3.226 \(\int x^2 \cot ^{-1}(a+b f^{c+d x}) \, dx\)

Optimal. Leaf size=313 \[ -\frac {i \text {Li}_4\left (\frac {b f^{c+d x}}{i-a}\right )}{d^3 \log ^3(f)}+\frac {i \text {Li}_4\left (-\frac {b f^{c+d x}}{a+i}\right )}{d^3 \log ^3(f)}+\frac {i x \text {Li}_3\left (\frac {b f^{c+d x}}{i-a}\right )}{d^2 \log ^2(f)}-\frac {i x \text {Li}_3\left (-\frac {b f^{c+d x}}{a+i}\right )}{d^2 \log ^2(f)}-\frac {i x^2 \text {Li}_2\left (\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x^2 \text {Li}_2\left (-\frac {b f^{c+d x}}{a+i}\right )}{2 d \log (f)}-\frac {1}{6} i x^3 \log \left (1-\frac {b f^{c+d x}}{-a+i}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {b f^{c+d x}}{a+i}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} i x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right ) \]

[Out]

-1/6*I*x^3*ln(1-b*f^(d*x+c)/(I-a))+1/6*I*x^3*ln(1+b*f^(d*x+c)/(I+a))+1/6*I*x^3*ln(1-I/(a+b*f^(d*x+c)))-1/6*I*x
^3*ln(1+I/(a+b*f^(d*x+c)))-1/2*I*x^2*polylog(2,b*f^(d*x+c)/(I-a))/d/ln(f)+1/2*I*x^2*polylog(2,-b*f^(d*x+c)/(I+
a))/d/ln(f)+I*x*polylog(3,b*f^(d*x+c)/(I-a))/d^2/ln(f)^2-I*x*polylog(3,-b*f^(d*x+c)/(I+a))/d^2/ln(f)^2-I*polyl
og(4,b*f^(d*x+c)/(I-a))/d^3/ln(f)^3+I*polylog(4,-b*f^(d*x+c)/(I+a))/d^3/ln(f)^3

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Rubi [A]  time = 2.45, antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 29, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {5144, 2551, 12, 6742, 2190, 2531, 6609, 2282, 6589} \[ \frac {i x \text {PolyLog}\left (3,\frac {b f^{c+d x}}{-a+i}\right )}{d^2 \log ^2(f)}-\frac {i x \text {PolyLog}\left (3,-\frac {b f^{c+d x}}{a+i}\right )}{d^2 \log ^2(f)}-\frac {i \text {PolyLog}\left (4,\frac {b f^{c+d x}}{-a+i}\right )}{d^3 \log ^3(f)}+\frac {i \text {PolyLog}\left (4,-\frac {b f^{c+d x}}{a+i}\right )}{d^3 \log ^3(f)}-\frac {i x^2 \text {PolyLog}\left (2,\frac {b f^{c+d x}}{-a+i}\right )}{2 d \log (f)}+\frac {i x^2 \text {PolyLog}\left (2,-\frac {b f^{c+d x}}{a+i}\right )}{2 d \log (f)}-\frac {1}{6} i x^3 \log \left (1-\frac {b f^{c+d x}}{-a+i}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {b f^{c+d x}}{a+i}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} i x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCot[a + b*f^(c + d*x)],x]

[Out]

(-I/6)*x^3*Log[1 - (b*f^(c + d*x))/(I - a)] + (I/6)*x^3*Log[1 + (b*f^(c + d*x))/(I + a)] + (I/6)*x^3*Log[1 - I
/(a + b*f^(c + d*x))] - (I/6)*x^3*Log[1 + I/(a + b*f^(c + d*x))] - ((I/2)*x^2*PolyLog[2, (b*f^(c + d*x))/(I -
a)])/(d*Log[f]) + ((I/2)*x^2*PolyLog[2, -((b*f^(c + d*x))/(I + a))])/(d*Log[f]) + (I*x*PolyLog[3, (b*f^(c + d*
x))/(I - a)])/(d^2*Log[f]^2) - (I*x*PolyLog[3, -((b*f^(c + d*x))/(I + a))])/(d^2*Log[f]^2) - (I*PolyLog[4, (b*
f^(c + d*x))/(I - a)])/(d^3*Log[f]^3) + (I*PolyLog[4, -((b*f^(c + d*x))/(I + a))])/(d^3*Log[f]^3)

