3.2.41 \(\int \cos (c+d x) \sqrt [3]{b \cos (c+d x)} (A+C \cos ^2(c+d x)) \, dx\) [141]

3.2.41.1 Optimal result
3.2.41.2 Mathematica [A] (verified)
3.2.41.3 Rubi [A] (verified)
3.2.41.4 Maple [F]
3.2.41.5 Fricas [F]
3.2.41.6 Sympy [F(-1)]
3.2.41.7 Maxima [F]
3.2.41.8 Giac [F]
3.2.41.9 Mupad [F(-1)]

3.2.41.1 Optimal result

Integrand size = 31, antiderivative size = 95 \[ \int \cos (c+d x) \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C (b \cos (c+d x))^{7/3} \sin (c+d x)}{10 b^2 d}-\frac {3 (10 A+7 C) (b \cos (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{70 b^2 d \sqrt {\sin ^2(c+d x)}} \]

output
3/10*C*(b*cos(d*x+c))^(7/3)*sin(d*x+c)/b^2/d-3/70*(10*A+7*C)*(b*cos(d*x+c) 
)^(7/3)*hypergeom([1/2, 7/6],[13/6],cos(d*x+c)^2)*sin(d*x+c)/b^2/d/(sin(d* 
x+c)^2)^(1/2)
 
3.2.41.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.96 \[ \int \cos (c+d x) \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 (b \cos (c+d x))^{4/3} \cot (c+d x) \left (13 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )+7 C \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {13}{6},\frac {19}{6},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{91 b d} \]

input
Integrate[Cos[c + d*x]*(b*Cos[c + d*x])^(1/3)*(A + C*Cos[c + d*x]^2),x]
 
output
(-3*(b*Cos[c + d*x])^(4/3)*Cot[c + d*x]*(13*A*Hypergeometric2F1[1/2, 7/6, 
13/6, Cos[c + d*x]^2] + 7*C*Cos[c + d*x]^2*Hypergeometric2F1[1/2, 13/6, 19 
/6, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(91*b*d)
 
3.2.41.3 Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2030, 3042, 3493, 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.

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

\(\Big \downarrow \) 2030

\(\displaystyle \frac {\int (b \cos (c+d x))^{4/3} \left (C \cos ^2(c+d x)+A\right )dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx}{b}\)

\(\Big \downarrow \) 3493

\(\displaystyle \frac {\frac {1}{10} (10 A+7 C) \int (b \cos (c+d x))^{4/3}dx+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{7/3}}{10 b d}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{10} (10 A+7 C) \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}dx+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{7/3}}{10 b d}}{b}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\frac {3 C \sin (c+d x) (b \cos (c+d x))^{7/3}}{10 b d}-\frac {3 (10 A+7 C) \sin (c+d x) (b \cos (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )}{70 b d \sqrt {\sin ^2(c+d x)}}}{b}\)

input
Int[Cos[c + d*x]*(b*Cos[c + d*x])^(1/3)*(A + C*Cos[c + d*x]^2),x]
 
output
((3*C*(b*Cos[c + d*x])^(7/3)*Sin[c + d*x])/(10*b*d) - (3*(10*A + 7*C)*(b*C 
os[c + d*x])^(7/3)*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c 
 + d*x])/(70*b*d*Sqrt[Sin[c + d*x]^2]))/b
 

3.2.41.3.1 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 3493
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*( 
x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f 
*(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2)   Int[(b*Sin[e + f*x] 
)^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 
3.2.41.4 Maple [F]

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

input
int(cos(d*x+c)*(cos(d*x+c)*b)^(1/3)*(A+C*cos(d*x+c)^2),x)
 
output
int(cos(d*x+c)*(cos(d*x+c)*b)^(1/3)*(A+C*cos(d*x+c)^2),x)
 
3.2.41.5 Fricas [F]

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

input
integrate(cos(d*x+c)*(b*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x, algorithm= 
"fricas")
 
output
integral((C*cos(d*x + c)^3 + A*cos(d*x + c))*(b*cos(d*x + c))^(1/3), x)
 
3.2.41.6 Sympy [F(-1)]

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

input
integrate(cos(d*x+c)*(b*cos(d*x+c))**(1/3)*(A+C*cos(d*x+c)**2),x)
 
output
Timed out
 
3.2.41.7 Maxima [F]

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

input
integrate(cos(d*x+c)*(b*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x, algorithm= 
"maxima")
 
output
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(1/3)*cos(d*x + c), x)
 
3.2.41.8 Giac [F]

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

input
integrate(cos(d*x+c)*(b*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x, algorithm= 
"giac")
 
output
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(1/3)*cos(d*x + c), x)
 
3.2.41.9 Mupad [F(-1)]

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

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