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

Optimal. Leaf size=145 \[ -\frac{3 (A+4 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )}{8 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 A b \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}+\frac{3 B \sin (c+d x) \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right )}{d \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}} \]

[Out]

(3*A*b*Sin[c + d*x])/(4*d*(b*Cos[c + d*x])^(4/3)) + (3*B*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]*Sin
[c + d*x])/(d*(b*Cos[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2]) - (3*(A + 4*C)*(b*Cos[c + d*x])^(2/3)*Hypergeometri
c2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*b*d*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.174072, antiderivative size = 145, normalized size of antiderivative = 1., 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, 3021, 2748, 2643} \[ -\frac{3 (A+4 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )}{8 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 A b \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}+\frac{3 B \sin (c+d x) \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right )}{d \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*b*Sin[c + d*x])/(4*d*(b*Cos[c + d*x])^(4/3)) + (3*B*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]*Sin
[c + d*x])/(d*(b*Cos[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2]) - (3*(A + 4*C)*(b*Cos[c + d*x])^(2/3)*Hypergeometri
c2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*b*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 ^2(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx &=b^2 \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{7/3}} \, dx\\ &=\frac{3 A b \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}+\frac{3 \int \frac{\frac{4 b^2 B}{3}+\frac{1}{3} b^2 (A+4 C) \cos (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx}{4 b}\\ &=\frac{3 A b \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}+(b B) \int \frac{1}{(b \cos (c+d x))^{4/3}} \, dx+\frac{1}{4} (A+4 C) \int \frac{1}{\sqrt [3]{b \cos (c+d x)}} \, dx\\ &=\frac{3 A b \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}+\frac{3 B \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)} \sqrt{\sin ^2(c+d x)}}-\frac{3 (A+4 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)}{8 b d \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [B]  time = 6.30457, size = 699, normalized size = 4.82 \[ \frac{4 B \csc (c) \cos ^{\frac{7}{3}}(c+d x) \left (A \sec ^2(c+d x)+B \sec (c+d x)+C\right ) \left (\frac{\tan (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{\tan ^2(c)+1} \sqrt{1-\cos \left (\tan ^{-1}(\tan (c))+d x\right )} \sqrt{\cos \left (\tan ^{-1}(\tan (c))+d x\right )+1} \sqrt [3]{\cos (c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}-\frac{\frac{\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right )}{\sqrt{\tan ^2(c)+1}}+\frac{3 \cos ^2(c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}{2 \left (\sin ^2(c)+\cos ^2(c)\right )}}{\sqrt [3]{\cos (c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}\right )}{d \sqrt [3]{b \cos (c+d x)} (2 A+2 B \cos (c+d x)+C \cos (2 c+2 d x)+C)}-\frac{A \cos ^{\frac{7}{3}}(c+d x) \cos \left (d x-\tan ^{-1}(\cot (c))\right ) \sin \left (d x-\tan ^{-1}(\cot (c))\right ) \left (A \sec ^2(c+d x)+B \sec (c+d x)+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 )}{2 d \sqrt [3]{b \cos (c+d x)} \sqrt [3]{\cos (c) \cos (d x)-\sin (c) \sin (d x)} \sqrt [3]{\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )} (2 A+2 B \cos (c+d x)+C \cos (2 c+2 d x)+C)}-\frac{2 C \cos ^{\frac{7}{3}}(c+d x) \cos \left (d x-\tan ^{-1}(\cot (c))\right ) \sin \left (d x-\tan ^{-1}(\cot (c))\right ) \left (A \sec ^2(c+d x)+B \sec (c+d x)+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 )}{d \sqrt [3]{b \cos (c+d x)} \sqrt [3]{\cos (c) \cos (d x)-\sin (c) \sin (d x)} \sqrt [3]{\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )} (2 A+2 B \cos (c+d x)+C \cos (2 c+2 d x)+C)}+\frac{\cos ^3(c+d x) \left (A \sec ^2(c+d x)+B \sec (c+d x)+C\right ) \left (\frac{3 \sec (c) \sec (c+d x) (A \sin (c)+4 B \sin (d x))}{2 d}+\frac{3 A \sec (c) \sin (d x) \sec ^2(c+d x)}{2 d}+\frac{6 B \csc (c) \sec (c)}{d}\right )}{\sqrt [3]{b \cos (c+d x)} (2 A+2 B \cos (c+d x)+C \cos (2 c+2 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]^2)/(b*Cos[c + d*x])^(1/3),x]

[Out]

(Cos[c + d*x]^3*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*((6*B*Csc[c]*Sec[c])/d + (3*A*Sec[c]*Sec[c + d*x]^2*Si
n[d*x])/(2*d) + (3*Sec[c]*Sec[c + d*x]*(A*Sin[c] + 4*B*Sin[d*x]))/(2*d)))/((b*Cos[c + d*x])^(1/3)*(2*A + C + 2
*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])) - (A*Cos[c + d*x]^(7/3)*Cos[d*x - ArcTan[Cot[c]]]*Hypergeometric2F1[1/2
, 2/3, 3/2, Cos[d*x - ArcTan[Cot[c]]]^2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*Sin[d*x - ArcTan[Cot[c]]])/(2
*d*(b*Cos[c + d*x])^(1/3)*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])*(Cos[c]*Cos[d*x] - Sin[c]*Sin[d*x]
)^(1/3)*(Sin[d*x - ArcTan[Cot[c]]]^2)^(1/3)) - (2*C*Cos[c + d*x]^(7/3)*Cos[d*x - ArcTan[Cot[c]]]*Hypergeometri
c2F1[1/2, 2/3, 3/2, Cos[d*x - ArcTan[Cot[c]]]^2]*(C + B*Sec[c + d*x] + A*Sec[c + d*x]^2)*Sin[d*x - ArcTan[Cot[
c]]])/(d*(b*Cos[c + d*x])^(1/3)*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x])*(Cos[c]*Cos[d*x] - Sin[c]*Si
n[d*x])^(1/3)*(Sin[d*x - ArcTan[Cot[c]]]^2)^(1/3)) + (4*B*Cos[c + d*x]^(7/3)*Csc[c]*(C + B*Sec[c + d*x] + A*Se
c[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/6}, {5/6}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*
Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*(Cos[c]*Cos[d*x + ArcTan[Tan[
c]]]*Sqrt[1 + Tan[c]^2])^(1/3)*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] +
(3*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(2*(Cos[c]^2 + Sin[c]^2)))/(Cos[c]*Cos[d*x + ArcTan[
Tan[c]]]*Sqrt[1 + Tan[c]^2])^(1/3)))/(d*(b*Cos[c + d*x])^(1/3)*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*c + 2*d*x
]))

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Maple [F]  time = 0.375, 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 ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}{\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)^2/(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)^2/(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 )^{2}}{\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)^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)*sec(d*x + c)^2/(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 )^{2}}{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)^2/(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)^2/(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)**2/(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 )^{2}}{\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)^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)*sec(d*x + c)^2/(b*cos(d*x + c))^(1/3), x)