3.63 \(\int \sqrt {e \cot (c+d x)} (a+b \cot (c+d x))^3 \, dx\)

Optimal. Leaf size=342 \[ -\frac {2 b \left (3 a^2-b^2\right ) \sqrt {e \cot (c+d x)}}{d}-\frac {\sqrt {e} (a-b) \left (a^2+4 a b+b^2\right ) \log \left (\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}+\frac {\sqrt {e} (a-b) \left (a^2+4 a b+b^2\right ) \log \left (\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}+\frac {\sqrt {e} (a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {\sqrt {e} (a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d}-\frac {8 a b^2 (e \cot (c+d x))^{3/2}}{5 d e}-\frac {2 b^2 (e \cot (c+d x))^{3/2} (a+b \cot (c+d x))}{5 d e} \]

[Out]

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

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Rubi [A]  time = 0.48, antiderivative size = 342, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3566, 3630, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {2 b \left (3 a^2-b^2\right ) \sqrt {e \cot (c+d x)}}{d}-\frac {\sqrt {e} (a-b) \left (a^2+4 a b+b^2\right ) \log \left (\sqrt {e} \cot (c+d x)-\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}+\frac {\sqrt {e} (a-b) \left (a^2+4 a b+b^2\right ) \log \left (\sqrt {e} \cot (c+d x)+\sqrt {2} \sqrt {e \cot (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}+\frac {\sqrt {e} (a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {\sqrt {e} (a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \cot (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d}-\frac {8 a b^2 (e \cot (c+d x))^{3/2}}{5 d e}-\frac {2 b^2 (e \cot (c+d x))^{3/2} (a+b \cot (c+d x))}{5 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)*(a^2 - 4*a*b + b^2)*Sqrt[e]*ArcTan[1 - (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d) - ((a + b
)*(a^2 - 4*a*b + b^2)*Sqrt[e]*ArcTan[1 + (Sqrt[2]*Sqrt[e*Cot[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d) - (2*b*(3*a^2 -
b^2)*Sqrt[e*Cot[c + d*x]])/d - (8*a*b^2*(e*Cot[c + d*x])^(3/2))/(5*d*e) - (2*b^2*(e*Cot[c + d*x])^(3/2)*(a + b
*Cot[c + d*x]))/(5*d*e) - ((a - b)*(a^2 + 4*a*b + b^2)*Sqrt[e]*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] - Sqrt[2]*Sq
rt[e*Cot[c + d*x]]])/(2*Sqrt[2]*d) + ((a - b)*(a^2 + 4*a*b + b^2)*Sqrt[e]*Log[Sqrt[e] + Sqrt[e]*Cot[c + d*x] +
Sqrt[2]*Sqrt[e*Cot[c + d*x]]])/(2*Sqrt[2]*d)

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 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 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 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 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 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3630

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

Rubi steps

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

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Mathematica [C]  time = 2.59, size = 247, normalized size = 0.72 \[ -\frac {\sqrt {e \cot (c+d x)} \left (\frac {2}{3} a \left (a^2-3 b^2\right ) \cot ^{\frac {3}{2}}(c+d x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\cot ^2(c+d x)\right )-\frac {1}{4} b \left (b^2-3 a^2\right ) \left (8 \sqrt {\cot (c+d x)}+\sqrt {2} \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-\sqrt {2} \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+2 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-2 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )\right )+2 a b^2 \cot ^{\frac {3}{2}}(c+d x)+\frac {2}{5} b^3 \cot ^{\frac {5}{2}}(c+d x)\right )}{d \sqrt {\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[e*Cot[c + d*x]]*(2*a*b^2*Cot[c + d*x]^(3/2) + (2*b^3*Cot[c + d*x]^(5/2))/5 + (2*a*(a^2 - 3*b^2)*Cot[c
+ d*x]^(3/2)*Hypergeometric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2])/3 - (b*(-3*a^2 + b^2)*(2*Sqrt[2]*ArcTan[1 - Sqrt
[2]*Sqrt[Cot[c + d*x]]] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]] + 8*Sqrt[Cot[c + d*x]] + Sqrt[2]*Lo
g[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])
)/4))/(d*Sqrt[Cot[c + d*x]]))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))^(1/2)*(a+b*cot(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))^(1/2)*(a+b*cot(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((b*cot(d*x + c) + a)^3*sqrt(e*cot(d*x + c)), x)

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maple [B]  time = 0.72, size = 750, normalized size = 2.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cot(d*x+c))^(1/2)*(a+b*cot(d*x+c))^3,x)

[Out]

