3.66 \(\int \frac{(a+b \cot (c+d x))^3}{(e \cot (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=313 \[ -\frac{(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\sqrt{e} \cot (c+d x)-\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2}}+\frac{(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\sqrt{e} \cot (c+d x)+\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{5/2}}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} d e^{5/2}}+\frac{16 a^2 b}{3 d e^2 \sqrt{e \cot (c+d x)}}+\frac{2 a^2 (a+b \cot (c+d x))}{3 d e (e \cot (c+d x))^{3/2}} \]

[Out]

-(((a - b)*(a^2 + 4*a*b + b^2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(5/2))) + ((a
- b)*(a^2 + 4*a*b + b^2)*ArcTan[1 + (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(5/2)) + (16*a^2*b)/
(3*d*e^2*Sqrt[e*Cot[c + d*x]]) + (2*a^2*(a + b*Cot[c + d*x]))/(3*d*e*(e*Cot[c + d*x])^(3/2)) - ((a + b)*(a^2 -
4*a*b + b^2)*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] - Sqrt[2]*Sqrt[e*Cot[c + d*x]]])/(2*Sqrt[2]*d*e^(5/2)) + ((a
+ b)*(a^2 - 4*a*b + b^2)*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] + Sqrt[2]*Sqrt[e*Cot[c + d*x]]])/(2*Sqrt[2]*d*e^(5
/2))

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Rubi [A]  time = 0.457541, antiderivative size = 313, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {3565, 3628, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\sqrt{e} \cot (c+d x)-\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2}}+\frac{(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\sqrt{e} \cot (c+d x)+\sqrt{2} \sqrt{e \cot (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{5/2}}+\frac{(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \cot (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} d e^{5/2}}+\frac{16 a^2 b}{3 d e^2 \sqrt{e \cot (c+d x)}}+\frac{2 a^2 (a+b \cot (c+d x))}{3 d e (e \cot (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cot[c + d*x])^3/(e*Cot[c + d*x])^(5/2),x]

[Out]

-(((a - b)*(a^2 + 4*a*b + b^2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(5/2))) + ((a
- b)*(a^2 + 4*a*b + b^2)*ArcTan[1 + (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(5/2)) + (16*a^2*b)/
(3*d*e^2*Sqrt[e*Cot[c + d*x]]) + (2*a^2*(a + b*Cot[c + d*x]))/(3*d*e*(e*Cot[c + d*x])^(3/2)) - ((a + b)*(a^2 -
4*a*b + b^2)*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] - Sqrt[2]*Sqrt[e*Cot[c + d*x]]])/(2*Sqrt[2]*d*e^(5/2)) + ((a
+ b)*(a^2 - 4*a*b + b^2)*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] + Sqrt[2]*Sqrt[e*Cot[c + d*x]]])/(2*Sqrt[2]*d*e^(5
/2))

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*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] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3628

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*(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*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
+ b^2, 0]

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 1168

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

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

Mathematica [C]  time = 0.358201, size = 104, normalized size = 0.33 \[ \frac{-6 b \left (b^2-3 a^2\right ) \text{Hypergeometric2F1}\left (-\frac{1}{4},1,\frac{3}{4},-\cot ^2(c+d x)\right )+2 a \left (a^2-3 b^2\right ) \tan (c+d x) \text{Hypergeometric2F1}\left (-\frac{3}{4},1,\frac{1}{4},-\cot ^2(c+d x)\right )+6 b^2 (a \tan (c+d x)+b)}{3 d e^2 \sqrt{e \cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cot[c + d*x])^3/(e*Cot[c + d*x])^(5/2),x]

[Out]

(-6*b*(-3*a^2 + b^2)*Hypergeometric2F1[-1/4, 1, 3/4, -Cot[c + d*x]^2] + 2*a*(a^2 - 3*b^2)*Hypergeometric2F1[-3
/4, 1, 1/4, -Cot[c + d*x]^2]*Tan[c + d*x] + 6*b^2*(b + a*Tan[c + d*x]))/(3*d*e^2*Sqrt[e*Cot[c + d*x]])

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Maple [B]  time = 0.028, size = 743, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cot(d*x+c))^3/(e*cot(d*x+c))^(5/2),x)

[Out]

1/2/d*a^3/e^3*(e^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)-3/2/d/e^3*(e^2)^(1/4)*2^(
1/2)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)*a*b^2-1/2/d*a^3/e^3*(e^2)^(1/4)*2^(1/2)*arctan(-2^(1/2
)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+3/2/d/e^3*(e^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))
^(1/2)+1)*a*b^2+1/4/d*a^3/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)))-3/4/d/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)))*a*b^2+3/4/d/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)))*a^2*b-1/4/d/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)))*b^3+3/2/d/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)*a^2*b-1/2/d/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)*b^3-3/2/d/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)*a^2*b+1/2/d/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)*b^3+2/3/d*a^3/e/(e*cot(d*x+c))^(3/2)+6*a^2
*b/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+b*cot(d*x+c))^3/(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+b*cot(d*x+c))^3/(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*} \int \frac{\left (a + b \cot{\left (c + d x \right )}\right )^{3}}{\left (e \cot{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cot(d*x+c))**3/(e*cot(d*x+c))**(5/2),x)

[Out]

Integral((a + b*cot(c + d*x))**3/(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 (b \cot \left (d x + c\right ) + a\right )}^{3}}{\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+b*cot(d*x+c))^3/(e*cot(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*cot(d*x + c) + a)^3/(e*cot(d*x + c))^(5/2), x)