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

Optimal. Leaf size=95 \[ \frac {3 C \sin (c+d x) (b \cos (c+d x))^{2/3}}{5 b^2 d}-\frac {3 (5 A+2 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right )}{10 b^2 d \sqrt {\sin ^2(c+d x)}} \]

[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)

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Rubi [A]  time = 0.07, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {16, 3014, 2643} \[ \frac {3 C \sin (c+d x) (b \cos (c+d x))^{2/3}}{5 b^2 d}-\frac {3 (5 A+2 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right )}{10 b^2 d \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*(A + 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])

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 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 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[
e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*
x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx &=\frac {\int \frac {A+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 {(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)}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 90, normalized size = 0.95 \[ -\frac {3 \sqrt {\sin ^2(c+d x)} \cot (c+d x) \left (4 A \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right )+C \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {4}{3};\frac {7}{3};\cos ^2(c+d x)\right )\right )}{8 b d \sqrt [3]{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}}{b^{2} \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.43, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x +c \right ) \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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