3.8 \(\int (e \cot (c+d x))^{5/2} (a+a \cot (c+d x))^2 \, dx\)

Optimal. Leaf size=269 \[ \frac {a^2 e^{5/2} \log \left (\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{\sqrt {2} d}-\frac {a^2 e^{5/2} \log \left (\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{\sqrt {2} d}+\frac {\sqrt {2} a^2 e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{d}-\frac {\sqrt {2} a^2 e^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{d}+\frac {4 a^2 e^2 \sqrt {e \cot (c+d x)}}{d}-\frac {2 a^2 (e \cot (c+d x))^{7/2}}{7 d e}-\frac {4 a^2 (e \cot (c+d x))^{5/2}}{5 d} \]

[Out]

-4/5*a^2*(e*cot(d*x+c))^(5/2)/d-2/7*a^2*(e*cot(d*x+c))^(7/2)/d/e+1/2*a^2*e^(5/2)*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^(5/2)*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^(5/2)*arctan(1-2^(1/2)*(e*cot(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/d-a^2*e^(5/2)*arctan(1+2^(
1/2)*(e*cot(d*x+c))^(1/2)/e^(1/2))*2^(1/2)/d+4*a^2*e^2*(e*cot(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.29, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3543, 12, 16, 3473, 3476, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac {4 a^2 e^2 \sqrt {e \cot (c+d x)}}{d}+\frac {a^2 e^{5/2} \log \left (\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{\sqrt {2} d}-\frac {a^2 e^{5/2} \log \left (\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{\sqrt {2} d}+\frac {\sqrt {2} a^2 e^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{d}-\frac {\sqrt {2} a^2 e^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{d}-\frac {2 a^2 (e \cot (c+d x))^{7/2}}{7 d e}-\frac {4 a^2 (e \cot (c+d x))^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^2,x]

[Out]

(Sqrt[2]*a^2*e^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/d - (Sqrt[2]*a^2*e^(5/2)*ArcTan[1 + (
Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/d + (4*a^2*e^2*Sqrt[e*Cot[c + d*x]])/d - (4*a^2*(e*Cot[c + d*x])^(5/2)
)/(5*d) - (2*a^2*(e*Cot[c + d*x])^(7/2))/(7*d*e) + (a^2*e^(5/2)*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] - Sqrt[2]*S
qrt[e*Cot[c + d*x]]])/(Sqrt[2]*d) - (a^2*e^(5/2)*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 204

