Integrand size = 25, antiderivative size = 390 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+b \sin (c+d x))^2} \, dx=\frac {3 a e^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{5/2} \sqrt [4]{-a^2+b^2} d}-\frac {3 a e^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 b^{5/2} \sqrt [4]{-a^2+b^2} d}-\frac {3 e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)}}+\frac {3 a^2 e^3 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^3 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {3 a^2 e^3 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{2 b^3 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))} \]
3/2*a*e^(5/2)*arctan(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2) )/b^(5/2)/(-a^2+b^2)^(1/4)/d-3/2*a*e^(5/2)*arctanh(b^(1/2)*(e*cos(d*x+c))^ (1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(5/2)/(-a^2+b^2)^(1/4)/d-e*(e*cos(d*x+c) )^(3/2)/b/d/(a+b*sin(d*x+c))+3/2*a^2*e^3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos( 1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(b-(-a^2+b^2)^(1/2)),2^(1 /2))*cos(d*x+c)^(1/2)/b^3/d/(b-(-a^2+b^2)^(1/2))/(e*cos(d*x+c))^(1/2)+3/2* a^2*e^3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2 *d*x+1/2*c),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*cos(d*x+c)^(1/2)/b^3/d/(b+(- a^2+b^2)^(1/2))/(e*cos(d*x+c))^(1/2)-3*e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/co s(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1/2 )/b^2/d/cos(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 15.18 (sec) , antiderivative size = 371, normalized size of antiderivative = 0.95 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+b \sin (c+d x))^2} \, dx=\frac {(e \cos (c+d x))^{5/2} \left (-8 b^{3/2} \cos ^{\frac {3}{2}}(c+d x)-\frac {\left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(c+d x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+b \cos (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+b \cos (c+d x)\right )\right )\right ) \left (a+b \sqrt {\sin ^2(c+d x)}\right )}{a^2-b^2}\right )}{8 b^{5/2} d \cos ^{\frac {5}{2}}(c+d x) (a+b \sin (c+d x))} \]
((e*Cos[c + d*x])^(5/2)*(-8*b^(3/2)*Cos[c + d*x]^(3/2) - ((8*b^(5/2)*Appel lF1[3/4, -1/2, 1, 7/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)]* Cos[c + d*x]^(3/2) + 3*Sqrt[2]*a*(a^2 - b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]* Sqrt[b]*Sqrt[Cos[c + d*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqr t[b]*Sqrt[Cos[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[Cos[c + d*x]] + b*Cos[c + d*x]] + Log[Sqr t[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[c + d*x]] + b*Co s[c + d*x]]))*(a + b*Sqrt[Sin[c + d*x]^2]))/(a^2 - b^2)))/(8*b^(5/2)*d*Cos [c + d*x]^(5/2)*(a + b*Sin[c + d*x]))
Time = 1.77 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.97, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3042, 3172, 3042, 3346, 3042, 3121, 3042, 3119, 3180, 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.
\(\displaystyle \int \frac {(e \cos (c+d x))^{5/2}}{(a+b \sin (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(e \cos (c+d x))^{5/2}}{(a+b \sin (c+d x))^2}dx\) |
\(\Big \downarrow \) 3172 |
\(\displaystyle -\frac {3 e^2 \int \frac {\sqrt {e \cos (c+d x)} \sin (c+d x)}{a+b \sin (c+d x)}dx}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3 e^2 \int \frac {\sqrt {e \cos (c+d x)} \sin (c+d x)}{a+b \sin (c+d x)}dx}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3346 |
\(\displaystyle -\frac {3 e^2 \left (\frac {\int \sqrt {e \cos (c+d x)}dx}{b}-\frac {a \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)}dx}{b}\right )}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3 e^2 \left (\frac {\int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {a \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)}dx}{b}\right )}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle -\frac {3 e^2 \left (\frac {\sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b \sqrt {\cos (c+d x)}}-\frac {a \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)}dx}{b}\right )}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3 e^2 \left (\frac {\sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b \sqrt {\cos (c+d x)}}-\frac {a \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)}dx}{b}\right )}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle -\frac {3 e^2 \left (\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)}dx}{b}\right )}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3180 |
\(\displaystyle -\frac {3 e^2 \left (\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a \left (\frac {b e \int \frac {\sqrt {e \cos (c+d x)}}{b^2 \cos ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2}d(e \cos (c+d x))}{d}-\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {3 e^2 \left (\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a \left (\frac {2 b e \int \frac {e^2 \cos ^2(c+d x)}{b^2 e^4 \cos ^4(c+d x)+\left (a^2-b^2\right ) e^2}d\sqrt {e \cos (c+d x)}}{d}-\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle -\frac {3 e^2 \left (\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a \left (\frac {2 b e \left (\frac {\int \frac {1}{b e^2 \cos ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \cos (c+d x)}}{2 b}-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \cos ^2(c+d x)}d\sqrt {e \cos (c+d x)}}{2 