-2/5/d/e^2*b^3*(e*cot(d*x+c))^(5/2)-2*a*b^2*(e*cot(d*x+c))^(3/2)/d/e-6/d*(e*cot(d*x+c))^(1/2)*a^2*b+2/d*b^3*(e
*cot(d*x+c))^(1/2)+3/4/d*(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)^(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)^(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)^(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)^(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)^(1/4)*2^(1/2)*arctan(2^(1/2)/(e^2)^(1/
4)*(e*cot(d*x+c))^(1/2)+1)*b^3-1/4/d*e*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)))*a^3+3/4/d*e*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)))*a*b^2+1/2/d*e*2^(1/2)/(e^2)^(1/4)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(
d*x+c))^(1/2)+1)*a^3-3/2/d*e*2^(1/2)/(e^2)^(1/4)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)*a*b^2-1/2
/d*e*2^(1/2)/(e^2)^(1/4)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)*a^3+3/2/d*e*2^(1/2)/(e^2)^(1/4)*ar
ctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)*a*b^2

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maxima [A]  time = 0.44, size = 316, normalized size = 0.92 \[ -\frac {{\left (\frac {10 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {10 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} - \frac {5 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}} + \frac {5 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}} + \frac {8 \, {\left (5 \, a b^{2} e \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {3}{2}} + b^{3} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {5}{2}} + 5 \, {\left (3 \, a^{2} b - b^{3}\right )} e^{2} \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{e^{3}}\right )} e}{20 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))^(1/2)*(a+b*cot(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/20*(10*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(e) + 2*sqrt(e/tan(d*x + c))
)/sqrt(e))/sqrt(e) + 10*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(e) - 2*sqrt(
e/tan(d*x + c)))/sqrt(e))/sqrt(e) - 5*sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*log(sqrt(2)*sqrt(e)*sqrt(e/tan(d
*x + c)) + e + e/tan(d*x + c))/sqrt(e) + 5*sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*log(-sqrt(2)*sqrt(e)*sqrt(e
/tan(d*x + c)) + e + e/tan(d*x + c))/sqrt(e) + 8*(5*a*b^2*e*(e/tan(d*x + c))^(3/2) + b^3*(e/tan(d*x + c))^(5/2
) + 5*(3*a^2*b - b^3)*e^2*sqrt(e/tan(d*x + c)))/e^3)*e/d

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mupad [B]  time = 2.55, size = 2071, normalized size = 6.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cot(c + d*x))^(1/2)*(a + b*cot(c + d*x))^3,x)

[Out]

