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

Optimal. Leaf size=96 \[ \frac{2 b^2 \sqrt{\sin (e+f x)} F\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{d \sec (e+f x)}}{3 d^2 f \sqrt{b \tan (e+f x)}}-\frac{2 b \sqrt{b \tan (e+f x)}}{3 f (d \sec (e+f x))^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12114, antiderivative size = 96, 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 = {2610, 2616, 2642, 2641} \[ \frac{2 b^2 \sqrt{\sin (e+f x)} F\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{d \sec (e+f x)}}{3 d^2 f \sqrt{b \tan (e+f x)}}-\frac{2 b \sqrt{b \tan (e+f x)}}{3 f (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2610

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(n - 1))/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan
[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, 3/2]
)) && 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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

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

Rubi steps

\begin{align*} \int \frac{(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{3/2}} \, dx &=-\frac{2 b \sqrt{b \tan (e+f x)}}{3 f (d \sec (e+f x))^{3/2}}+\frac{b^2 \int \frac{\sqrt{d \sec (e+f x)}}{\sqrt{b \tan (e+f x)}} \, dx}{3 d^2}\\ &=-\frac{2 b \sqrt{b \tan (e+f x)}}{3 f (d \sec (e+f x))^{3/2}}+\frac{\left (b^2 \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}\right ) \int \frac{1}{\sqrt{b \sin (e+f x)}} \, dx}{3 d^2 \sqrt{b \tan (e+f x)}}\\ &=-\frac{2 b \sqrt{b \tan (e+f x)}}{3 f (d \sec (e+f x))^{3/2}}+\frac{\left (b^2 \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}\right ) \int \frac{1}{\sqrt{\sin (e+f x)}} \, dx}{3 d^2 \sqrt{b \tan (e+f x)}}\\ &=\frac{2 b^2 F\left (\left .\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )\right |2\right ) \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}}{3 d^2 f \sqrt{b \tan (e+f x)}}-\frac{2 b \sqrt{b \tan (e+f x)}}{3 f (d \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.595159, size = 98, normalized size = 1.02 \[ -\frac{2 b \sqrt{b \tan (e+f x)} \left (\sqrt{\sec (e+f x)+1}-\sqrt{2} \sec ^{\frac{3}{2}}(e+f x) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )}{3 f \sqrt{\sec (e+f x)+1} (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.202, size = 207, normalized size = 2.2 \begin{align*} -{\frac{\sqrt{2}}{3\,f \left ( \cos \left ( fx+e \right ) -1 \right ) \sin \left ( fx+e \right ) } \left ( i\sin \left ( fx+e \right ) \sqrt{{\frac{-i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{-{\frac{i\cos \left ( fx+e \right ) -i-\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{\sqrt{2}}{2}} \right ) + \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sqrt{2}-\cos \left ( fx+e \right ) \sqrt{2} \right ) \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}} \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/f*2^(1/2)*(I*sin(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*2^(1/2)-cos(f*x+e)*2^(1/2))*(b*sin(f*x+e)/cos(f*x+e))^(3/2)/(cos(f*x+e)-1)/(d/cos(f*x+
e))^(3/2)/sin(f*x+e)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e))^(3/2)/(d*sec(f*x + e))^(3/2), 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 )} b \tan \left (f x + e\right )}{d^{2} \sec \left (f x + e\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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