Integrand size = 25, antiderivative size = 397 \[ \int \frac {(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)} \, dx=-\frac {\sqrt [4]{-a^2+b^2} e^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{3/2} d}-\frac {\sqrt [4]{-a^2+b^2} e^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{3/2} d}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}+\frac {2 a e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 d \sqrt {e \cos (c+d x)}}-\frac {a \left (a^2-b^2\right ) e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b^2 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}}-\frac {a \left (a^2-b^2\right ) e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b^2 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \cos (c+d x)}} \]
-(-a^2+b^2)^(1/4)*e^(3/2)*arctan(b^(1/2)*(e*cos(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*cos(d* x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(3/2)/d+2*a*e^2*(cos(1/2*d*x+1/2*c )^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d* x+c)^(1/2)/b^2/d/(e*cos(d*x+c))^(1/2)-a*(a^2-b^2)*e^2*(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^2/d/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(e*c os(d*x+c))^(1/2)-a*(a^2-b^2)*e^2*(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^2/d/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(e*cos(d*x+c))^(1/2)+2*e* (e*cos(d*x+c))^(1/2)/b/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 11.92 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.55 \[ \int \frac {(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)} \, dx=-\frac {e \sqrt {e \cos (c+d x)} \left (a^2-b^2+b^2 \cos ^2(c+d x)\right ) \csc ^2(c+d x) \left (2 b \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},\left (1-\frac {a^2}{b^2}\right ) \sec ^2(c+d x)\right ) \tan (c+d x)+a \left (\operatorname {EllipticPi}\left (-\frac {\sqrt {-a^2+b^2}}{b},\arcsin \left (\sqrt [4]{\sec ^2(c+d x)}\right ),-1\right )+\operatorname {EllipticPi}\left (\frac {\sqrt {-a^2+b^2}}{b},\arcsin \left (\sqrt [4]{\sec ^2(c+d x)}\right ),-1\right )\right ) \sqrt [4]{\sec ^2(c+d x)} \sqrt {-\tan ^2(c+d x)}\right )}{b^2 d (b+a \csc (c+d x)) \left (-a \sqrt {\sec ^2(c+d x)}+b \tan (c+d x)\right )} \]
-((e*Sqrt[e*Cos[c + d*x]]*(a^2 - b^2 + b^2*Cos[c + d*x]^2)*Csc[c + d*x]^2* (2*b*Hypergeometric2F1[-1/4, 1, 3/4, (1 - a^2/b^2)*Sec[c + d*x]^2]*Tan[c + d*x] + a*(EllipticPi[-(Sqrt[-a^2 + b^2]/b), ArcSin[(Sec[c + d*x]^2)^(1/4) ], -1] + EllipticPi[Sqrt[-a^2 + b^2]/b, ArcSin[(Sec[c + d*x]^2)^(1/4)], -1 ])*(Sec[c + d*x]^2)^(1/4)*Sqrt[-Tan[c + d*x]^2]))/(b^2*d*(b + a*Csc[c + d* x])*(-(a*Sqrt[Sec[c + d*x]^2]) + b*Tan[c + d*x])))
Time = 1.84 (sec) , antiderivative size = 392, normalized size of antiderivative = 0.99, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3042, 3174, 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.
