3.26 \(\int \frac {1}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))} \, dx\)

Optimal. Leaf size=83 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{a d \sqrt {e}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} (\cot (c+d x)+1)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d \sqrt {e}} \]

[Out]

-arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a/d/e^(1/2)-1/2*arctanh(1/2*(1+cot(d*x+c))*e^(1/2)*2^(1/2)/(e*cot(d*x+c)
)^(1/2))/a/d*2^(1/2)/e^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3574, 3532, 208, 3634, 63, 205} \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{a d \sqrt {e}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} (\cot (c+d x)+1)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d \sqrt {e}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[e*Cot[c + d*x]]*(a + a*Cot[c + d*x])),x]

[Out]

-(ArcTan[Sqrt[e*Cot[c + d*x]]/Sqrt[e]]/(a*d*Sqrt[e])) - ArcTanh[(Sqrt[e]*(1 + Cot[c + d*x]))/(Sqrt[2]*Sqrt[e*C
ot[c + d*x]])]/(Sqrt[2]*a*d*Sqrt[e])

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 205

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

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]

Rule 3532

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-2*d^2)/f,
Subst[Int[1/(2*c*d + b*x^2), x], x, (c - d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x
] && EqQ[c^2 - d^2, 0]

Rule 3574

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/
(c^2 + d^2), Int[(a + b*Tan[e + f*x])^m*(c - d*Tan[e + f*x]), x], x] + Dist[d^2/(c^2 + d^2), Int[((a + b*Tan[e
+ f*x])^m*(1 + Tan[e + f*x]^2))/(c + d*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rubi steps

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

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Mathematica [A]  time = 0.52, size = 107, normalized size = 1.29 \[ -\frac {\sqrt {\cot (c+d x)} \left (\sqrt {2} \left (\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-\log \left (-\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}-1\right )\right )+4 \tan ^{-1}\left (\sqrt {\cot (c+d x)}\right )\right )}{4 a d \sqrt {e \cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[e*Cot[c + d*x]]*(a + a*Cot[c + d*x])),x]

[Out]

-1/4*(Sqrt[Cot[c + d*x]]*(4*ArcTan[Sqrt[Cot[c + d*x]]] + Sqrt[2]*(-Log[-1 + Sqrt[2]*Sqrt[Cot[c + d*x]] - Cot[c
+ d*x]] + Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])))/(a*d*Sqrt[e*Cot[c + d*x]])

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fricas [A]  time = 0.62, size = 321, normalized size = 3.87 \[ \left [\frac {\sqrt {2} \sqrt {-e} \arctan \left (\frac {\sqrt {2} \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1\right )}}{2 \, {\left (e \cos \left (2 \, d x + 2 \, c\right ) + e\right )}}\right ) - \sqrt {-e} \log \left (\frac {e \cos \left (2 \, d x + 2 \, c\right ) - e \sin \left (2 \, d x + 2 \, c\right ) + 2 \, \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + e}{\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1}\right )}{2 \, a d e}, \frac {\sqrt {2} \sqrt {e} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) - \sin \left (2 \, d x + 2 \, c\right ) - 1\right )} + 2 \, e \sin \left (2 \, d x + 2 \, c\right ) + e\right ) - 4 \, \sqrt {e} \arctan \left (\frac {\sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{\sqrt {e}}\right )}{4 \, a d e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))^(1/2)/(a+a*cot(d*x+c)),x, algorithm="fricas")

[Out]

