3.26 \(\int \sqrt{b \tan ^2(c+d x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0166386, 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} \[ -\frac{\cot (c+d x) \sqrt{b \tan ^2(c+d x)} \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Tan[c + d*x]^2],x]

[Out]

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

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

Mathematica [A]  time = 0.0389083, size = 32, normalized size = 1. \[ -\frac{\cot (c+d x) \sqrt{b \tan ^2(c+d x)} \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Tan[c + d*x]^2],x]

[Out]

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

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Maple [A]  time = 0.026, size = 37, normalized size = 1.2 \begin{align*}{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d\tan \left ( dx+c \right ) }\sqrt{b \left ( \tan \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(d*x+c)^2)^(1/2),x)

[Out]

1/2/d*(b*tan(d*x+c)^2)^(1/2)/tan(d*x+c)*ln(1+tan(d*x+c)^2)

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Maxima [A]  time = 1.57723, size = 26, normalized size = 0.81 \begin{align*} \frac{\sqrt{b} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(b)*log(tan(d*x + c)^2 + 1)/d

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Fricas [A]  time = 1.32182, size = 100, normalized size = 3.12 \begin{align*} -\frac{\sqrt{b \tan \left (d x + c\right )^{2}} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, d \tan \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(b*tan(d*x + c)^2)*log(1/(tan(d*x + c)^2 + 1))/(d*tan(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(d*x+c)**2)**(1/2),x)

[Out]

Integral(sqrt(b*tan(c + d*x)**2), x)

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Giac [A]  time = 1.44285, size = 35, normalized size = 1.09 \begin{align*} \frac{\sqrt{b} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) \mathrm{sgn}\left (\tan \left (d x + c\right )\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(b)*log(tan(d*x + c)^2 + 1)*sgn(tan(d*x + c))/d