\(\int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [210]

Optimal result

Integrand size = 40, antiderivative size = 167 \[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 B \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (7+3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 C \cos ^{3+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (10+3 m),\frac {1}{6} (16+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (10+3 m) \sqrt {\sin ^2(c+d x)}} \] Output:

-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],cos(d*x+c)^2)*sin(d*x+c)/d/(7+3*m)/(sin(d*x+c)^2)^(1/2)-3*C*cos(d 
*x+c)^(3+m)*(b*cos(d*x+c))^(1/3)*hypergeom([1/2, 5/3+1/2*m],[8/3+1/2*m],co 
s(d*x+c)^2)*sin(d*x+c)/d/(10+3*m)/(sin(d*x+c)^2)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.84 \[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \csc (c+d x) \left (C (7+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3}+\frac {m}{2},\frac {8}{3}+\frac {m}{2},\cos ^2(c+d x)\right )+B (10+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (7+3 m) (10+3 m)} \] Input:

Integrate[Cos[c + d*x]^m*(b*Cos[c + d*x])^(1/3)*(B*Cos[c + d*x] + C*Cos[c 
+ d*x]^2),x]
 

Output:

(-3*Cos[c + d*x]^(2 + m)*(b*Cos[c + d*x])^(1/3)*Csc[c + d*x]*(C*(7 + 3*m)* 
Cos[c + d*x]*Hypergeometric2F1[1/2, 5/3 + m/2, 8/3 + m/2, Cos[c + d*x]^2] 
+ B*(10 + 3*m)*Hypergeometric2F1[1/2, (7 + 3*m)/6, (13 + 3*m)/6, Cos[c + d 
*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(7 + 3*m)*(10 + 3*m))
 

Rubi [A] (verified)

Time = 0.54 (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.

\(\displaystyle \int \sqrt [3]{b \cos (c+d x)} \cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 2034

\(\displaystyle \frac {\sqrt [3]{b \cos (c+d x)} \int \cos ^{m+\frac {1}{3}}(c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)\right )dx}{\sqrt [3]{\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt [3]{b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {1}{3}} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{\sqrt [3]{\cos (c+d x)}}\)

\(\Big \downarrow \) 3489

\(\displaystyle \frac {\sqrt [3]{b \cos (c+d x)} \int \cos ^{m+\frac {4}{3}}(c+d x) (B+C \cos (c+d x))dx}{\sqrt [3]{\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt [3]{b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {4}{3}} \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{\sqrt [3]{\cos (c+d x)}}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\sqrt [3]{b \cos (c+d x)} \left (B \int \cos ^{m+\frac {4}{3}}(c+d x)dx+C \int \cos ^{m+\frac {7}{3}}(c+d x)dx\right )}{\sqrt [3]{\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt [3]{b \cos (c+d x)} \left (B \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {4}{3}}dx+C \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {7}{3}}dx\right )}{\sqrt [3]{\cos (c+d x)}}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\sqrt [3]{b \cos (c+d x)} \left (-\frac {3 B \sin (c+d x) \cos ^{m+\frac {7}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+7),\frac {1}{6} (3 m+13),\cos ^2(c+d x)\right )}{d (3 m+7) \sqrt {\sin ^2(c+d x)}}-\frac {3 C \sin (c+d x) \cos ^{m+\frac {10}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+10),\frac {1}{6} (3 m+16),\cos ^2(c+d x)\right )}{d (3 m+10) \sqrt {\sin ^2(c+d x)}}\right )}{\sqrt [3]{\cos (c+d x)}}\)

Input:

Int[Cos[c + d*x]^m*(b*Cos[c + d*x])^(1/3)*(B*Cos[c + d*x] + C*Cos[c + d*x] 
^2),x]
 

Output:

((b*Cos[c + d*x])^(1/3)*((-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)*S 
qrt[Sin[c + d*x]^2]) - (3*C*Cos[c + d*x]^(10/3 + m)*Hypergeometric2F1[1/2, 
 (10 + 3*m)/6, (16 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(10 + 3*m)*S 
qrt[Sin[c + d*x]^2])))/Cos[c + d*x]^(1/3)
 

Defintions of rubi rules used

rule 2034

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3122

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]
 

rule 3227

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]
 

rule 3489

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]
 

Maple [F]

\[\int \cos \left (d x +c \right )^{m} \left (b \cos \left (d x +c \right )\right )^{\frac {1}{3}} \left (B \cos \left (d x +c \right )+C \cos \left (d x +c \right )^{2}\right )d x\]

Input:

int(cos(d*x+c)^m*(b*cos(d*x+c))^(1/3)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)
 

Output:

int(cos(d*x+c)^m*(b*cos(d*x+c))^(1/3)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)
 

Fricas [F]

\[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right )^{m} \,d x } \] Input:

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(1/3)*(B*cos(d*x+c)+C*cos(d*x+c)^2), 
x, algorithm="fricas")
 

Output:

integral((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c))^(1/3)*cos(d* 
x + c)^m, x)
 

Sympy [F]

\[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int \sqrt [3]{b \cos {\left (c + d x \right )}} \left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \cos ^{m}{\left (c + d x \right )}\, dx \] Input:

integrate(cos(d*x+c)**m*(b*cos(d*x+c))**(1/3)*(B*cos(d*x+c)+C*cos(d*x+c)** 
2),x)
 

Output:

Integral((b*cos(c + d*x))**(1/3)*(B + C*cos(c + d*x))*cos(c + d*x)*cos(c + 
 d*x)**m, x)
 

Maxima [F]

\[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right )^{m} \,d x } \] Input:

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(1/3)*(B*cos(d*x+c)+C*cos(d*x+c)^2), 
x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c))^(1/3)*cos(d 
*x + c)^m, x)
 

Giac [F]

\[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right )^{m} \,d x } \] Input:

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(1/3)*(B*cos(d*x+c)+C*cos(d*x+c)^2), 
x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c))^(1/3)*cos(d 
*x + c)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^m\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{1/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right ) \,d x \] Input:

int(cos(c + d*x)^m*(b*cos(c + d*x))^(1/3)*(B*cos(c + d*x) + C*cos(c + d*x) 
^2),x)
 

Output:

int(cos(c + d*x)^m*(b*cos(c + d*x))^(1/3)*(B*cos(c + d*x) + C*cos(c + d*x) 
^2), x)
 

Reduce [F]

\[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=b^{\frac {1}{3}} \left (\left (\int \cos \left (d x +c \right )^{m +\frac {1}{3}} \cos \left (d x +c \right )d x \right ) b +\left (\int \cos \left (d x +c \right )^{m +\frac {1}{3}} \cos \left (d x +c \right )^{2}d x \right ) c \right ) \] Input:

int(cos(d*x+c)^m*(b*cos(d*x+c))^(1/3)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)
 

Output:

b**(1/3)*(int(cos(c + d*x)**((3*m + 1)/3)*cos(c + d*x),x)*b + int(cos(c + 
d*x)**((3*m + 1)/3)*cos(c + d*x)**2,x)*c)