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

Optimal. Leaf size=95 \[ \frac{3 (8 A+5 C) \sin (c+d x) (b \sec (c+d x))^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{2},\frac{2}{3},\cos ^2(c+d x)\right )}{16 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{5/3}}{8 b^2 d} \]

[Out]

(3*(8*A + 5*C)*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(2/3)*Sin[c + d*x])/(16*b*d*
Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(5/3)*Tan[c + d*x])/(8*b^2*d)

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Rubi [A]  time = 0.0868797, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {16, 4046, 3772, 2643} \[ \frac{3 (8 A+5 C) \sin (c+d x) (b \sec (c+d x))^{2/3} \, _2F_1\left (-\frac{1}{3},\frac{1}{2};\frac{2}{3};\cos ^2(c+d x)\right )}{16 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \tan (c+d x) (b \sec (c+d x))^{5/3}}{8 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(8*A + 5*C)*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(2/3)*Sin[c + d*x])/(16*b*d*
Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(5/3)*Tan[c + d*x])/(8*b^2*d)

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 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

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{\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx &=\frac{\int (b \sec (c+d x))^{5/3} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b^2}\\ &=\frac{3 C (b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b^2 d}+\frac{(8 A+5 C) \int (b \sec (c+d x))^{5/3} \, dx}{8 b^2}\\ &=\frac{3 C (b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b^2 d}+\frac{\left ((8 A+5 C) \left (\frac{\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \frac{1}{\left (\frac{\cos (c+d x)}{b}\right )^{5/3}} \, dx}{8 b^2}\\ &=\frac{3 (8 A+5 C) \, _2F_1\left (-\frac{1}{3},\frac{1}{2};\frac{2}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{16 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 C (b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b^2 d}\\ \end{align*}

Mathematica [C]  time = 2.72516, size = 207, normalized size = 2.18 \[ \frac{3 i e^{-i (c+d x)} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{8/3} \left (A+C \sec ^2(c+d x)\right ) \left ((8 A+5 C) \left (1+e^{2 i (c+d x)}\right )^{8/3} \text{Hypergeometric2F1}\left (\frac{2}{3},\frac{5}{6},\frac{11}{6},-e^{2 i (c+d x)}\right )-5 \left (8 A \left (1+e^{2 i (c+d x)}\right )^2+C \left (14 e^{2 i (c+d x)}+5 e^{4 i (c+d x)}+1\right )\right )\right )}{20 \sqrt [3]{2} d \sec ^{\frac{5}{3}}(c+d x) \sqrt [3]{b \sec (c+d x)} (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)/20)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(8/3)*(-5*(8*A*(1 + E^((2*I)*(c + d*x)))^2 + C*(1 + 14
*E^((2*I)*(c + d*x)) + 5*E^((4*I)*(c + d*x)))) + (8*A + 5*C)*(1 + E^((2*I)*(c + d*x)))^(8/3)*Hypergeometric2F1
[2/3, 5/6, 11/6, -E^((2*I)*(c + d*x))])*(A + C*Sec[c + d*x]^2))/(2^(1/3)*d*E^(I*(c + d*x))*(A + 2*C + A*Cos[2*
(c + d*x)])*Sec[c + d*x]^(5/3)*(b*Sec[c + d*x])^(1/3))

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Maple [F]  time = 0.141, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( dx+c \right ) \right ) ^{2} \left ( A+C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ){\frac{1}{\sqrt [3]{b\sec \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(b*sec(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 \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \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(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)*sec(d*x + c)^2/(b*sec(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 \sec \left (d x + c\right )^{3} + A \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{\frac{2}{3}}}{b}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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