Integrand size = 40, antiderivative size = 167 \[ \int \frac {\cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=-\frac {3 B \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 C \cos ^{3+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (8+3 m),\frac {1}{6} (14+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (8+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}} \]
-3*B*cos(d*x+c)^(2+m)*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)/(b*cos(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)-3*C*cos(d *x+c)^(3+m)*hypergeom([1/2, 4/3+1/2*m],[7/3+1/2*m],cos(d*x+c)^2)*sin(d*x+c )/d/(8+3*m)/(b*cos(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)
Time = 0.29 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=-\frac {3 \cos ^{2+m}(c+d x) \csc (c+d x) \left (B (8+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\cos ^2(c+d x)\right )+C (5+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (8+3 m),\frac {7}{3}+\frac {m}{2},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (5+3 m) (8+3 m) \sqrt [3]{b \cos (c+d x)}} \]
(-3*Cos[c + d*x]^(2 + m)*Csc[c + d*x]*(B*(8 + 3*m)*Hypergeometric2F1[1/2, (5 + 3*m)/6, (11 + 3*m)/6, Cos[c + d*x]^2] + C*(5 + 3*m)*Cos[c + d*x]*Hype rgeometric2F1[1/2, (8 + 3*m)/6, 7/3 + m/2, Cos[c + d*x]^2])*Sqrt[Sin[c + d *x]^2])/(d*(5 + 3*m)*(8 + 3*m)*(b*Cos[c + d*x])^(1/3))
Time = 0.51 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {2034, 3042, 3489, 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.
\(\displaystyle \int \frac {\cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \int \cos ^{m-\frac {1}{3}}(c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)\right )dx}{\sqrt [3]{b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {1}{3}} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{\sqrt [3]{b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3489 |
\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \int \cos ^{m+\frac {2}{3}}(c+d x) (B+C \cos (c+d x))dx}{\sqrt [3]{b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {2}{3}} \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{\sqrt [3]{b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (B \int \cos ^{m+\frac {2}{3}}(c+d x)dx+C \int \cos ^{m+\frac {5}{3}}(c+d x)dx\right )}{\sqrt [3]{b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (B \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {2}{3}}dx+C \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {5}{3}}dx\right )}{\sqrt [3]{b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (-\frac {3 B \sin (c+d x) \cos ^{m+\frac {5}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+5),\frac {1}{6} (3 m+11),\cos ^2(c+d x)\right )}{d (3 m+5) \sqrt {\sin ^2(c+d x)}}-\frac {3 C \sin (c+d x) \cos ^{m+\frac {8}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+8),\frac {1}{6} (3 m+14),\cos ^2(c+d x)\right )}{d (3 m+8) \sqrt {\sin ^2(c+d x)}}\right )}{\sqrt [3]{b \cos (c+d x)}}\) |
(Cos[c + d*x]^(1/3)*((-3*B*Cos[c + d*x]^(5/3 + m)*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[ Sin[c + d*x]^2]) - (3*C*Cos[c + d*x]^(8/3 + m)*Hypergeometric2F1[1/2, (8 + 3*m)/6, (14 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(8 + 3*m)*Sqrt[Sin [c + d*x]^2])))/(b*Cos[c + d*x])^(1/3)
3.3.13.3.1 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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b Int[(b*Sin[e + f* x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x]
\[\int \frac {\left (\cos ^{m}\left (d x +c \right )\right ) \left (B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\left (\cos \left (d x +c \right ) b \right )^{\frac {1}{3}}}d x\]
\[ \int \frac {\cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \cos \left (d x + c\right )^{m}}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]
integrate(cos(d*x+c)^m*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/3), x, algorithm="fricas")
\[ \int \frac {\cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int \frac {\left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \cos ^{m}{\left (c + d x \right )}}{\sqrt [3]{b \cos {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \cos \left (d x + c\right )^{m}}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]
integrate(cos(d*x+c)^m*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/3), x, algorithm="maxima")
\[ \int \frac {\cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \cos \left (d x + c\right )^{m}}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]
integrate(cos(d*x+c)^m*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/3), x, algorithm="giac")
Timed out. \[ \int \frac {\cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^m\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{1/3}} \,d x \]