3.4.59 \(\int \frac {\cos (c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{(b \cos (c+d x))^{4/3}} \, dx\) [359]

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

[Out]

3/5*C*(b*cos(d*x+c))^(2/3)*sin(d*x+c)/b^2/d-3/10*(5*A+2*C)*(b*cos(d*x+c))^(2/3)*hypergeom([1/3, 1/2],[4/3],cos
(d*x+c)^2)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)-3/5*B*(b*cos(d*x+c))^(5/3)*hypergeom([1/2, 5/6],[11/6],cos(d*
x+c)^2)*sin(d*x+c)/b^3/d/(sin(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {16, 3102, 2827, 2722} \begin {gather*} -\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^2 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^3 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{2/3}}{5 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*(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])^(2/3)*Sin[c + d*x])/(5*b^2*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^2*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(5/3)*Hyp
ergeometric2F1[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 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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2827

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 3102

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

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Mathematica [A]
time = 0.24, size = 111, normalized size = 0.72 \begin {gather*} -\frac {3 \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 ) \sin (2 (c+d x))}{20 b d \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(4/3),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*b*d*(b*Cos[c + d*x])^(1/3)*Sqrt[Sin
[c + d*x]^2])

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Maple [F]
time = 0.21, size = 0, normalized size = 0.00 \[\int \frac {\cos \left (d x +c \right ) \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(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)*(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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)), x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 5990 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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