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

Optimal. Leaf size=91 \[ \frac{3 (A-2 C) \sin (c+d x) (b \cos (c+d x))^{4/3} \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right )}{8 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 A b \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3}} \]

[Out]

(3*A*b*Sin[c + d*x])/(2*d*(b*Cos[c + d*x])^(2/3)) + (3*(A - 2*C)*(b*Cos[c + d*x])^(4/3)*Hypergeometric2F1[1/2,
2/3, 5/3, 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 = 0.0933749, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {16, 3012, 2643} \[ \frac{3 (A-2 C) \sin (c+d x) (b \cos (c+d x))^{4/3} \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right )}{8 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 A b \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*b*Sin[c + d*x])/(2*d*(b*Cos[c + d*x])^(2/3)) + (3*(A - 2*C)*(b*Cos[c + d*x])^(4/3)*Hypergeometric2F1[1/2,
2/3, 5/3, 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 3012

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

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 \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=b^2 \int \frac{A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{5/3}} \, dx\\ &=\frac{3 A b \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3}}+\frac{1}{2} (-A+2 C) \int \sqrt [3]{b \cos (c+d x)} \, dx\\ &=\frac{3 A b \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3}}+\frac{3 (A-2 C) (b \cos (c+d x))^{4/3} \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{8 b d \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.163227, size = 88, normalized size = 0.97 \[ -\frac{3 b \sqrt{\sin ^2(c+d x)} \csc (c+d x) \left (C \cos ^2(c+d x) \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right )-2 A \, _2F_1\left (-\frac{1}{3},\frac{1}{2};\frac{2}{3};\cos ^2(c+d x)\right )\right )}{4 d (b \cos (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*Csc[c + d*x]*(-2*A*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2] + C*Cos[c + d*x]^2*Hypergeometric2F
1[1/2, 2/3, 5/3, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(4*d*(b*Cos[c + d*x])^(2/3))

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Maple [F]  time = 0.346, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{b\cos \left ( dx+c \right ) } \left ( A+C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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