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

Optimal. Leaf size=154 \[ -\frac{3 (5 A+2 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )}{10 b d \sqrt{\sin ^2(c+d x)}}-\frac{3 B \sin (c+d x) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )}{5 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \sin (c+d x) (b \cos (c+d x))^{2/3}}{5 b d} \]

[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]) - (3*B*(b*Cos[c + d*x])^(5/3)*Hyperge
ometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*b^2*d*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.122842, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3023, 2748, 2643} \[ -\frac{3 (5 A+2 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )}{10 b d \sqrt{\sin ^2(c+d x)}}-\frac{3 B \sin (c+d x) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )}{5 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \sin (c+d x) (b \cos (c+d x))^{2/3}}{5 b d} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x] + 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]) - (3*B*(b*Cos[c + d*x])^(5/3)*Hyperge
ometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*b^2*d*Sqrt[Sin[c + d*x]^2])

Rule 3023

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin{align*} \int \frac{A+B \cos (c+d x)+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 \int \frac{\frac{1}{3} b (5 A+2 C)+\frac{5}{3} b B \cos (c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx}{5 b}\\ &=\frac{3 C (b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b d}+\frac{B \int (b \cos (c+d x))^{2/3} \, dx}{b}+\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} \, _2F_1\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)}}-\frac{3 B (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{5 b^2 d \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.141897, size = 108, normalized size = 0.7 \[ -\frac{3 \sin (2 (c+d x)) \left ((5 A+2 C) \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )+2 B \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )-2 C \sqrt{\sin ^2(c+d x)}\right )}{20 d \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*((5*A + 2*C)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2] + 2*B*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/
6, 11/6, Cos[c + d*x]^2] - 2*C*Sqrt[Sin[c + d*x]^2])*Sin[2*(c + d*x)])/(20*d*(b*Cos[c + d*x])^(1/3)*Sqrt[Sin[c
+ d*x]^2])

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Maple [F]  time = 0.246, size = 0, normalized size = 0. \begin{align*} \int{(A+B\cos \left ( dx+c \right ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2}){\frac{1}{\sqrt [3]{b\cos \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\left (b \cos \left (d x + c\right )\right )^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{2}{3}}}{b \cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\left (b \cos \left (d x + c\right )\right )^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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