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

Optimal. Leaf size=164 \[ \frac {5 e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{8 a^3 d}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{2 \sqrt {2} a^3 d}+\frac {e \sqrt {e \cot (c+d x)}}{8 d \left (a^3 \cot (c+d x)+a^3\right )}-\frac {e \sqrt {e \cot (c+d x)}}{4 a d (a \cot (c+d x)+a)^2} \]

[Out]

5/8*e^(3/2)*arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a^3/d+1/4*e^(3/2)*arctan(1/2*(e^(1/2)-cot(d*x+c)*e^(1/2))*2^(
1/2)/(e*cot(d*x+c))^(1/2))/a^3/d*2^(1/2)-1/4*e*(e*cot(d*x+c))^(1/2)/a/d/(a+a*cot(d*x+c))^2+1/8*e*(e*cot(d*x+c)
)^(1/2)/d/(a^3+a^3*cot(d*x+c))

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Rubi [A]  time = 0.66, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3567, 3649, 3653, 3532, 205, 3634, 63} \[ \frac {5 e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{8 a^3 d}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{2 \sqrt {2} a^3 d}+\frac {e \sqrt {e \cot (c+d x)}}{8 d \left (a^3 \cot (c+d x)+a^3\right )}-\frac {e \sqrt {e \cot (c+d x)}}{4 a d (a \cot (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*e^(3/2)*ArcTan[Sqrt[e*Cot[c + d*x]]/Sqrt[e]])/(8*a^3*d) + (e^(3/2)*ArcTan[(Sqrt[e] - Sqrt[e]*Cot[c + d*x])/
(Sqrt[2]*Sqrt[e*Cot[c + d*x]])])/(2*Sqrt[2]*a^3*d) - (e*Sqrt[e*Cot[c + d*x]])/(4*a*d*(a + a*Cot[c + d*x])^2) +
(e*Sqrt[e*Cot[c + d*x]])/(8*d*(a^3 + a^3*Cot[c + d*x]))

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 3567

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[
1/((m + 1)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[a*c^2*(m + 1) + a*
d^2*(n - 1) + b*c*d*(m - n + 2) - (b*c^2 - 2*a*c*d - b*d^2)*(m + 1)*Tan[e + f*x] - d*(b*c - a*d)*(m + n)*Tan[e
+ f*x]^2, 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] && LtQ[m, -1] && LtQ[1, n, 2] && IntegerQ[2*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]

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3653

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[((c + d*Tan[e + f*x])^n*(1 + Tan[e + f*x]^2))/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rubi steps

