3.24 \(\int \frac {(e \cot (c+d x))^{3/2}}{a+a \cot (c+d x)} \, dx\)

Optimal. Leaf size=87 \[ \frac {e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} \cot (c+d x)+\sqrt {e}}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d}-\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{a d} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 87, 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 = {3573, 3532, 208, 3634, 63, 205} \[ \frac {e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} \cot (c+d x)+\sqrt {e}}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d}-\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cot[c + d*x])^(3/2)/(a + a*Cot[c + d*x]),x]

[Out]

-((e^(3/2)*ArcTan[Sqrt[e*Cot[c + d*x]]/Sqrt[e]])/(a*d)) + (e^(3/2)*ArcTanh[(Sqrt[e] + Sqrt[e]*Cot[c + d*x])/(S
qrt[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 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 3573

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1
/(c^2 + d^2), Int[Simp[a^2*c - b^2*c + 2*a*b*d + (2*a*b*c - a^2*d + b^2*d)*Tan[e + f*x], x]/Sqrt[a + b*Tan[e +
f*x]], x], x] + Dist[(b*c - a*d)^2/(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, e, f}, 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 {(e \cot (c+d x))^{3/2}}{a+a \cot (c+d x)} \, dx &=\frac {\int \frac {-a e^2+a e^2 \cot (c+d x)}{\sqrt {e \cot (c+d x)}} \, dx}{2 a^2}+\frac {1}{2} e^2 \int \frac {1+\cot ^2(c+d x)}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))} \, dx\\ &=\frac {e^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {-e x} (a-a x)} \, dx,x,-\cot (c+d x)\right )}{2 d}-\frac {e^4 \operatorname {Subst}\left (\int \frac {1}{2 a^2 e^4-e x^2} \, dx,x,\frac {-a e^2-a e^2 \cot (c+d x)}{\sqrt {e \cot (c+d x)}}\right )}{d}\\ &=\frac {e^{3/2} \tanh ^{-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 {e \operatorname {Subst}\left (\int \frac {1}{a+\frac {a x^2}{e}} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{d}\\ &=-\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{a d}+\frac {e^{3/2} \tanh ^{-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 = 4.05, size = 107, normalized size = 1.23 \[ -\frac {(e \cot (c+d x))^{3/2} \left (\sqrt {2} \left (\log \left (-\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}-1\right )-\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )\right )+4 \tan ^{-1}\left (\sqrt {\cot (c+d x)}\right )\right )}{4 a d \cot ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cot[c + d*x])^(3/2)/(a + a*Cot[c + d*x]),x]

[Out]

-1/4*((e*Cot[c + d*x])^(3/2)*(4*ArcTan[Sqrt[Cot[c + d*x]]] + Sqrt[2]*(Log[-1 + Sqrt[2]*Sqrt[Cot[c + d*x]] - Co
t[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])))/(a*d*Cot[c + d*x]^(3/2))

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fricas [A]  time = 0.79, size = 333, normalized size = 3.83 \[ \left [-\frac {\sqrt {2} \sqrt {-e} e \arctan \left (\frac {{\left (\sqrt {2} \cos \left (2 \, d x + 2 \, c\right ) + \sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) + \sqrt {2}\right )} \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{2 \, {\left (e \cos \left (2 \, d x + 2 \, c\right ) + e\right )}}\right ) - \sqrt {-e} e \log \left (\frac {e \cos \left (2 \, d x + 2 \, c\right ) - e \sin \left (2 \, d x + 2 \, c\right ) - 2 \, \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + e}{\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1}\right )}{2 \, a d}, \frac {\sqrt {2} e^{\frac {3}{2}} \log \left (-{\left (\sqrt {2} \cos \left (2 \, d x + 2 \, c\right ) - \sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) - \sqrt {2}\right )} \sqrt {e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} + 2 \, e \sin \left (2 \, d x + 2 \, c\right ) + e\right ) - 4 \, e^{\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{\sqrt {e}}\right )}{4 \, a d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))^(3/2)/(a+a*cot(d*x+c)),x, algorithm="fricas")

[Out]

[-1/2*(sqrt(2)*sqrt(-e)*e*arctan(1/2*(sqrt(2)*cos(2*d*x + 2*c) + sqrt(2)*sin(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)) - sqrt(-e)*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), 1/4*(sqrt(2)*e^(3/2)*log(-(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)) + 2*e*sin(2*d*x + 2*c) + e)
- 4*e^(3/2)*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 \[ \int \frac {\left (e \cot \left (d x + c\right )\right )^{\frac {3}{2}}}{a \cot \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))^(3/2)/(a+a*cot(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*cot(d*x + c))^(3/2)/(a*cot(d*x + c) + a), x)

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maple [B]  time = 0.73, size = 368, normalized size = 4.23 \[ -\frac {e^{\frac {3}{2}} \arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{d a}+\frac {e \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{8 d a}+\frac {e \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a}-\frac {e \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a}-\frac {e^{2} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{8 d a \left (e^{2}\right )^{\frac {1}{4}}}-\frac {e^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a \left (e^{2}\right )^{\frac {1}{4}}}+\frac {e^{2} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{4 d a \left (e^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cot(d*x+c))^(3/2)/(a+cot(d*x+c)*a),x)

[Out]

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

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maxima [A]  time = 0.70, size = 118, normalized size = 1.36 \[ \frac {e {\left (\frac {e {\left (\frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}} - \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}}\right )}}{a} - \frac {4 \, \sqrt {e} \arctan \left (\frac {\sqrt {\frac {e}{\tan \left (d x + c\right )}}}{\sqrt {e}}\right )}{a}\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))^(3/2)/(a+a*cot(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.49, size = 79, normalized size = 0.91 \[ \frac {\sqrt {2}\,e^{3/2}\,\mathrm {atanh}\left (\frac {12\,\sqrt {2}\,e^{25/2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{12\,e^{13}\,\mathrm {cot}\left (c+d\,x\right )+12\,e^{13}}\right )}{2\,a\,d}-\frac {e^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cot(c + d*x))^(3/2)/(a + a*cot(c + d*x)),x)

[Out]

(2^(1/2)*e^(3/2)*atanh((12*2^(1/2)*e^(25/2)*(e*cot(c + d*x))^(1/2))/(12*e^13*cot(c + d*x) + 12*e^13)))/(2*a*d)
- (e^(3/2)*atan((e*cot(c + d*x))^(1/2)/e^(1/2)))/(a*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\left (e \cot {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\cot {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))**(3/2)/(a+a*cot(d*x+c)),x)

[Out]

Integral((e*cot(c + d*x))**(3/2)/(cot(c + d*x) + 1), x)/a

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