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

Optimal. Leaf size=88 \[ -\frac {3 b (4 A+C) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{8 d (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{4 d} \]

[Out]

-3/8*b*(4*A+C)*hypergeom([1/3, 1/2],[4/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(2/3)/(sin(d*x+c)^2)^(1/2)
+3/4*C*(b*sec(d*x+c))^(1/3)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4131, 3857, 2722} \begin {gather*} \frac {3 C \tan (c+d x) \sqrt [3]{b \sec (c+d x)}}{4 d}-\frac {3 b (4 A+C) \sin (c+d x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right )}{8 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(4*A + C)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*d*(b*Sec[c + d*x])^(2/3)*Sqr
t[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(4*d)

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3857

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 4131

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

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 0.77, size = 162, normalized size = 1.84 \begin {gather*} \frac {3 \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (-i \sqrt [3]{2} (4 A+C) \sqrt [3]{\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt [3]{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-e^{2 i (c+d x)}\right )+C \sec ^{\frac {4}{3}}(c+d x) \sin (c+d x)\right )}{2 d (A+2 C+A \cos (2 (c+d x))) \sec ^{\frac {7}{3}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*Sec[c + d*x])^(1/3)*(A + C*Sec[c + d*x]^2)*((-I)*2^(1/3)*(4*A + C)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d
*x))))^(1/3)*(1 + E^((2*I)*(c + d*x)))^(1/3)*Hypergeometric2F1[1/6, 1/3, 7/6, -E^((2*I)*(c + d*x))] + C*Sec[c
+ d*x]^(4/3)*Sin[c + d*x]))/(2*d*(A + 2*C + A*Cos[2*(c + d*x)])*Sec[c + d*x]^(7/3))

________________________________________________________________________________________

Maple [F]
time = 0.33, size = 0, normalized size = 0.00 \[\int \left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________