Integrand size = 21, antiderivative size = 82 \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{2/3} \, dx=-\frac {3 \cos ^{1+m}(c+d x) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5+3 m) \sqrt {\sin ^2(c+d x)}} \] Output:
-3*cos(d*x+c)^(1+m)*(b*cos(d*x+c))^(2/3)*hypergeom([1/2, 5/6+1/2*m],[11/6+ 1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(5+3*m)/(sin(d*x+c)^2)^(1/2)
Time = 0.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{2/3} \, dx=-\frac {\cos ^{1+m}(c+d x) (b \cos (c+d x))^{2/3} \csc (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} \left (\frac {5}{3}+m\right ),\frac {1}{2} \left (\frac {11}{3}+m\right ),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}}{d \left (\frac {5}{3}+m\right )} \] Input:
Integrate[Cos[c + d*x]^m*(b*Cos[c + d*x])^(2/3),x]
Output:
-((Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(2/3)*Csc[c + d*x]*Hypergeometric 2F1[1/2, (5/3 + m)/2, (11/3 + m)/2, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2])/ (d*(5/3 + m)))
Time = 0.26 (sec) , antiderivative size = 82, 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 = {2034, 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.
\(\displaystyle \int (b \cos (c+d x))^{2/3} \cos ^m(c+d x) \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \frac {(b \cos (c+d x))^{2/3} \int \cos ^{m+\frac {2}{3}}(c+d x)dx}{\cos ^{\frac {2}{3}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(b \cos (c+d x))^{2/3} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {2}{3}}dx}{\cos ^{\frac {2}{3}}(c+d x)}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle -\frac {3 \sin (c+d x) (b \cos (c+d x))^{2/3} \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+5),\frac {1}{6} (3 m+11),\cos ^2(c+d x)\right )}{d (3 m+5) \sqrt {\sin ^2(c+d x)}}\) |
Input:
Int[Cos[c + d*x]^m*(b*Cos[c + d*x])^(2/3),x]
Output:
(-3*Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(2/3)*Hypergeometric2F1[1/2, (5 + 3*m)/6, (11 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(5 + 3*m)*Sqrt[Si n[c + d*x]^2])
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[b^IntPart [n]*((b*v)^FracPart[n]/(a^IntPart[n]*(a*v)^FracPart[n])) Int[(a*v)^(m + n )*Fx, x], x] /; FreeQ[{a, b, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[m + n]
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]
\[\int \cos \left (d x +c \right )^{m} \left (\cos \left (d x +c \right ) b \right )^{\frac {2}{3}}d x\]
Input:
int(cos(d*x+c)^m*(cos(d*x+c)*b)^(2/3),x)
Output:
int(cos(d*x+c)^m*(cos(d*x+c)*b)^(2/3),x)
\[ \int \cos ^m(c+d x) (b \cos (c+d x))^{2/3} \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}} \cos \left (d x + c\right )^{m} \,d x } \] Input:
integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(2/3),x, algorithm="fricas")
Output:
integral((b*cos(d*x + c))^(2/3)*cos(d*x + c)^m, x)
\[ \int \cos ^m(c+d x) (b \cos (c+d x))^{2/3} \, dx=\int \left (b \cos {\left (c + d x \right )}\right )^{\frac {2}{3}} \cos ^{m}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)**m*(b*cos(d*x+c))**(2/3),x)
Output:
Integral((b*cos(c + d*x))**(2/3)*cos(c + d*x)**m, x)
\[ \int \cos ^m(c+d x) (b \cos (c+d x))^{2/3} \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}} \cos \left (d x + c\right )^{m} \,d x } \] Input:
integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(2/3),x, algorithm="maxima")
Output:
integrate((b*cos(d*x + c))^(2/3)*cos(d*x + c)^m, x)
\[ \int \cos ^m(c+d x) (b \cos (c+d x))^{2/3} \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}} \cos \left (d x + c\right )^{m} \,d x } \] Input:
integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(2/3),x, algorithm="giac")
Output:
integrate((b*cos(d*x + c))^(2/3)*cos(d*x + c)^m, x)
Timed out. \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{2/3} \, dx=\int {\cos \left (c+d\,x\right )}^m\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{2/3} \,d x \] Input:
int(cos(c + d*x)^m*(b*cos(c + d*x))^(2/3),x)
Output:
int(cos(c + d*x)^m*(b*cos(c + d*x))^(2/3), x)
\[ \int \cos ^m(c+d x) (b \cos (c+d x))^{2/3} \, dx=b^{\frac {2}{3}} \left (\int \cos \left (d x +c \right )^{m +\frac {2}{3}}d x \right ) \] Input:
int(cos(d*x+c)^m*(b*cos(d*x+c))^(2/3),x)
Output:
b**(2/3)*int(cos(c + d*x)**((3*m + 2)/3),x)