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

Optimal. Leaf size=119 \[ -\frac{3 A \sin (c+d x) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )}{5 b^3 d \sqrt{\sin ^2(c+d x)}}-\frac{3 B \sin (c+d x) (b \cos (c+d x))^{8/3} \, _2F_1\left (\frac{1}{2},\frac{4}{3};\frac{7}{3};\cos ^2(c+d x)\right )}{8 b^4 d \sqrt{\sin ^2(c+d x)}} \]

[Out]

(-3*A*(b*Cos[c + d*x])^(5/3)*Hypergeometric2F1[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]) - (3*B*(b*Cos[c + d*x])^(8/3)*Hypergeometric2F1[1/2, 4/3, 7/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*
b^4*d*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.0734252, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {16, 2748, 2643} \[ -\frac{3 A \sin (c+d x) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )}{5 b^3 d \sqrt{\sin ^2(c+d x)}}-\frac{3 B \sin (c+d x) (b \cos (c+d x))^{8/3} \, _2F_1\left (\frac{1}{2},\frac{4}{3};\frac{7}{3};\cos ^2(c+d x)\right )}{8 b^4 d \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*A*(b*Cos[c + d*x])^(5/3)*Hypergeometric2F1[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]) - (3*B*(b*Cos[c + d*x])^(8/3)*Hypergeometric2F1[1/2, 4/3, 7/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*
b^4*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 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{\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx &=\frac{\int (b \cos (c+d x))^{2/3} (A+B \cos (c+d x)) \, dx}{b^2}\\ &=\frac{A \int (b \cos (c+d x))^{2/3} \, dx}{b^2}+\frac{B \int (b \cos (c+d x))^{5/3} \, dx}{b^3}\\ &=-\frac{3 A (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)}}-\frac{3 B (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)}{8 b^4 d \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.166257, size = 94, normalized size = 0.79 \[ -\frac{3 \sqrt{\sin ^2(c+d x)} \cos ^2(c+d x) \cot (c+d x) \left (8 A \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )+5 B \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{4}{3};\frac{7}{3};\cos ^2(c+d x)\right )\right )}{40 d (b \cos (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Cos[c + d*x]^2*Cot[c + d*x]*(8*A*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2] + 5*B*Cos[c + d*x]*Hype
rgeometric2F1[1/2, 4/3, 7/3, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(40*d*(b*Cos[c + d*x])^(4/3))

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Maple [F]  time = 0.327, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ( A+B\cos \left ( dx+c \right ) \right ) \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(cos(d*x+c)^2*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(4/3),x)

[Out]

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

[Out]

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^2/(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 (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{2}{3}}}{b^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

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

[Out]

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