\(\int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sin (e+f x)}} \, dx\) [151]

Optimal result

Integrand size = 25, antiderivative size = 64 \[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sin (e+f x)}} \, dx=\frac {6 \cos ^2(e+f x)^{5/4} \operatorname {Hypergeometric2F1}\left (\frac {13}{12},\frac {5}{4},\frac {25}{12},\sin ^2(e+f x)\right ) (d \tan (e+f x))^{5/2}}{13 d f \sqrt [3]{b \sin (e+f x)}} \] Output:

6/13*(cos(f*x+e)^2)^(5/4)*hypergeom([13/12, 5/4],[25/12],sin(f*x+e)^2)*(d* 
tan(f*x+e))^(5/2)/d/f/(b*sin(f*x+e))^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.82 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98 \[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sin (e+f x)}} \, dx=-\frac {2 d \left (-1+\sqrt [4]{\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{12},\frac {1}{4},\frac {13}{12},\sin ^2(e+f x)\right )\right ) \sqrt {d \tan (e+f x)}}{f \sqrt [3]{b \sin (e+f x)}} \] Input:

Integrate[(d*Tan[e + f*x])^(3/2)/(b*Sin[e + f*x])^(1/3),x]
 

Output:

(-2*d*(-1 + (Cos[e + f*x]^2)^(1/4)*Hypergeometric2F1[1/12, 1/4, 13/12, Sin 
[e + f*x]^2])*Sqrt[d*Tan[e + f*x]])/(f*(b*Sin[e + f*x])^(1/3))
 

Rubi [A] (verified)

Time = 0.32 (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.

\(\displaystyle \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sin (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sin (e+f x)}}dx\)

\(\Big \downarrow \) 3082

\(\displaystyle \frac {b \cos ^{\frac {5}{2}}(e+f x) (d \tan (e+f x))^{5/2} \int \frac {(b \sin (e+f x))^{7/6}}{\cos ^{\frac {3}{2}}(e+f x)}dx}{d (b \sin (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b \cos ^{\frac {5}{2}}(e+f x) (d \tan (e+f x))^{5/2} \int \frac {(b \sin (e+f x))^{7/6}}{\cos (e+f x)^{3/2}}dx}{d (b \sin (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3057

\(\displaystyle \frac {6 \cos ^2(e+f x)^{5/4} (d \tan (e+f x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {13}{12},\frac {5}{4},\frac {25}{12},\sin ^2(e+f x)\right )}{13 d f \sqrt [3]{b \sin (e+f x)}}\)

Input:

Int[(d*Tan[e + f*x])^(3/2)/(b*Sin[e + f*x])^(1/3),x]
 

Output:

(6*(Cos[e + f*x]^2)^(5/4)*Hypergeometric2F1[13/12, 5/4, 25/12, Sin[e + f*x 
]^2]*(d*Tan[e + f*x])^(5/2))/(13*d*f*(b*Sin[e + f*x])^(1/3))
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3057

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]
 

rule 3082

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]
 

Maple [F]

\[\int \frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{\left (b \sin \left (f x +e \right )\right )^{\frac {1}{3}}}d x\]

Input:

int((d*tan(f*x+e))^(3/2)/(b*sin(f*x+e))^(1/3),x)
 

Output:

int((d*tan(f*x+e))^(3/2)/(b*sin(f*x+e))^(1/3),x)
 

Fricas [F]

\[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int { \frac {\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:

integrate((d*tan(f*x+e))^(3/2)/(b*sin(f*x+e))^(1/3),x, algorithm="fricas")
 

Output:

integral((b*sin(f*x + e))^(2/3)*sqrt(d*tan(f*x + e))*d*tan(f*x + e)/(b*sin 
(f*x + e)), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sin (e+f x)}} \, dx=\text {Timed out} \] Input:

integrate((d*tan(f*x+e))**(3/2)/(b*sin(f*x+e))**(1/3),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int { \frac {\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:

integrate((d*tan(f*x+e))^(3/2)/(b*sin(f*x+e))^(1/3),x, algorithm="maxima")
 

Output:

integrate((d*tan(f*x + e))^(3/2)/(b*sin(f*x + e))^(1/3), x)
 

Giac [F]

\[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int { \frac {\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \sin \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:

integrate((d*tan(f*x+e))^(3/2)/(b*sin(f*x+e))^(1/3),x, algorithm="giac")
 

Output:

integrate((d*tan(f*x + e))^(3/2)/(b*sin(f*x + e))^(1/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sin (e+f x)}} \, dx=\int \frac {{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{{\left (b\,\sin \left (e+f\,x\right )\right )}^{1/3}} \,d x \] Input:

int((d*tan(e + f*x))^(3/2)/(b*sin(e + f*x))^(1/3),x)
 

Output:

int((d*tan(e + f*x))^(3/2)/(b*sin(e + f*x))^(1/3), x)
 

Reduce [F]

\[ \int \frac {(d \tan (e+f x))^{3/2}}{\sqrt [3]{b \sin (e+f x)}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )}{\sin \left (f x +e \right )^{\frac {1}{3}}}d x \right ) d}{b^{\frac {1}{3}}} \] Input:

int((d*tan(f*x+e))^(3/2)/(b*sin(f*x+e))^(1/3),x)
 

Output:

(sqrt(d)*int((sqrt(tan(e + f*x))*tan(e + f*x))/sin(e + f*x)**(1/3),x)*d)/b 
**(1/3)