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

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Rubi [A]  time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.00, 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 3475

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

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
]])

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*}

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Mathematica [A]  time = 0.04, size = 32, normalized size = 1.00 \[ -\frac {\cot (e+f x) \sqrt {b \tan ^2(e+f x)} \log (\cos (e+f x))}{f} \]

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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fricas [A]  time = 0.41, size = 38, normalized size = 1.19 \[ -\frac {\sqrt {b \tan \left (f x + e\right )^{2}} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, f \tan \left (f x + e\right )} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)-sqrt(b)*sign(tan(f*x+exp(1)))*ln(abs(cos(f*x+exp(1))))/f

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maple [A]  time = 0.32, size = 37, normalized size = 1.16 \[ \frac {\sqrt {b \left (\tan ^{2}\left (f x +e \right )\right )}\, \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \tan \left (f x +e \right )} \]

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 = 0.69, size = 19, normalized size = 0.59 \[ \frac {\sqrt {b} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{2 \, f} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x)^2)^(1/2),x)

[Out]

int((b*tan(e + f*x)^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \tan ^{2}{\left (e + f x \right )}}\, dx \]

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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