3.624 \(\int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x)) \, dx\)

Optimal. Leaf size=78 \[ \frac {3 a d \sin (e+f x) (d \sec (e+f x))^{2/3} \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(e+f x)\right )}{2 f \sqrt {\sin ^2(e+f x)}}+\frac {3 b (d \sec (e+f x))^{5/3}}{5 f} \]

[Out]

3/5*b*(d*sec(f*x+e))^(5/3)/f+3/2*a*d*hypergeom([-1/3, 1/2],[2/3],cos(f*x+e)^2)*(d*sec(f*x+e))^(2/3)*sin(f*x+e)
/f/(sin(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 78, 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} \[ \frac {3 a d \sin (e+f x) (d \sec (e+f x))^{2/3} \text {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(e+f x)\right )}{2 f \sqrt {\sin ^2(e+f x)}}+\frac {3 b (d \sec (e+f x))^{5/3}}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(5/3)*(a + b*Tan[e + f*x]),x]

[Out]

(3*b*(d*Sec[e + f*x])^(5/3))/(5*f) + (3*a*d*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[e + f*x]^2]*(d*Sec[e + f*x])
^(2/3)*Sin[e + f*x])/(2*f*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 (d \sec (e+f x))^{5/3} (a+b \tan (e+f x)) \, dx &=\frac {3 b (d \sec (e+f x))^{5/3}}{5 f}+a \int (d \sec (e+f x))^{5/3} \, dx\\ &=\frac {3 b (d \sec (e+f x))^{5/3}}{5 f}+\left (a \left (\frac {\cos (e+f x)}{d}\right )^{2/3} (d \sec (e+f x))^{2/3}\right ) \int \frac {1}{\left (\frac {\cos (e+f x)}{d}\right )^{5/3}} \, dx\\ &=\frac {3 b (d \sec (e+f x))^{5/3}}{5 f}+\frac {3 a d \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(e+f x)\right ) (d \sec (e+f x))^{2/3} \sin (e+f x)}{2 f \sqrt {\sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.51, size = 126, normalized size = 1.62 \[ \frac {d (d \sec (e+f x))^{2/3} (a+b \tan (e+f x)) \left (3 \cos ^2(e+f x)^{2/3} (5 a \sin (2 (e+f x))+4 b)-10 a \sin (e+f x) \cos ^3(e+f x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {3}{2};\sin ^2(e+f x)\right )\right )}{20 f \cos ^2(e+f x)^{2/3} (a \cos (e+f x)+b \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(5/3)*(a + b*Tan[e + f*x]),x]

[Out]

(d*(d*Sec[e + f*x])^(2/3)*(-10*a*Cos[e + f*x]^3*Hypergeometric2F1[1/3, 1/2, 3/2, Sin[e + f*x]^2]*Sin[e + f*x]
+ 3*(Cos[e + f*x]^2)^(2/3)*(4*b + 5*a*Sin[2*(e + f*x)]))*(a + b*Tan[e + f*x]))/(20*f*(Cos[e + f*x]^2)^(2/3)*(a
*Cos[e + f*x] + b*Sin[e + f*x]))

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b d \sec \left (f x + e\right ) \tan \left (f x + e\right ) + a d \sec \left (f x + e\right )\right )} \left (d \sec \left (f x + e\right )\right )^{\frac {2}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e)),x, algorithm="fricas")

[Out]

integral((b*d*sec(f*x + e)*tan(f*x + e) + a*d*sec(f*x + e))*(d*sec(f*x + e))^(2/3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (b \tan \left (f x + e\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(5/3)*(b*tan(f*x + e) + a), x)

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maple [F]  time = 0.58, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}} \left (a +b \tan \left (f x +e \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e)),x)

[Out]

int((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (b \tan \left (f x + e\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/3)*(a+b*tan(f*x+e)),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(5/3)*(b*tan(f*x + e) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}\,\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(5/3)*(a + b*tan(e + f*x)),x)

[Out]

int((d/cos(e + f*x))^(5/3)*(a + b*tan(e + f*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}} \left (a + b \tan {\left (e + f x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(5/3)*(a+b*tan(f*x+e)),x)

[Out]

Integral((d*sec(e + f*x))**(5/3)*(a + b*tan(e + f*x)), x)

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