Rule 12

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

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2282

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 2531

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 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 5144

Int[ArcCot[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Dist[I/2, Int[x^m*Log[1 - I/(a +
b*f^(c + d*x))], x], x] - Dist[I/2, Int[x^m*Log[1 + I/(a + b*f^(c + d*x))], x], x] /; FreeQ[{a, b, c, d, f}, x
] && IntegerQ[m] && m > 0

Rule 6589

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 6609

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]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int x^2 \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx &=\frac {1}{2} i \int x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right ) \, dx-\frac {1}{2} i \int x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right ) \, dx\\ &=\frac {1}{6} i x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} i x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )+\frac {1}{6} \int \frac {b d f^{c+d x} x^3 \log (f)}{\left (i (1-i a)+b f^{c+d x}\right ) \left (a+b f^{c+d x}\right )} \, dx+\frac {1}{6} \int \frac {b d f^{c+d x} x^3 \log (f)}{\left (-i (1+i a)+b f^{c+d x}\right ) \left (a+b f^{c+d x}\right )} \, dx\\ &=\frac {1}{6} i x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} i x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )+\frac {1}{6} (b d \log (f)) \int \frac {f^{c+d x} x^3}{\left (i (1-i a)+b f^{c+d x}\right ) \left (a+b f^{c+d x}\right )} \, dx+\frac {1}{6} (b d \log (f)) \int \frac {f^{c+d x} x^3}{\left (-i (1+i a)+b f^{c+d x}\right ) \left (a+b f^{c+d x}\right )} \, dx\\ &=\frac {1}{6} i x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} i x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )+\frac {1}{6} (b d \log (f)) \int \left (\frac {i f^{c+d x} x^3}{a+b f^{c+d x}}-\frac {i f^{c+d x} x^3}{-i+a+b f^{c+d x}}\right ) \, dx+\frac {1}{6} (b d \log (f)) \int \left (-\frac {i f^{c+d x} x^3}{a+b f^{c+d x}}+\frac {i f^{c+d x} x^3}{i+a+b f^{c+d x}}\right ) \, dx\\ &=\frac {1}{6} i x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} i x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} (i b d \log (f)) \int \frac {f^{c+d x} x^3}{-i+a+b f^{c+d x}} \, dx+\frac {1}{6} (i b d \log (f)) \int \frac {f^{c+d x} x^3}{i+a+b f^{c+d x}} \, dx\\ &=-\frac {1}{6} i x^3 \log \left (1-\frac {b f^{c+d x}}{i-a}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {b f^{c+d x}}{i+a}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} i x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )+\frac {1}{2} i \int x^2 \log \left (1+\frac {b f^{c+d x}}{-i+a}\right ) \, dx-\frac {1}{2} i \int x^2 \log \left (1+\frac {b f^{c+d x}}{i+a}\right ) \, dx\\ &=-\frac {1}{6} i x^3 \log \left (1-\frac {b f^{c+d x}}{i-a}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {b f^{c+d x}}{i+a}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} i x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )-\frac {i x^2 \text {Li}_2\left (\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x^2 \text {Li}_2\left (-\frac {b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac {i \int x \text {Li}_2\left (-\frac {b f^{c+d x}}{-i+a}\right ) \, dx}{d \log (f)}-\frac {i \int x \text {Li}_2\left (-\frac {b f^{c+d x}}{i+a}\right ) \, dx}{d \log (f)}\\ &=-\frac {1}{6} i x^3 \log \left (1-\frac {b f^{c+d x}}{i-a}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {b f^{c+d x}}{i+a}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} i x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )-\frac {i x^2 \text {Li}_2\left (\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x^2 \text {Li}_2\left (-\frac {b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac {i x \text {Li}_3\left (\frac {b f^{c+d x}}{i-a}\right )}{d^2 \log ^2(f)}-\frac {i x \text {Li}_3\left (-\frac {b f^{c+d x}}{i+a}\right )}{d^2 \log ^2(f)}-\frac {i \int \text {Li}_3\left (-\frac {b f^{c+d x}}{-i+a}\right ) \, dx}{d^2 \log ^2(f)}+\frac {i \int \text {Li}_3\left (-\frac {b f^{c+d x}}{i+a}\right ) \, dx}{d^2 \log ^2(f)}\\ &=-\frac {1}{6} i x^3 \log \left (1-\frac {b f^{c+d x}}{i-a}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {b f^{c+d x}}{i+a}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} i x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )-\frac {i x^2 \text {Li}_2\left (\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x^2 \text {Li}_2\left (-\frac {b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac {i x \text {Li}_3\left (\frac {b f^{c+d x}}{i-a}\right )}{d^2 \log ^2(f)}-\frac {i x \text {Li}_3\left (-\frac {b f^{c+d x}}{i+a}\right )}{d^2 \log ^2(f)}-\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {b x}{i-a}\right )}{x} \, dx,x,f^{c+d x}\right )}{d^3 \log ^3(f)}+\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {b x}{i+a}\right )}{x} \, dx,x,f^{c+d x}\right )}{d^3 \log ^3(f)}\\ &=-\frac {1}{6} i x^3 \log \left (1-\frac {b f^{c+d x}}{i-a}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {b f^{c+d x}}{i+a}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} i x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )-\frac {i x^2 \text {Li}_2\left (\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x^2 \text {Li}_2\left (-\frac {b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac {i x \text {Li}_3\left (\frac {b f^{c+d x}}{i-a}\right )}{d^2 \log ^2(f)}-\frac {i x \text {Li}_3\left (-\frac {b f^{c+d x}}{i+a}\right )}{d^2 \log ^2(f)}-\frac {i \text {Li}_4\left (\frac {b f^{c+d x}}{i-a}\right )}{d^3 \log ^3(f)}+\frac {i \text {Li}_4\left (-\frac {b f^{c+d x}}{i+a}\right )}{d^3 \log ^3(f)}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 313, normalized size = 1.00 \[ -\frac {i \text {Li}_4\left (\frac {b f^{c+d x}}{i-a}\right )}{d^3 \log ^3(f)}+\frac {i \text {Li}_4\left (-\frac {b f^{c+d x}}{a+i}\right )}{d^3 \log ^3(f)}+\frac {i x \text {Li}_3\left (\frac {b f^{c+d x}}{i-a}\right )}{d^2 \log ^2(f)}-\frac {i x \text {Li}_3\left (-\frac {b f^{c+d x}}{a+i}\right )}{d^2 \log ^2(f)}-\frac {i x^2 \text {Li}_2\left (\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x^2 \text {Li}_2\left (-\frac {b f^{c+d x}}{a+i}\right )}{2 d \log (f)}-\frac {1}{6} i x^3 \log \left (1-\frac {b f^{c+d x}}{-a+i}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {b f^{c+d x}}{a+i}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} i x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCot[a + b*f^(c + d*x)],x]

