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

Optimal. Leaf size=90 \[ \frac {3 C \tan (c+d x)}{2 d \sqrt [3]{b \sec (c+d x)}}-\frac {3 b (2 A-C) \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right )}{8 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}} \]

[Out]

-3/8*b*(2*A-C)*hypergeom([1/2, 2/3],[5/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(4/3)/(sin(d*x+c)^2)^(1/2)
+3/2*C*tan(d*x+c)/d/(b*sec(d*x+c))^(1/3)

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Rubi [A]  time = 0.07, antiderivative size = 90, 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 = {4046, 3772, 2643} \[ \frac {3 C \tan (c+d x)}{2 d \sqrt [3]{b \sec (c+d x)}}-\frac {3 b (2 A-C) \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right )}{8 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(2*A - C)*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*d*(b*Sec[c + d*x])^(4/3)*Sqr
t[Sin[c + d*x]^2]) + (3*C*Tan[c + d*x])/(2*d*(b*Sec[c + d*x])^(1/3))

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

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Mathematica [C]  time = 0.78, size = 127, normalized size = 1.41 \[ -\frac {3 i \left ((2 A-C) e^{2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {5}{6};\frac {11}{6};-e^{2 i (c+d x)}\right )-5 \left (A e^{2 i (c+d x)}+A-C e^{2 i (c+d x)}\right )\right )}{5 d \left (1+e^{2 i (c+d x)}\right ) \sqrt [3]{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}}{b \sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [F]  time = 0.81, size = 0, normalized size = 0.00 \[ \int \frac {A +C \left (\sec ^{2}\left (d x +c \right )\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/(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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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