3.360 \(\int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\)

Optimal. Leaf size=152 \[ \frac{3 (2 A-C) \sin (c+d x) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )}{5 b^3 d \sqrt{\sin ^2(c+d x)}}+\frac{3 A \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)}}-\frac{3 B \sin (c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )}{2 b^2 d \sqrt{\sin ^2(c+d x)}} \]

[Out]

(3*A*Sin[c + d*x])/(b*d*(b*Cos[c + d*x])^(1/3)) - (3*B*(b*Cos[c + d*x])^(2/3)*Hypergeometric2F1[1/3, 1/2, 4/3,
Cos[c + d*x]^2]*Sin[c + d*x])/(2*b^2*d*Sqrt[Sin[c + d*x]^2]) + (3*(2*A - C)*(b*Cos[c + d*x])^(5/3)*Hypergeome
tric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*b^3*d*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.139211, antiderivative size = 152, 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 = {3021, 2748, 2643} \[ \frac{3 (2 A-C) \sin (c+d x) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )}{5 b^3 d \sqrt{\sin ^2(c+d x)}}+\frac{3 A \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)}}-\frac{3 B \sin (c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )}{2 b^2 d \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(b*Cos[c + d*x])^(4/3),x]

[Out]

(3*A*Sin[c + d*x])/(b*d*(b*Cos[c + d*x])^(1/3)) - (3*B*(b*Cos[c + d*x])^(2/3)*Hypergeometric2F1[1/3, 1/2, 4/3,
Cos[c + d*x]^2]*Sin[c + d*x])/(2*b^2*d*Sqrt[Sin[c + d*x]^2]) + (3*(2*A - C)*(b*Cos[c + d*x])^(5/3)*Hypergeome
tric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*b^3*d*Sqrt[Sin[c + d*x]^2])

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

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)}{(b \cos (c+d x))^{4/3}} \, dx &=\frac{3 A \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)}}+\frac{3 \int \frac{\frac{b^2 B}{3}-\frac{1}{3} b^2 (2 A-C) \cos (c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx}{b^3}\\ &=\frac{3 A \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)}}+\frac{B \int \frac{1}{\sqrt [3]{b \cos (c+d x)}} \, dx}{b}-\frac{(2 A-C) \int (b \cos (c+d x))^{2/3} \, dx}{b^2}\\ &=\frac{3 A \sin (c+d x)}{b d \sqrt [3]{b \cos (c+d x)}}-\frac{3 B (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)}{2 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 (2 A-C) (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^3 d \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.224296, size = 115, normalized size = 0.76 \[ -\frac{3 \sqrt{\sin ^2(c+d x)} \cot (c+d x) \left (\cos (c+d x) \left (5 B \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )+2 C \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )\right )-10 A \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right )\right )}{10 d (b \cos (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(b*Cos[c + d*x])^(4/3),x]

[Out]

(-3*Cot[c + d*x]*(-10*A*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2] + Cos[c + d*x]*(5*B*Hypergeometric2F
1[1/3, 1/2, 4/3, Cos[c + d*x]^2] + 2*C*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]))*Sqrt[S
in[c + d*x]^2])/(10*d*(b*Cos[c + d*x])^(4/3))

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Maple [F]  time = 0.249, size = 0, normalized size = 0. \begin{align*} \int{(A+B\cos \left ( dx+c \right ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2}) \left ( b\cos \left ( dx+c \right ) \right ) ^{-{\frac{4}{3}}}}\, 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))^(4/3),x)

[Out]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3),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{4}{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))^(4/3),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/(b*cos(d*x + c))^(4/3), 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^{2} \cos \left (d x + c\right )^{2}}, 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))^(4/3),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^2*cos(d*x + c)^2), 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))**(4/3),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{4}{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))^(4/3),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/(b*cos(d*x + c))^(4/3), x)