Integrand size = 25, antiderivative size = 90 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{3/2}} \, dx=-\frac {2 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{a^2 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}+\frac {2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{a^2 f} \]
-2*b^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f *x+1/2*e),2^(1/2))*(a*sin(f*x+e))^(1/2)/a^2/f/cos(f*x+e)^(1/2)/(b*tan(f*x+ e))^(1/2)+2*b*(a*sin(f*x+e))^(1/2)*(b*tan(f*x+e))^(1/2)/a^2/f
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.55 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{3/2}} \, dx=\frac {\left (2 \cos (e+f x) \cos ^2(e+f x)^{3/4}-\cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\sin ^2(e+f x)\right )\right ) (b \tan (e+f x))^{3/2}}{a f \cos ^2(e+f x)^{3/4} \sqrt {a \sin (e+f x)}} \]
((2*Cos[e + f*x]*(Cos[e + f*x]^2)^(3/4) - Cos[e + f*x]^3*Hypergeometric2F1 [1/4, 1/2, 3/2, Sin[e + f*x]^2])*(b*Tan[e + f*x])^(3/2))/(a*f*(Cos[e + f*x ]^2)^(3/4)*Sqrt[a*Sin[e + f*x]])
Time = 0.41 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3073, 3042, 3081, 3042, 3119}
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.
\(\displaystyle \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 3073 |
\(\displaystyle \frac {2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{a^2 f}-\frac {b^2 \int \frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \tan (e+f x)}}dx}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{a^2 f}-\frac {b^2 \int \frac {\sqrt {a \sin (e+f x)}}{\sqrt {b \tan (e+f x)}}dx}{a^2}\) |
\(\Big \downarrow \) 3081 |
\(\displaystyle \frac {2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{a^2 f}-\frac {b^2 \sqrt {a \sin (e+f x)} \int \sqrt {\cos (e+f x)}dx}{a^2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{a^2 f}-\frac {b^2 \sqrt {a \sin (e+f x)} \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{a^2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2 b \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}{a^2 f}-\frac {2 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {a \sin (e+f x)}}{a^2 f \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}\) |
(-2*b^2*EllipticE[(e + f*x)/2, 2]*Sqrt[a*Sin[e + f*x]])/(a^2*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[e + f*x]]) + (2*b*Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f *x]])/(a^2*f)
3.2.24.3.1 Defintions of rubi rules used
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[b*(a*Sin[e + f*x])^(m + 2)*((b*Tan[e + f*x])^(n - 1)/ (a^2*f*(n - 1))), x] - Simp[b^2*((m + 2)/(a^2*(n - 1))) Int[(a*Sin[e + f* x])^(m + 2)*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && G tQ[n, 1] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, 3/2])) && IntegersQ[2*m, 2 *n]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[Cos[e + f*x]^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^ n) Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(- 1)]) || IntegersQ[m - 1/2, n - 1/2])
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]
Result contains complex when optimal does not.
Time = 2.13 (sec) , antiderivative size = 329, normalized size of antiderivative = 3.66
method | result | size |
default | \(-\frac {2 \left (i \sqrt {\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}+1}\, \sqrt {-\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}+1}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )-i \sqrt {\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}+1}\, \sqrt {-\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}+1}\, E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+\left (\csc ^{3}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{3}+\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ) \left (\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}-1\right ) {\left (-\frac {b \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}-1}\right )}^{\frac {3}{2}}}{f {\left (\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}+1\right )}^{2} {\left (\frac {a \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}+1}\right )}^{\frac {3}{2}}}\) | \(329\) |
-2/f*(I*(csc(f*x+e)^2*(1-cos(f*x+e))^2+1)^(1/2)*(-csc(f*x+e)^2*(1-cos(f*x+ e))^2+1)^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)-I*(csc(f*x+e)^2*(1-c os(f*x+e))^2+1)^(1/2)*(-csc(f*x+e)^2*(1-cos(f*x+e))^2+1)^(1/2)*EllipticE(I *(csc(f*x+e)-cot(f*x+e)),I)+csc(f*x+e)^3*(1-cos(f*x+e))^3+csc(f*x+e)-cot(f *x+e))*(csc(f*x+e)^2*(1-cos(f*x+e))^2-1)*(-b/(csc(f*x+e)^2*(1-cos(f*x+e))^ 2-1)*(csc(f*x+e)-cot(f*x+e)))^(3/2)/(csc(f*x+e)^2*(1-cos(f*x+e))^2+1)^2/(1 /(csc(f*x+e)^2*(1-cos(f*x+e))^2+1)*a*(csc(f*x+e)-cot(f*x+e)))^(3/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.14 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {2} \sqrt {-a b} b {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + \sqrt {2} \sqrt {-a b} b {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, \sqrt {a \sin \left (f x + e\right )} b \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{a^{2} f} \]
(sqrt(2)*sqrt(-a*b)*b*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co s(f*x + e) + I*sin(f*x + e))) + sqrt(2)*sqrt(-a*b)*b*weierstrassZeta(-4, 0 , weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) + 2*sqrt(a*si n(f*x + e))*b*sqrt(b*sin(f*x + e)/cos(f*x + e)))/(a^2*f)
Timed out. \[ \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{3/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]