Integrand size = 31, antiderivative size = 167 \[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=-\frac {3 A \cos ^{1+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4+3 m),\frac {1}{6} (10+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (4+3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 B \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (7+3 m) \sqrt {\sin ^2(c+d x)}} \]
-3*A*cos(d*x+c)^(1+m)*(b*cos(d*x+c))^(1/3)*hypergeom([1/2, 2/3+1/2*m],[5/3 +1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(4+3*m)/(sin(d*x+c)^2)^(1/2)-3*B*cos(d* x+c)^(2+m)*(b*cos(d*x+c))^(1/3)*hypergeom([1/2, 7/6+1/2*m],[13/6+1/2*m],co s(d*x+c)^2)*sin(d*x+c)/d/(7+3*m)/(sin(d*x+c)^2)^(1/2)
Time = 0.20 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.84 \[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=-\frac {3 \cos ^{1+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \csc (c+d x) \left (A (7+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4+3 m),\frac {5}{3}+\frac {m}{2},\cos ^2(c+d x)\right )+B (4+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (4+3 m) (7+3 m)} \]
(-3*Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(1/3)*Csc[c + d*x]*(A*(7 + 3*m)* Hypergeometric2F1[1/2, (4 + 3*m)/6, 5/3 + m/2, Cos[c + d*x]^2] + B*(4 + 3* m)*Cos[c + d*x]*Hypergeometric2F1[1/2, (7 + 3*m)/6, (13 + 3*m)/6, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(4 + 3*m)*(7 + 3*m))
Time = 0.42 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2034, 3042, 3227, 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.
\(\displaystyle \int \sqrt [3]{b \cos (c+d x)} \cos ^m(c+d x) (A+B \cos (c+d x)) \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \frac {\sqrt [3]{b \cos (c+d x)} \int \cos ^{m+\frac {1}{3}}(c+d x) (A+B \cos (c+d x))dx}{\sqrt [3]{\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt [3]{b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {1}{3}} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{\sqrt [3]{\cos (c+d x)}}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\sqrt [3]{b \cos (c+d x)} \left (A \int \cos ^{m+\frac {1}{3}}(c+d x)dx+B \int \cos ^{m+\frac {4}{3}}(c+d x)dx\right )}{\sqrt [3]{\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt [3]{b \cos (c+d x)} \left (A \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {1}{3}}dx+B \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {4}{3}}dx\right )}{\sqrt [3]{\cos (c+d x)}}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\sqrt [3]{b \cos (c+d x)} \left (-\frac {3 A \sin (c+d x) \cos ^{m+\frac {4}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+4),\frac {1}{6} (3 m+10),\cos ^2(c+d x)\right )}{d (3 m+4) \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) \cos ^{m+\frac {7}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+7),\frac {1}{6} (3 m+13),\cos ^2(c+d x)\right )}{d (3 m+7) \sqrt {\sin ^2(c+d x)}}\right )}{\sqrt [3]{\cos (c+d x)}}\) |
((b*Cos[c + d*x])^(1/3)*((-3*A*Cos[c + d*x]^(4/3 + m)*Hypergeometric2F1[1/ 2, (4 + 3*m)/6, (10 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(4 + 3*m)*S qrt[Sin[c + d*x]^2]) - (3*B*Cos[c + d*x]^(7/3 + m)*Hypergeometric2F1[1/2, (7 + 3*m)/6, (13 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(7 + 3*m)*Sqrt [Sin[c + d*x]^2])))/Cos[c + d*x]^(1/3)
3.10.29.3.1 Defintions of rubi rules used
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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
\[\int \left (\cos ^{m}\left (d x +c \right )\right ) \left (\cos \left (d x +c \right ) b \right )^{\frac {1}{3}} \left (A +B \cos \left (d x +c \right )\right )d x\]
\[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right )^{m} \,d x } \]
\[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int \sqrt [3]{b \cos {\left (c + d x \right )}} \left (A + B \cos {\left (c + d x \right )}\right ) \cos ^{m}{\left (c + d x \right )}\, dx \]
\[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right )^{m} \,d x } \]
\[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right )^{m} \,d x } \]
Timed out. \[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^m\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{1/3}\,\left (A+B\,\cos \left (c+d\,x\right )\right ) \,d x \]