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

Optimal. Leaf size=32 \[ -\frac{\cot (e+f x) \sqrt{b \tan ^2(e+f x)} \log (\cos (e+f x))}{f} \]

[Out]

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

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Rubi [A]  time = 0.0172923, antiderivative size = 32, 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} \[ -\frac{\cot (e+f x) \sqrt{b \tan ^2(e+f x)} \log (\cos (e+f x))}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Tan[e + f*x]^2],x]

[Out]

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

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 \sqrt{b \tan ^2(e+f x)} \, dx &=\left (\cot (e+f x) \sqrt{b \tan ^2(e+f x)}\right ) \int \tan (e+f x) \, dx\\ &=-\frac{\cot (e+f x) \log (\cos (e+f x)) \sqrt{b \tan ^2(e+f x)}}{f}\\ \end{align*}

Mathematica [A]  time = 0.0392099, size = 32, normalized size = 1. \[ -\frac{\cot (e+f x) \sqrt{b \tan ^2(e+f x)} \log (\cos (e+f x))}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Tan[e + f*x]^2],x]

[Out]

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

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Maple [A]  time = 0.024, size = 37, normalized size = 1.2 \begin{align*}{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f\tan \left ( fx+e \right ) }\sqrt{b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.63487, size = 26, normalized size = 0.81 \begin{align*} \frac{\sqrt{b} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(b)*log(tan(f*x + e)^2 + 1)/f

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Fricas [A]  time = 1.9592, size = 100, normalized size = 3.12 \begin{align*} -\frac{\sqrt{b \tan \left (f x + e\right )^{2}} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, f \tan \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(b*tan(e + f*x)**2), x)

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Giac [A]  time = 1.25947, size = 38, normalized size = 1.19 \begin{align*} \frac{\sqrt{b} \log \left (\tan \left (f x + e\right )^{2} + 1\right ) \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((b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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