[1/2*(sqrt(2)*sqrt(-e)*arctan(1/2*sqrt(2)*sqrt(-e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))*(cos(2*d*x
+ 2*c) + sin(2*d*x + 2*c) + 1)/(e*cos(2*d*x + 2*c) + e)) - sqrt(-e)*log((e*cos(2*d*x + 2*c) - e*sin(2*d*x + 2*
c) + 2*sqrt(-e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))*sin(2*d*x + 2*c) + e)/(cos(2*d*x + 2*c) + sin(
2*d*x + 2*c) + 1)))/(a*d*e), 1/4*(sqrt(2)*sqrt(e)*log(sqrt(2)*sqrt(e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x
+ 2*c))*(cos(2*d*x + 2*c) - sin(2*d*x + 2*c) - 1) + 2*e*sin(2*d*x + 2*c) + e) - 4*sqrt(e)*arctan(sqrt((e*cos(2
*d*x + 2*c) + e)/sin(2*d*x + 2*c))/sqrt(e)))/(a*d*e)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \cot \left (d x + c\right ) + a\right )} \sqrt {e \cot \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))^(1/2)/(a+a*cot(d*x+c)),x, algorithm="giac")

[Out]

integrate(1/((a*cot(d*x + c) + a)*sqrt(e*cot(d*x + c))), x)

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maple [B]  time = 0.88, size = 365, normalized size = 4.40 \[ -\frac {\arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{a d \sqrt {e}}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{8 d a e}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a e}+\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a e}+\frac {\sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{8 d a \left (e^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a \left (e^{2}\right )^{\frac {1}{4}}}-\frac {\sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a \left (e^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*cot(d*x+c))^(1/2)/(a+cot(d*x+c)*a),x)

[Out]

-arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a/d/e^(1/2)-1/8/d/a/e*(e^2)^(1/4)*2^(1/2)*ln((e*cot(d*x+c)+(e^2)^(1/4)*(
e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*cot(d*x+c)-(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))-
1/4/d/a/e*(e^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+1/4/d/a/e*(e^2)^(1/4)*2^(1/2)
*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+1/8/d/a*2^(1/2)/(e^2)^(1/4)*ln((e*cot(d*x+c)-(e^2)^(1/4)*
(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*cot(d*x+c)+(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))
+1/4/d/a*2^(1/2)/(e^2)^(1/4)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)-1/4/d/a*2^(1/2)/(e^2)^(1/4)*ar
ctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)

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maxima [A]  time = 0.63, size = 120, normalized size = 1.45 \[ -\frac {e {\left (\frac {\frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}} - \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}}}{a e} + \frac {4 \, \arctan \left (\frac {\sqrt {\frac {e}{\tan \left (d x + c\right )}}}{\sqrt {e}}\right )}{a e^{\frac {3}{2}}}\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))^(1/2)/(a+a*cot(d*x+c)),x, algorithm="maxima")

[Out]

-1/4*e*((sqrt(2)*log(sqrt(2)*sqrt(e)*sqrt(e/tan(d*x + c)) + e + e/tan(d*x + c))/sqrt(e) - sqrt(2)*log(-sqrt(2)
*sqrt(e)*sqrt(e/tan(d*x + c)) + e + e/tan(d*x + c))/sqrt(e))/(a*e) + 4*arctan(sqrt(e/tan(d*x + c))/sqrt(e))/(a
*e^(3/2)))/d

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mupad [B]  time = 0.52, size = 79, normalized size = 0.95 \[ -\frac {\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{a\,d\,\sqrt {e}}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {12\,\sqrt {2}\,e^{9/2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{12\,e^5\,\mathrm {cot}\left (c+d\,x\right )+12\,e^5}\right )}{2\,a\,d\,\sqrt {e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*cot(c + d*x))^(1/2)*(a + a*cot(c + d*x))),x)

[Out]

- atan((e*cot(c + d*x))^(1/2)/e^(1/2))/(a*d*e^(1/2)) - (2^(1/2)*atanh((12*2^(1/2)*e^(9/2)*(e*cot(c + d*x))^(1/
2))/(12*e^5*cot(c + d*x) + 12*e^5)))/(2*a*d*e^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{\sqrt {e \cot {\left (c + d x \right )}} \cot {\left (c + d x \right )} + \sqrt {e \cot {\left (c + d x \right )}}}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))**(1/2)/(a+a*cot(d*x+c)),x)

[Out]

Integral(1/(sqrt(e*cot(c + d*x))*cot(c + d*x) + sqrt(e*cot(c + d*x))), x)/a

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