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

Optimal. Leaf size=101 \[ \frac{2 d^2 \sqrt{\sin (e+f x)} F\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{d \sec (e+f x)}}{3 b^2 f \sqrt{b \tan (e+f x)}}-\frac{2 d^2 \sqrt{d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.126031, antiderivative size = 101, 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 = {2608, 2616, 2642, 2641} \[ \frac{2 d^2 \sqrt{\sin (e+f x)} F\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{d \sec (e+f x)}}{3 b^2 f \sqrt{b \tan (e+f x)}}-\frac{2 d^2 \sqrt{d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2608

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*(n + 1)), x] - Dist[(a^2*(m - 2))/(b^2*(n + 1)), Int[(a*Sec[e
+ f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && (GtQ[m, 1] || (Eq
Q[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{(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{5/2}} \, dx &=-\frac{2 d^2 \sqrt{d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}+\frac{d^2 \int \frac{\sqrt{d \sec (e+f x)}}{\sqrt{b \tan (e+f x)}} \, dx}{3 b^2}\\ &=-\frac{2 d^2 \sqrt{d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}+\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}{3 b^2 \sqrt{b \tan (e+f x)}}\\ &=-\frac{2 d^2 \sqrt{d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}+\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}{3 b^2 \sqrt{b \tan (e+f x)}}\\ &=-\frac{2 d^2 \sqrt{d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}+\frac{2 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)}}{3 b^2 f \sqrt{b \tan (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.443671, size = 116, normalized size = 1.15 \[ \frac{2 d^3 \sqrt{b \tan (e+f x)} \left (\sqrt{2} \sqrt{\sec (e+f x)} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-\cot (e+f x) \csc (e+f x) \sqrt{\sec (e+f x)+1}\right )}{3 b^3 f \sqrt{\sec (e+f x)+1} \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.205, size = 315, normalized size = 3.1 \begin{align*}{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{3\,f} \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 ) }}}+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 ) -\cos \left ( fx+e \right ) \sqrt{2} \right ) \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}} \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}}} \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))^(5/2),x)

[Out]

1/3/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)+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^(1/2))*(d/cos(f*x+e))^(5/2)*sin(f*x+e)/(b*sin(f*x+e)/cos(f*x+e))^(5/2)

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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}}}{\left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}}\,{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))^(5/2),x, algorithm="maxima")

[Out]

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

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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^{3} \tan \left (f x + e\right )^{3}}, 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))^(5/2),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^3*tan(f*x + e)^3), 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))**(5/2)/(b*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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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}}}{\left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}}\,{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))^(5/2),x, algorithm="giac")

[Out]

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