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

Optimal. Leaf size=154 \[ -\frac{3 (11 A+8 C) \sin (c+d x) (b \cos (c+d x))^{8/3} \, _2F_1\left (\frac{1}{2},\frac{4}{3};\frac{7}{3};\cos ^2(c+d x)\right )}{88 b^4 d \sqrt{\sin ^2(c+d x)}}-\frac{3 B \sin (c+d x) (b \cos (c+d x))^{11/3} \, _2F_1\left (\frac{1}{2},\frac{11}{6};\frac{17}{6};\cos ^2(c+d x)\right )}{11 b^5 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \sin (c+d x) (b \cos (c+d x))^{8/3}}{11 b^4 d} \]

[Out]

(3*C*(b*Cos[c + d*x])^(8/3)*Sin[c + d*x])/(11*b^4*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^4*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(11/3)*
Hypergeometric2F1[1/2, 11/6, 17/6, Cos[c + d*x]^2]*Sin[c + d*x])/(11*b^5*d*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.142863, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {16, 3023, 2748, 2643} \[ -\frac{3 (11 A+8 C) \sin (c+d x) (b \cos (c+d x))^{8/3} \, _2F_1\left (\frac{1}{2},\frac{4}{3};\frac{7}{3};\cos ^2(c+d x)\right )}{88 b^4 d \sqrt{\sin ^2(c+d x)}}-\frac{3 B \sin (c+d x) (b \cos (c+d x))^{11/3} \, _2F_1\left (\frac{1}{2},\frac{11}{6};\frac{17}{6};\cos ^2(c+d x)\right )}{11 b^5 d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \sin (c+d x) (b \cos (c+d x))^{8/3}}{11 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*(b*Cos[c + d*x])^(8/3)*Sin[c + d*x])/(11*b^4*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^4*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(11/3)*
Hypergeometric2F1[1/2, 11/6, 17/6, Cos[c + d*x]^2]*Sin[c + d*x])/(11*b^5*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 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{\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx &=\frac{\int (b \cos (c+d x))^{5/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{b^3}\\ &=\frac{3 C (b \cos (c+d x))^{8/3} \sin (c+d x)}{11 b^4 d}+\frac{3 \int (b \cos (c+d x))^{5/3} \left (\frac{1}{3} b (11 A+8 C)+\frac{11}{3} b B \cos (c+d x)\right ) \, dx}{11 b^4}\\ &=\frac{3 C (b \cos (c+d x))^{8/3} \sin (c+d x)}{11 b^4 d}+\frac{B \int (b \cos (c+d x))^{8/3} \, dx}{b^4}+\frac{(11 A+8 C) \int (b \cos (c+d x))^{5/3} \, dx}{11 b^3}\\ &=\frac{3 C (b \cos (c+d x))^{8/3} \sin (c+d x)}{11 b^4 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^4 d \sqrt{\sin ^2(c+d x)}}-\frac{3 B (b \cos (c+d x))^{11/3} \, _2F_1\left (\frac{1}{2},\frac{11}{6};\frac{17}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{11 b^5 d \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.335602, size = 114, normalized size = 0.74 \[ -\frac{3 \sin (c+d x) \cos ^4(c+d x) \left ((11 A+8 C) \, _2F_1\left (\frac{1}{2},\frac{4}{3};\frac{7}{3};\cos ^2(c+d x)\right )+8 B \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{11}{6};\frac{17}{6};\cos ^2(c+d x)\right )-8 C \sqrt{\sin ^2(c+d x)}\right )}{88 d \sqrt{\sin ^2(c+d x)} (b \cos (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Cos[c + d*x]^4*Sin[c + d*x]*((11*A + 8*C)*Hypergeometric2F1[1/2, 4/3, 7/3, Cos[c + d*x]^2] + 8*B*Cos[c + d
*x]*Hypergeometric2F1[1/2, 11/6, 17/6, Cos[c + d*x]^2] - 8*C*Sqrt[Sin[c + d*x]^2]))/(88*d*(b*Cos[c + d*x])^(4/
3)*Sqrt[Sin[c + d*x]^2])

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Maple [F]  time = 0.427, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cos \left ( dx+c \right ) \right ) ^{3} \left ( A+B\cos \left ( dx+c \right ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \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)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3),x)

[Out]

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