3.2.58 \(\int x^2 \cot ^{-1}(c+d \tan (a+b x)) \, dx\) [158]
Optimal. Leaf size=403 \[ \frac {1}{3} x^3 \cot ^{-1}(c+d \tan (a+b x))-\frac {1}{6} i x^3 \log \left (1+\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )-\frac {x^2 \text {PolyLog}\left (2,-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{4 b}+\frac {x^2 \text {PolyLog}\left (2,-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{4 b}-\frac {i x \text {PolyLog}\left (3,-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{4 b^2}+\frac {i x \text {PolyLog}\left (3,-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{4 b^2}+\frac {\text {PolyLog}\left (4,-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{8 b^3}-\frac {\text {PolyLog}\left (4,-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{8 b^3} \]
[Out]
1/3*x^3*arccot(c+d*tan(b*x+a))-1/6*I*x^3*ln(1+(1+I*c+d)*exp(2*I*a+2*I*b*x)/(1+I*c-d))+1/6*I*x^3*ln(1+(c+I*(1-d
))*exp(2*I*a+2*I*b*x)/(c+I*(1+d)))-1/4*x^2*polylog(2,-(1+I*c+d)*exp(2*I*a+2*I*b*x)/(1+I*c-d))/b+1/4*x^2*polylo
g(2,-(c+I*(1-d))*exp(2*I*a+2*I*b*x)/(c+I*(1+d)))/b-1/4*I*x*polylog(3,-(1+I*c+d)*exp(2*I*a+2*I*b*x)/(1+I*c-d))/
b^2+1/4*I*x*polylog(3,-(c+I*(1-d))*exp(2*I*a+2*I*b*x)/(c+I*(1+d)))/b^2+1/8*polylog(4,-(1+I*c+d)*exp(2*I*a+2*I*
b*x)/(1+I*c-d))/b^3-1/8*polylog(4,-(c+I*(1-d))*exp(2*I*a+2*I*b*x)/(c+I*(1+d)))/b^3
________________________________________________________________________________________
Rubi [A]
time = 0.39, antiderivative size = 403, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5284, 2221,
2611, 6744, 2320, 6724} \begin {gather*} \frac {\text {Li}_4\left (-\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )}{8 b^3}-\frac {\text {Li}_4\left (-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (d+1)}\right )}{8 b^3}-\frac {i x \text {Li}_3\left (-\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )}{4 b^2}+\frac {i x \text {Li}_3\left (-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (d+1)}\right )}{4 b^2}-\frac {x^2 \text {Li}_2\left (-\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )}{4 b}+\frac {x^2 \text {Li}_2\left (-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (d+1)}\right )}{4 b}-\frac {1}{6} i x^3 \log \left (1+\frac {(i c+d+1) e^{2 i a+2 i b x}}{i c-d+1}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (d+1)}\right )+\frac {1}{3} x^3 \cot ^{-1}(d \tan (a+b x)+c) \end {gather*}
Antiderivative was successfully verified.
[In]
Int[x^2*ArcCot[c + d*Tan[a + b*x]],x]
[Out]
(x^3*ArcCot[c + d*Tan[a + b*x]])/3 - (I/6)*x^3*Log[1 + ((1 + I*c + d)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c - d)]
+ (I/6)*x^3*Log[1 + ((c + I*(1 - d))*E^((2*I)*a + (2*I)*b*x))/(c + I*(1 + d))] - (x^2*PolyLog[2, -(((1 + I*c +
d)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c - d))])/(4*b) + (x^2*PolyLog[2, -(((c + I*(1 - d))*E^((2*I)*a + (2*I)*b*
x))/(c + I*(1 + d)))])/(4*b) - ((I/4)*x*PolyLog[3, -(((1 + I*c + d)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c - d))])/
b^2 + ((I/4)*x*PolyLog[3, -(((c + I*(1 - d))*E^((2*I)*a + (2*I)*b*x))/(c + I*(1 + d)))])/b^2 + PolyLog[4, -(((
