3.4 \(\int \frac{1}{\sqrt{b \tan ^2(e+f x)}} \, dx\)

Optimal. Leaf size=31 \[ \frac{\tan (e+f x) \log (\sin (e+f x))}{f \sqrt{b \tan ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.0230007, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3658, 3475} \[ \frac{\tan (e+f x) \log (\sin (e+f x))}{f \sqrt{b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*Tan[e + f*x]^2],x]

[Out]

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

Mathematica [A]  time = 0.0816531, size = 39, normalized size = 1.26 \[ \frac{\tan (e+f x) (\log (\tan (e+f x))+\log (\cos (e+f x)))}{f \sqrt{b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*Tan[e + f*x]^2],x]

[Out]

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

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Maple [A]  time = 0.043, size = 47, normalized size = 1.5 \begin{align*}{\frac{\tan \left ( fx+e \right ) \left ( 2\,\ln \left ( \tan \left ( fx+e \right ) \right ) -\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \right ) }{2\,f}{\frac{1}{\sqrt{b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.63721, size = 45, normalized size = 1.45 \begin{align*} -\frac{\frac{\log \left (\tan \left (f x + e\right )^{2} + 1\right )}{\sqrt{b}} - \frac{2 \, \log \left (\tan \left (f x + e\right )\right )}{\sqrt{b}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.85147, size = 119, normalized size = 3.84 \begin{align*} \frac{\sqrt{b \tan \left (f x + e\right )^{2}} \log \left (\frac{\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, b f \tan \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(b*tan(e + f*x)**2), x)

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Giac [B]  time = 1.32646, size = 113, normalized size = 3.65 \begin{align*} -\frac{\frac{2 \, \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1\right )}{\sqrt{b} \mathrm{sgn}\left (\tan \left (f x + e\right )\right )} - \frac{\log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}{\sqrt{b} \mathrm{sgn}\left (\tan \left (f x + e\right )\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1)/(sqrt(b)*sgn(tan(f*x + e))) - log(-(cos(f*x + e) - 1)/
(cos(f*x + e) + 1))/(sqrt(b)*sgn(tan(f*x + e))))/f