Integrand size = 21, antiderivative size = 58 \[ \int \sqrt [3]{b \cos (c+d x)} \sec ^3(c+d x) \, dx=\frac {3 b^2 \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3} \sqrt {\sin ^2(c+d x)}} \] Output:
3/5*b^2*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 = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09 \[ \int \sqrt [3]{b \cos (c+d x)} \sec ^3(c+d x) \, dx=\frac {3 \sqrt [3]{b \cos (c+d x)} \csc (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right ) \sec ^2(c+d x) \sqrt {\sin ^2(c+d x)}}{5 d} \] Input:
Integrate[(b*Cos[c + d*x])^(1/3)*Sec[c + d*x]^3,x]
Output:
(3*(b*Cos[c + d*x])^(1/3)*Csc[c + d*x]*Hypergeometric2F1[-5/6, 1/2, 1/6, C os[c + d*x]^2]*Sec[c + d*x]^2*Sqrt[Sin[c + d*x]^2])/(5*d)
Time = 0.23 (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.
\(\displaystyle \int \sec ^3(c+d x) \sqrt [3]{b \cos (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt [3]{b \sin \left (c+d x+\frac {\pi }{2}\right )}}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b^3 \int \frac {1}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{8/3}}dx\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {3 b^2 \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right )}{5 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{5/3}}\) |
Input:
Int[(b*Cos[c + d*x])^(1/3)*Sec[c + d*x]^3,x]
Output:
(3*b^2*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]
\[\int \left (\cos \left (d x +c \right ) b \right )^{\frac {1}{3}} \sec \left (d x +c \right )^{3}d x\]
Input:
int((cos(d*x+c)*b)^(1/3)*sec(d*x+c)^3,x)
Output:
int((cos(d*x+c)*b)^(1/3)*sec(d*x+c)^3,x)
\[ \int \sqrt [3]{b \cos (c+d x)} \sec ^3(c+d x) \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{3} \,d x } \] Input:
integrate((b*cos(d*x+c))^(1/3)*sec(d*x+c)^3,x, algorithm="fricas")
Output:
integral((b*cos(d*x + c))^(1/3)*sec(d*x + c)^3, x)
\[ \int \sqrt [3]{b \cos (c+d x)} \sec ^3(c+d x) \, dx=\int \sqrt [3]{b \cos {\left (c + d x \right )}} \sec ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate((b*cos(d*x+c))**(1/3)*sec(d*x+c)**3,x)
Output:
Integral((b*cos(c + d*x))**(1/3)*sec(c + d*x)**3, x)
\[ \int \sqrt [3]{b \cos (c+d x)} \sec ^3(c+d x) \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{3} \,d x } \] Input:
integrate((b*cos(d*x+c))^(1/3)*sec(d*x+c)^3,x, algorithm="maxima")
Output:
integrate((b*cos(d*x + c))^(1/3)*sec(d*x + c)^3, x)
\[ \int \sqrt [3]{b \cos (c+d x)} \sec ^3(c+d x) \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{3} \,d x } \] Input:
integrate((b*cos(d*x+c))^(1/3)*sec(d*x+c)^3,x, algorithm="giac")
Output:
integrate((b*cos(d*x + c))^(1/3)*sec(d*x + c)^3, x)
Timed out. \[ \int \sqrt [3]{b \cos (c+d x)} \sec ^3(c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{1/3}}{{\cos \left (c+d\,x\right )}^3} \,d x \] Input:
int((b*cos(c + d*x))^(1/3)/cos(c + d*x)^3,x)
Output:
int((b*cos(c + d*x))^(1/3)/cos(c + d*x)^3, x)
\[ \int \sqrt [3]{b \cos (c+d x)} \sec ^3(c+d x) \, dx=b^{\frac {1}{3}} \left (\int \cos \left (d x +c \right )^{\frac {1}{3}} \sec \left (d x +c \right )^{3}d x \right ) \] Input:
int((b*cos(d*x+c))^(1/3)*sec(d*x+c)^3,x)
Output:
b**(1/3)*int(cos(c + d*x)**(1/3)*sec(c + d*x)**3,x)