Integrand size = 31, antiderivative size = 167 \[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{\sqrt [3]{b \cos (c+d x)}} \, dx=-\frac {3 A \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (2+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}} \] Output:
-3*A*cos(d*x+c)^(1+m)*hypergeom([1/2, 1/3+1/2*m],[4/3+1/2*m],cos(d*x+c)^2) *sin(d*x+c)/d/(2+3*m)/(b*cos(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)-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)
Time = 0.23 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{\sqrt [3]{b \cos (c+d x)}} \, dx=-\frac {3 \cos ^{1+m}(c+d x) \csc (c+d x) \left (A (5+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\cos ^2(c+d x)\right )+B (2+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (2+3 m) (5+3 m) \sqrt [3]{b \cos (c+d x)}} \] Input:
Integrate[(Cos[c + d*x]^m*(A + B*Cos[c + d*x]))/(b*Cos[c + d*x])^(1/3),x]
Output:
(-3*Cos[c + d*x]^(1 + m)*Csc[c + d*x]*(A*(5 + 3*m)*Hypergeometric2F1[1/2, (2 + 3*m)/6, (8 + 3*m)/6, Cos[c + d*x]^2] + B*(2 + 3*m)*Cos[c + d*x]*Hyper geometric2F1[1/2, (5 + 3*m)/6, (11 + 3*m)/6, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(2 + 3*m)*(5 + 3*m)*(b*Cos[c + d*x])^(1/3))
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.
\(\displaystyle \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{\sqrt [3]{b \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 2034 |
\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \int \cos ^{m-\frac {1}{3}}(c+d x) (A+B \cos (c+d x))dx}{\sqrt [3]{b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {1}{3}} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{\sqrt [3]{b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (A \int \cos ^{m-\frac {1}{3}}(c+d x)dx+B \int \cos ^{m+\frac {2}{3}}(c+d x)dx\right )}{\sqrt [3]{b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (A \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {1}{3}}dx+B \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {2}{3}}dx\right )}{\sqrt [3]{b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (-\frac {3 A \sin (c+d x) \cos ^{m+\frac {2}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+2),\frac {1}{6} (3 m+8),\cos ^2(c+d x)\right )}{d (3 m+2) \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) \cos ^{m+\frac {5}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+5),\frac {1}{6} (3 m+11),\cos ^2(c+d x)\right )}{d (3 m+5) \sqrt {\sin ^2(c+d x)}}\right )}{\sqrt [3]{b \cos (c+d x)}}\) |
Input:
Int[(Cos[c + d*x]^m*(A + B*Cos[c + d*x]))/(b*Cos[c + d*x])^(1/3),x]
Output:
(Cos[c + d*x]^(1/3)*((-3*A*Cos[c + d*x]^(2/3 + m)*Hypergeometric2F1[1/2, ( 2 + 3*m)/6, (8 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(2 + 3*m)*Sqrt[S in[c + d*x]^2]) - (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])))/(b*Cos[c + d*x])^(1/3)
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]
\[\int \frac {\cos \left (d x +c \right )^{m} \left (A +B \cos \left (d x +c \right )\right )}{\left (\cos \left (d x +c \right ) b \right )^{\frac {1}{3}}}d x\]
Input:
int(cos(d*x+c)^m*(A+B*cos(d*x+c))/(cos(d*x+c)*b)^(1/3),x)
Output:
int(cos(d*x+c)^m*(A+B*cos(d*x+c))/(cos(d*x+c)*b)^(1/3),x)
\[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{m}}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(cos(d*x+c)^m*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(1/3),x, algorithm= "fricas")
Output:
integral((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(2/3)*cos(d*x + c)^m/(b*cos (d*x + c)), x)
\[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \cos ^{m}{\left (c + d x \right )}}{\sqrt [3]{b \cos {\left (c + d x \right )}}}\, dx \] Input:
integrate(cos(d*x+c)**m*(A+B*cos(d*x+c))/(b*cos(d*x+c))**(1/3),x)
Output:
Integral((A + B*cos(c + d*x))*cos(c + d*x)**m/(b*cos(c + d*x))**(1/3), x)
\[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{m}}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(cos(d*x+c)^m*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(1/3),x, algorithm= "maxima")
Output:
integrate((B*cos(d*x + c) + A)*cos(d*x + c)^m/(b*cos(d*x + c))^(1/3), x)
\[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{m}}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(cos(d*x+c)^m*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(1/3),x, algorithm= "giac")
Output:
integrate((B*cos(d*x + c) + A)*cos(d*x + c)^m/(b*cos(d*x + c))^(1/3), x)
Timed out. \[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^m\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{1/3}} \,d x \] Input:
int((cos(c + d*x)^m*(A + B*cos(c + d*x)))/(b*cos(c + d*x))^(1/3),x)
Output:
int((cos(c + d*x)^m*(A + B*cos(c + d*x)))/(b*cos(c + d*x))^(1/3), x)
\[ \int \frac {\cos ^m(c+d x) (A+B \cos (c+d x))}{\sqrt [3]{b \cos (c+d x)}} \, dx=\frac {\left (\int \frac {\cos \left (d x +c \right )^{m}}{\cos \left (d x +c \right )^{\frac {1}{3}}}d x \right ) a +\left (\int \cos \left (d x +c \right )^{m} \cos \left (d x +c \right )^{\frac {2}{3}}d x \right ) b}{b^{\frac {1}{3}}} \] Input:
int(cos(d*x+c)^m*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(1/3),x)
Output:
(int(cos(c + d*x)**m/cos(c + d*x)**(1/3),x)*a + int((cos(c + d*x)**m*cos(c + d*x))/cos(c + d*x)**(1/3),x)*b)/b**(1/3)