Integrand size = 21, antiderivative size = 58 \[ \int \frac {\cos ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=-\frac {3 b^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 d (b \sec (c+d x))^{10/3} \sqrt {\sin ^2(c+d x)}} \] Output:
-3/10*b^3*hypergeom([1/2, 5/3],[8/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x +c))^(10/3)/(sin(d*x+c)^2)^(1/2)
Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=-\frac {3 b^2 \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{7 d (b \sec (c+d x))^{7/3}} \] Input:
Integrate[Cos[c + d*x]^2/(b*Sec[c + d*x])^(1/3),x]
Output:
(-3*b^2*Cot[c + d*x]*Hypergeometric2F1[-7/6, 1/2, -1/6, Sec[c + d*x]^2]*Sq rt[-Tan[c + d*x]^2])/(7*d*(b*Sec[c + d*x])^(7/3))
Time = 0.37 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 2030, 4259, 3042, 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.
\(\displaystyle \int \frac {\cos ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt [3]{b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b^2 \int \frac {1}{\left (b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{7/3}}dx\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle b^2 \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \left (\frac {\cos (c+d x)}{b}\right )^{7/3}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{7/3}dx\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle -\frac {3 b^3 \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{10/3}}\) |
Input:
Int[Cos[c + d*x]^2/(b*Sec[c + d*x])^(1/3),x]
Output:
(-3*b^3*Hypergeometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2]*Sin[c + d*x])/(10 *d*(b*Sec[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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^(n - 1)*((Sin[c + d*x]/b)^(n - 1) Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
\[\int \frac {\cos \left (d x +c \right )^{2}}{\left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
Input:
int(cos(d*x+c)^2/(b*sec(d*x+c))^(1/3),x)
Output:
int(cos(d*x+c)^2/(b*sec(d*x+c))^(1/3),x)
\[ \int \frac {\cos ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(cos(d*x+c)^2/(b*sec(d*x+c))^(1/3),x, algorithm="fricas")
Output:
integral((b*sec(d*x + c))^(2/3)*cos(d*x + c)^2/(b*sec(d*x + c)), x)
\[ \int \frac {\cos ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{\sqrt [3]{b \sec {\left (c + d x \right )}}}\, dx \] Input:
integrate(cos(d*x+c)**2/(b*sec(d*x+c))**(1/3),x)
Output:
Integral(cos(c + d*x)**2/(b*sec(c + d*x))**(1/3), x)
\[ \int \frac {\cos ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(cos(d*x+c)^2/(b*sec(d*x+c))^(1/3),x, algorithm="maxima")
Output:
integrate(cos(d*x + c)^2/(b*sec(d*x + c))^(1/3), x)
\[ \int \frac {\cos ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(cos(d*x+c)^2/(b*sec(d*x+c))^(1/3),x, algorithm="giac")
Output:
integrate(cos(d*x + c)^2/(b*sec(d*x + c))^(1/3), x)
Timed out. \[ \int \frac {\cos ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \] Input:
int(cos(c + d*x)^2/(b/cos(c + d*x))^(1/3),x)
Output:
int(cos(c + d*x)^2/(b/cos(c + d*x))^(1/3), x)
\[ \int \frac {\cos ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=\frac {\int \frac {\cos \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{\frac {1}{3}}}d x}{b^{\frac {1}{3}}} \] Input:
int(cos(d*x+c)^2/(b*sec(d*x+c))^(1/3),x)
Output:
int(cos(c + d*x)**2/sec(c + d*x)**(1/3),x)/b**(1/3)