[Out]

(-1/6*I)*x^3*Log[1 - (b*f^(c + d*x))/(I - a)] + (I/6)*x^3*Log[1 + (b*f^(c + d*x))/(I + a)] + (I/6)*x^3*Log[1 -
I/(a + b*f^(c + d*x))] - (I/6)*x^3*Log[1 + I/(a + b*f^(c + d*x))] - ((I/2)*x^2*PolyLog[2, (b*f^(c + d*x))/(I
- a)])/(d*Log[f]) + ((I/2)*x^2*PolyLog[2, -((b*f^(c + d*x))/(I + a))])/(d*Log[f]) + (I*x*PolyLog[3, (b*f^(c +
d*x))/(I - a)])/(d^2*Log[f]^2) - (I*x*PolyLog[3, -((b*f^(c + d*x))/(I + a))])/(d^2*Log[f]^2) - (I*PolyLog[4, (
b*f^(c + d*x))/(I - a)])/(d^3*Log[f]^3) + (I*PolyLog[4, -((b*f^(c + d*x))/(I + a))])/(d^3*Log[f]^3)

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fricas [C]  time = 0.79, size = 378, normalized size = 1.21 \[ \frac {2 \, d^{3} x^{3} \operatorname {arccot}\left (b f^{d x + c} + a\right ) \log \relax (f)^{3} - 3 i \, d^{2} x^{2} {\rm Li}_2\left (-\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \relax (f)^{2} + 3 i \, d^{2} x^{2} {\rm Li}_2\left (-\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \relax (f)^{2} - i \, c^{3} \log \left (b f^{d x + c} + a + i\right ) \log \relax (f)^{3} + i \, c^{3} \log \left (b f^{d x + c} + a - i\right ) \log \relax (f)^{3} + {\left (-i \, d^{3} x^{3} - i \, c^{3}\right )} \log \relax (f)^{3} \log \left (\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + {\left (i \, d^{3} x^{3} + i \, c^{3}\right )} \log \relax (f)^{3} \log \left (\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + 6 i \, d x \log \relax (f) {\rm polylog}\left (3, -\frac {{\left (a b + i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) - 6 i \, d x \log \relax (f) {\rm polylog}\left (3, -\frac {{\left (a b - i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) - 6 i \, {\rm polylog}\left (4, -\frac {{\left (a b + i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) + 6 i \, {\rm polylog}\left (4, -\frac {{\left (a b - i \, b\right )} f^{d x + c}}{a^{2} + 1}\right )}{6 \, d^{3} \log \relax (f)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccot(a+b*f^(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(2*d^3*x^3*arccot(b*f^(d*x + c) + a)*log(f)^3 - 3*I*d^2*x^2*dilog(-(a^2 + (a*b + I*b)*f^(d*x + c) + 1)/(a^
2 + 1) + 1)*log(f)^2 + 3*I*d^2*x^2*dilog(-(a^2 + (a*b - I*b)*f^(d*x + c) + 1)/(a^2 + 1) + 1)*log(f)^2 - I*c^3*
log(b*f^(d*x + c) + a + I)*log(f)^3 + I*c^3*log(b*f^(d*x + c) + a - I)*log(f)^3 + (-I*d^3*x^3 - I*c^3)*log(f)^
3*log((a^2 + (a*b + I*b)*f^(d*x + c) + 1)/(a^2 + 1)) + (I*d^3*x^3 + I*c^3)*log(f)^3*log((a^2 + (a*b - I*b)*f^(
d*x + c) + 1)/(a^2 + 1)) + 6*I*d*x*log(f)*polylog(3, -(a*b + I*b)*f^(d*x + c)/(a^2 + 1)) - 6*I*d*x*log(f)*poly
log(3, -(a*b - I*b)*f^(d*x + c)/(a^2 + 1)) - 6*I*polylog(4, -(a*b + I*b)*f^(d*x + c)/(a^2 + 1)) + 6*I*polylog(
4, -(a*b - I*b)*f^(d*x + c)/(a^2 + 1)))/(d^3*log(f)^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {arccot}\left (b f^{d x + c} + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccot(a+b*f^(d*x+c)),x, algorithm="giac")

[Out]

integrate(x^2*arccot(b*f^(d*x + c) + a), x)

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maple [B]  time = 0.48, size = 764, normalized size = 2.44 \[ -\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x \,c^{2}}{2 d^{2}}-\frac {i c^{2} \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{-i+a}\right ) x}{2 d^{2}}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x \,c^{2}}{2 d^{2}}+\frac {i c^{2} \ln \left (\frac {b \,f^{d x} f^{c}+i+a}{i+a}\right ) x}{2 d^{2}}+\frac {i \polylog \left (2, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) c^{2}}{2 d^{3} \ln \relax (f )}-\frac {i c^{2} \dilog \left (\frac {b \,f^{d x} f^{c}+a -i}{-i+a}\right )}{2 d^{3} \ln \relax (f )}+\frac {i \polylog \left (2, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x^{2}}{2 d \ln \relax (f )}-\frac {i \polylog \left (2, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) c^{2}}{2 d^{3} \ln \relax (f )}+\frac {i c^{2} \dilog \left (\frac {b \,f^{d x} f^{c}+i+a}{i+a}\right )}{2 d^{3} \ln \relax (f )}-\frac {i \polylog \left (4, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right )}{d^{3} \ln \relax (f )^{3}}+\frac {i \polylog \left (4, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right )}{d^{3} \ln \relax (f )^{3}}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) c^{3}}{3 d^{3}}-\frac {i c^{3} \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{-i+a}\right )}{2 d^{3}}+\frac {i c^{3} \ln \left (\frac {b \,f^{d x} f^{c}+i+a}{i+a}\right )}{2 d^{3}}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) c^{3}}{3 d^{3}}-\frac {i c^{3} \ln \left (1-i a -i f^{d x} f^{c} b \right )}{6 d^{3}}+\frac {i c^{3} \ln \left (i f^{d x} f^{c} b +i a +1\right )}{6 d^{3}}+\frac {\pi \,x^{3}}{6}-\frac {i \polylog \left (3, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x}{d^{2} \ln \relax (f )^{2}}+\frac {i \polylog \left (3, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x}{d^{2} \ln \relax (f )^{2}}-\frac {i \polylog \left (2, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x^{2}}{2 d \ln \relax (f )}+\frac {i x^{3} \ln \left (1+i \left (a +b \,f^{d x +c}\right )\right )}{6}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x^{3}}{6}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x^{3}}{6}-\frac {i x^{3} \ln \left (1-i \left (a +b \,f^{d x +c}\right )\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccot(a+b*f^(d*x+c)),x)

