3.292 \(\int (d \sec (e+f x))^{3/2} \sqrt{b \tan (e+f x)} \, dx\)

Optimal. Leaf size=93 \[ \frac{d^2 (b \tan (e+f x))^{3/2}}{b f \sqrt{d \sec (e+f x)}}-\frac{d^2 E\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{f \sqrt{\sin (e+f x)} \sqrt{d \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.11169, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2613, 2616, 2640, 2639} \[ \frac{d^2 (b \tan (e+f x))^{3/2}}{b f \sqrt{d \sec (e+f x)}}-\frac{d^2 E\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{f \sqrt{\sin (e+f x)} \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]],x]

[Out]

-((d^2*EllipticE[(e - Pi/2 + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(f*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[e + f*x]])) + (
d^2*(b*Tan[e + f*x])^(3/2))/(b*f*Sqrt[d*Sec[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 2616

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^(m + n)*(b
*Tan[e + f*x])^n)/((a*Sec[e + f*x])^n*(b*Sin[e + f*x])^n), Int[(b*Sin[e + f*x])^n/Cos[e + f*x]^(m + n), x], x]
/; FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[n + 1/2] && IntegerQ[m + 1/2]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, 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 (d \sec (e+f x))^{3/2} \sqrt{b \tan (e+f x)} \, dx &=\frac{d^2 (b \tan (e+f x))^{3/2}}{b f \sqrt{d \sec (e+f x)}}-\frac{1}{2} d^2 \int \frac{\sqrt{b \tan (e+f x)}}{\sqrt{d \sec (e+f x)}} \, dx\\ &=\frac{d^2 (b \tan (e+f x))^{3/2}}{b f \sqrt{d \sec (e+f x)}}-\frac{\left (d^2 \sqrt{b \tan (e+f x)}\right ) \int \sqrt{b \sin (e+f x)} \, dx}{2 \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=\frac{d^2 (b \tan (e+f x))^{3/2}}{b f \sqrt{d \sec (e+f x)}}-\frac{\left (d^2 \sqrt{b \tan (e+f x)}\right ) \int \sqrt{\sin (e+f x)} \, dx}{2 \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}}\\ &=-\frac{d^2 E\left (\left .\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{f \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}}+\frac{d^2 (b \tan (e+f x))^{3/2}}{b f \sqrt{d \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.00158, size = 71, normalized size = 0.76 \[ \frac{d \sin (e+f x) \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)} \left (1-\frac{\, _2F_1\left (-\frac{1}{4},\frac{1}{4};\frac{3}{4};\sec ^2(e+f x)\right )}{\left (-\tan ^2(e+f x)\right )^{3/4}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]],x]

[Out]

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

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Maple [C]  time = 0.277, size = 572, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*2^(1/2)*(b*sin(f*x+e)/cos(f*x+e))^(1/2)*(d/cos(f*x+e))^(3/2)*cos(f*x+e)*(2*cos(f*x+e)^2*(-I*(cos(f*x+e)-
1)/sin(f*x+e))^(1/2)*EllipticE(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*((I*cos(f*x+e)-I+si
n(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)-cos(f*x+e)^2*(-I*(cos(f*x+e)-1)/si
n(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)
*EllipticF(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))+2*cos(f*x+e)*(-I*(cos(f*x+e)-1)/sin(f*x
+e))^(1/2)*EllipticE(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*((I*cos(f*x+e)-I+sin(f*x+e))/
sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)-cos(f*x+e)*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1
/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*cos(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((
(I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))-cos(f*x+e)*2^(1/2)+2^(1/2))/sin(f*x+e)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}} \sqrt{b \tan \left (f x + e\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{d \sec \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )} d \sec \left (f x + e\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}} \sqrt{b \tan \left (f x + e\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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