\(\int \frac {\sqrt {d \tan (e+f x)}}{(b \sin (e+f x))^{4/3}} \, dx\) [148]

Optimal result

Integrand size = 25, antiderivative size = 62 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(b \sin (e+f x))^{4/3}} \, dx=\frac {6 \cos ^2(e+f x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{12},\frac {3}{4},\frac {13}{12},\sin ^2(e+f x)\right ) (d \tan (e+f x))^{3/2}}{d f (b \sin (e+f x))^{4/3}} \] Output:

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

Mathematica [A] (verified)

Time = 10.87 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(b \sin (e+f x))^{4/3}} \, dx=\frac {3 \operatorname {Hypergeometric2F1}\left (\frac {1}{12},\frac {3}{4},\frac {13}{12},\sin ^2(e+f x)\right ) \sin (2 (e+f x)) \sqrt {d \tan (e+f x)}}{f \sqrt [4]{\cos ^2(e+f x)} (b \sin (e+f x))^{4/3}} \] Input:

Integrate[Sqrt[d*Tan[e + f*x]]/(b*Sin[e + f*x])^(4/3),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 62, 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.

\(\displaystyle \int \frac {\sqrt {d \tan (e+f x)}}{(b \sin (e+f x))^{4/3}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {d \tan (e+f x)}}{(b \sin (e+f x))^{4/3}}dx\)

\(\Big \downarrow \) 3082

\(\displaystyle \frac {b \cos ^{\frac {3}{2}}(e+f x) (d \tan (e+f x))^{3/2} \int \frac {1}{\sqrt {\cos (e+f x)} (b \sin (e+f x))^{5/6}}dx}{d (b \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b \cos ^{\frac {3}{2}}(e+f x) (d \tan (e+f x))^{3/2} \int \frac {1}{\sqrt {\cos (e+f x)} (b \sin (e+f x))^{5/6}}dx}{d (b \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3057

\(\displaystyle \frac {6 \cos ^2(e+f x)^{3/4} (d \tan (e+f x))^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{12},\frac {3}{4},\frac {13}{12},\sin ^2(e+f x)\right )}{d f (b \sin (e+f x))^{4/3}}\)

Input:

Int[Sqrt[d*Tan[e + f*x]]/(b*Sin[e + f*x])^(4/3),x]
 

Output:

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

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]

\[\int \frac {\sqrt {d \tan \left (f x +e \right )}}{\left (b \sin \left (f x +e \right )\right )^{\frac {4}{3}}}d x\]

Input:

int((d*tan(f*x+e))^(1/2)/(b*sin(f*x+e))^(4/3),x)
 

Output:

int((d*tan(f*x+e))^(1/2)/(b*sin(f*x+e))^(4/3),x)
 

Fricas [F]

\[ \int \frac {\sqrt {d \tan (e+f x)}}{(b \sin (e+f x))^{4/3}} \, dx=\int { \frac {\sqrt {d \tan \left (f x + e\right )}}{\left (b \sin \left (f x + e\right )\right )^{\frac {4}{3}}} \,d x } \] Input:

integrate((d*tan(f*x+e))^(1/2)/(b*sin(f*x+e))^(4/3),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {d \tan (e+f x)}}{(b \sin (e+f x))^{4/3}} \, dx=\text {Timed out} \] Input:

integrate((d*tan(f*x+e))**(1/2)/(b*sin(f*x+e))**(4/3),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sqrt {d \tan (e+f x)}}{(b \sin (e+f x))^{4/3}} \, dx=\int { \frac {\sqrt {d \tan \left (f x + e\right )}}{\left (b \sin \left (f x + e\right )\right )^{\frac {4}{3}}} \,d x } \] Input:

integrate((d*tan(f*x+e))^(1/2)/(b*sin(f*x+e))^(4/3),x, algorithm="maxima")
 

Output:

integrate(sqrt(d*tan(f*x + e))/(b*sin(f*x + e))^(4/3), x)
 

Giac [F]

\[ \int \frac {\sqrt {d \tan (e+f x)}}{(b \sin (e+f x))^{4/3}} \, dx=\int { \frac {\sqrt {d \tan \left (f x + e\right )}}{\left (b \sin \left (f x + e\right )\right )^{\frac {4}{3}}} \,d x } \] Input:

integrate((d*tan(f*x+e))^(1/2)/(b*sin(f*x+e))^(4/3),x, algorithm="giac")
 

Output:

integrate(sqrt(d*tan(f*x + e))/(b*sin(f*x + e))^(4/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {d \tan (e+f x)}}{(b \sin (e+f x))^{4/3}} \, dx=\int \frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{{\left (b\,\sin \left (e+f\,x\right )\right )}^{4/3}} \,d x \] Input:

int((d*tan(e + f*x))^(1/2)/(b*sin(e + f*x))^(4/3),x)
 

Output:

int((d*tan(e + f*x))^(1/2)/(b*sin(e + f*x))^(4/3), x)
 

Reduce [F]

\[ \int \frac {\sqrt {d \tan (e+f x)}}{(b \sin (e+f x))^{4/3}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\sin \left (f x +e \right )^{\frac {4}{3}}}d x \right )}{b^{\frac {4}{3}}} \] Input:

int((d*tan(f*x+e))^(1/2)/(b*sin(f*x+e))^(4/3),x)
 

Output:

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