3.405 \(\int \frac{\cot (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx\)

Optimal. Leaf size=70 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} f}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a+b}}\right )}{f \sqrt{a+b}} \]

[Out]

ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]]/(Sqrt[a]*f) - ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a + b]]/(Sqr
t[a + b]*f)

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Rubi [A]  time = 0.106442, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4139, 446, 86, 63, 208} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} f}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a+b}}\right )}{f \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]]/(Sqrt[a]*f) - ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a + b]]/(Sqr
t[a + b]*f)

Rule 4139

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sec[e + f*x], x]}, Dist[1/f, Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p)/x
, x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[
n, 2] || EqQ[n, 4] || IGtQ[p, 0] || IntegersQ[2*n, p])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 86

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), In
t[(e + f*x)^p/(a + b*x), x], x] - Dist[d/(b*c - a*d), Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] &&  !IntegerQ[p]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\cot (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x \left (-1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{(-1+x) x \sqrt{a+b x}} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a+b x}} \, dx,x,\sec ^2(e+f x)\right )}{2 f}-\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{-1-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sec ^2(e+f x)}\right )}{b f}-\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sec ^2(e+f x)}\right )}{b f}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} f}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a+b}}\right )}{\sqrt{a+b} f}\\ \end{align*}

Mathematica [F]  time = 2.35647, size = 0, normalized size = 0. \[ \int \frac{\cot (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Cot[e + f*x]/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

Integrate[Cot[e + f*x]/Sqrt[a + b*Sec[e + f*x]^2], x]

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Maple [B]  time = 0.407, size = 376, normalized size = 5.4 \begin{align*}{\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{2\,f\cos \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) }\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}} \left ( \ln \left ( -4\,{\frac{1}{-1+\cos \left ( fx+e \right ) } \left ( \cos \left ( fx+e \right ) \sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}}\sqrt{a+b}+a\cos \left ( fx+e \right ) +\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}}\sqrt{a+b}+b \right ) } \right ) \sqrt{a}-\ln \left ( -2\,{\frac{-1+\cos \left ( fx+e \right ) }{\sqrt{a+b} \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( \cos \left ( fx+e \right ) \sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}}\sqrt{a+b}-a\cos \left ( fx+e \right ) +\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}}\sqrt{a+b}+b \right ) } \right ) \sqrt{a}-2\,\ln \left ( 4\,\cos \left ( fx+e \right ) \sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}}\sqrt{a}+4\,a\cos \left ( fx+e \right ) +4\,\sqrt{a}\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}} \right ) \sqrt{a+b} \right ){\frac{1}{\sqrt{a+b}}}{\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

1/2/f/(a+b)^(1/2)/a^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(ln(-4*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(
1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/(-1
+cos(f*x+e)))*a^(1/2)-ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2
)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a^(1/2)-2*
ln(4*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)+4*a*cos(f*x+e)+4*a^(1/2)*((b+a*cos(f*x+e)^
2)/(1+cos(f*x+e))^2)^(1/2))*(a+b)^(1/2))*sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(1/2)/cos(f*x+e)/(-1+c
os(f*x+e))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )}{\sqrt{b \sec \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cot(f*x + e)/sqrt(b*sec(f*x + e)^2 + a), x)

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Fricas [B]  time = 1.1719, size = 2527, normalized size = 36.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*((a + b)*sqrt(a)*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*cos(f*x + e)^6 + 160*a^2*b^2*cos(f*x + e)^4 + 32*
a*b^3*cos(f*x + e)^2 + b^4 + 8*(16*a^3*cos(f*x + e)^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x + e)^4 + b^
3*cos(f*x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) + 2*sqrt(a + b)*a*log(2*((8*a^2 + 8*a*b
+ b^2)*cos(f*x + e)^4 + 2*(4*a*b + 3*b^2)*cos(f*x + e)^2 + b^2 - 4*((2*a + b)*cos(f*x + e)^4 + b*cos(f*x + e)
^2)*sqrt(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)))/((a^2 +
a*b)*f), 1/8*(4*a*sqrt(-a - b)*arctan(1/2*((2*a + b)*cos(f*x + e)^2 + b)*sqrt(-a - b)*sqrt((a*cos(f*x + e)^2
+ b)/cos(f*x + e)^2)/((a^2 + a*b)*cos(f*x + e)^2 + a*b + b^2)) + (a + b)*sqrt(a)*log(128*a^4*cos(f*x + e)^8 +
256*a^3*b*cos(f*x + e)^6 + 160*a^2*b^2*cos(f*x + e)^4 + 32*a*b^3*cos(f*x + e)^2 + b^4 + 8*(16*a^3*cos(f*x + e)
^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x + e)^4 + b^3*cos(f*x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 +
b)/cos(f*x + e)^2)))/((a^2 + a*b)*f), -1/4*(sqrt(-a)*(a + b)*arctan(1/4*(8*a^2*cos(f*x + e)^4 + 8*a*b*cos(f*x
+ e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(2*a^3*cos(f*x + e)^4 + 3*a^2*b*cos(f*x + e
)^2 + a*b^2)) - sqrt(a + b)*a*log(2*((8*a^2 + 8*a*b + b^2)*cos(f*x + e)^4 + 2*(4*a*b + 3*b^2)*cos(f*x + e)^2 +
b^2 - 4*((2*a + b)*cos(f*x + e)^4 + b*cos(f*x + e)^2)*sqrt(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)
)/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)))/((a^2 + a*b)*f), -1/4*(sqrt(-a)*(a + b)*arctan(1/4*(8*a^2*cos(f*x
+ e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(2*a^3*cos(f*x + e)^
4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2)) - 2*a*sqrt(-a - b)*arctan(1/2*((2*a + b)*cos(f*x + e)^2 + b)*sqrt(-a - b)
*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a^2 + a*b)*cos(f*x + e)^2 + a*b + b^2)))/((a^2 + a*b)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Integral(cot(e + f*x)/sqrt(a + b*sec(e + f*x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )}{\sqrt{b \sec \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cot(f*x + e)/sqrt(b*sec(f*x + e)^2 + a), x)