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

Optimal. Leaf size=149 \[ \frac{3 (2 A-C) \sin (c+d x) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )}{5 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 A \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}}-\frac{3 B \sin (c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )}{2 b d \sqrt{\sin ^2(c+d x)}} \]

[Out]

(3*A*Sin[c + d*x])/(d*(b*Cos[c + d*x])^(1/3)) - (3*B*(b*Cos[c + d*x])^(2/3)*Hypergeometric2F1[1/3, 1/2, 4/3, C
os[c + d*x]^2]*Sin[c + d*x])/(2*b*d*Sqrt[Sin[c + d*x]^2]) + (3*(2*A - C)*(b*Cos[c + d*x])^(5/3)*Hypergeometric
2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*b^2*d*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.163131, antiderivative size = 149, normalized size of antiderivative = 1., 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, 3021, 2748, 2643} \[ \frac{3 (2 A-C) \sin (c+d x) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )}{5 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 A \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}}-\frac{3 B \sin (c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )}{2 b d \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*Sin[c + d*x])/(d*(b*Cos[c + d*x])^(1/3)) - (3*B*(b*Cos[c + d*x])^(2/3)*Hypergeometric2F1[1/3, 1/2, 4/3, C
os[c + d*x]^2]*Sin[c + d*x])/(2*b*d*Sqrt[Sin[c + d*x]^2]) + (3*(2*A - C)*(b*Cos[c + d*x])^(5/3)*Hypergeometric
2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*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 3021

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]

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 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]

Rubi steps

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

Mathematica [A]  time = 6.24293, size = 268, normalized size = 1.8 \[ \frac{(b \cos (c+d x))^{2/3} (A \sec (c+d x)+B+C \cos (c+d x)) \left (\frac{4 (2 A-C) \sec (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \, _2F_1\left (-\frac{1}{2},-\frac{1}{6};\frac{5}{6};\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right )}{\sqrt{\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )}}+\csc (c) \left (3 \sqrt{\sec ^2(c)} ((4 A-C) \cos (d x)-C \cos (2 c+d x))-5 (2 A-C) \sec (c) \cos \left (c-\tan ^{-1}(\tan (c))-d x\right )+(C-2 A) \sec (c) \cos \left (c+\tan ^{-1}(\tan (c))+d x\right )\right )-\frac{2 B \sqrt{\sec ^2(c)} \sin \left (2 d x-2 \tan ^{-1}(\cot (c))\right ) \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{3}{2};\cos ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )}{\sqrt [3]{\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )}}\right )}{2 b d \sqrt{\sec ^2(c)} (2 A+2 B \cos (c+d x)+C \cos (2 (c+d x))+C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b*Cos[c + d*x])^(2/3)*(B + C*Cos[c + d*x] + A*Sec[c + d*x])*(Csc[c]*(-5*(2*A - C)*Cos[c - d*x - ArcTan[Tan[c
]]]*Sec[c] + (-2*A + C)*Cos[c + d*x + ArcTan[Tan[c]]]*Sec[c] + 3*((4*A - C)*Cos[d*x] - C*Cos[2*c + d*x])*Sqrt[
Sec[c]^2]) - (2*B*Hypergeometric2F1[1/2, 2/3, 3/2, Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Sec[c]^2]*Sin[2*d*x - 2*A
rcTan[Cot[c]]])/(Sin[d*x - ArcTan[Cot[c]]]^2)^(1/3) + (4*(2*A - C)*HypergeometricPFQ[{-1/2, -1/6}, {5/6}, Cos[
d*x + ArcTan[Tan[c]]]^2]*Sec[c]*Sin[d*x + ArcTan[Tan[c]]])/Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/(2*b*d*(2*A + C
+ 2*B*Cos[c + d*x] + C*Cos[2*(c + d*x)])*Sqrt[Sec[c]^2])

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Maple [F]  time = 0.345, size = 0, normalized size = 0. \begin{align*} \int{ \left ( A+B\cos \left ( dx+c \right ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sec \left ( dx+c \right ){\frac{1}{\sqrt [3]{b\cos \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\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{2}{3}} \sec \left (d x + c\right )}{b \cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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