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

Rule 211

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

Rule 329

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 617

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 628

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 1162

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 1165

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 3473

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 3476

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 3543

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))^{5/2} (a+a \cot (c+d x))^2 \, dx &=-\frac {2 a^2 (e \cot (c+d x))^{7/2}}{7 d e}+\int 2 a^2 \cot (c+d x) (e \cot (c+d x))^{5/2} \, dx\\ &=-\frac {2 a^2 (e \cot (c+d x))^{7/2}}{7 d e}+\left (2 a^2\right ) \int \cot (c+d x) (e \cot (c+d x))^{5/2} \, dx\\ &=-\frac {2 a^2 (e \cot (c+d x))^{7/2}}{7 d e}+\frac {\left (2 a^2\right ) \int (e \cot (c+d x))^{7/2} \, dx}{e}\\ &=-\frac {4 a^2 (e \cot (c+d x))^{5/2}}{5 d}-\frac {2 a^2 (e \cot (c+d x))^{7/2}}{7 d e}-\left (2 a^2 e\right ) \int (e \cot (c+d x))^{3/2} \, dx\\ &=\frac {4 a^2 e^2 \sqrt {e \cot (c+d x)}}{d}-\frac {4 a^2 (e \cot (c+d x))^{5/2}}{5 d}-\frac {2 a^2 (e \cot (c+d x))^{7/2}}{7 d e}+\left (2 a^2 e^3\right ) \int \frac {1}{\sqrt {e \cot (c+d x)}} \, dx\\ &=\frac {4 a^2 e^2 \sqrt {e \cot (c+d x)}}{d}-\frac {4 a^2 (e \cot (c+d x))^{5/2}}{5 d}-\frac {2 a^2 (e \cot (c+d x))^{7/2}}{7 d e}-\frac {\left (2 a^2 e^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (e^2+x^2\right )} \, dx,x,e \cot (c+d x)\right )}{d}\\ &=\frac {4 a^2 e^2 \sqrt {e \cot (c+d x)}}{d}-\frac {4 a^2 (e \cot (c+d x))^{5/2}}{5 d}-\frac {2 a^2 (e \cot (c+d x))^{7/2}}{7 d e}-\frac {\left (4 a^2 e^4\right ) \operatorname {Subst}\left (\int \frac {1}{e^2+x^4} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{d}\\ &=\frac {4 a^2 e^2 \sqrt {e \cot (c+d x)}}{d}-\frac {4 a^2 (e \cot (c+d x))^{5/2}}{5 d}-\frac {2 a^2 (e \cot (c+d x))^{7/2}}{7 d e}-\frac {\left (2 a^2 e^3\right ) \operatorname {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^3\right ) \operatorname {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^2 \sqrt {e \cot (c+d x)}}{d}-\frac {4 a^2 (e \cot (c+d x))^{5/2}}{5 d}-\frac {2 a^2 (e \cot (c+d x))^{7/2}}{7 d e}+\frac {\left (a^2 e^{5/2}\right ) \operatorname {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^{5/2}\right ) \operatorname {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\right ) \operatorname {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^3\right ) \operatorname {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^2 \sqrt {e \cot (c+d x)}}{d}-\frac {4 a^2 (e \cot (c+d x))^{5/2}}{5 d}-\frac {2 a^2 (e \cot (c+d x))^{7/2}}{7 d e}+\frac {a^2 e^{5/2} \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^{5/2} \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^{5/2}\right ) \operatorname {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^{5/2}\right ) \operatorname {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^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{d}-\frac {\sqrt {2} a^2 e^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {4 a^2 e^2 \sqrt {e \cot (c+d x)}}{d}-\frac {4 a^2 (e \cot (c+d x))^{5/2}}{5 d}-\frac {2 a^2 (e \cot (c+d x))^{7/2}}{7 d e}+\frac {a^2 e^{5/2} \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^{5/2} \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 [A]  time = 1.19, size = 187, normalized size = 0.70 \[ -\frac {a^2 (e \cot (c+d x))^{5/2} \left (20 \cot ^{\frac {7}{2}}(c+d x)+56 \cot ^{\frac {5}{2}}(c+d x)-280 \sqrt {\cot (c+d x)}-35 \sqrt {2} \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+35 \sqrt {2} \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-70 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )+70 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )\right )}{70 d \cot ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^2,x]

[Out]

-1/70*(a^2*(e*Cot[c + d*x])^(5/2)*(-70*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] + 70*Sqrt[2]*ArcTan[1 +
Sqrt[2]*Sqrt[Cot[c + d*x]]] - 280*Sqrt[Cot[c + d*x]] + 56*Cot[c + d*x]^(5/2) + 20*Cot[c + d*x]^(7/2) - 35*Sqrt
[2]*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] + 35*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c
+ d*x]]))/(d*Cot[c + d*x]^(5/2))

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))^(5/2)*(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \cot \left (d x + c\right ) + a\right )}^{2} \left (e \cot \left (d x + c\right )\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))^(5/2)*(a+a*cot(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((a*cot(d*x + c) + a)^2*(e*cot(d*x + c))^(5/2), x)

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maple [A]  time = 0.61, size = 234, normalized size = 0.87 \[ -\frac {2 a^{2} \left (e \cot \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d e}-\frac {4 a^{2} \left (e \cot \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d}+\frac {4 a^{2} e^{2} \sqrt {e \cot \left (d x +c \right )}}{d}+\frac {a^{2} e^{2} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d}-\frac {a^{2} e^{2} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{2 d}-\frac {a^{2} e^{2} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cot(d*x+c))^(5/2)*(a+cot(d*x+c)*a)^2,x)

