3.6 \(\int \frac{1}{(b \tan ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=97 \[ -\frac{\cot ^3(e+f x)}{4 b^2 f \sqrt{b \tan ^2(e+f x)}}+\frac{\cot (e+f x)}{2 b^2 f \sqrt{b \tan ^2(e+f x)}}+\frac{\tan (e+f x) \log (\sin (e+f x))}{b^2 f \sqrt{b \tan ^2(e+f x)}} \]

[Out]

Cot[e + f*x]/(2*b^2*f*Sqrt[b*Tan[e + f*x]^2]) - Cot[e + f*x]^3/(4*b^2*f*Sqrt[b*Tan[e + f*x]^2]) + (Log[Sin[e +
f*x]]*Tan[e + f*x])/(b^2*f*Sqrt[b*Tan[e + f*x]^2])

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Rubi [A]  time = 0.0390056, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3658, 3473, 3475} \[ -\frac{\cot ^3(e+f x)}{4 b^2 f \sqrt{b \tan ^2(e+f x)}}+\frac{\cot (e+f x)}{2 b^2 f \sqrt{b \tan ^2(e+f x)}}+\frac{\tan (e+f x) \log (\sin (e+f x))}{b^2 f \sqrt{b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^2)^(-5/2),x]

[Out]

Cot[e + f*x]/(2*b^2*f*Sqrt[b*Tan[e + f*x]^2]) - Cot[e + f*x]^3/(4*b^2*f*Sqrt[b*Tan[e + f*x]^2]) + (Log[Sin[e +
f*x]]*Tan[e + f*x])/(b^2*f*Sqrt[b*Tan[e + f*x]^2])

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{1}{\left (b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\tan (e+f x) \int \cot ^5(e+f x) \, dx}{b^2 \sqrt{b \tan ^2(e+f x)}}\\ &=-\frac{\cot ^3(e+f x)}{4 b^2 f \sqrt{b \tan ^2(e+f x)}}-\frac{\tan (e+f x) \int \cot ^3(e+f x) \, dx}{b^2 \sqrt{b \tan ^2(e+f x)}}\\ &=\frac{\cot (e+f x)}{2 b^2 f \sqrt{b \tan ^2(e+f x)}}-\frac{\cot ^3(e+f x)}{4 b^2 f \sqrt{b \tan ^2(e+f x)}}+\frac{\tan (e+f x) \int \cot (e+f x) \, dx}{b^2 \sqrt{b \tan ^2(e+f x)}}\\ &=\frac{\cot (e+f x)}{2 b^2 f \sqrt{b \tan ^2(e+f x)}}-\frac{\cot ^3(e+f x)}{4 b^2 f \sqrt{b \tan ^2(e+f x)}}+\frac{\log (\sin (e+f x)) \tan (e+f x)}{b^2 f \sqrt{b \tan ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.246315, size = 68, normalized size = 0.7 \[ \frac{\tan ^5(e+f x) \left (-\cot ^4(e+f x)+2 \cot ^2(e+f x)+4 \log (\tan (e+f x))+4 \log (\cos (e+f x))\right )}{4 f \left (b \tan ^2(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Cot[e + f*x]^2 - Cot[e + f*x]^4 + 4*Log[Cos[e + f*x]] + 4*Log[Tan[e + f*x]])*Tan[e + f*x]^5)/(4*f*(b*Tan[e
+ f*x]^2)^(5/2))

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Maple [A]  time = 0.023, size = 74, normalized size = 0.8 \begin{align*}{\frac{\tan \left ( fx+e \right ) \left ( 4\,\ln \left ( \tan \left ( fx+e \right ) \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{4}-2\,\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{4}+2\, \left ( \tan \left ( fx+e \right ) \right ) ^{2}-1 \right ) }{4\,f} \left ( b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/f*tan(f*x+e)*(4*ln(tan(f*x+e))*tan(f*x+e)^4-2*ln(1+tan(f*x+e)^2)*tan(f*x+e)^4+2*tan(f*x+e)^2-1)/(b*tan(f*x
+e)^2)^(5/2)

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Maxima [A]  time = 1.68978, size = 89, normalized size = 0.92 \begin{align*} -\frac{\frac{2 \, \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{b^{\frac{5}{2}}} - \frac{4 \, \log \left (\tan \left (f x + e\right )\right )}{b^{\frac{5}{2}}} - \frac{2 \, \sqrt{b} \tan \left (f x + e\right )^{2} - \sqrt{b}}{b^{3} \tan \left (f x + e\right )^{4}}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

-1/4*(2*log(tan(f*x + e)^2 + 1)/b^(5/2) - 4*log(tan(f*x + e))/b^(5/2) - (2*sqrt(b)*tan(f*x + e)^2 - sqrt(b))/(
b^3*tan(f*x + e)^4))/f

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Fricas [A]  time = 2.05207, size = 207, normalized size = 2.13 \begin{align*} \frac{{\left (2 \, \log \left (\frac{\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{4} + 3 \, \tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} - 1\right )} \sqrt{b \tan \left (f x + e\right )^{2}}}{4 \, b^{3} f \tan \left (f x + e\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

1/4*(2*log(tan(f*x + e)^2/(tan(f*x + e)^2 + 1))*tan(f*x + e)^4 + 3*tan(f*x + e)^4 + 2*tan(f*x + e)^2 - 1)*sqrt
(b*tan(f*x + e)^2)/(b^3*f*tan(f*x + e)^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)**2)**(5/2),x)

[Out]

Integral((b*tan(e + f*x)**2)**(-5/2), x)

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Giac [B]  time = 1.99926, size = 392, normalized size = 4.04 \begin{align*} -\frac{\mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 12 \, \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + \frac{64 \, \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right ) \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )} - \frac{32 \, \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2}\right ) \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )} + \frac{48 \, \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 12 \, \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4}}}{64 \, b^{\frac{5}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

-1/64*(sgn(-tan(1/2*f*x + 1/2*e)^2 + 1)*sgn(tan(1/2*f*x + 1/2*e))*tan(1/2*f*x + 1/2*e)^4 - 12*sgn(-tan(1/2*f*x
+ 1/2*e)^2 + 1)*sgn(tan(1/2*f*x + 1/2*e))*tan(1/2*f*x + 1/2*e)^2 + 64*log(tan(1/2*f*x + 1/2*e)^2 + 1)*sgn(-ta
n(1/2*f*x + 1/2*e)^2 + 1)/sgn(tan(1/2*f*x + 1/2*e)) - 32*log(tan(1/2*f*x + 1/2*e)^2)*sgn(-tan(1/2*f*x + 1/2*e)
^2 + 1)/sgn(tan(1/2*f*x + 1/2*e)) + (48*sgn(-tan(1/2*f*x + 1/2*e)^2 + 1)*tan(1/2*f*x + 1/2*e)^4 - 12*sgn(-tan(
1/2*f*x + 1/2*e)^2 + 1)*tan(1/2*f*x + 1/2*e)^2 + sgn(-tan(1/2*f*x + 1/2*e)^2 + 1))/(sgn(tan(1/2*f*x + 1/2*e))*
tan(1/2*f*x + 1/2*e)^4))/(b^(5/2)*f)