\(\int \frac {\sqrt {b \sin (e+f x)}}{\sqrt [3]{d \tan (e+f x)}} \, dx\) [155]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 25, antiderivative size = 64 \[ \int \frac {\sqrt {b \sin (e+f x)}}{\sqrt [3]{d \tan (e+f x)}} \, dx=\frac {6 \sqrt [3]{\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {7}{12},\frac {19}{12},\sin ^2(e+f x)\right ) \sqrt {b \sin (e+f x)} (d \tan (e+f x))^{2/3}}{7 d f} \] Output:

6/7*(cos(f*x+e)^2)^(1/3)*hypergeom([1/3, 7/12],[19/12],sin(f*x+e)^2)*(b*si 
n(f*x+e))^(1/2)*(d*tan(f*x+e))^(2/3)/d/f
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.87 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {b \sin (e+f x)}}{\sqrt [3]{d \tan (e+f x)}} \, dx=\frac {6 \operatorname {Hypergeometric2F1}\left (\frac {7}{12},\frac {5}{4},\frac {19}{12},-\tan ^2(e+f x)\right ) \sqrt [4]{\sec ^2(e+f x)} \sqrt {b \sin (e+f x)} (d \tan (e+f x))^{2/3}}{7 d f} \] Input:

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

Output:

(6*Hypergeometric2F1[7/12, 5/4, 19/12, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^( 
1/4)*Sqrt[b*Sin[e + f*x]]*(d*Tan[e + f*x])^(2/3))/(7*d*f)
 

Rubi [A] (verified)

Time = 0.30 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3082

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3057

\(\displaystyle \frac {6 \sqrt [3]{\cos ^2(e+f x)} \sqrt {b \sin (e+f x)} (d \tan (e+f x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {7}{12},\frac {19}{12},\sin ^2(e+f x)\right )}{7 d f}\)

Input:

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

Output:

(6*(Cos[e + f*x]^2)^(1/3)*Hypergeometric2F1[1/3, 7/12, 19/12, Sin[e + f*x] 
^2]*Sqrt[b*Sin[e + f*x]]*(d*Tan[e + f*x])^(2/3))/(7*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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