[Out]

-2/7*a^2*(e*cot(d*x+c))^(7/2)/d/e-4/5*a^2*(e*cot(d*x+c))^(5/2)/d+4*a^2*e^2*(e*cot(d*x+c))^(1/2)/d+1/d*a^2*e^2*
(e^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)-1/2/d*a^2*e^2*(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/d*a^2*e^2*(e^2)^(1/4)*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 = 0.59, size = 232, normalized size = 0.86 \[ -\frac {{\left (35 \, {\left (2 \, \sqrt {2} e^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right ) + 2 \, \sqrt {2} e^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right ) + \sqrt {2} e^{\frac {3}{2}} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right ) - \sqrt {2} e^{\frac {3}{2}} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )\right )} a^{2} - \frac {4 \, {\left (70 \, a^{2} e^{3} \sqrt {\frac {e}{\tan \left (d x + c\right )}} - 14 \, a^{2} e \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {5}{2}} - 5 \, a^{2} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {7}{2}}\right )}}{e^{2}}\right )} e}{70 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))^(5/2)*(a+a*cot(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/70*(35*(2*sqrt(2)*e^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(e) + 2*sqrt(e/tan(d*x + c)))/sqrt(e)) + 2*sqrt(2
)*e^(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(e) - 2*sqrt(e/tan(d*x + c)))/sqrt(e)) + sqrt(2)*e^(3/2)*log(sqrt(2
)*sqrt(e)*sqrt(e/tan(d*x + c)) + e + e/tan(d*x + c)) - sqrt(2)*e^(3/2)*log(-sqrt(2)*sqrt(e)*sqrt(e/tan(d*x + c
)) + e + e/tan(d*x + c)))*a^2 - 4*(70*a^2*e^3*sqrt(e/tan(d*x + c)) - 14*a^2*e*(e/tan(d*x + c))^(5/2) - 5*a^2*(
e/tan(d*x + c))^(7/2))/e^2)*e/d

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mupad [B]  time = 1.73, size = 125, normalized size = 0.46 \[ \frac {4\,a^2\,e^2\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{d}-\frac {4\,a^2\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{5/2}}{5\,d}-\frac {2\,a^2\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{7/2}}{7\,d\,e}+\frac {{\left (-1\right )}^{1/4}\,a^2\,e^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )\,2{}\mathrm {i}}{d}+\frac {2\,{\left (-1\right )}^{1/4}\,a^2\,e^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,1{}\mathrm {i}}{\sqrt {e}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cot(c + d*x))^(5/2)*(a + a*cot(c + d*x))^2,x)

[Out]

(4*a^2*e^2*(e*cot(c + d*x))^(1/2))/d - (4*a^2*(e*cot(c + d*x))^(5/2))/(5*d) - (2*a^2*(e*cot(c + d*x))^(7/2))/(
7*d*e) + ((-1)^(1/4)*a^2*e^(5/2)*atan(((-1)^(1/4)*(e*cot(c + d*x))^(1/2))/e^(1/2))*2i)/d + (2*(-1)^(1/4)*a^2*e
^(5/2)*atan(((-1)^(1/4)*(e*cot(c + d*x))^(1/2)*1i)/e^(1/2)))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int \left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx + \int 2 \left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}} \cot {\left (c + d x \right )}\, dx + \int \left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}} \cot ^{2}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))**(5/2)*(a+a*cot(d*x+c))**2,x)

[Out]

a**2*(Integral((e*cot(c + d*x))**(5/2), x) + Integral(2*(e*cot(c + d*x))**(5/2)*cot(c + d*x), x) + Integral((e
*cot(c + d*x))**(5/2)*cot(c + d*x)**2, x))

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