Integrand size = 25, antiderivative size = 64 \[ \int \sqrt [3]{b \sec (e+f x)} \sqrt {d \tan (e+f x)} \, dx=\frac {2 \cos ^2(e+f x)^{11/12} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {11}{12},\frac {7}{4},\sin ^2(e+f x)\right ) \sqrt [3]{b \sec (e+f x)} (d \tan (e+f x))^{3/2}}{3 d f} \] Output:
2/3*(cos(f*x+e)^2)^(11/12)*hypergeom([3/4, 11/12],[7/4],sin(f*x+e)^2)*(b*s ec(f*x+e))^(1/3)*(d*tan(f*x+e))^(3/2)/d/f
Time = 0.06 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.97 \[ \int \sqrt [3]{b \sec (e+f x)} \sqrt {d \tan (e+f x)} \, dx=\frac {3 d \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{4},\frac {7}{6},\sec ^2(e+f x)\right ) \sqrt [3]{b \sec (e+f x)} \sqrt [4]{-\tan ^2(e+f x)}}{f \sqrt {d \tan (e+f x)}} \] Input:
Integrate[(b*Sec[e + f*x])^(1/3)*Sqrt[d*Tan[e + f*x]],x]
Output:
(3*d*Hypergeometric2F1[1/6, 1/4, 7/6, Sec[e + f*x]^2]*(b*Sec[e + f*x])^(1/ 3)*(-Tan[e + f*x]^2)^(1/4))/(f*Sqrt[d*Tan[e + f*x]])
Time = 0.23 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {3042, 3097}
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.
\(\displaystyle \int \sqrt [3]{b \sec (e+f x)} \sqrt {d \tan (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt [3]{b \sec (e+f x)} \sqrt {d \tan (e+f x)}dx\) |
\(\Big \downarrow \) 3097 |
\(\displaystyle \frac {2 \cos ^2(e+f x)^{11/12} \sqrt [3]{b \sec (e+f x)} (d \tan (e+f x))^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {11}{12},\frac {7}{4},\sin ^2(e+f x)\right )}{3 d f}\) |
Input:
Int[(b*Sec[e + f*x])^(1/3)*Sqrt[d*Tan[e + f*x]],x]
Output:
(2*(Cos[e + f*x]^2)^(11/12)*Hypergeometric2F1[3/4, 11/12, 7/4, Sin[e + f*x ]^2]*(b*Sec[e + f*x])^(1/3)*(d*Tan[e + f*x])^(3/2))/(3*d*f)
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[(a*Sec[e + f*x])^m*(b*Tan[e + f*x])^(n + 1)*((Cos[e + f*x]^2)^((m + n + 1)/2)/(b*f*(n + 1)))*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2, (n + 3)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[(n - 1)/2] && !IntegerQ[m/2]
\[\int \left (b \sec \left (f x +e \right )\right )^{\frac {1}{3}} \sqrt {d \tan \left (f x +e \right )}d x\]
Input:
int((b*sec(f*x+e))^(1/3)*(d*tan(f*x+e))^(1/2),x)
Output:
int((b*sec(f*x+e))^(1/3)*(d*tan(f*x+e))^(1/2),x)
\[ \int \sqrt [3]{b \sec (e+f x)} \sqrt {d \tan (e+f x)} \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {1}{3}} \sqrt {d \tan \left (f x + e\right )} \,d x } \] Input:
integrate((b*sec(f*x+e))^(1/3)*(d*tan(f*x+e))^(1/2),x, algorithm="fricas")
Output:
integral((b*sec(f*x + e))^(1/3)*sqrt(d*tan(f*x + e)), x)
\[ \int \sqrt [3]{b \sec (e+f x)} \sqrt {d \tan (e+f x)} \, dx=\int \sqrt [3]{b \sec {\left (e + f x \right )}} \sqrt {d \tan {\left (e + f x \right )}}\, dx \] Input:
integrate((b*sec(f*x+e))**(1/3)*(d*tan(f*x+e))**(1/2),x)
Output:
Integral((b*sec(e + f*x))**(1/3)*sqrt(d*tan(e + f*x)), x)
\[ \int \sqrt [3]{b \sec (e+f x)} \sqrt {d \tan (e+f x)} \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {1}{3}} \sqrt {d \tan \left (f x + e\right )} \,d x } \] Input:
integrate((b*sec(f*x+e))^(1/3)*(d*tan(f*x+e))^(1/2),x, algorithm="maxima")
Output:
integrate((b*sec(f*x + e))^(1/3)*sqrt(d*tan(f*x + e)), x)
\[ \int \sqrt [3]{b \sec (e+f x)} \sqrt {d \tan (e+f x)} \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {1}{3}} \sqrt {d \tan \left (f x + e\right )} \,d x } \] Input:
integrate((b*sec(f*x+e))^(1/3)*(d*tan(f*x+e))^(1/2),x, algorithm="giac")
Output:
integrate((b*sec(f*x + e))^(1/3)*sqrt(d*tan(f*x + e)), x)
Timed out. \[ \int \sqrt [3]{b \sec (e+f x)} \sqrt {d \tan (e+f x)} \, dx=\int \sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{1/3} \,d x \] Input:
int((d*tan(e + f*x))^(1/2)*(b/cos(e + f*x))^(1/3),x)
Output:
int((d*tan(e + f*x))^(1/2)*(b/cos(e + f*x))^(1/3), x)
\[ \int \sqrt [3]{b \sec (e+f x)} \sqrt {d \tan (e+f x)} \, dx=\sqrt {d}\, b^{\frac {1}{3}} \left (\int \sqrt {\tan \left (f x +e \right )}\, \sec \left (f x +e \right )^{\frac {1}{3}}d x \right ) \] Input:
int((b*sec(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))*sec(e + f*x)**(1/3),x)