3.4.60 \(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\) [360]

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

[Out]

3*A*sin(d*x+c)/b/d/(b*cos(d*x+c))^(1/3)-3/2*B*(b*cos(d*x+c))^(2/3)*hypergeom([1/3, 1/2],[4/3],cos(d*x+c)^2)*si
n(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)+3/5*(2*A-C)*(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 = 152, 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 = {3100, 2827, 2722} \begin {gather*} \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)}} \end {gather*}

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 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 3100

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]

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*}

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Mathematica [A]
time = 0.25, size = 115, normalized size = 0.76 \begin {gather*} -\frac {3 \cot (c+d x) \left (-10 A \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right )+\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 )\right ) \sqrt {\sin ^2(c+d x)}}{10 d (b \cos (c+d x))^{4/3}} \end {gather*}

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 = 0.16, size = 0, normalized size = 0.00 \[\int \frac {A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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 = 0.00, 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((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 = 0.00, 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((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(-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((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 3881 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((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)

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

[Out]

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

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