3.298 \(\int \frac{\sqrt{b \tan (e+f x)}}{(d \sec (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=132 \[ \frac{4 (b \tan (e+f x))^{3/2}}{15 b d^2 f (d \sec (e+f x))^{5/2}}+\frac{8 E\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{15 d^4 f \sqrt{\sin (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}} \]

[Out]

(8*EllipticE[(e - Pi/2 + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(15*d^4*f*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[e + f*x]]) +
(2*(b*Tan[e + f*x])^(3/2))/(9*b*f*(d*Sec[e + f*x])^(9/2)) + (4*(b*Tan[e + f*x])^(3/2))/(15*b*d^2*f*(d*Sec[e +
f*x])^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.167954, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2612, 2616, 2640, 2639} \[ \frac{4 (b \tan (e+f x))^{3/2}}{15 b d^2 f (d \sec (e+f x))^{5/2}}+\frac{8 E\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{15 d^4 f \sqrt{\sin (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Tan[e + f*x]]/(d*Sec[e + f*x])^(9/2),x]

[Out]

(8*EllipticE[(e - Pi/2 + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(15*d^4*f*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[e + f*x]]) +
(2*(b*Tan[e + f*x])^(3/2))/(9*b*f*(d*Sec[e + f*x])^(9/2)) + (4*(b*Tan[e + f*x])^(3/2))/(15*b*d^2*f*(d*Sec[e +
f*x])^(5/2))

Rule 2612

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*m), x] + Dist[(m + n + 1)/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[
e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integer
sQ[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 \frac{\sqrt{b \tan (e+f x)}}{(d \sec (e+f x))^{9/2}} \, dx &=\frac{2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}+\frac{2 \int \frac{\sqrt{b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx}{3 d^2}\\ &=\frac{2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}+\frac{4 (b \tan (e+f x))^{3/2}}{15 b d^2 f (d \sec (e+f x))^{5/2}}+\frac{4 \int \frac{\sqrt{b \tan (e+f x)}}{\sqrt{d \sec (e+f x)}} \, dx}{15 d^4}\\ &=\frac{2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}+\frac{4 (b \tan (e+f x))^{3/2}}{15 b d^2 f (d \sec (e+f x))^{5/2}}+\frac{\left (4 \sqrt{b \tan (e+f x)}\right ) \int \sqrt{b \sin (e+f x)} \, dx}{15 d^4 \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=\frac{2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}+\frac{4 (b \tan (e+f x))^{3/2}}{15 b d^2 f (d \sec (e+f x))^{5/2}}+\frac{\left (4 \sqrt{b \tan (e+f x)}\right ) \int \sqrt{\sin (e+f x)} \, dx}{15 d^4 \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}}\\ &=\frac{8 E\left (\left .\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{15 d^4 f \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}}+\frac{2 (b \tan (e+f x))^{3/2}}{9 b f (d \sec (e+f x))^{9/2}}+\frac{4 (b \tan (e+f x))^{3/2}}{15 b d^2 f (d \sec (e+f x))^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.931741, size = 92, normalized size = 0.7 \[ \frac{b \sin ^2(e+f x) (5 \cos (2 (e+f x))+17)-24 b \sqrt [4]{-\tan ^2(e+f x)} \, _2F_1\left (-\frac{1}{4},\frac{1}{4};\frac{3}{4};\sec ^2(e+f x)\right )}{45 d^4 f \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[b*Tan[e + f*x]]/(d*Sec[e + f*x])^(9/2),x]

[Out]

(b*(17 + 5*Cos[2*(e + f*x)])*Sin[e + f*x]^2 - 24*b*Hypergeometric2F1[-1/4, 1/4, 3/4, Sec[e + f*x]^2]*(-Tan[e +
f*x]^2)^(1/4))/(45*d^4*f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]])

________________________________________________________________________________________

Maple [C]  time = 0.216, size = 585, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(9/2),x)

[Out]

-1/45/f*2^(1/2)*(5*cos(f*x+e)^5*2^(1/2)-12*cos(f*x+e)*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(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)*EllipticF(((I*cos(f*x+e)-I+sin(f*x
+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))+24*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)^3*2^(1/2)-12*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*co
s(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))
*(-I*(cos(f*x+e)-1)/sin(f*x+e))^(1/2)+24*((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)*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
)-1)/sin(f*x+e))^(1/2)+6*cos(f*x+e)*2^(1/2)-12*2^(1/2))*(b*sin(f*x+e)/cos(f*x+e))^(1/2)/(d/cos(f*x+e))^(9/2)/c
os(f*x+e)^4/sin(f*x+e)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \tan \left (f x + e\right )}}{\left (d \sec \left (f x + e\right )\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(f*x + e))/(d*sec(f*x + e))^(9/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \sec \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )}}{d^{5} \sec \left (f x + e\right )^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*sec(f*x + e))*sqrt(b*tan(f*x + e))/(d^5*sec(f*x + e)^5), x)

________________________________________________________________________________________

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((b*tan(f*x+e))**(1/2)/(d*sec(f*x+e))**(9/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b \tan \left (f x + e\right )}}{\left (d \sec \left (f x + e\right )\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(9/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*tan(f*x + e))/(d*sec(f*x + e))^(9/2), x)