Integrand size = 25, antiderivative size = 64 \[ \int \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{4/3} \, dx=\frac {3 \cos ^2(e+f x)^{17/12} \operatorname {Hypergeometric2F1}\left (\frac {7}{6},\frac {17}{12},\frac {13}{6},\sin ^2(e+f x)\right ) \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{7/3}}{7 d f} \] Output:
3/7*(cos(f*x+e)^2)^(17/12)*hypergeom([7/6, 17/12],[13/6],sin(f*x+e)^2)*(b* sec(f*x+e))^(1/2)*(d*tan(f*x+e))^(7/3)/d/f
Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.97 \[ \int \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{4/3} \, dx=\frac {2 d \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{4},\frac {5}{4},\sec ^2(e+f x)\right ) \sqrt {b \sec (e+f x)} \sqrt [3]{d \tan (e+f x)}}{f \sqrt [6]{-\tan ^2(e+f x)}} \] Input:
Integrate[Sqrt[b*Sec[e + f*x]]*(d*Tan[e + f*x])^(4/3),x]
Output:
(2*d*Hypergeometric2F1[-1/6, 1/4, 5/4, Sec[e + f*x]^2]*Sqrt[b*Sec[e + f*x] ]*(d*Tan[e + f*x])^(1/3))/(f*(-Tan[e + f*x]^2)^(1/6))
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.
\(\displaystyle \int \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{4/3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{4/3}dx\) |
\(\Big \downarrow \) 3097 |
\(\displaystyle \frac {3 \cos ^2(e+f x)^{17/12} \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {7}{6},\frac {17}{12},\frac {13}{6},\sin ^2(e+f x)\right )}{7 d f}\) |
Input:
Int[Sqrt[b*Sec[e + f*x]]*(d*Tan[e + f*x])^(4/3),x]
Output:
(3*(Cos[e + f*x]^2)^(17/12)*Hypergeometric2F1[7/6, 17/12, 13/6, Sin[e + f* x]^2]*Sqrt[b*Sec[e + f*x]]*(d*Tan[e + f*x])^(7/3))/(7*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]
\[\int \sqrt {b \sec \left (f x +e \right )}\, \left (d \tan \left (f x +e \right )\right )^{\frac {4}{3}}d x\]
Input:
int((b*sec(f*x+e))^(1/2)*(d*tan(f*x+e))^(4/3),x)
Output:
int((b*sec(f*x+e))^(1/2)*(d*tan(f*x+e))^(4/3),x)
\[ \int \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{4/3} \, dx=\int { \sqrt {b \sec \left (f x + e\right )} \left (d \tan \left (f x + e\right )\right )^{\frac {4}{3}} \,d x } \] Input:
integrate((b*sec(f*x+e))^(1/2)*(d*tan(f*x+e))^(4/3),x, algorithm="fricas")
Output:
integral(sqrt(b*sec(f*x + e))*(d*tan(f*x + e))^(1/3)*d*tan(f*x + e), x)
Timed out. \[ \int \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{4/3} \, dx=\text {Timed out} \] Input:
integrate((b*sec(f*x+e))**(1/2)*(d*tan(f*x+e))**(4/3),x)
Output:
Timed out
\[ \int \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{4/3} \, dx=\int { \sqrt {b \sec \left (f x + e\right )} \left (d \tan \left (f x + e\right )\right )^{\frac {4}{3}} \,d x } \] Input:
integrate((b*sec(f*x+e))^(1/2)*(d*tan(f*x+e))^(4/3),x, algorithm="maxima")
Output:
integrate(sqrt(b*sec(f*x + e))*(d*tan(f*x + e))^(4/3), x)
\[ \int \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{4/3} \, dx=\int { \sqrt {b \sec \left (f x + e\right )} \left (d \tan \left (f x + e\right )\right )^{\frac {4}{3}} \,d x } \] Input:
integrate((b*sec(f*x+e))^(1/2)*(d*tan(f*x+e))^(4/3),x, algorithm="giac")
Output:
integrate(sqrt(b*sec(f*x + e))*(d*tan(f*x + e))^(4/3), x)
Timed out. \[ \int \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{4/3} \, dx=\int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{4/3}\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}} \,d x \] Input:
int((d*tan(e + f*x))^(4/3)*(b/cos(e + f*x))^(1/2),x)
Output:
int((d*tan(e + f*x))^(4/3)*(b/cos(e + f*x))^(1/2), x)
\[ \int \sqrt {b \sec (e+f x)} (d \tan (e+f x))^{4/3} \, dx=\frac {2 d^{\frac {4}{3}} \sqrt {b}\, \left (3 \tan \left (f x +e \right )^{\frac {1}{3}} \sqrt {\sec \left (f x +e \right )}-\left (\int \frac {\sqrt {\sec \left (f x +e \right )}}{\tan \left (f x +e \right )^{\frac {2}{3}}}d x \right ) f \right )}{5 f} \] Input:
int((b*sec(f*x+e))^(1/2)*(d*tan(f*x+e))^(4/3),x)
Output:
(2*d**(1/3)*sqrt(b)*d*(3*tan(e + f*x)**(1/3)*sqrt(sec(e + f*x)) - int((tan (e + f*x)**(1/3)*sqrt(sec(e + f*x)))/tan(e + f*x),x)*f))/(5*f)