Integrand size = 21, antiderivative size = 58 \[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 b^2 \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {1}{2},-\frac {2}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 d (b \cos (c+d x))^{10/3} \sqrt {\sin ^2(c+d x)}} \] Output:
3/10*b^2*hypergeom([-5/3, 1/2],[-2/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*cos(d* x+c))^(10/3)/(sin(d*x+c)^2)^(1/2)
Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 b^2 \csc (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {1}{2},-\frac {2}{3},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}}{10 d (b \cos (c+d x))^{10/3}} \] Input:
Integrate[Sec[c + d*x]^3/(b*Cos[c + d*x])^(4/3),x]
Output:
(3*b^2*Csc[c + d*x]*Hypergeometric2F1[-5/3, 1/2, -2/3, Cos[c + d*x]^2]*Sqr t[Sin[c + d*x]^2])/(10*d*(b*Cos[c + d*x])^(10/3))
Time = 0.24 (sec) , antiderivative size = 58, 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.
\(\displaystyle \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b^3 \int \frac {1}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{13/3}}dx\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {3 b^2 \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {1}{2},-\frac {2}{3},\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{10/3}}\) |
Input:
Int[Sec[c + d*x]^3/(b*Cos[c + d*x])^(4/3),x]
Output:
(3*b^2*Hypergeometric2F1[-5/3, 1/2, -2/3, Cos[c + d*x]^2]*Sin[c + d*x])/(1 0*d*(b*Cos[c + d*x])^(10/3)*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]
\[\int \frac {\sec \left (d x +c \right )^{3}}{\left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}}}d x\]
Input:
int(sec(d*x+c)^3/(cos(d*x+c)*b)^(4/3),x)
Output:
int(sec(d*x+c)^3/(cos(d*x+c)*b)^(4/3),x)
\[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^3/(b*cos(d*x+c))^(4/3),x, algorithm="fricas")
Output:
integral((b*cos(d*x + c))^(2/3)*sec(d*x + c)^3/(b^2*cos(d*x + c)^2), x)
\[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (b \cos {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \] Input:
integrate(sec(d*x+c)**3/(b*cos(d*x+c))**(4/3),x)
Output:
Integral(sec(c + d*x)**3/(b*cos(c + d*x))**(4/3), x)
\[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^3/(b*cos(d*x+c))^(4/3),x, algorithm="maxima")
Output:
integrate(sec(d*x + c)^3/(b*cos(d*x + c))^(4/3), x)
\[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^3/(b*cos(d*x+c))^(4/3),x, algorithm="giac")
Output:
integrate(sec(d*x + c)^3/(b*cos(d*x + c))^(4/3), x)
Timed out. \[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}} \,d x \] Input:
int(1/(cos(c + d*x)^3*(b*cos(c + d*x))^(4/3)),x)
Output:
int(1/(cos(c + d*x)^3*(b*cos(c + d*x))^(4/3)), x)
\[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\frac {\int \frac {\sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{\frac {4}{3}}}d x}{b^{\frac {4}{3}}} \] Input:
int(sec(d*x+c)^3/(b*cos(d*x+c))^(4/3),x)
Output:
int(sec(c + d*x)**3/(cos(c + d*x)**(1/3)*cos(c + d*x)),x)/(b**(1/3)*b)