1 + I*c + d)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c - d))]/(8*b^3) - PolyLog[4, -(((c + I*(1 - d))*E^((2*I)*a + (2*
I)*b*x))/(c + I*(1 + d)))]/(8*b^3)
Rule 2221
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Rule 2320
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]
Rule 2611
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]
Rule 5284
Int[ArcCot[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m +
1)*(ArcCot[c + d*Tan[a + b*x]]/(f*(m + 1))), x] + (-Dist[b*((1 - I*c - d)/(f*(m + 1))), Int[(e + f*x)^(m + 1)
*(E^(2*I*a + 2*I*b*x)/(1 - I*c + d + (1 - I*c - d)*E^(2*I*a + 2*I*b*x))), x], x] + Dist[b*((1 + I*c + d)/(f*(m
+ 1))), Int[(e + f*x)^(m + 1)*(E^(2*I*a + 2*I*b*x)/(1 + I*c - d + (1 + I*c + d)*E^(2*I*a + 2*I*b*x))), x], x]
) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && NeQ[(c + I*d)^2, -1]
Rule 6724
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]
Rule 6744
Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Dist[f*(m/(b*c*p*Log[F])), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]
Rubi steps
\begin {align*} \int x^2 \cot ^{-1}(c+d \tan (a+b x)) \, dx &=\frac {1}{3} x^3 \cot ^{-1}(c+d \tan (a+b x))-\frac {1}{3} (b (1-i c-d)) \int \frac {e^{2 i a+2 i b x} x^3}{1-i c+d+(1-i c-d) e^{2 i a+2 i b x}} \, dx+\frac {1}{3} (b (1+i c+d)) \int \frac {e^{2 i a+2 i b x} x^3}{1+i c-d+(1+i c+d) e^{2 i a+2 i b x}} \, dx\\ &=\frac {1}{3} x^3 \cot ^{-1}(c+d \tan (a+b x))-\frac {1}{6} i x^3 \log \left (1+\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )-\frac {1}{2} i \int x^2 \log \left (1+\frac {(1-i c-d) e^{2 i a+2 i b x}}{1-i c+d}\right ) \, dx+\frac {1}{2} i \int x^2 \log \left (1+\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right ) \, dx\\ &=\frac {1}{3} x^3 \cot ^{-1}(c+d \tan (a+b x))-\frac {1}{6} i x^3 \log \left (1+\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )-\frac {x^2 \text {Li}_2\left (-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{4 b}+\frac {x^2 \text {Li}_2\left (-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{4 b}-\frac {\int x \text {Li}_2\left (-\frac {(1-i c-d) e^{2 i a+2 i b x}}{1-i c+d}\right ) \, dx}{2 b}+\frac {\int x \text {Li}_2\left (-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right ) \, dx}{2 b}\\ &=\frac {1}{3} x^3 \cot ^{-1}(c+d \tan (a+b x))-\frac {1}{6} i x^3 \log \left (1+\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )-\frac {x^2 \text {Li}_2\left (-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{4 b}+\frac {x^2 \text {Li}_2\left (-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{4 b}-\frac {i x \text {Li}_3\left (-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{4 b^2}+\frac {i x \text {Li}_3\left (-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{4 b^2}-\frac {i \int \text {Li}_3\left (-\frac {(1-i c-d) e^{2 i a+2 i b x}}{1-i c+d}\right ) \, dx}{4 b^2}+\frac {i \int \text {Li}_3\left (-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right ) \, dx}{4 b^2}\\ &=\frac {1}{3} x^3 \cot ^{-1}(c+d \tan (a+b x))-\frac {1}{6} i x^3 \log \left (1+\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )-\frac {x^2 \text {Li}_2\left (-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{4 