\(\int \sqrt [3]{b \cos (c+d x)} (A+C \cos ^2(c+d x)) \, dx\) [142]

Optimal result

Integrand size = 25, antiderivative size = 95 \[ \int \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C (b \cos (c+d x))^{4/3} \sin (c+d x)}{7 b d}-\frac {3 (7 A+4 C) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{28 b d \sqrt {\sin ^2(c+d x)}} \] Output:

3/7*C*(b*cos(d*x+c))^(4/3)*sin(d*x+c)/b/d-3/28*(7*A+4*C)*(b*cos(d*x+c))^(4 
/3)*hypergeom([1/2, 2/3],[5/3],cos(d*x+c)^2)*sin(d*x+c)/b/d/(sin(d*x+c)^2) 
^(1/2)
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93 \[ \int \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 \sqrt [3]{b \cos (c+d x)} \cot (c+d x) \left (5 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )+2 C \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{20 d} \] Input:

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

Output:

(-3*(b*Cos[c + d*x])^(1/3)*Cot[c + d*x]*(5*A*Hypergeometric2F1[1/2, 2/3, 5 
/3, Cos[c + d*x]^2] + 2*C*Cos[c + d*x]^2*Hypergeometric2F1[1/2, 5/3, 8/3, 
Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(20*d)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3493

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle \frac {3 C \sin (c+d x) (b \cos (c+d x))^{4/3}}{7 b d}-\frac {3 (7 A+4 C) \sin (c+d x) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{28 b d \sqrt {\sin ^2(c+d x)}}\)

Input:

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

Output:

(3*C*(b*Cos[c + d*x])^(4/3)*Sin[c + d*x])/(7*b*d) - (3*(7*A + 4*C)*(b*Cos[ 
c + d*x])^(4/3)*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d 
*x])/(28*b*d*Sqrt[Sin[c + d*x]^2])
 

Defintions of rubi rules used

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate((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), x)
 

Giac [F]

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

integrate((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), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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