\(\int \cos ^2(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx\) [887]

Optimal result

Integrand size = 31, antiderivative size = 119 \[ \int \cos ^2(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=-\frac {3 A (b \cos (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 b^3 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{13/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {13}{6},\frac {19}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{13 b^4 d \sqrt {\sin ^2(c+d x)}} \] Output:

-3/10*A*(b*cos(d*x+c))^(10/3)*hypergeom([1/2, 5/3],[8/3],cos(d*x+c)^2)*sin 
(d*x+c)/b^3/d/(sin(d*x+c)^2)^(1/2)-3/13*B*(b*cos(d*x+c))^(13/3)*hypergeom( 
[1/2, 13/6],[19/6],cos(d*x+c)^2)*sin(d*x+c)/b^4/d/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.79 \[ \int \cos ^2(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=-\frac {3 \cos ^2(c+d x) \sqrt [3]{b \cos (c+d x)} \cot (c+d x) \left (13 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right )+10 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {13}{6},\frac {19}{6},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{130 d} \] Input:

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

Output:

(-3*Cos[c + d*x]^2*(b*Cos[c + d*x])^(1/3)*Cot[c + d*x]*(13*A*Hypergeometri 
c2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2] + 10*B*Cos[c + d*x]*Hypergeometric2F1[ 
1/2, 13/6, 19/6, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(130*d)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2030, 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.

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

\(\Big \downarrow \) 2030

\(\displaystyle \frac {\int (b \cos (c+d x))^{7/3} (A+B \cos (c+d x))dx}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{b^2}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {A \int (b \cos (c+d x))^{7/3}dx+\frac {B \int (b \cos (c+d x))^{10/3}dx}{b}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3}dx+\frac {B \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{10/3}dx}{b}}{b^2}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {-\frac {3 A \sin (c+d x) (b \cos (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right )}{10 b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{13/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {13}{6},\frac {19}{6},\cos ^2(c+d x)\right )}{13 b^2 d \sqrt {\sin ^2(c+d x)}}}{b^2}\)

Input:

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

Output:

((-3*A*(b*Cos[c + d*x])^(10/3)*Hypergeometric2F1[1/2, 5/3, 8/3, Cos[c + d* 
x]^2]*Sin[c + d*x])/(10*b*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^ 
(13/3)*Hypergeometric2F1[1/2, 13/6, 19/6, Cos[c + d*x]^2]*Sin[c + d*x])/(1 
3*b^2*d*Sqrt[Sin[c + d*x]^2]))/b^2
 

Defintions of rubi rules used

rule 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \cos ^2(c+d x) \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=b^{\frac {1}{3}} \left (\left (\int \cos \left (d x +c \right )^{\frac {10}{3}}d x \right ) b +\left (\int \cos \left (d x +c \right )^{\frac {7}{3}}d x \right ) a \right ) \] Input:

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

Output:

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