b}+\frac {x^2 \text {Li}_2\left (-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{4 b}-\frac {i x \text {Li}_3\left (-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{4 b^2}+\frac {i x \text {Li}_3\left (-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{4 b^2}+\frac {\text {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {(1+i c+d) x}{1+i c-d}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}-\frac {\text {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {(c-i (-1+d)) x}{c+i (1+d)}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}\\ &=\frac {1}{3} x^3 \cot ^{-1}(c+d \tan (a+b x))-\frac {1}{6} i x^3 \log \left (1+\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )-\frac {x^2 \text {Li}_2\left (-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{4 b}+\frac {x^2 \text {Li}_2\left (-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{4 b}-\frac {i x \text {Li}_3\left (-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{4 b^2}+\frac {i x \text {Li}_3\left (-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{4 b^2}+\frac {\text {Li}_4\left (-\frac {(1+i c+d) e^{2 i a+2 i b x}}{1+i c-d}\right )}{8 b^3}-\frac {\text {Li}_4\left (-\frac {(c+i (1-d)) e^{2 i a+2 i b x}}{c+i (1+d)}\right )}{8 b^3}\\ \end {align*}
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Mathematica [A]
time = 0.63, size = 363, normalized size = 0.90 \begin {gather*} \frac {1}{3} x^3 \cot ^{-1}(c+d \tan (a+b x))+\frac {-4 i b^3 x^3 \log \left (1+\frac {(c-i (1+d)) e^{2 i (a+b x)}}{c+i (-1+d)}\right )+4 i b^3 x^3 \log \left (1+\frac {(i+c-i d) e^{2 i (a+b x)}}{c+i (1+d)}\right )-6 b^2 x^2 \text {PolyLog}\left (2,-\frac {(c-i (1+d)) e^{2 i (a+b x)}}{c+i (-1+d)}\right )+6 b^2 x^2 \text {PolyLog}\left (2,-\frac {(i+c-i d) e^{2 i (a+b x)}}{c+i (1+d)}\right )-6 i b x \text {PolyLog}\left (3,-\frac {(c-i (1+d)) e^{2 i (a+b x)}}{c+i (-1+d)}\right )+6 i b x \text {PolyLog}\left (3,-\frac {(i+c-i d) e^{2 i (a+b x)}}{c+i (1+d)}\right )+3 \text {PolyLog}\left (4,-\frac {(c-i (1+d)) e^{2 i (a+b x)}}{c+i (-1+d)}\right )-3 \text {PolyLog}\left (4,-\frac {(i+c-i d) e^{2 i (a+b x)}}{c+i (1+d)}\right )}{24 b^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[x^2*ArcCot[c + d*Tan[a + b*x]],x]
[Out]
(x^3*ArcCot[c + d*Tan[a + b*x]])/3 + ((-4*I)*b^3*x^3*Log[1 + ((c - I*(1 + d))*E^((2*I)*(a + b*x)))/(c + I*(-1
+ d))] + (4*I)*b^3*x^3*Log[1 + ((I + c - I*d)*E^((2*I)*(a + b*x)))/(c + I*(1 + d))] - 6*b^2*x^2*PolyLog[2, -((
(c - I*(1 + d))*E^((2*I)*(a + b*x)))/(c + I*(-1 + d)))] + 6*b^2*x^2*PolyLog[2, -(((I + c - I*d)*E^((2*I)*(a +
b*x)))/(c + I*(1 + d)))] - (6*I)*b*x*PolyLog[3, -(((c - I*(1 + d))*E^((2*I)*(a + b*x)))/(c + I*(-1 + d)))] + (
6*I)*b*x*PolyLog[3, -(((I + c - I*d)*E^((2*I)*(a + b*x)))/(c + I*(1 + d)))] + 3*PolyLog[4, -(((c - I*(1 + d))*
E^((2*I)*(a + b*x)))/(c + I*(-1 + d)))] - 3*PolyLog[4, -(((I + c - I*d)*E^((2*I)*(a + b*x)))/(c + I*(1 + d)))]
)/(24*b^3)
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 28.52, size = 8070, normalized size = 20.02
| | |
method |
result |
size |
| | |
risch |
\(\text {Expression too large to display}\) |
\(8070\) |
| | |
