3.2.58 \(\int \frac {\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{\sqrt [3]{b \cos (c+d x)}} \, dx\) [158]

Optimal. Leaf size=95 \[ \frac {3 C (b \cos (c+d x))^{8/3} \sin (c+d x)}{11 b^3 d}-\frac {3 (11 A+8 C) (b \cos (c+d x))^{8/3} \, _2F_1\left (\frac {1}{2},\frac {4}{3};\frac {7}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{88 b^3 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

3/11*C*(b*cos(d*x+c))^(8/3)*sin(d*x+c)/b^3/d-3/88*(11*A+8*C)*(b*cos(d*x+c))^(8/3)*hypergeom([1/2, 4/3],[7/3],c
os(d*x+c)^2)*sin(d*x+c)/b^3/d/(sin(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {16, 3093, 2722} \begin {gather*} \frac {3 C \sin (c+d x) (b \cos (c+d x))^{8/3}}{11 b^3 d}-\frac {3 (11 A+8 C) \sin (c+d x) (b \cos (c+d x))^{8/3} \, _2F_1\left (\frac {1}{2},\frac {4}{3};\frac {7}{3};\cos ^2(c+d x)\right )}{88 b^3 d \sqrt {\sin ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*(b*Cos[c + d*x])^(8/3)*Sin[c + d*x])/(11*b^3*d) - (3*(11*A + 8*C)*(b*Cos[c + d*x])^(8/3)*Hypergeometric2F
1[1/2, 4/3, 7/3, Cos[c + d*x]^2]*Sin[c + d*x])/(88*b^3*d*Sqrt[Sin[c + d*x]^2])

Rule 16

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

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*} \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx &=\frac {\int (b \cos (c+d x))^{5/3} \left (A+C \cos ^2(c+d x)\right ) \, dx}{b^2}\\ &=\frac {3 C (b \cos (c+d x))^{8/3} \sin (c+d x)}{11 b^3 d}+\frac {(11 A+8 C) \int (b \cos (c+d x))^{5/3} \, dx}{11 b^2}\\ &=\frac {3 C (b \cos (c+d x))^{8/3} \sin (c+d x)}{11 b^3 d}-\frac {3 (11 A+8 C) (b \cos (c+d x))^{8/3} \, _2F_1\left (\frac {1}{2},\frac {4}{3};\frac {7}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{88 b^3 d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 96, normalized size = 1.01 \begin {gather*} -\frac {3 \cot (c+d x) \left (7 A \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {4}{3};\frac {7}{3};\cos ^2(c+d x)\right )+4 C \cos ^4(c+d x) \, _2F_1\left (\frac {1}{2},\frac {7}{3};\frac {10}{3};\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{56 d \sqrt [3]{b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Cot[c + d*x]*(7*A*Cos[c + d*x]^2*Hypergeometric2F1[1/2, 4/3, 7/3, Cos[c + d*x]^2] + 4*C*Cos[c + d*x]^4*Hyp
ergeometric2F1[1/2, 7/3, 10/3, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(56*d*(b*Cos[c + d*x])^(1/3))

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Maple [F]
time = 0.23, size = 0, normalized size = 0.00 \[\int \frac {\left (\cos ^{2}\left (d x +c \right )\right ) \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3065 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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