3.351 \(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt [3]{b \cos (c+d x)}} \, dx\)

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

[Out]

3/11*C*(b*cos(d*x+c))^(8/3)*sin(d*x+c)/b^3/d-3/88*(11*A+8*C)*(b*cos(d*x+c))^(8/3)*hypergeom([1/2, 4/3],[7/3],c
os(d*x+c)^2)*sin(d*x+c)/b^3/d/(sin(d*x+c)^2)^(1/2)-3/11*B*(b*cos(d*x+c))^(11/3)*hypergeom([1/2, 11/6],[17/6],c
os(d*x+c)^2)*sin(d*x+c)/b^4/d/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 154, normalized size of antiderivative = 1.00, 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} \[ -\frac {3 (11 A+8 C) \sin (c+d x) (b \cos (c+d x))^{8/3} \, _2F_1\left (\frac {1}{2},\frac {4}{3};\frac {7}{3};\cos ^2(c+d x)\right )}{88 b^3 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{11/3} \, _2F_1\left (\frac {1}{2},\frac {11}{6};\frac {17}{6};\cos ^2(c+d x)\right )}{11 b^4 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{8/3}}{11 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.28, size = 114, normalized size = 0.74 \[ -\frac {3 \sin (c+d x) \cos ^3(c+d x) \left ((11 A+8 C) \, _2F_1\left (\frac {1}{2},\frac {4}{3};\frac {7}{3};\cos ^2(c+d x)\right )+8 B \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {11}{6};\frac {17}{6};\cos ^2(c+d x)\right )-8 C \sqrt {\sin ^2(c+d x)}\right )}{88 d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Cos[c + d*x]^3*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])^(1/
3)*Sqrt[Sin[c + d*x]^2])

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fricas [F]  time = 1.29, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{3} + B \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}}{b}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/3),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.44, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{2}\left (d x +c \right )\right ) \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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