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*arccot(c+d*tan(b*x+a)),x,method=_RETURNVERBOSE)
[Out]
result too large to display
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*arccot(c+d*tan(b*x+a)),x, algorithm="maxima")
[Out]
1/6*x^3*arctan2(-(d + 1)*cos(2*b*x + 2*a) + c*sin(2*b*x + 2*a) + d - 1, -c*cos(2*b*x + 2*a) - (d + 1)*sin(2*b*
x + 2*a) - c) - 1/6*x^3*arctan2(-(d - 1)*cos(2*b*x + 2*a) + c*sin(2*b*x + 2*a) + d + 1, -c*cos(2*b*x + 2*a) -
(d - 1)*sin(2*b*x + 2*a) - c) - 4*b*d*integrate(-1/3*(2*(c^2 + d^2 + 1)*x^3*cos(2*b*x + 2*a)^2 + 2*c*d*x^3*sin
(2*b*x + 2*a) + 2*(c^2 + d^2 + 1)*x^3*sin(2*b*x + 2*a)^2 + (c^2 - d^2 + 1)*x^3*cos(2*b*x + 2*a) - (2*c*d*x^3*s
in(2*b*x + 2*a) - (c^2 - d^2 + 1)*x^3*cos(2*b*x + 2*a))*cos(4*b*x + 4*a) + (2*c*d*x^3*cos(2*b*x + 2*a) + (c^2
- d^2 + 1)*x^3*sin(2*b*x + 2*a))*sin(4*b*x + 4*a))/(c^4 + d^4 + 2*(c^2 - 1)*d^2 + (c^4 + d^4 + 2*(c^2 - 1)*d^2
+ 2*c^2 + 1)*cos(4*b*x + 4*a)^2 + 4*(c^4 + d^4 + 2*(c^2 + 1)*d^2 + 2*c^2 + 1)*cos(2*b*x + 2*a)^2 + (c^4 + d^4
+ 2*(c^2 - 1)*d^2 + 2*c^2 + 1)*sin(4*b*x + 4*a)^2 + 4*(c^4 + d^4 + 2*(c^2 + 1)*d^2 + 2*c^2 + 1)*sin(2*b*x + 2
*a)^2 + 2*c^2 + 2*(c^4 + d^4 - 2*(3*c^2 + 1)*d^2 + 2*c^2 + 2*(c^4 - d^4 + 2*c^2 + 1)*cos(2*b*x + 2*a) - 4*(c*d
^3 + (c^3 + c)*d)*sin(2*b*x + 2*a) + 1)*cos(4*b*x + 4*a) + 4*(c^4 - d^4 + 2*c^2 + 1)*cos(2*b*x + 2*a) - 4*(2*c
*d^3 - 2*(c^3 + c)*d - 2*(c*d^3 + (c^3 + c)*d)*cos(2*b*x + 2*a) - (c^4 - d^4 + 2*c^2 + 1)*sin(2*b*x + 2*a))*si
n(4*b*x + 4*a) + 8*(c*d^3 + (c^3 + c)*d)*sin(2*b*x + 2*a) + 1), x)
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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 1965 vs. \(2 (290) = 580\).
time = 3.96, size = 1965, normalized size = 4.88 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*arccot(c+d*tan(b*x+a)),x, algorithm="fricas")
[Out]
1/48*(16*b^3*x^3*arccot(d*tan(b*x + a) + c) - 6*b^2*x^2*dilog(2*((I*c*d - d^2 + d)*tan(b*x + a)^2 - c^2 - I*c*
d + (I*c^2 - 2*c*d - I*d^2 + I)*tan(b*x + a) + d - 1)/((c^2 + d^2 - 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*d
+ 1) + 1) + 6*b^2*x^2*dilog(2*((I*c*d - d^2 - d)*tan(b*x + a)^2 - c^2 - I*c*d + (I*c^2 - 2*c*d - I*d^2 + I)*ta
n(b*x + a) - d - 1)/((c^2 + d^2 + 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*d + 1) + 1) - 6*b^2*x^2*dilog(2*((-I
*c*d - d^2 + d)*tan(b*x + a)^2 - c^2 + I*c*d + (-I*c^2 - 2*c*d + I*d^2 - I)*tan(b*x + a) + d - 1)/((c^2 + d^2
- 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*d + 1) + 1) + 6*b^2*x^2*dilog(2*((-I*c*d - d^2 - d)*tan(b*x + a)^2 -
c^2 + I*c*d + (-I*c^2 - 2*c*d + I*d^2 - I)*tan(b*x + a) - d - 1)/((c^2 + d^2 + 2*d + 1)*tan(b*x + a)^2 + c^2
+ d^2 + 2*d + 1) + 1) - 4*I*a^3*log(((I*c*d + d^2 + d)*tan(b*x + a)^2 - c^2 + I*c*d + (I*c^2 + I*d^2 + 2*I*d +
I)*tan(b*x + a) - d - 1)/(tan(b*x + a)^2 + 1)) + 4*I*a^3*log(((I*c*d + d^2 - d)*tan(b*x + a)^2 - c^2 + I*c*d
+ (I*c^2 + I*d^2 - 2*I*d + I)*tan(b*x + a) + d - 1)/(tan(b*x + a)^2 + 1)) - 4*I*a^3*log(((I*c*d - d^2 + d)*tan
(b*x + a)^2 + c^2 + I*c*d + (I*c^2 + I*d^2 - 2*I*d + I)*tan(b*x + a) - d + 1)/(tan(b*x + a)^2 + 1)) + 4*I*a^3*
log(((I*c*d - d^2 - d)*tan(b*x + a)^2 + c^2 + I*c*d + (I*c^2 + I*d^2 + 2*I*d + I)*tan(b*x + a) + d + 1)/(tan(b
