\(\int \sqrt [3]{b \sin (e+f x)} \sqrt {d \tan (e+f x)} \, dx\) [146]

Optimal result

Integrand size = 25, antiderivative size = 64 \[ \int \sqrt [3]{b \sin (e+f x)} \sqrt {d \tan (e+f x)} \, dx=\frac {6 \cos ^2(e+f x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {11}{12},\frac {23}{12},\sin ^2(e+f x)\right ) \sqrt [3]{b \sin (e+f x)} (d \tan (e+f x))^{3/2}}{11 d f} \] Output:

6/11*(cos(f*x+e)^2)^(3/4)*hypergeom([3/4, 11/12],[23/12],sin(f*x+e)^2)*(b* 
sin(f*x+e))^(1/3)*(d*tan(f*x+e))^(3/2)/d/f
 

Mathematica [A] (verified)

Time = 10.82 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.08 \[ \int \sqrt [3]{b \sin (e+f x)} \sqrt {d \tan (e+f x)} \, dx=\frac {3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {11}{12},\frac {23}{12},\sin ^2(e+f x)\right ) \sqrt [3]{b \sin (e+f x)} \sin (2 (e+f x)) \sqrt {d \tan (e+f x)}}{11 f \sqrt [4]{\cos ^2(e+f x)}} \] Input:

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

Output:

(3*Hypergeometric2F1[3/4, 11/12, 23/12, Sin[e + f*x]^2]*(b*Sin[e + f*x])^( 
1/3)*Sin[2*(e + f*x)]*Sqrt[d*Tan[e + f*x]])/(11*f*(Cos[e + f*x]^2)^(1/4))
 

Rubi [A] (verified)

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3082

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3057

\(\displaystyle \frac {6 \cos ^2(e+f x)^{3/4} \sqrt [3]{b \sin (e+f x)} (d \tan (e+f x))^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {11}{12},\frac {23}{12},\sin ^2(e+f x)\right )}{11 d f}\)

Input:

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

Output:

(6*(Cos[e + f*x]^2)^(3/4)*Hypergeometric2F1[3/4, 11/12, 23/12, Sin[e + f*x 
]^2]*(b*Sin[e + f*x])^(1/3)*(d*Tan[e + f*x])^(3/2))/(11*d*f)
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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