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