*x + a)^2 + 1)) - 6*I*b*x*polylog(3, ((c^2 + 2*I*c*d - d^2 + 1)*tan(b*x + a)^2 - c^2 - 2*I*c*d + d^2 - 2*(-I*c
^2 + 2*c*d + I*d^2 - I)*tan(b*x + a) - 1)/((c^2 + d^2 + 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*d + 1)) + 6*I*
b*x*polylog(3, ((c^2 - 2*I*c*d - d^2 + 1)*tan(b*x + a)^2 - c^2 + 2*I*c*d + d^2 - 2*(I*c^2 + 2*c*d - I*d^2 + I)
*tan(b*x + a) - 1)/((c^2 + d^2 + 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*d + 1)) + 6*I*b*x*polylog(3, ((c^2 +
2*I*c*d - d^2 + 1)*tan(b*x + a)^2 - c^2 - 2*I*c*d + d^2 - 2*(-I*c^2 + 2*c*d + I*d^2 - I)*tan(b*x + a) - 1)/((c
^2 + d^2 - 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*d + 1)) - 6*I*b*x*polylog(3, ((c^2 - 2*I*c*d - d^2 + 1)*tan
(b*x + a)^2 - c^2 + 2*I*c*d + d^2 - 2*(I*c^2 + 2*c*d - I*d^2 + I)*tan(b*x + a) - 1)/((c^2 + d^2 - 2*d + 1)*tan
(b*x + a)^2 + c^2 + d^2 - 2*d + 1)) - 4*(-I*b^3*x^3 - I*a^3)*log(-2*((I*c*d - d^2 + d)*tan(b*x + a)^2 - c^2 -
I*c*d + (I*c^2 - 2*c*d - I*d^2 + I)*tan(b*x + a) + d - 1)/((c^2 + d^2 - 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 -
2*d + 1)) - 4*(I*b^3*x^3 + I*a^3)*log(-2*((I*c*d - d^2 - d)*tan(b*x + a)^2 - c^2 - I*c*d + (I*c^2 - 2*c*d - I*
d^2 + I)*tan(b*x + a) - d - 1)/((c^2 + d^2 + 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*d + 1)) - 4*(I*b^3*x^3 +
I*a^3)*log(-2*((-I*c*d - d^2 + d)*tan(b*x + a)^2 - c^2 + I*c*d + (-I*c^2 - 2*c*d + I*d^2 - I)*tan(b*x + a) + d
- 1)/((c^2 + d^2 - 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*d + 1)) - 4*(-I*b^3*x^3 - I*a^3)*log(-2*((-I*c*d -
d^2 - d)*tan(b*x + a)^2 - c^2 + I*c*d + (-I*c^2 - 2*c*d + I*d^2 - I)*tan(b*x + a) - d - 1)/((c^2 + d^2 + 2*d
+ 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*d + 1)) - 3*polylog(4, ((c^2 + 2*I*c*d - d^2 + 1)*tan(b*x + a)^2 - c^2 - 2
*I*c*d + d^2 - 2*(-I*c^2 + 2*c*d + I*d^2 - I)*tan(b*x + a) - 1)/((c^2 + d^2 + 2*d + 1)*tan(b*x + a)^2 + c^2 +
d^2 + 2*d + 1)) - 3*polylog(4, ((c^2 - 2*I*c*d - d^2 + 1)*tan(b*x + a)^2 - c^2 + 2*I*c*d + d^2 - 2*(I*c^2 + 2*
c*d - I*d^2 + I)*tan(b*x + a) - 1)/((c^2 + d^2 + 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*d + 1)) + 3*polylog(4
, ((c^2 + 2*I*c*d - d^2 + 1)*tan(b*x + a)^2 - c^2 - 2*I*c*d + d^2 - 2*(-I*c^2 + 2*c*d + I*d^2 - I)*tan(b*x + a
) - 1)/((c^2 + d^2 - 2*d + 1)*tan(b*x + a)^2 + c^2 + d^2 - 2*d + 1)) + 3*polylog(4, ((c^2 - 2*I*c*d - d^2 + 1)
*tan(b*x + a)^2 - c^2 + 2*I*c*d + d^2 - 2*(I*c^2 + 2*c*d - I*d^2 + I)*tan(b*x + a) - 1)/((c^2 + d^2 - 2*d + 1)
*tan(b*x + a)^2 + c^2 + d^2 - 2*d + 1)))/b^3
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2*acot(c+d*tan(b*x+a)),x)
[Out]
Timed out
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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(x^2*arccot(c+d*tan(b*x+a)),x, algorithm="giac")
[Out]
integrate(x^2*arccot(d*tan(b*x + a) + c), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,\mathrm {acot}\left (c+d\,\mathrm {tan}\left (a+b\,x\right )\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*acot(c + d*tan(a + b*x)),x)
[Out]
int(x^2*acot(c + d*tan(a + b*x)), x)
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