3.1.9 \(\int (e \cot (c+d x))^{3/2} (a+a \cot (c+d x))^2 \, dx\) [9]

Optimal. Leaf size=246 \[ -\frac {\sqrt {2} a^2 e^{3/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {\sqrt {2} a^2 e^{3/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{d}-\frac {4 a^2 (e \cot (c+d x))^{3/2}}{3 d}-\frac {2 a^2 (e \cot (c+d x))^{5/2}}{5 d e}+\frac {a^2 e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{\sqrt {2} d}-\frac {a^2 e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}\right )}{\sqrt {2} d} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3624, 12, 16, 3554, 3557, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {\sqrt {2} a^2 e^{3/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {\sqrt {2} a^2 e^{3/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{d}+\frac {a^2 e^{3/2} \log \left (\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{\sqrt {2} d}-\frac {a^2 e^{3/2} \log \left (\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{\sqrt {2} d}-\frac {2 a^2 (e \cot (c+d x))^{5/2}}{5 d e}-\frac {4 a^2 (e \cot (c+d x))^{3/2}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*a^2*e^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/d) + (Sqrt[2]*a^2*e^(3/2)*ArcTan[1
+ (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/d - (4*a^2*(e*Cot[c + d*x])^(3/2))/(3*d) - (2*a^2*(e*Cot[c + d*x])^
(5/2))/(5*d*e) + (a^2*e^(3/2)*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] - Sqrt[2]*Sqrt[e*Cot[c + d*x]]])/(Sqrt[2]*d)
- (a^2*e^(3/2)*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] + Sqrt[2]*Sqrt[e*Cot[c + d*x]]])/(Sqrt[2]*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3624

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

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.40, size = 52, normalized size = 0.21 \begin {gather*} -\frac {2 a^2 (e \cot (c+d x))^{3/2} \left (10+3 \cot (c+d x)-10 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\cot ^2(c+d x)\right )\right )}{15 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*(e*Cot[c + d*x])^(3/2)*(10 + 3*Cot[c + d*x] - 10*Hypergeometric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2]))/(15
*d)

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Maple [A]
time = 0.44, size = 172, normalized size = 0.70

method	result	size
derivativedivides	\(-\frac {2 a^{2} \left (\frac {\left (e \cot \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 e \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\frac {e^{3} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (e^{2}\right )^{\frac {1}{4}}}\right )}{d e}\)	\(172\)
default	\(-\frac {2 a^{2} \left (\frac {\left (e \cot \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 e \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\frac {e^{3} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (e^{2}\right )^{\frac {1}{4}}}\right )}{d e}\)	\(172\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.52, size = 146, normalized size = 0.59 \begin {gather*} \frac {{\left (15 \, {\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{2} - \frac {40 \, a^{2}}{\tan \left (d x + c\right )^{\frac {3}{2}}} - \frac {12 \, a^{2}}{\tan \left (d x + c\right )^{\frac {5}{2}}}\right )} e^{\frac {3}{2}}}{30 \, d} \end {gather*}

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))^2,x, algorithm="maxima")

[Out]

1/30*(15*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt
(2) - 2/sqrt(tan(d*x + c)))) - sqrt(2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + sqrt(2)*log(-sqr
t(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^2 - 40*a^2/tan(d*x + c)^(3/2) - 12*a^2/tan(d*x + c)^(5/2))*e^
(3/2)/d

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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))^2,x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   catdef: division by zero

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

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))**2,x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.95, size = 104, normalized size = 0.42 \begin {gather*} \frac {2\,{\left (-1\right )}^{1/4}\,a^2\,e^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{d}-\frac {2\,a^2\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{5/2}}{5\,d\,e}-\frac {4\,a^2\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}{3\,d}+\frac {{\left (-1\right )}^{1/4}\,a^2\,e^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,1{}\mathrm {i}}{\sqrt {e}}\right )\,2{}\mathrm {i}}{d} \end {gather*}

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))^2,x)

[Out]

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

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