Integrand size = 21, antiderivative size = 56 \[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 b \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3} \sqrt {\sin ^2(c+d x)}} \] Output:
3/5*b*hypergeom([-5/6, 1/2],[1/6],cos(d*x+c)^2)*sin(d*x+c)/d/(b*cos(d*x+c) )^(5/3)/(sin(d*x+c)^2)^(1/2)
Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.04 \[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 b^2 \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}}{5 d (b \cos (c+d x))^{8/3}} \] Input:
Integrate[Sec[c + d*x]^2/(b*Cos[c + d*x])^(2/3),x]
Output:
(3*b^2*Cot[c + d*x]*Hypergeometric2F1[-5/6, 1/2, 1/6, Cos[c + d*x]^2]*Sqrt [Sin[c + d*x]^2])/(5*d*(b*Cos[c + d*x])^(8/3))
Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 2030, 3122}
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.
\(\displaystyle \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b^2 \int \frac {1}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{8/3}}dx\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {3 b \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right )}{5 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{5/3}}\) |
Input:
Int[Sec[c + d*x]^2/(b*Cos[c + d*x])^(2/3),x]
Output:
(3*b*Hypergeometric2F1[-5/6, 1/2, 1/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*d* (b*Cos[c + d*x])^(5/3)*Sqrt[Sin[c + d*x]^2])
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[2*n]
\[\int \frac {\sec \left (d x +c \right )^{2}}{\left (\cos \left (d x +c \right ) b \right )^{\frac {2}{3}}}d x\]
Input:
int(sec(d*x+c)^2/(cos(d*x+c)*b)^(2/3),x)
Output:
int(sec(d*x+c)^2/(cos(d*x+c)*b)^(2/3),x)
\[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^2/(b*cos(d*x+c))^(2/3),x, algorithm="fricas")
Output:
integral((b*cos(d*x + c))^(1/3)*sec(d*x + c)^2/(b*cos(d*x + c)), x)
\[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (b \cos {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \] Input:
integrate(sec(d*x+c)**2/(b*cos(d*x+c))**(2/3),x)
Output:
Integral(sec(c + d*x)**2/(b*cos(c + d*x))**(2/3), x)
\[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^2/(b*cos(d*x+c))^(2/3),x, algorithm="maxima")
Output:
integrate(sec(d*x + c)^2/(b*cos(d*x + c))^(2/3), x)
\[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^2/(b*cos(d*x+c))^(2/3),x, algorithm="giac")
Output:
integrate(sec(d*x + c)^2/(b*cos(d*x + c))^(2/3), x)
Timed out. \[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^2\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \] Input:
int(1/(cos(c + d*x)^2*(b*cos(c + d*x))^(2/3)),x)
Output:
int(1/(cos(c + d*x)^2*(b*cos(c + d*x))^(2/3)), x)
\[ \int \frac {\sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {\int \frac {\sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{\frac {2}{3}}}d x}{b^{\frac {2}{3}}} \] Input:
int(sec(d*x+c)^2/(b*cos(d*x+c))^(2/3),x)
Output:
int(sec(c + d*x)**2/cos(c + d*x)**(2/3),x)/b**(2/3)