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

Optimal. Leaf size=154 \[ -\frac {3 (10 A+7 C) \sin (c+d x) (b \cos (c+d x))^{7/3} \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right )}{70 b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{10/3} \, _2F_1\left (\frac {1}{2},\frac {5}{3};\frac {8}{3};\cos ^2(c+d x)\right )}{10 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{7/3}}{10 b d} \]

[Out]

3/10*C*(b*cos(d*x+c))^(7/3)*sin(d*x+c)/b/d-3/70*(10*A+7*C)*(b*cos(d*x+c))^(7/3)*hypergeom([1/2, 7/6],[13/6],co
s(d*x+c)^2)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)-3/10*B*(b*cos(d*x+c))^(10/3)*hypergeom([1/2, 5/3],[8/3],cos(d*
x+c)^2)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 154, normalized size of antiderivative = 1.00, 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} \[ -\frac {3 (10 A+7 C) \sin (c+d x) (b \cos (c+d x))^{7/3} \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right )}{70 b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{10/3} \, _2F_1\left (\frac {1}{2},\frac {5}{3};\frac {8}{3};\cos ^2(c+d x)\right )}{10 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{7/3}}{10 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*(b*Cos[c + d*x])^(7/3)*Sin[c + d*x])/(10*b*d) - (3*(10*A + 7*C)*(b*Cos[c + d*x])^(7/3)*Hypergeometric2F1[
1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(70*b*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(10/3)*Hyp
ergeometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2]*Sin[c + d*x])/(10*b^2*d*Sqrt[Sin[c + d*x]^2])

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]

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 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]

Rubi steps

\begin {align*} \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {3 C (b \cos (c+d x))^{7/3} \sin (c+d x)}{10 b d}+\frac {3 \int (b \cos (c+d x))^{4/3} \left (\frac {1}{3} b (10 A+7 C)+\frac {10}{3} b B \cos (c+d x)\right ) \, dx}{10 b}\\ &=\frac {3 C (b \cos (c+d x))^{7/3} \sin (c+d x)}{10 b d}+\frac {B \int (b \cos (c+d x))^{7/3} \, dx}{b}+\frac {1}{10} (10 A+7 C) \int (b \cos (c+d x))^{4/3} \, dx\\ &=\frac {3 C (b \cos (c+d x))^{7/3} \sin (c+d x)}{10 b d}-\frac {3 (10 A+7 C) (b \cos (c+d x))^{7/3} \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{70 b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{10/3} \, _2F_1\left (\frac {1}{2},\frac {5}{3};\frac {8}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{10 b^2 d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 109, normalized size = 0.71 \[ -\frac {3 \sin (c+d x) (b \cos (c+d x))^{7/3} \left ((10 A+7 C) \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right )+7 B \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{3};\frac {8}{3};\cos ^2(c+d x)\right )-7 C \sqrt {\sin ^2(c+d x)}\right )}{70 b d \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b*Cos[c + d*x])^(7/3)*Sin[c + d*x]*((10*A + 7*C)*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2] + 7*B*
Cos[c + d*x]*Hypergeometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2] - 7*C*Sqrt[Sin[c + d*x]^2]))/(70*b*d*Sqrt[Sin[c
+ d*x]^2])

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b \cos \left (d x + c\right )^{3} + B b \cos \left (d x + c\right )^{2} + A b \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [F]  time = 0.39, size = 0, normalized size = 0.00 \[ \int \left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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