Integrand size = 19, antiderivative size = 58 \[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} \, dx=-\frac {3 b^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}} \] Output:
-3/2*b^2*hypergeom([1/3, 1/2],[4/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+ c))^(2/3)/(sin(d*x+c)^2)^(1/2)
Time = 0.00 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.97 \[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} \, dx=\frac {3 b \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sec ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sqrt {-\tan ^2(c+d x)}}{d} \] Input:
Integrate[Cos[c + d*x]*(b*Sec[c + d*x])^(4/3),x]
Output:
(3*b*Cot[c + d*x]*Hypergeometric2F1[1/6, 1/2, 7/6, Sec[c + d*x]^2]*(b*Sec[ c + d*x])^(1/3)*Sqrt[-Tan[c + d*x]^2])/d
Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, 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.
\(\displaystyle \int \cos (c+d x) (b \sec (c+d x))^{4/3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b \int \sqrt [3]{b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle b \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\sqrt [3]{\frac {\cos (c+d x)}{b}}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\sqrt [3]{\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}}}dx\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle -\frac {3 b^2 \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}\) |
Input:
Int[Cos[c + d*x]*(b*Sec[c + d*x])^(4/3),x]
Output:
(-3*b^2*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(2* d*(b*Sec[c + d*x])^(2/3)*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]
\[\int \cos \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}d x\]
Input:
int(cos(d*x+c)*(b*sec(d*x+c))^(4/3),x)
Output:
int(cos(d*x+c)*(b*sec(d*x+c))^(4/3),x)
\[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right ) \,d x } \] Input:
integrate(cos(d*x+c)*(b*sec(d*x+c))^(4/3),x, algorithm="fricas")
Output:
integral((b*sec(d*x + c))^(1/3)*b*cos(d*x + c)*sec(d*x + c), x)
Timed out. \[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)*(b*sec(d*x+c))**(4/3),x)
Output:
Timed out
\[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right ) \,d x } \] Input:
integrate(cos(d*x+c)*(b*sec(d*x+c))^(4/3),x, algorithm="maxima")
Output:
integrate((b*sec(d*x + c))^(4/3)*cos(d*x + c), x)
\[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} \, dx=\int { \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right ) \,d x } \] Input:
integrate(cos(d*x+c)*(b*sec(d*x+c))^(4/3),x, algorithm="giac")
Output:
integrate((b*sec(d*x + c))^(4/3)*cos(d*x + c), x)
Timed out. \[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} \, dx=\int \cos \left (c+d\,x\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3} \,d x \] Input:
int(cos(c + d*x)*(b/cos(c + d*x))^(4/3),x)
Output:
int(cos(c + d*x)*(b/cos(c + d*x))^(4/3), x)
\[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} \, dx=b^{\frac {4}{3}} \left (\int \sec \left (d x +c \right )^{\frac {4}{3}} \cos \left (d x +c \right )d x \right ) \] Input:
int(cos(d*x+c)*(b*sec(d*x+c))^(4/3),x)
Output:
b**(1/3)*int(sec(c + d*x)**(1/3)*cos(c + d*x)*sec(c + d*x),x)*b