3.231 \(\int \sec ^3(e+f x) \sqrt {d \tan (e+f x)} \, dx\)

Optimal. Leaf size=107 \[ \frac {4 \cos (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac {2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}-\frac {4 \cos (e+f x) E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \tan (e+f x)}}{5 f \sqrt {\sin (2 e+2 f x)}} \]

[Out]

4/5*cos(f*x+e)*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f*x)*EllipticE(cos(e+1/4*Pi+f*x),2^(1/2))*(d*tan(f*x+e
))^(1/2)/f/sin(2*f*x+2*e)^(1/2)+4/5*cos(f*x+e)*(d*tan(f*x+e))^(3/2)/d/f+2/5*sec(f*x+e)*(d*tan(f*x+e))^(3/2)/d/
f

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Rubi [A]  time = 0.13, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2613, 2615, 2572, 2639} \[ \frac {4 \cos (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac {2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}-\frac {4 \cos (e+f x) E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \tan (e+f x)}}{5 f \sqrt {\sin (2 e+2 f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^3*Sqrt[d*Tan[e + f*x]],x]

[Out]

(-4*Cos[e + f*x]*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[d*Tan[e + f*x]])/(5*f*Sqrt[Sin[2*e + 2*f*x]]) + (4*Cos[e +
f*x]*(d*Tan[e + f*x])^(3/2))/(5*d*f) + (2*Sec[e + f*x]*(d*Tan[e + f*x])^(3/2))/(5*d*f)

Rule 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +
f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
x]

Rule 2613

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(a^2*(m - 2))/(m + n - 1), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 2615

Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[(Sqrt[Cos[e + f*x]]*Sqrt[b*
Tan[e + f*x]])/Sqrt[Sin[e + f*x]], Int[Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin {align*} \int \sec ^3(e+f x) \sqrt {d \tan (e+f x)} \, dx &=\frac {2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac {2}{5} \int \sec (e+f x) \sqrt {d \tan (e+f x)} \, dx\\ &=\frac {4 \cos (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac {2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}-\frac {4}{5} \int \cos (e+f x) \sqrt {d \tan (e+f x)} \, dx\\ &=\frac {4 \cos (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac {2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}-\frac {\left (4 \sqrt {\cos (e+f x)} \sqrt {d \tan (e+f x)}\right ) \int \sqrt {\cos (e+f x)} \sqrt {\sin (e+f x)} \, dx}{5 \sqrt {\sin (e+f x)}}\\ &=\frac {4 \cos (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac {2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}-\frac {\left (4 \cos (e+f x) \sqrt {d \tan (e+f x)}\right ) \int \sqrt {\sin (2 e+2 f x)} \, dx}{5 \sqrt {\sin (2 e+2 f x)}}\\ &=-\frac {4 \cos (e+f x) E\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {d \tan (e+f x)}}{5 f \sqrt {\sin (2 e+2 f x)}}+\frac {4 \cos (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}+\frac {2 \sec (e+f x) (d \tan (e+f x))^{3/2}}{5 d f}\\ \end {align*}

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Mathematica [C]  time = 0.44, size = 102, normalized size = 0.95 \[ \frac {2 \sqrt {d \tan (e+f x)} \left (3 \sqrt {\sec ^2(e+f x)} (2 \sin (e+f x)+\tan (e+f x) \sec (e+f x))-4 \tan (e+f x) \sec (e+f x) \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\tan ^2(e+f x)\right )\right )}{15 f \sqrt {\sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^3*Sqrt[d*Tan[e + f*x]],x]

[Out]

(2*Sqrt[d*Tan[e + f*x]]*(-4*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[e + f*x]^2]*Sec[e + f*x]*Tan[e + f*x] + 3*Sq
rt[Sec[e + f*x]^2]*(2*Sin[e + f*x] + Sec[e + f*x]*Tan[e + f*x])))/(15*f*Sqrt[Sec[e + f*x]^2])

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \tan \left (f x + e\right )} \sec \left (f x + e\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3*(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(f*x + e))*sec(f*x + e)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3*(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(d*tan(f*x + e))*sec(f*x + e)^3, x)

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maple [B]  time = 0.57, size = 559, normalized size = 5.22 \[ -\frac {\left (-1+\cos \left (f x +e \right )\right )^{2} \left (2 \left (\cos ^{3}\left (f x +e \right )\right ) \EllipticF \left (\sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}-4 \left (\cos ^{3}\left (f x +e \right )\right ) \EllipticE \left (\sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}+2 \left (\cos ^{2}\left (f x +e \right )\right ) \EllipticF \left (\sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}-4 \left (\cos ^{2}\left (f x +e \right )\right ) \EllipticE \left (\sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}+2 \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {2}-\left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}-\sqrt {2}\right ) \sqrt {\frac {d \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \left (1+\cos \left (f x +e \right )\right )^{2} \sqrt {2}}{5 f \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^3*(d*tan(f*x+e))^(1/2),x)

[Out]

-1/5/f*(-1+cos(f*x+e))^2*(2*cos(f*x+e)^3*EllipticF((-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))
*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(-sin(f*x+e)-1+cos(f*x+e))
/sin(f*x+e))^(1/2)-4*cos(f*x+e)^3*EllipticE((-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*((-1+c
os(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*
x+e))^(1/2)+2*cos(f*x+e)^2*EllipticF((-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*((-1+cos(f*x+
e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(
1/2)-4*cos(f*x+e)^2*EllipticE((-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*((-1+cos(f*x+e))/sin
(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2)+2*
cos(f*x+e)^3*2^(1/2)-cos(f*x+e)^2*2^(1/2)-2^(1/2))*(d*sin(f*x+e)/cos(f*x+e))^(1/2)*(1+cos(f*x+e))^2/cos(f*x+e)
^2/sin(f*x+e)^5*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3*(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(d*tan(f*x + e))*sec(f*x + e)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^(1/2)/cos(e + f*x)^3,x)

[Out]

int((d*tan(e + f*x))^(1/2)/cos(e + f*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**3*(d*tan(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(d*tan(e + f*x))*sec(e + f*x)**3, x)

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