\begin {align*} \int \frac {(e \cot (c+d x))^{3/2}}{(a+a \cot (c+d x))^3} \, dx &=-\frac {e \sqrt {e \cot (c+d x)}}{4 a d (a+a \cot (c+d x))^2}-\frac {\int \frac {\frac {a e^2}{2}-2 a e^2 \cot (c+d x)-\frac {3}{2} a e^2 \cot ^2(c+d x)}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))^2} \, dx}{4 a^2}\\ &=-\frac {e \sqrt {e \cot (c+d x)}}{4 a d (a+a \cot (c+d x))^2}+\frac {e \sqrt {e \cot (c+d x)}}{8 d \left (a^3+a^3 \cot (c+d x)\right )}+\frac {\int \frac {-\frac {1}{2} a^3 e^3+4 a^3 e^3 \cot (c+d x)-\frac {1}{2} a^3 e^3 \cot ^2(c+d x)}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))} \, dx}{8 a^5 e}\\ &=-\frac {e \sqrt {e \cot (c+d x)}}{4 a d (a+a \cot (c+d x))^2}+\frac {e \sqrt {e \cot (c+d x)}}{8 d \left (a^3+a^3 \cot (c+d x)\right )}+\frac {\int \frac {4 a^4 e^3+4 a^4 e^3 \cot (c+d x)}{\sqrt {e \cot (c+d x)}} \, dx}{16 a^7 e}-\frac {\left (5 e^2\right ) \int \frac {1+\cot ^2(c+d x)}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))} \, dx}{16 a^2}\\ &=-\frac {e \sqrt {e \cot (c+d x)}}{4 a d (a+a \cot (c+d x))^2}+\frac {e \sqrt {e \cot (c+d x)}}{8 d \left (a^3+a^3 \cot (c+d x)\right )}-\frac {\left (5 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-e x} (a-a x)} \, dx,x,-\cot (c+d x)\right )}{16 a^2 d}-\frac {\left (2 a e^5\right ) \operatorname {Subst}\left (\int \frac {1}{-32 a^8 e^6-e x^2} \, dx,x,\frac {4 a^4 e^3-4 a^4 e^3 \cot (c+d x)}{\sqrt {e \cot (c+d x)}}\right )}{d}\\ &=\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{2 \sqrt {2} a^3 d}-\frac {e \sqrt {e \cot (c+d x)}}{4 a d (a+a \cot (c+d x))^2}+\frac {e \sqrt {e \cot (c+d x)}}{8 d \left (a^3+a^3 \cot (c+d x)\right )}+\frac {(5 e) \operatorname {Subst}\left (\int \frac {1}{a+\frac {a x^2}{e}} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{8 a^2 d}\\ &=\frac {5 e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{8 a^3 d}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{2 \sqrt {2} a^3 d}-\frac {e \sqrt {e \cot (c+d x)}}{4 a d (a+a \cot (c+d x))^2}+\frac {e \sqrt {e \cot (c+d x)}}{8 d \left (a^3+a^3 \cot (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 2.02, size = 131, normalized size = 0.80 \[ \frac {e \sqrt {e \cot (c+d x)} \left (\frac {2 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-2 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+5 \tan ^{-1}\left (\sqrt {\cot (c+d x)}\right )}{\sqrt {\cot (c+d x)}}+\frac {\tan (c+d x)-\sec ^2(c+d x)+1}{(\tan (c+d x)+1)^2}\right )}{8 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[e*Cot[c + d*x]]*((2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt
[Cot[c + d*x]]] + 5*ArcTan[Sqrt[Cot[c + d*x]]])/Sqrt[Cot[c + d*x]] + (1 - Sec[c + d*x]^2 + Tan[c + d*x])/(1 +
Tan[c + d*x])^2))/(8*a^3*d)

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fricas [A]  time = 0.54, size = 533, normalized size = 3.25 \[ \left [\frac {2 \, {\left (\sqrt {2} e \sin \left (2 \, d x + 2 \, c\right ) + \sqrt {2} e\right )} \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 ) + 5 \, {\left (e \sin \left (2 \, d x + 2 \, c\right ) + e\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 ) + {\left (e \cos \left (2 \, d x + 2 \, c\right ) + e \sin \left (2 \, d x + 2 \, c\right ) - e\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{16 \, {\left (a^{3} d \sin \left (2 \, d x + 2 \, c\right ) + a^{3} d\right )}}, \frac {4 \, {\left (\sqrt {2} e \sin \left (2 \, d x + 2 \, c\right ) + \sqrt {2} e\right )} \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 ) + 10 \, {\left (e \sin \left (2 \, d x + 2 \, c\right ) + e\right )} \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 ) + {\left (e \cos \left (2 \, d x + 2 \, c\right ) + e \sin \left (2 \, d x + 2 \, c\right ) - e\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{16 \, {\left (a^{3} d \sin \left (2 \, d x + 2 \, c\right ) + a^{3} d\right )}}\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))^3,x, algorithm="fricas")

[Out]

[1/16*(2*(sqrt(2)*e*sin(2*d*x + 2*c) + sqrt(2)*e)*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*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)) - 2*e*sin(2*d*x + 2*c) + e) + 5*(e*
sin(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)) + (e*cos(2*d*x + 2
*c) + e*sin(2*d*x + 2*c) - e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(a^3*d*sin(2*d*x + 2*c) + a^3*d
), 1/16*(4*(sqrt(2)*e*sin(2*d*x + 2*c) + sqrt(2)*e)*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)) +
10*(e*sin(2*d*x + 2*c) + e)*sqrt(e)*arctan(sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))/sqrt(e)) + (e*cos(2
*d*x + 2*c) + e*sin(2*d*x + 2*c) - e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(a^3*d*sin(2*d*x + 2*c)
+ a^3*d)]