(e*cot(c + d*x))^(1/2)*((2*b^3)/d - (6*a^2*b)/d) + atan((((16*(e*cot(c + d*x))^(1/2)*(a^6*e^4 - b^6*e^4 + 15*a
^2*b^4*e^4 - 15*a^4*b^2*e^4))/d^2 - (8*(4*b^3*d^2*e^4 - 12*a^2*b*d^2*e^4)*((b^6*e*1i - a^6*e*1i - a^2*b^4*e*15
i - 20*a^3*b^3*e + a^4*b^2*e*15i + 6*a*b^5*e + 6*a^5*b*e)/(4*d^2))^(1/2))/d^3)*((b^6*e*1i - a^6*e*1i - a^2*b^4
*e*15i - 20*a^3*b^3*e + a^4*b^2*e*15i + 6*a*b^5*e + 6*a^5*b*e)/(4*d^2))^(1/2)*1i + ((16*(e*cot(c + d*x))^(1/2)
*(a^6*e^4 - b^6*e^4 + 15*a^2*b^4*e^4 - 15*a^4*b^2*e^4))/d^2 + (8*(4*b^3*d^2*e^4 - 12*a^2*b*d^2*e^4)*((b^6*e*1i
- a^6*e*1i - a^2*b^4*e*15i - 20*a^3*b^3*e + a^4*b^2*e*15i + 6*a*b^5*e + 6*a^5*b*e)/(4*d^2))^(1/2))/d^3)*((b^6
*e*1i - a^6*e*1i - a^2*b^4*e*15i - 20*a^3*b^3*e + a^4*b^2*e*15i + 6*a*b^5*e + 6*a^5*b*e)/(4*d^2))^(1/2)*1i)/((
(16*(e*cot(c + d*x))^(1/2)*(a^6*e^4 - b^6*e^4 + 15*a^2*b^4*e^4 - 15*a^4*b^2*e^4))/d^2 - (8*(4*b^3*d^2*e^4 - 12
*a^2*b*d^2*e^4)*((b^6*e*1i - a^6*e*1i - a^2*b^4*e*15i - 20*a^3*b^3*e + a^4*b^2*e*15i + 6*a*b^5*e + 6*a^5*b*e)/
(4*d^2))^(1/2))/d^3)*((b^6*e*1i - a^6*e*1i - a^2*b^4*e*15i - 20*a^3*b^3*e + a^4*b^2*e*15i + 6*a*b^5*e + 6*a^5*
b*e)/(4*d^2))^(1/2) - ((16*(e*cot(c + d*x))^(1/2)*(a^6*e^4 - b^6*e^4 + 15*a^2*b^4*e^4 - 15*a^4*b^2*e^4))/d^2 +
(8*(4*b^3*d^2*e^4 - 12*a^2*b*d^2*e^4)*((b^6*e*1i - a^6*e*1i - a^2*b^4*e*15i - 20*a^3*b^3*e + a^4*b^2*e*15i +
6*a*b^5*e + 6*a^5*b*e)/(4*d^2))^(1/2))/d^3)*((b^6*e*1i - a^6*e*1i - a^2*b^4*e*15i - 20*a^3*b^3*e + a^4*b^2*e*1
5i + 6*a*b^5*e + 6*a^5*b*e)/(4*d^2))^(1/2) + (16*(3*a*b^8*e^5 - a^9*e^5 + 8*a^3*b^6*e^5 + 6*a^5*b^4*e^5))/d^3)
)*((b^6*e*1i - a^6*e*1i - a^2*b^4*e*15i - 20*a^3*b^3*e + a^4*b^2*e*15i + 6*a*b^5*e + 6*a^5*b*e)/(4*d^2))^(1/2)
*2i + atan((((16*(e*cot(c + d*x))^(1/2)*(a^6*e^4 - b^6*e^4 + 15*a^2*b^4*e^4 - 15*a^4*b^2*e^4))/d^2 - (8*(4*b^3
*d^2*e^4 - 12*a^2*b*d^2*e^4)*((a^6*e*1i - b^6*e*1i + a^2*b^4*e*15i - 20*a^3*b^3*e - a^4*b^2*e*15i + 6*a*b^5*e
+ 6*a^5*b*e)/(4*d^2))^(1/2))/d^3)*((a^6*e*1i - b^6*e*1i + a^2*b^4*e*15i - 20*a^3*b^3*e - a^4*b^2*e*15i + 6*a*b
^5*e + 6*a^5*b*e)/(4*d^2))^(1/2)*1i + ((16*(e*cot(c + d*x))^(1/2)*(a^6*e^4 - b^6*e^4 + 15*a^2*b^4*e^4 - 15*a^4
*b^2*e^4))/d^2 + (8*(4*b^3*d^2*e^4 - 12*a^2*b*d^2*e^4)*((a^6*e*1i - b^6*e*1i + a^2*b^4*e*15i - 20*a^3*b^3*e -
a^4*b^2*e*15i + 6*a*b^5*e + 6*a^5*b*e)/(4*d^2))^(1/2))/d^3)*((a^6*e*1i - b^6*e*1i + a^2*b^4*e*15i - 20*a^3*b^3
*e - a^4*b^2*e*15i + 6*a*b^5*e + 6*a^5*b*e)/(4*d^2))^(1/2)*1i)/(((16*(e*cot(c + d*x))^(1/2)*(a^6*e^4 - b^6*e^4
+ 15*a^2*b^4*e^4 - 15*a^4*b^2*e^4))/d^2 - (8*(4*b^3*d^2*e^4 - 12*a^2*b*d^2*e^4)*((a^6*e*1i - b^6*e*1i + a^2*b
^4*e*15i - 20*a^3*b^3*e - a^4*b^2*e*15i + 6*a*b^5*e + 6*a^5*b*e)/(4*d^2))^(1/2))/d^3)*((a^6*e*1i - b^6*e*1i +
a^2*b^4*e*15i - 20*a^3*b^3*e - a^4*b^2*e*15i + 6*a*b^5*e + 6*a^5*b*e)/(4*d^2))^(1/2) - ((16*(e*cot(c + d*x))^(
1/2)*(a^6*e^4 - b^6*e^4 + 15*a^2*b^4*e^4 - 15*a^4*b^2*e^4))/d^2 + (8*(4*b^3*d^2*e^4 - 12*a^2*b*d^2*e^4)*((a^6*
e*1i - b^6*e*1i + a^2*b^4*e*15i - 20*a^3*b^3*e - a^4*b^2*e*15i + 6*a*b^5*e + 6*a^5*b*e)/(4*d^2))^(1/2))/d^3)*(
(a^6*e*1i - b^6*e*1i + a^2*b^4*e*15i - 20*a^3*b^3*e - a^4*b^2*e*15i + 6*a*b^5*e + 6*a^5*b*e)/(4*d^2))^(1/2) +
(16*(3*a*b^8*e^5 - a^9*e^5 + 8*a^3*b^6*e^5 + 6*a^5*b^4*e^5))/d^3))*((a^6*e*1i - b^6*e*1i + a^2*b^4*e*15i - 20*
a^3*b^3*e - a^4*b^2*e*15i + 6*a*b^5*e + 6*a^5*b*e)/(4*d^2))^(1/2)*2i - (2*b^3*(e*cot(c + d*x))^(5/2))/(5*d*e^2
) - (2*a*b^2*(e*cot(c + d*x))^(3/2))/(d*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(e*cot(c + d*x))*(a + b*cot(c + d*x))**3, x)

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