\(\int \frac {A+C \cos ^2(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx\) [160]

Optimal result

Integrand size = 25, antiderivative size = 95 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx=\frac {3 C (b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b d}-\frac {3 (5 A+2 C) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 b d \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {3093, 2722} \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx=\frac {3 C \sin (c+d x) (b \cos (c+d x))^{2/3}}{5 b d}-\frac {3 (5 A+2 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{10 b d \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

(3*C*(b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(5*b*d) - (3*(5*A + 2*C)*(b*Cos[c + d*x])^(2/3)*Hypergeometric2F1[1/
3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(10*b*d*Sqrt[Sin[c + d*x]^2])

Rule 2722

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]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3093

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] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {3 C (b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b d}+\frac {1}{5} (5 A+2 C) \int \frac {1}{\sqrt [3]{b \cos (c+d x)}} \, dx \\ & = \frac {3 C (b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b d}-\frac {3 (5 A+2 C) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 b d \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.92 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx=-\frac {3 \cot (c+d x) \left (4 A \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )+C \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{8 d \sqrt [3]{b \cos (c+d x)}} \]

[In]

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

[Out]

(-3*Cot[c + d*x]*(4*A*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2] + C*Cos[c + d*x]^2*Hypergeometric2F1[1/
2, 4/3, 7/3, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(8*d*(b*Cos[c + d*x])^(1/3))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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