Integrand size = 25, antiderivative size = 430 \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+b \sec (c+d x)} \, dx=\frac {b \left (a^2-b^2\right )^{3/4} e^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{7/2} d}-\frac {b \left (a^2-b^2\right )^{3/4} e^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{a^{7/2} d}-\frac {b^2 \left (a^2-b^2\right ) e^3 \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{a^4 \left (a-\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}-\frac {b^2 \left (a^2-b^2\right ) e^3 \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{a^4 \left (a+\sqrt {a^2-b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {2 \left (3 a^2-5 b^2\right ) e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a^3 d \sqrt {\sin (c+d x)}}+\frac {2 e (5 b-3 a \cos (c+d x)) (e \sin (c+d x))^{3/2}}{15 a^2 d} \]
b*(a^2-b^2)^(3/4)*e^(5/2)*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1 /4)/e^(1/2))/a^(7/2)/d-b*(a^2-b^2)^(3/4)*e^(5/2)*arctanh(a^(1/2)*(e*sin(d* x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))/a^(7/2)/d+2/15*e*(5*b-3*a*cos(d*x+c)) *(e*sin(d*x+c))^(3/2)/a^2/d+b^2*(a^2-b^2)*e^3*(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(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^4/d/(a-(a^2-b^2)^(1/2))/( e*sin(d*x+c))^(1/2)+b^2*(a^2-b^2)*e^3*(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(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^4/d/(a+(a^2-b^2)^(1/2))/(e*sin(d* x+c))^(1/2)-2/5*(3*a^2-5*b^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)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin( d*x+c))^(1/2)/a^3/d/sin(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 36.32 (sec) , antiderivative size = 853, normalized size of antiderivative = 1.98 \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+b \sec (c+d x)} \, dx=-\frac {(b+a \cos (c+d x)) \sec (c+d x) (e \sin (c+d x))^{5/2} \left (\frac {\left (-3 a^2+5 b^2\right ) \cos ^2(c+d x) \left (3 \sqrt {2} b \left (-a^2+b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )\right )+8 a^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)\right ) \left (b+a \sqrt {1-\sin ^2(c+d x)}\right )}{12 a^{3/2} \left (a^2-b^2\right ) (b+a \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac {4 a b \cos (c+d x) \left (\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+i a \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+i a \sin (c+d x)\right )\right )}{\sqrt {a} \sqrt [4]{a^2-b^2}}+\frac {b \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)}{3 \left (-a^2+b^2\right )}\right ) \left (b+a \sqrt {1-\sin ^2(c+d x)}\right )}{(b+a \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}\right )}{5 a^2 d (a+b \sec (c+d x)) \sin ^{\frac {5}{2}}(c+d x)}+\frac {(b+a \cos (c+d x)) \csc ^2(c+d x) \sec (c+d x) (e \sin (c+d x))^{5/2} \left (\frac {2 b \sin (c+d x)}{3 a^2}-\frac {\sin (2 (c+d x))}{5 a}\right )}{d (a+b \sec (c+d x))} \]
-1/5*((b + a*Cos[c + d*x])*Sec[c + d*x]*(e*Sin[c + d*x])^(5/2)*(((-3*a^2 + 5*b^2)*Cos[c + d*x]^2*(3*Sqrt[2]*b*(-a^2 + b^2)^(3/4)*(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[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]]) + 8*a^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Sin[c + d*x ]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)]*Sin[c + d*x]^(3/2))*(b + a*Sqrt[1 - Sin[c + d*x]^2]))/(12*a^(3/2)*(a^2 - b^2)*(b + a*Cos[c + d*x])*(1 - Sin[c + d*x]^2)) + (4*a*b*Cos[c + d*x]*(((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sqr t[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)*S qrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*a*Sin[c + d*x]] + Log[Sqrt [a^2 - b^2] + (1 + I)*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*a*S in[c + d*x]]))/(Sqrt[a]*(a^2 - b^2)^(1/4)) + (b*AppellF1[3/4, 1/2, 1, 7/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)]*Sin[c + d*x]^(3/2))/(3* (-a^2 + b^2)))*(b + a*Sqrt[1 - Sin[c + d*x]^2]))/((b + a*Cos[c + d*x])*Sqr t[1 - Sin[c + d*x]^2])))/(a^2*d*(a + b*Sec[c + d*x])*Sin[c + d*x]^(5/2)) + ((b + a*Cos[c + d*x])*Csc[c + d*x]^2*Sec[c + d*x]*(e*Sin[c + d*x])^(5/2)* ((2*b*Sin[c + d*x])/(3*a^2) - Sin[2*(c + d*x)]/(5*a)))/(d*(a + b*Sec[c ...
