3.27 \(\int \frac{1}{\sqrt{b \tan ^2(c+d x)}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0160449, 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} \[ \frac{\tan (c+d x) \log (\sin (c+d x))}{d \sqrt{b \tan ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*Tan[c + d*x]^2],x]

[Out]

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

Mathematica [A]  time = 0.0835878, size = 39, normalized size = 1.26 \[ \frac{\tan (c+d x) (\log (\tan (c+d x))+\log (\cos (c+d x)))}{d \sqrt{b \tan ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*Tan[c + d*x]^2],x]

[Out]

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

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Maple [A]  time = 0.025, size = 45, normalized size = 1.5 \begin{align*} -{\frac{\tan \left ( dx+c \right ) \left ( \ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) -2\,\ln \left ( \tan \left ( dx+c \right ) \right ) \right ) }{2\,d}{\frac{1}{\sqrt{b \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.61233, size = 45, normalized size = 1.45 \begin{align*} -\frac{\frac{\log \left (\tan \left (d x + c\right )^{2} + 1\right )}{\sqrt{b}} - \frac{2 \, \log \left (\tan \left (d x + c\right )\right )}{\sqrt{b}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.31158, size = 119, normalized size = 3.84 \begin{align*} \frac{\sqrt{b \tan \left (d x + c\right )^{2}} \log \left (\frac{\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, b d \tan \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.61994, size = 109, normalized size = 3.52 \begin{align*} \frac{\frac{\log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{\sqrt{b} \mathrm{sgn}\left (\tan \left (d x + c\right )\right )} - \frac{2 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{\sqrt{b} \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(1/(b*tan(d*x+c)^2)^(1/2),x, algorithm="giac")

[Out]

1/2*(log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(sqrt(b)*sgn(tan(d*x + c))) - 2*log(abs(-(cos(d*x + c)
- 1)/(cos(d*x + c) + 1) + 1))/(sqrt(b)*sgn(tan(d*x + c))))/d