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

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))^3,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.84, size = 434, normalized size = 2.65 \[ \frac {e^{2} \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}{8 d \,a^{3} \left (e \cot \left (d x +c \right )+e \right )^{2}}-\frac {e^{3} \sqrt {e \cot \left (d x +c \right )}}{8 d \,a^{3} \left (e \cot \left (d x +c \right )+e \right )^{2}}+\frac {5 e^{\frac {3}{2}} \arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{8 a^{3} d}-\frac {e \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{16 d \,a^{3}}-\frac {e \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{8 d \,a^{3}}+\frac {e \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{8 d \,a^{3}}-\frac {e^{2} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{16 d \,a^{3} \left (e^{2}\right )^{\frac {1}{4}}}-\frac {e^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{8 d \,a^{3} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {e^{2} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{8 d \,a^{3} \left (e^{2}\right )^{\frac {1}{4}}} \]

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)^3,x)

[Out]

1/8/d/a^3*e^2/(e*cot(d*x+c)+e)^2*(e*cot(d*x+c))^(3/2)-1/8/d/a^3*e^3/(e*cot(d*x+c)+e)^2*(e*cot(d*x+c))^(1/2)+5/
8*e^(3/2)*arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a^3/d-1/16/d/a^3*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)))-1/8/d/a^3*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^3*e*(e^2)^(1/
4)*2^(1/2)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)-1/16/d/a^3*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)))-1/8/d/a^3*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)+1/8/d/a
^3*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 = 0.92, size = 189, normalized size = 1.15 \[ -\frac {e {\left (\frac {e^{2} \sqrt {\frac {e}{\tan \left (d x + c\right )}} - e \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {3}{2}}}{a^{3} e^{2} + \frac {2 \, a^{3} e^{2}}{\tan \left (d x + c\right )} + \frac {a^{3} e^{2}}{\tan \left (d x + c\right )^{2}}} + \frac {2 \, e {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}}\right )}}{a^{3}} - \frac {5 \, \sqrt {e} \arctan \left (\frac {\sqrt {\frac {e}{\tan \left (d x + c\right )}}}{\sqrt {e}}\right )}{a^{3}}\right )}}{8 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

-1/8*e*((e^2*sqrt(e/tan(d*x + c)) - e*(e/tan(d*x + c))^(3/2))/(a^3*e^2 + 2*a^3*e^2/tan(d*x + c) + a^3*e^2/tan(
d*x + c)^2) + 2*e*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(e) + 2*sqrt(e/tan(d*x + c)))/sqrt(e))/sqrt(e) + sq
rt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(e) - 2*sqrt(e/tan(d*x + c)))/sqrt(e))/sqrt(e))/a^3 - 5*sqrt(e)*arctan(
sqrt(e/tan(d*x + c))/sqrt(e))/a^3)/d

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mupad [B]  time = 0.94, size = 178, normalized size = 1.09 \[ \frac {5\,e^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{8\,a^3\,d}-\frac {\frac {e^3\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{8}-\frac {e^2\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}{8}}{d\,a^3\,e^2\,{\mathrm {cot}\left (c+d\,x\right )}^2+2\,d\,a^3\,e^2\,\mathrm {cot}\left (c+d\,x\right )+d\,a^3\,e^2}-\frac {\sqrt {2}\,e^{3/2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{2\,\sqrt {e}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{2\,\sqrt {e}}+\frac {\sqrt {2}\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}{2\,e^{3/2}}\right )\right )}{8\,a^3\,d} \]

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))^3,x)

[Out]

(5*e^(3/2)*atan((e*cot(c + d*x))^(1/2)/e^(1/2)))/(8*a^3*d) - ((e^3*(e*cot(c + d*x))^(1/2))/8 - (e^2*(e*cot(c +
d*x))^(3/2))/8)/(a^3*d*e^2 + a^3*d*e^2*cot(c + d*x)^2 + 2*a^3*d*e^2*cot(c + d*x)) - (2^(1/2)*e^(3/2)*(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)))))/(8*a^3*d)

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

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))**3,x)

[Out]

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

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