Time = 1.95 (sec) , antiderivative size = 417, normalized size of antiderivative = 0.97, number of steps used = 24, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.920, Rules used = {3042, 4360, 25, 25, 3042, 3344, 27, 3042, 3346, 3042, 3121, 3042, 3119, 3180, 25, 266, 827, 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.
| \(\displaystyle \int \frac {(e \sin (c+d x))^{5/2}}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}{a-b \csc \left (c+d x-\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 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 |
| \(\displaystyle -\int -\frac {\cos (c+d x) (e \sin (c+d x))^{5/2}}{b+a \cos (c+d x)}dx\) |
| \(\Big \downarrow \) 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 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}{a \sin \left (c+d x+\frac {\pi }{2}\right )+b}dx\) |
| \(\Big \downarrow \) 3344 |
| \(\displaystyle \frac {2 e^2 \int -\frac {\left (2 a b-\left (3 a^2-5 b^2\right ) \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{2 (b+a \cos (c+d x))}dx}{5 a^2}+\frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \int \frac {\left (2 a b-\left (3 a^2-5 b^2\right ) \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{b+a \cos (c+d x)}dx}{5 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \int \frac {\sqrt {-e \cos \left (c+d x+\frac {\pi }{2}\right )} \left (2 a b+\left (5 b^2-3 a^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 a^2}\) |
| \(\Big \downarrow \) 3346 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \int \frac {\sqrt {e \sin (c+d x)}}{b+a \cos (c+d x)}dx}{a}-\frac {\left (3 a^2-5 b^2\right ) \int \sqrt {e \sin (c+d x)}dx}{a}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{b-a \sin \left (c+d x-\frac {\pi }{2}\right )}dx}{a}-\frac {\left (3 a^2-5 b^2\right ) \int \sqrt {e \sin (c+d x)}dx}{a}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{b-a \sin \left (c+d x-\frac {\pi }{2}\right )}dx}{a}-\frac {\left (3 a^2-5 b^2\right ) \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{a \sqrt {\sin (c+d x)}}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{b-a \sin \left (c+d x-\frac {\pi }{2}\right )}dx}{a}-\frac {\left (3 a^2-5 b^2\right ) \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{a \sqrt {\sin (c+d x)}}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \int \frac {\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}{b-a \sin \left (c+d x-\frac {\pi }{2}\right )}dx}{a}-\frac {2 \left (3 a^2-5 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3180 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \left (-\frac {a e \int -\frac {\sqrt {e \sin (c+d x)}}{\left (a^2-b^2\right ) e^2-a^2 e^2 \sin ^2(c+d x)}d(e \sin (c+d x))}{d}-\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 a}+\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 a}\right )}{a}-\frac {2 \left (3 a^2-5 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \left (\frac {a e \int \frac {\sqrt {e \sin (c+d x)}}{\left (a^2-b^2\right ) e^2-a^2 e^2 \sin ^2(c+d x)}d(e \sin (c+d x))}{d}-\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 a}+\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 a}\right )}{a}-\frac {2 \left (3 a^2-5 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \left (\frac {2 a e \int \frac {e^2 \sin ^2(c+d x)}{\left (a^2-b^2\right ) e^2-a^2 e^4 \sin ^4(c+d x)}d\sqrt {e \sin (c+d x)}}{d}-\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 a}+\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 a}\right )}{a}-\frac {2 \left (3 a^2-5 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \left (\frac {2 a e \left (\frac {\int \frac {1}{\sqrt {a^2-b^2} e-a e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 a}-\frac {\int \frac {1}{a e^2 \sin ^2(c+d x)+\sqrt {a^2-b^2} e}d\sqrt {e \sin (c+d x)}}{2 a}\right )}{d}-\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 a}+\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 a}\right )}{a}-\frac {2 \left (3 a^2-5 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \left (-\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 a}+\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 a}+\frac {2 a e \left (\frac {\int \frac {1}{\sqrt {a^2-b^2} e-a e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 a}-\frac {\arctan \left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 a^{3/2} \sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d}\right )}{a}-\frac {2 \left (3 a^2-5 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \left (-\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 a}+\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 a}+\frac {2 a e \left (\frac {\text {arctanh}\left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 a^{3/2} \sqrt {e} \sqrt [4]{a^2-b^2}}-\frac {\arctan \left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 a^{3/2} \sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d}\right )}{a}-\frac {2 \left (3 a^2-5 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \left (-\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 a}+\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 a}+\frac {2 a e \left (\frac {\text {arctanh}\left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 a^{3/2} \sqrt {e} \sqrt [4]{a^2-b^2}}-\frac {\arctan \left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 a^{3/2} \sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d}\right )}{a}-\frac {2 \left (3 a^2-5 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \left (-\frac {b e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 a \sqrt {e \sin (c+d x)}}+\frac {b e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 a \sqrt {e \sin (c+d x)}}+\frac {2 a e \left (\frac {\text {arctanh}\left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 a^{3/2} \sqrt {e} \sqrt [4]{a^2-b^2}}-\frac {\arctan \left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 a^{3/2} \sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d}\right )}{a}-\frac {2 \left (3 a^2-5 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \left (-\frac {b e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {a^2-b^2}-a \sin (c+d x)\right )}dx}{2 a \sqrt {e \sin (c+d x)}}+\frac {b e \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (a \sin (c+d x)+\sqrt {a^2-b^2}\right )}dx}{2 a \sqrt {e \sin (c+d x)}}+\frac {2 a e \left (\frac {\text {arctanh}\left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 a^{3/2} \sqrt {e} \sqrt [4]{a^2-b^2}}-\frac {\arctan \left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 a^{3/2} \sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d}\right )}{a}-\frac {2 \left (3 a^2-5 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}}\right )}{5 a^2}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {2 e (e \sin (c+d x))^{3/2} (5 b-3 a \cos (c+d x))}{15 a^2 d}-\frac {e^2 \left (\frac {5 b \left (a^2-b^2\right ) \left (\frac {b e \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a-\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{a d \left (a-\sqrt {a^2-b^2}\right ) \sqrt {e \sin (c+d x)}}+\frac {b e \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+\sqrt {a^2-b^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{a d \left (\sqrt {a^2-b^2}+a\right ) \sqrt {e \sin (c+d x)}}+\frac {2 a e \left (\frac {\text {arctanh}\left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 a^{3/2} \sqrt {e} \sqrt [4]{a^2-b^2}}-\frac {\arctan \left (\frac {\sqrt {a} \sqrt {e} \sin (c+d x)}{\sqrt [4]{a^2-b^2}}\right )}{2 a^{3/2} \sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d}\right )}{a}-\frac {2 \left (3 a^2-5 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{a d \sqrt {\sin (c+d x)}}\right )}{5 a^2}\) |
(2*e*(5*b - 3*a*Cos[c + d*x])*(e*Sin[c + d*x])^(3/2))/(15*a^2*d) - (e^2*(( -2*(3*a^2 - 5*b^2)*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/ (a*d*Sqrt[Sin[c + d*x]]) + (5*b*(a^2 - b^2)*((2*a*e*(-1/2*ArcTan[(Sqrt[a]* Sqrt[e]*Sin[c + d*x])/(a^2 - b^2)^(1/4)]/(a^(3/2)*(a^2 - b^2)^(1/4)*Sqrt[e ]) + ArcTanh[(Sqrt[a]*Sqrt[e]*Sin[c + d*x])/(a^2 - b^2)^(1/4)]/(2*a^(3/2)* (a^2 - b^2)^(1/4)*Sqrt[e])))/d + (b*e*EllipticPi[(2*a)/(a - Sqrt[a^2 - b^2 ]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(a*(a - Sqrt[a^2 - b^2])*d* Sqrt[e*Sin[c + d*x]]) + (b*e*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(a*(a + Sqrt[a^2 - b^2])*d*Sqrt[e*Si n[c + d*x]])))/a))/(5*a^2)
3.3.34.3.1 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[((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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_ )]), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Simp[a*(g/(2*b)) Int[1/(S qrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (-Simp[a*(g/(2*b)) In t[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x] + Simp[b*(g/f) Su bst[Int[Sqrt[x]/(g^2*(a^2 - b^2) + b^2*x^2), x], x, g*Cos[e + f*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*(g* Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d* p + b*d*(m + p)*Sin[e + f*x])/(b^2*f*(m + p)*(m + p + 1))), x] + Simp[g^2*( (p - 1)/(b^2*(m + p)*(m + p + 1))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Si n[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1) - d*(a^ 2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1 , 0] && 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 = 13.67 (sec) , antiderivative size = 690, normalized size of antiderivative = 1.60
| method | result | size |
| default | \(\frac {2 e b \left (\frac {\left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3 a^{2}}+\frac {e^{2} \left (a^{2}-b^{2}\right ) \left (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 )-\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 )\right )}{4 a^{4} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )+\frac {\sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, a \,e^{3} \left (-\frac {6 \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 {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \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 {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \cos \left (d x +c \right )^{4}+2 \cos \left (d x +c \right )^{2}}{5 a^{2} \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}}+\frac {b^{2} \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \left (2 \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right )}{a^{4} \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}}-\frac {b^{2} \left (a^{2}-b^{2}\right ) \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 a^{2} \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 a^{2} \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 )}{a^{4}}\right )}{\cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(690\) |
(2*e*b*(1/3*(e*sin(d*x+c))^(3/2)/a^2+1/4*e^2*(a^2-b^2)/a^4/(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))-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)))))+(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*a*e^3*(-1/5/a^2 /(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*(6*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+ 2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-3*( -sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((-s in(d*x+c)+1)^(1/2),1/2*2^(1/2))-2*cos(d*x+c)^4+2*cos(d*x+c)^2)+b^2/a^4*(-s in(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)*(2*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-Ellipti cF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2)))-b^2*(a^2-b^2)/a^4*(-1/2/a^2*(-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*(-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))))/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d
Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+b \sec (c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+b \sec (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(e \sin (c+d x))^{5/2}}{a+b \sec (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{b \sec \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {(e \sin (c+d x))^{5/2}}{a+b \sec (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{b \sec \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{a+b \sec (c+d x)} \, dx=\int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}}{b+a\,\cos \left (c+d\,x\right )} \,d x \]