[Out]

-I/d^2/ln(f)^2*polylog(3,I*b/(1-I*a)*f^(d*x)*f^c)*x+I/d^2/ln(f)^2*polylog(3,I*b/(-I*a-1)*f^(d*x)*f^c)*x+1/3*I/
d^3*ln(1-I*b/(-I*a-1)*f^(d*x)*f^c)*c^3-1/2*I/d^3*c^3*ln((b*f^(d*x)*f^c+a-I)/(-I+a))+1/2*I/d^3*c^3*ln((b*f^(d*x
)*f^c+I+a)/(I+a))-1/3*I/d^3*ln(1-I*b/(1-I*a)*f^(d*x)*f^c)*c^3-1/6*I/d^3*c^3*ln(1-I*a-I*f^(d*x)*f^c*b)-I/d^3/ln
(f)^3*polylog(4,I*b/(-I*a-1)*f^(d*x)*f^c)+1/6*I/d^3*c^3*ln(I*f^(d*x)*f^c*b+I*a+1)+I/d^3/ln(f)^3*polylog(4,I*b/
(1-I*a)*f^(d*x)*f^c)+1/6*Pi*x^3+1/6*I*x^3*ln(1+I*(a+b*f^(d*x+c)))-1/6*I*ln(1-I*b/(-I*a-1)*f^(d*x)*f^c)*x^3+1/6
*I*ln(1-I*b/(1-I*a)*f^(d*x)*f^c)*x^3-1/6*I*x^3*ln(1-I*(a+b*f^(d*x+c)))-1/2*I/d^2*ln(1-I*b/(1-I*a)*f^(d*x)*f^c)
*x*c^2-1/2*I/d/ln(f)*polylog(2,I*b/(-I*a-1)*f^(d*x)*f^c)*x^2+1/2*I/d^3/ln(f)*polylog(2,I*b/(-I*a-1)*f^(d*x)*f^
c)*c^2-1/2*I/d^3/ln(f)*c^2*dilog((b*f^(d*x)*f^c+a-I)/(-I+a))-1/2*I/d^2*c^2*ln((b*f^(d*x)*f^c+a-I)/(-I+a))*x+1/
2*I/d^2*ln(1-I*b/(-I*a-1)*f^(d*x)*f^c)*x*c^2+1/2*I/d/ln(f)*polylog(2,I*b/(1-I*a)*f^(d*x)*f^c)*x^2-1/2*I/d^3/ln
(f)*polylog(2,I*b/(1-I*a)*f^(d*x)*f^c)*c^2+1/2*I/d^3/ln(f)*c^2*dilog((b*f^(d*x)*f^c+I+a)/(I+a))+1/2*I/d^2*c^2*
ln((b*f^(d*x)*f^c+I+a)/(I+a))*x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b d f^{c} \int \frac {f^{d x} x^{3}}{3 \, {\left (b^{2} f^{2 \, d x} f^{2 \, c} + 2 \, a b f^{d x} f^{c} + a^{2} + 1\right )}}\,{d x} \log \relax (f) + \frac {1}{3} \, x^{3} \arctan \left (\frac {1}{b f^{d x} f^{c} + a}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccot(a+b*f^(d*x+c)),x, algorithm="maxima")

[Out]

b*d*f^c*integrate(1/3*f^(d*x)*x^3/(b^2*f^(2*d*x)*f^(2*c) + 2*a*b*f^(d*x)*f^c + a^2 + 1), x)*log(f) + 1/3*x^3*a
rctan(1/(b*f^(d*x)*f^c + a))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,\mathrm {acot}\left (a+b\,f^{c+d\,x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acot(a + b*f^(c + d*x)),x)

[Out]

int(x^2*acot(a + b*f^(c + d*x)), x)

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sympy [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(x**2*acot(a+b*f**(d*x+c)),x)

[Out]

Timed out

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