b}\right )}{d}-\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {3 e^2 \left (\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a \left (\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \cos ^2(c+d x)}d\sqrt {e \cos (c+d x)}}{2 b}\right )}{d}-\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}\right )}{b}\right )}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {3 e^2 \left (\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a \left (-\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b}+\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )}{b}\right )}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3 e^2 \left (\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a \left (-\frac {a e \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b}+\frac {a e \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )} \left (b \sin \left (c+d x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 b}+\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )}{b}\right )}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle -\frac {3 e^2 \left (\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a \left (-\frac {a e \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 b \sqrt {e \cos (c+d x)}}+\frac {a e \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {e \cos (c+d x)}}+\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )}{b}\right )}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3 e^2 \left (\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a \left (-\frac {a e \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b \sqrt {e \cos (c+d x)}}+\frac {a e \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b \sin \left (c+d x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 b \sqrt {e \cos (c+d x)}}+\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}\right )}{b}\right )}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle -\frac {3 e^2 \left (\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a \left (\frac {2 b e \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} \sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{b d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \cos (c+d x)}}+\frac {a e \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{b d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \cos (c+d x)}}\right )}{b}\right )}{2 b}-\frac {e (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))}\) |
(-3*e^2*((2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(b*d*Sqrt[Cos[ c + d*x]]) - (a*((2*b*e*(ArcTan[(Sqrt[b]*Sqrt[e]*Cos[c + d*x])/(-a^2 + b^2 )^(1/4)]/(2*b^(3/2)*(-a^2 + b^2)^(1/4)*Sqrt[e]) - ArcTanh[(Sqrt[b]*Sqrt[e] *Cos[c + d*x])/(-a^2 + b^2)^(1/4)]/(2*b^(3/2)*(-a^2 + b^2)^(1/4)*Sqrt[e])) )/d + (a*e*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(b*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*x]]) + (a*e*Sqr t[Cos[c + d*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/ (b*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*x]])))/b))/(2*b) - (e*(e*Cos[ c + d*x])^(3/2))/(b*d*(a + b*Sin[c + d*x]))
3.6.87.3.1 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[(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[(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[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_)*((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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 15.57 (sec) , antiderivative size = 1869, normalized size of antiderivative = 4.79
(8*e^3*a*b*(1/16/b^4/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(ln((2*e*cos(1/2*d* x+1/2*c)^2-e-(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)* 2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(2*e*cos(1/2*d*x+1/2*c)^2-e+(e^2*(a^2-b ^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b ^2)^(1/2)))+2*arctan((2^(1/2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)+(e^2*(a^2 -b^2)/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4))+2*arctan((2^(1/2)*(2*e*cos(1/ 2*d*x+1/2*c)^2-e)^(1/2)-(e^2*(a^2-b^2)/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/ 4)))-1/64*(a^2-b^2)/b^4/(e^2*(a^2-b^2)/b^2)^(1/4)*(16*(e^2*(a^2-b^2)/b^2)^ (1/4)*(cos(1/2*d*x+1/2*c)^2-1/2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*b^2+(4 *cos(1/2*d*x+1/2*c)^4*b^2-4*cos(1/2*d*x+1/2*c)^2*b^2+a^2)*e*2^(1/2)*(ln((2 *e*cos(1/2*d*x+1/2*c)^2-e-(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c )^2-e)^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(2*e*cos(1/2*d*x+1/2*c)^2- e+(e^2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^ 2*(a^2-b^2)/b^2)^(1/2)))+2*arctan((2^(1/2)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1 /2)+(e^2*(a^2-b^2)/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4))+2*arctan((2^(1/2 )*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)-(e^2*(a^2-b^2)/b^2)^(1/4))/(e^2*(a^2- b^2)/b^2)^(1/4))))/e/(a-b)/(a+b)/(4*cos(1/2*d*x+1/2*c)^4*b^2-4*cos(1/2*d*x +1/2*c)^2*b^2+a^2))-2*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^ (1/2)*e^3*(1/b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^( 1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))/(-2*sin(1/2*d*x+1/2*c)^4*e+s...
Timed out. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+b \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+b \sin (c+d x))^2} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]