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

Optimal. Leaf size=87 \[ \frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{a d}+\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3572, 3532, 205, 3634, 63} \[ \frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{a d}+\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 3572

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

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

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Mathematica [A]  time = 0.25, size = 98, normalized size = 1.13 \[ \frac {\sqrt {e \cot (c+d x)} \left (\sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-\sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+2 \tan ^{-1}\left (\sqrt {\cot (c+d x)}\right )\right )}{2 a d \sqrt {\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.84, size = 331, normalized size = 3.80 \[ \left [\frac {\sqrt {2} \sqrt {-e} \log \left (-{\left (\sqrt {2} \cos \left (2 \, d x + 2 \, c\right ) + \sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) - \sqrt {2}\right )} \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} - 2 \, e \sin \left (2 \, d x + 2 \, c\right ) + e\right ) + 2 \, \sqrt {-e} \log \left (\frac {e \cos \left (2 \, d x + 2 \, c\right ) - e \sin \left (2 \, d x + 2 \, c\right ) + 2 \, \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + e}{\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1}\right )}{4 \, a d}, \frac {\sqrt {2} \sqrt {e} \arctan \left (-\frac {{\left (\sqrt {2} \cos \left (2 \, d x + 2 \, c\right ) - \sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) + \sqrt {2}\right )} \sqrt {e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{2 \, {\left (e \cos \left (2 \, d x + 2 \, c\right ) + e\right )}}\right ) + 2 \, \sqrt {e} \arctan \left (\frac {\sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{\sqrt {e}}\right )}{2 \, a d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.81, size = 358, normalized size = 4.11 \[ \frac {\arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right ) \sqrt {e}}{d a}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{8 d a}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a}+\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a}-\frac {e \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{8 d a \left (e^{2}\right )^{\frac {1}{4}}}-\frac {e \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a \left (e^{2}\right )^{\frac {1}{4}}}+\frac {e \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a \left (e^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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))*e^(1/2)/d/a-1/8/d/a*(e^2)^(1/4)*2^(1/2)*ln((e*cot(d*x+c)+(e^2)^(1/4)*(e*c
ot(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^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^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*e*2^(1/2)/(e^2)^(1/4)*ln((e*cot(d*x+c)-(e^2)^(1/4)*(e*co
t(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*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*e*2^(1/2)/(e^2)^(1/4)*arc
tan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)

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maxima [A]  time = 0.81, size = 113, normalized size = 1.30 \[ -\frac {e {\left (\frac {\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}}}{a} - \frac {2 \, \arctan \left (\frac {\sqrt {\frac {e}{\tan \left (d x + c\right )}}}{\sqrt {e}}\right )}{a \sqrt {e}}\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*e*((sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(e) + 2*sqrt(e/tan(d*x + c)))/sqrt(e))/sqrt(e) + sqrt(2)*arct
an(-1/2*sqrt(2)*(sqrt(2)*sqrt(e) - 2*sqrt(e/tan(d*x + c)))/sqrt(e))/sqrt(e))/a - 2*arctan(sqrt(e/tan(d*x + c))
/sqrt(e))/(a*sqrt(e)))/d

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mupad [B]  time = 0.37, size = 102, normalized size = 1.17 \[ \frac {\sqrt {e}\,\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{a\,d}-\frac {\sqrt {2}\,\sqrt {e}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{2\,\sqrt {e}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{2\,\sqrt {e}}+\frac {\sqrt {2}\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}{2\,e^{3/2}}\right )\right )}{4\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^(1/2)*atan((e*cot(c + d*x))^(1/2)/e^(1/2)))/(a*d) - (2^(1/2)*e^(1/2)*(2*atan((2^(1/2)*(e*cot(c + d*x))^(1/2
))/(2*e^(1/2))) + 2*atan((2^(1/2)*(e*cot(c + d*x))^(1/2))/(2*e^(1/2)) + (2^(1/2)*(e*cot(c + d*x))^(3/2))/(2*e^
(3/2)))))/(4*a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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