3.5 \(\int \frac{1}{(b \tan ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=66 \[ -\frac{\cot (e+f x)}{2 b f \sqrt{b \tan ^2(e+f x)}}-\frac{\tan (e+f x) \log (\sin (e+f x))}{b f \sqrt{b \tan ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0368908, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3658, 3473, 3475} \[ -\frac{\cot (e+f x)}{2 b f \sqrt{b \tan ^2(e+f x)}}-\frac{\tan (e+f x) \log (\sin (e+f x))}{b f \sqrt{b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^2)^(-3/2),x]

[Out]

-Cot[e + f*x]/(2*b*f*Sqrt[b*Tan[e + f*x]^2]) - (Log[Sin[e + f*x]]*Tan[e + f*x])/(b*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 )^{3/2}} \, dx &=\frac{\tan (e+f x) \int \cot ^3(e+f x) \, dx}{b \sqrt{b \tan ^2(e+f x)}}\\ &=-\frac{\cot (e+f x)}{2 b f \sqrt{b \tan ^2(e+f x)}}-\frac{\tan (e+f x) \int \cot (e+f x) \, dx}{b \sqrt{b \tan ^2(e+f x)}}\\ &=-\frac{\cot (e+f x)}{2 b f \sqrt{b \tan ^2(e+f x)}}-\frac{\log (\sin (e+f x)) \tan (e+f x)}{b f \sqrt{b \tan ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.391674, size = 56, normalized size = 0.85 \[ -\frac{\tan ^3(e+f x) \left (\cot ^2(e+f x)+2 \log (\tan (e+f x))+2 \log (\cos (e+f x))\right )}{2 f \left (b \tan ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.02, size = 64, normalized size = 1. \begin{align*} -{\frac{\tan \left ( fx+e \right ) \left ( 2\,\ln \left ( \tan \left ( fx+e \right ) \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{2}-\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) }{2\,f} \left ( b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(f*x+e)^2)^(3/2),x)

[Out]

-1/2/f*tan(f*x+e)*(2*ln(tan(f*x+e))*tan(f*x+e)^2-ln(1+tan(f*x+e)^2)*tan(f*x+e)^2+1)/(b*tan(f*x+e)^2)^(3/2)

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Maxima [A]  time = 1.64679, size = 62, normalized size = 0.94 \begin{align*} \frac{\frac{\log \left (\tan \left (f x + e\right )^{2} + 1\right )}{b^{\frac{3}{2}}} - \frac{2 \, \log \left (\tan \left (f x + e\right )\right )}{b^{\frac{3}{2}}} - \frac{1}{b^{\frac{3}{2}} \tan \left (f x + e\right )^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.89531, size = 177, normalized size = 2.68 \begin{align*} -\frac{\sqrt{b \tan \left (f x + e\right )^{2}}{\left (\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 )^{2} + \tan \left (f x + e\right )^{2} + 1\right )}}{2 \, b^{2} f \tan \left (f x + e\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

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

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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{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)**2)**(3/2),x)

[Out]

Integral((b*tan(e + f*x)**2)**(-3/2), x)

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Giac [B]  time = 1.62848, size = 300, normalized size = 4.55 \begin{align*} -\frac{\frac{\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 )\right )} - \frac{8 \, \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{4 \, \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{4 \, \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 )^{2}}}{8 \, b^{\frac{3}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

-1/8*(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)) - 8*log(tan(1/2*f*x +
1/2*e)^2 + 1)*sgn(-tan(1/2*f*x + 1/2*e)^2 + 1)/sgn(tan(1/2*f*x + 1/2*e)) + 4*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)) - (4*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)^2))/(b^(3/2)*f)