3.316 \(\int \frac{(d \sec (e+f x))^{5/2}}{\sqrt{b \tan (e+f x)}} \, dx\)

Optimal. Leaf size=92 \[ \frac{d^2 \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)}}{b f}+\frac{d^2 \sqrt{\sin (e+f x)} F\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{d \sec (e+f x)}}{f \sqrt{b \tan (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.112805, antiderivative size = 92, 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, 2642, 2641} \[ \frac{d^2 \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)}}{b f}+\frac{d^2 \sqrt{\sin (e+f x)} F\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{d \sec (e+f x)}}{f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(5/2)/Sqrt[b*Tan[e + f*x]],x]

[Out]

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

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 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{(d \sec (e+f x))^{5/2}}{\sqrt{b \tan (e+f x)}} \, dx &=\frac{d^2 \sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}}{b f}+\frac{1}{2} d^2 \int \frac{\sqrt{d \sec (e+f x)}}{\sqrt{b \tan (e+f x)}} \, dx\\ &=\frac{d^2 \sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}}{b f}+\frac{\left (d^2 \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}\right ) \int \frac{1}{\sqrt{b \sin (e+f x)}} \, dx}{2 \sqrt{b \tan (e+f x)}}\\ &=\frac{d^2 \sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}}{b f}+\frac{\left (d^2 \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}\right ) \int \frac{1}{\sqrt{\sin (e+f x)}} \, dx}{2 \sqrt{b \tan (e+f x)}}\\ &=\frac{d^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)}}{f \sqrt{b \tan (e+f x)}}+\frac{d^2 \sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}}{b f}\\ \end{align*}

Mathematica [C]  time = 2.34878, size = 83, normalized size = 0.9 \[ \frac{d^2 \sqrt{d \sec (e+f x)} \left (\sin (e+f x) \cos (e+f x) \sec ^2(e+f x)^{3/4} \, _2F_1\left (\frac{1}{4},\frac{3}{4};\frac{5}{4};-\tan ^2(e+f x)\right )+\tan (e+f x)\right )}{f \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(5/2)/Sqrt[b*Tan[e + f*x]],x]

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.21, size = 208, normalized size = 2.3 \begin{align*} -{\frac{\sqrt{2}\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2\,f \left ( \cos \left ( fx+e \right ) -1 \right ) } \left ( i\cos \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 ) }}}\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 ) \sqrt{{\frac{-i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }}}-\cos \left ( fx+e \right ) \sqrt{2}+\sqrt{2} \right ) \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}}{\frac{1}{\sqrt{{\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(1/2),x)

[Out]

-1/2/f*2^(1/2)*(I*cos(f*x+e)*sin(f*x+e)*((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))*(-I*(cos(f*x+e)
-1)/sin(f*x+e))^(1/2)-cos(f*x+e)*2^(1/2)+2^(1/2))*(d/cos(f*x+e))^(5/2)*sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)-1)/(b
*sin(f*x+e)/cos(f*x+e))^(1/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \sec \left (f x + e\right )\right )^{\frac{5}{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))^(5/2)/(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(5/2)/sqrt(b*tan(f*x + e)), 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^{2} \sec \left (f x + e\right )^{2}}{b \tan \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))^(5/2)/(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^2*sec(f*x + e)^2/(b*tan(f*x + e)), 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((d*sec(f*x+e))**(5/2)/(b*tan(f*x+e))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \sec \left (f x + e\right )\right )^{\frac{5}{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))^(5/2)/(b*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(5/2)/sqrt(b*tan(f*x + e)), x)