Integrand size = 25, antiderivative size = 64 \[ \int \sqrt [3]{b \sin (e+f x)} (d \tan (e+f x))^{3/2} \, dx=\frac {6 \cos ^2(e+f x)^{5/4} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {17}{12},\frac {29}{12},\sin ^2(e+f x)\right ) \sqrt [3]{b \sin (e+f x)} (d \tan (e+f x))^{5/2}}{17 d f} \] Output:
6/17*(cos(f*x+e)^2)^(5/4)*hypergeom([5/4, 17/12],[29/12],sin(f*x+e)^2)*(b* sin(f*x+e))^(1/3)*(d*tan(f*x+e))^(5/2)/d/f
Time = 10.81 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98 \[ \int \sqrt [3]{b \sin (e+f x)} (d \tan (e+f x))^{3/2} \, dx=-\frac {2 d \left (-1+\sqrt [4]{\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {5}{12},\frac {17}{12},\sin ^2(e+f x)\right )\right ) \sqrt [3]{b \sin (e+f x)} \sqrt {d \tan (e+f x)}}{f} \] Input:
Integrate[(b*Sin[e + f*x])^(1/3)*(d*Tan[e + f*x])^(3/2),x]
Output:
(-2*d*(-1 + (Cos[e + f*x]^2)^(1/4)*Hypergeometric2F1[1/4, 5/12, 17/12, Sin [e + f*x]^2])*(b*Sin[e + f*x])^(1/3)*Sqrt[d*Tan[e + f*x]])/f
Time = 0.31 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 3082, 3042, 3057}
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.
\(\displaystyle \int \sqrt [3]{b \sin (e+f x)} (d \tan (e+f x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt [3]{b \sin (e+f x)} (d \tan (e+f x))^{3/2}dx\) |
\(\Big \downarrow \) 3082 |
\(\displaystyle \frac {b \cos ^{\frac {5}{2}}(e+f x) (d \tan (e+f x))^{5/2} \int \frac {(b \sin (e+f x))^{11/6}}{\cos ^{\frac {3}{2}}(e+f x)}dx}{d (b \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \cos ^{\frac {5}{2}}(e+f x) (d \tan (e+f x))^{5/2} \int \frac {(b \sin (e+f x))^{11/6}}{\cos (e+f x)^{3/2}}dx}{d (b \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle \frac {6 \cos ^2(e+f x)^{5/4} \sqrt [3]{b \sin (e+f x)} (d \tan (e+f x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {17}{12},\frac {29}{12},\sin ^2(e+f x)\right )}{17 d f}\) |
Input:
Int[(b*Sin[e + f*x])^(1/3)*(d*Tan[e + f*x])^(3/2),x]
Output:
(6*(Cos[e + f*x]^2)^(5/4)*Hypergeometric2F1[5/4, 17/12, 29/12, Sin[e + f*x ]^2]*(b*Sin[e + f*x])^(1/3)*(d*Tan[e + f*x])^(5/2))/(17*d*f)
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[b^(2*IntPart[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*Frac Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr acPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[ e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[a*Cos[e + f*x]^(n + 1)*((b*Tan[e + f*x])^(n + 1)/(b* (a*Sin[e + f*x])^(n + 1))) 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]
\[\int \left (b \sin \left (f x +e \right )\right )^{\frac {1}{3}} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}d x\]
Input:
int((b*sin(f*x+e))^(1/3)*(d*tan(f*x+e))^(3/2),x)
Output:
int((b*sin(f*x+e))^(1/3)*(d*tan(f*x+e))^(3/2),x)
\[ \int \sqrt [3]{b \sin (e+f x)} (d \tan (e+f x))^{3/2} \, dx=\int { \left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}} \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((b*sin(f*x+e))^(1/3)*(d*tan(f*x+e))^(3/2),x, algorithm="fricas")
Output:
integral((b*sin(f*x + e))^(1/3)*sqrt(d*tan(f*x + e))*d*tan(f*x + e), x)
Timed out. \[ \int \sqrt [3]{b \sin (e+f x)} (d \tan (e+f x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate((b*sin(f*x+e))**(1/3)*(d*tan(f*x+e))**(3/2),x)
Output:
Timed out
\[ \int \sqrt [3]{b \sin (e+f x)} (d \tan (e+f x))^{3/2} \, dx=\int { \left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}} \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((b*sin(f*x+e))^(1/3)*(d*tan(f*x+e))^(3/2),x, algorithm="maxima")
Output:
integrate((b*sin(f*x + e))^(1/3)*(d*tan(f*x + e))^(3/2), x)
Timed out. \[ \int \sqrt [3]{b \sin (e+f x)} (d \tan (e+f x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate((b*sin(f*x+e))^(1/3)*(d*tan(f*x+e))^(3/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \sqrt [3]{b \sin (e+f x)} (d \tan (e+f x))^{3/2} \, dx=\int {\left (b\,\sin \left (e+f\,x\right )\right )}^{1/3}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2} \,d x \] Input:
int((b*sin(e + f*x))^(1/3)*(d*tan(e + f*x))^(3/2),x)
Output:
int((b*sin(e + f*x))^(1/3)*(d*tan(e + f*x))^(3/2), x)
\[ \int \sqrt [3]{b \sin (e+f x)} (d \tan (e+f x))^{3/2} \, dx=\frac {\sqrt {d}\, b^{\frac {1}{3}} d \left (6 \sqrt {\tan \left (f x +e \right )}\, \sin \left (f x +e \right )^{\frac {1}{3}}-2 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \cos \left (f x +e \right )}{\sin \left (f x +e \right )^{\frac {2}{3}}}d x \right ) f -3 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \sin \left (f x +e \right )^{\frac {1}{3}}}{\tan \left (f x +e \right )}d x \right ) f \right )}{3 f} \] Input:
int((b*sin(f*x+e))^(1/3)*(d*tan(f*x+e))^(3/2),x)
Output:
(sqrt(d)*b**(1/3)*d*(6*sqrt(tan(e + f*x))*sin(e + f*x)**(1/3) - 2*int((sqr t(tan(e + f*x))*sin(e + f*x)**(1/3)*cos(e + f*x))/sin(e + f*x),x)*f - 3*in t((sqrt(tan(e + f*x))*sin(e + f*x)**(1/3))/tan(e + f*x),x)*f))/(3*f)