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

Optimal. Leaf size=247 \[ \frac{4 a^2}{d e^2 \sqrt{e \cot (c+d x)}}+\frac{a^2 \log \left (\sqrt{e} \cot (c+d x)-\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{\sqrt{2} d e^{5/2}}-\frac{a^2 \log \left (\sqrt{e} \cot (c+d x)+\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{\sqrt{2} d e^{5/2}}-\frac{\sqrt{2} a^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{d e^{5/2}}+\frac{\sqrt{2} a^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}+1\right )}{d e^{5/2}}+\frac{2 a^2}{3 d e (e \cot (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.237426, antiderivative size = 247, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {3542, 12, 3474, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{4 a^2}{d e^2 \sqrt{e \cot (c+d x)}}+\frac{a^2 \log \left (\sqrt{e} \cot (c+d x)-\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{\sqrt{2} d e^{5/2}}-\frac{a^2 \log \left (\sqrt{e} \cot (c+d x)+\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{\sqrt{2} d e^{5/2}}-\frac{\sqrt{2} a^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{d e^{5/2}}+\frac{\sqrt{2} a^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}+1\right )}{d e^{5/2}}+\frac{2 a^2}{3 d e (e \cot (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3542

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

Rule 12

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

Rule 3474

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 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 297

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 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 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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 1.24281, size = 233, normalized size = 0.94 \[ \frac{a^2 (\tan (c+d x)+1)^2 \left (48 \cos ^2(c+d x) \text{Hypergeometric2F1}\left (-\frac{1}{4},1,\frac{3}{4},-\cot ^2(c+d x)\right )+\sin (c+d x) \left (8 \cos (c+d x) \text{Hypergeometric2F1}\left (-\frac{3}{4},1,\frac{1}{4},-\cot ^2(c+d x)\right )+3 \sqrt{2} \sin (c+d x) \cot ^{\frac{5}{2}}(c+d x) \left (\log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )-\log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )+2 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )-2 \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )\right )\right )\right )}{12 d e^2 \sqrt{e \cot (c+d x)} (\sin (c+d x)+\cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(48*Cos[c + d*x]^2*Hypergeometric2F1[-1/4, 1, 3/4, -Cot[c + d*x]^2] + Sin[c + d*x]*(8*Cos[c + d*x]*Hyperg
eometric2F1[-3/4, 1, 1/4, -Cot[c + d*x]^2] + 3*Sqrt[2]*Cot[c + d*x]^(5/2)*(2*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d
*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]] + Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] - Log[1
+ Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])*Sin[c + d*x]))*(1 + Tan[c + d*x])^2)/(12*d*e^2*Sqrt[e*Cot[c + d*
x]]*(Cos[c + d*x] + Sin[c + d*x])^2)

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Maple [A]  time = 0.023, size = 216, normalized size = 0.9 \begin{align*}{\frac{{a}^{2}\sqrt{2}}{2\,d{e}^{2}}\ln \left ({ \left ( e\cot \left ( dx+c \right ) -\sqrt [4]{{e}^{2}}\sqrt{e\cot \left ( dx+c \right ) }\sqrt{2}+\sqrt{{e}^{2}} \right ) \left ( e\cot \left ( dx+c \right ) +\sqrt [4]{{e}^{2}}\sqrt{e\cot \left ( dx+c \right ) }\sqrt{2}+\sqrt{{e}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{e}^{2}}}}}+{\frac{{a}^{2}\sqrt{2}}{d{e}^{2}}\arctan \left ({\sqrt{2}\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{e}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{e}^{2}}}}}-{\frac{{a}^{2}\sqrt{2}}{d{e}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{e}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{e}^{2}}}}}+{\frac{2\,{a}^{2}}{3\,de} \left ( e\cot \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}+4\,{\frac{{a}^{2}}{d{e}^{2}\sqrt{e\cot \left ( dx+c \right ) }}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \frac{1}{\left (e \cot{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx + \int \frac{2 \cot{\left (c + d x \right )}}{\left (e \cot{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx + \int \frac{\cot ^{2}{\left (c + d x \right )}}{\left (e \cot{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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