/4)/(-a^2*e^2+b^2*e^ 2)*(ln(((e*sin(d*x+c))^(1/2)+(e^2*(a^2-b^2)/a^2)^(1/4))/((e*sin(d*x+c))^(1 /2)-(e^2*(a^2-b^2)/a^2)^(1/4)))+2*arctan((e*sin(d*x+c))^(1/2)/(e^2*(a^2-b^ 2)/a^2)^(1/4)))+1/3/e^2/(a^2-b^2)/(e*sin(d*x+c))^(3/2))+(cos(d*x+c)^2*e*si n(d*x+c))^(1/2)*a/e^2*(1/3/(a^2-b^2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(co s(d*x+c)^2-1)*((-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/ 2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+2*cos(d*x+c)^2*sin(d*x+c)) -1/(a+b)/(a-b)*b^2*(-1/2/(a^2-b^2)^(1/2)/a*(-sin(d*x+c)+1)^(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)/a)*EllipticPi((-sin(d*x+c)+1)^(1/2),1/(1-(a^2-b^2)^(1/2)/a),1/2* 2^(1/2))+1/2/(a^2-b^2)^(1/2)/a*(-sin(d*x+c)+1)^(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)/a) *EllipticPi((-sin(d*x+c)+1)^(1/2),1/(1+(a^2-b^2)^(1/2)/a),1/2*2^(1/2))))/c os(d*x+c)/(e*sin(d*x+c))^(1/2))/d
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b*sec(d*x+c))/(e*sin(d*x+c))**(5/2),x)
Output:
Timed out
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="maxima")
Output:
Timed out
\[ \int \frac {1}{(a+b \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(a+b*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="giac")
Output:
integrate(1/((b*sec(d*x + c) + a)*(e*sin(d*x + c))^(5/2)), x)
Timed out. \[ \int \frac {1}{(a+b \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,\left (b+a\,\cos \left (c+d\,x\right )\right )} \,d x \] Input:
int(1/((e*sin(c + d*x))^(5/2)*(a + b/cos(c + d*x))),x)
Output:
int(cos(c + d*x)/((e*sin(c + d*x))^(5/2)*(b + a*cos(c + d*x))), x)
\[ \int \frac {1}{(a+b \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )}}{\sec \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +\sin \left (d x +c \right )^{3} a}d x \right )}{e^{3}} \] Input:
int(1/(a+b*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x)
Output:
(sqrt(e)*int(sqrt(sin(c + d*x))/(sec(c + d*x)*sin(c + d*x)**3*b + sin(c + d*x)**3*a),x))/e**3