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

[Out]

3/16*(8*A+5*C)*hypergeom([-1/3, 1/2],[2/3],cos(d*x+c)^2)*(b*sec(d*x+c))^(2/3)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1
/2)+3/8*C*(b*sec(d*x+c))^(5/3)*tan(d*x+c)/b^2/d

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Rubi [A]  time = 0.09, antiderivative size = 95, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [C]  time = 3.11, size = 207, normalized size = 2.18 \[ \frac {3 i e^{-i (c+d x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{8/3} \left ((8 A+5 C) \left (1+e^{2 i (c+d x)}\right )^{8/3} \, _2F_1\left (\frac {2}{3},\frac {5}{6};\frac {11}{6};-e^{2 i (c+d x)}\right )-5 \left (8 A \left (1+e^{2 i (c+d x)}\right )^2+C \left (14 e^{2 i (c+d x)}+5 e^{4 i (c+d x)}+1\right )\right )\right ) \left (A+C \sec ^2(c+d x)\right )}{20 \sqrt [3]{2} d \sec ^{\frac {5}{3}}(c+d x) \sqrt [3]{b \sec (c+d x)} (A \cos (2 (c+d x))+A+2 C)} \]

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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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{3} + A \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}}{b}, x\right ) \]

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

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)

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maple [F]  time = 0.86, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sec ^{2}\left (d x +c \right )\right ) \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\sqrt [3]{b \sec {\left (c + d x \right )}}}\, dx \]

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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