| \(\displaystyle \int \frac {(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)}dx\) |
| \(\Big \downarrow \) 3174 |
| \(\displaystyle \frac {e^2 \int \frac {b+a \sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \int \frac {b+a \sin (c+d x)}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3346 |
| \(\displaystyle \frac {e^2 \left (\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)}}dx}{b}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{b}\right )}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \left (\frac {a \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{b}\right )}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {e^2 \left (\frac {a \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b \sqrt {e \cos (c+d x)}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{b}\right )}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \left (\frac {a \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \sqrt {e \cos (c+d x)}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{b}\right )}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} (a+b \sin (c+d x))}dx}{b}\right )}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3181 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (\frac {b e \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b^2 \cos ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2\right )}d(e \cos (c+d x))}{d}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (\frac {2 b e \int \frac {1}{b^2 e^4 \cos ^4(c+d x)+\left (a^2-b^2\right ) e^2}d\sqrt {e \cos (c+d x)}}{d}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \cos ^2(c+d x)}d\sqrt {e \cos (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\int \frac {1}{b e^2 \cos ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \cos (c+d x)}}{2 e \sqrt {b^2-a^2}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \cos ^2(c+d x)}d\sqrt {e \cos (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\right )}{b}\right )}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}+\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (-\frac {a \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )} \left (b \sin \left (c+d x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}+\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (-\frac {a \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {b^2-a^2}-b \cos (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \cos (c+d x)}}-\frac {a \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \left (b \cos (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \cos (c+d x)}}+\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (-\frac {a \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sqrt {b^2-a^2}-b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \cos (c+d x)}}-\frac {a \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b \sin \left (c+d x+\frac {\pi }{2}\right )+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \cos (c+d x)}}+\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\right )}{b}\right )}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {e^2 \left (\frac {2 a \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d \sqrt {e \cos (c+d x)}}-\frac {\left (a^2-b^2\right ) \left (\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \cos (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}+\frac {a \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{d \sqrt {b^2-a^2} \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \cos (c+d x)}}-\frac {a \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{d \sqrt {b^2-a^2} \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \cos (c+d x)}}\right )}{b}\right )}{b}+\frac {2 e \sqrt {e \cos (c+d x)}}{b d}\) |
(2*e*Sqrt[e*Cos[c + d*x]])/(b*d) + (e^2*((2*a*Sqrt[Cos[c + d*x]]*EllipticF [(c + d*x)/2, 2])/(b*d*Sqrt[e*Cos[c + d*x]]) - ((a^2 - b^2)*((2*b*e*(-1/2* ArcTan[(Sqrt[b]*Sqrt[e]*Cos[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]*Cos[c + d*x])/(-a^2 + b^2) ^(1/4)]/(2*Sqrt[b]*(-a^2 + b^2)^(3/4)*e^(3/2))))/d + (a*Sqrt[Cos[c + d*x]] *EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(Sqrt[-a^2 + b^ 2]*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*x]]) - (a*Sqrt[Cos[c + d*x]]* EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(Sqrt[-a^2 + b^2 ]*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*x]])))/b))/b
3.6.78.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + 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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.93 (sec) , antiderivative size = 844, normalized size of antiderivative = 2.13
(4*e^2*b*(1/2/b^2/e*(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)-(a^2-b^2)/b^2*(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)))/(16*a^2-16*b^2)/e) -2*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*e^2*(1/b^2* (sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-e*(2*sin( 1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c) ,2^(1/2))+1/16*(-a^2+b^2)/b^4*sum(1/_alpha/(2*_alpha^2-1)*(2^(1/2)/(e*(2*_ alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*arctanh(1/2*e*(4*_alpha^2-3)/(4*a^2-3*b^ 2)*(4*a^2*cos(1/2*d*x+1/2*c)^2-3*cos(1/2*d*x+1/2*c)^2*b^2+b^2*_alpha^2-3*a ^2+2*b^2)*2^(1/2)/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)/(-e*(2*sin(1/2* d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2))+8*b^2/a^2*_alpha*(_alpha^2-1)*( sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-e*sin(1/2* d*x+1/2*c)^2*(2*sin(1/2*d*x+1/2*c)^2-1))^(1/2)*EllipticPi(cos(1/2*d*x+1/2* c),-4*b^2/a^2*(_alpha^2-1),2^(1/2))),_alpha=RootOf(4*_Z^4*b^2-4*_Z^2*b^2+a ^2)))/sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2))/d
Timed out. \[ \int \frac {(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{b \sin \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{b \sin \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{3/2}}{a+b \sin (c+d x)} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}}{a+b\,\sin \left (c+d\,x\right )} \,d x \]