3.625 \(\int \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x)) \, dx\)

Optimal. Leaf size=76 \[ \frac {3 b \sqrt [3]{d \sec (e+f x)}}{f}-\frac {3 a d \sin (e+f x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(e+f x)\right )}{2 f \sqrt {\sin ^2(e+f x)} (d \sec (e+f x))^{2/3}} \]

[Out]

3*b*(d*sec(f*x+e))^(1/3)/f-3/2*a*d*hypergeom([1/3, 1/2],[4/3],cos(f*x+e)^2)*sin(f*x+e)/f/(d*sec(f*x+e))^(2/3)/
(sin(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3486, 3772, 2643} \[ \frac {3 b \sqrt [3]{d \sec (e+f x)}}{f}-\frac {3 a d \sin (e+f x) \text {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(e+f x)\right )}{2 f \sqrt {\sin ^2(e+f x)} (d \sec (e+f x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(1/3)*(a + b*Tan[e + f*x]),x]

[Out]

(3*b*(d*Sec[e + f*x])^(1/3))/f - (3*a*d*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[e + f*x]^2]*Sin[e + f*x])/(2*f*(d
*Sec[e + f*x])^(2/3)*Sqrt[Sin[e + f*x]^2])

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 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

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]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 58, normalized size = 0.76 \[ \frac {\sqrt [3]{d \sec (e+f x)} \left (a \cos ^2(e+f x)^{2/3} \tan (e+f x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {3}{2};\sin ^2(e+f x)\right )+3 b\right )}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(1/3)*(a + b*Tan[e + f*x]),x]

[Out]

((d*Sec[e + f*x])^(1/3)*(3*b + a*(Cos[e + f*x]^2)^(2/3)*Hypergeometric2F1[1/2, 2/3, 3/2, Sin[e + f*x]^2]*Tan[e
+ f*x]))/f

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}} {\left (b \tan \left (f x + e\right ) + a\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(1/3)*(a+b*tan(f*x+e)),x, algorithm="fricas")

[Out]

integral((d*sec(f*x + e))^(1/3)*(b*tan(f*x + e) + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}} {\left (b \tan \left (f x + e\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(1/3)*(a+b*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(1/3)*(b*tan(f*x + e) + a), x)

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maple [F]  time = 0.58, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}} \left (a +b \tan \left (f x +e \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(1/3)*(a+b*tan(f*x+e)),x)

[Out]

int((d*sec(f*x+e))^(1/3)*(a+b*tan(f*x+e)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}} {\left (b \tan \left (f x + e\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(1/3)*(a+b*tan(f*x+e)),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(1/3)*(b*tan(f*x + e) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{1/3}\,\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(1/3)*(a + b*tan(e + f*x)),x)

[Out]

int((d/cos(e + f*x))^(1/3)*(a + b*tan(e + f*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt [3]{d \sec {\left (e + f x \right )}} \left (a + b \tan {\left (e + f x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(1/3)*(a+b*tan(f*x+e)),x)

[Out]

Integral((d*sec(e + f*x))**(1/3)*(a + b*tan(e + f*x)), x)

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