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

Optimal. Leaf size=250 \[ \frac{i \text{PolyLog}\left (3,\frac{b f^{c+d x}}{-a+i}\right )}{2 d^2 \log ^2(f)}-\frac{i \text{PolyLog}\left (3,-\frac{b f^{c+d x}}{a+i}\right )}{2 d^2 \log ^2(f)}-\frac{i x \text{PolyLog}\left (2,\frac{b f^{c+d x}}{-a+i}\right )}{2 d \log (f)}+\frac{i x \text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a+i}\right )}{2 d \log (f)}-\frac{1}{4} i x^2 \log \left (1-\frac{b f^{c+d x}}{-a+i}\right )+\frac{1}{4} i x^2 \log \left (1+\frac{b f^{c+d x}}{a+i}\right )+\frac{1}{4} i x^2 \log \left (1-\frac{i}{a+b f^{c+d x}}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{i}{a+b f^{c+d x}}\right ) \]

[Out]

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

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Rubi [A]  time = 2.65375, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {5144, 2551, 12, 6742, 2190, 2531, 2282, 6589} \[ \frac{i \text{PolyLog}\left (3,\frac{b f^{c+d x}}{-a+i}\right )}{2 d^2 \log ^2(f)}-\frac{i \text{PolyLog}\left (3,-\frac{b f^{c+d x}}{a+i}\right )}{2 d^2 \log ^2(f)}-\frac{i x \text{PolyLog}\left (2,\frac{b f^{c+d x}}{-a+i}\right )}{2 d \log (f)}+\frac{i x \text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a+i}\right )}{2 d \log (f)}-\frac{1}{4} i x^2 \log \left (1-\frac{b f^{c+d x}}{-a+i}\right )+\frac{1}{4} i x^2 \log \left (1+\frac{b f^{c+d x}}{a+i}\right )+\frac{1}{4} i x^2 \log \left (1-\frac{i}{a+b f^{c+d x}}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{i}{a+b f^{c+d x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCot[a + b*f^(c + d*x)],x]

[Out]

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

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 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 12

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

Rule 6742

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

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 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 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 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]

Rubi steps

\begin{align*} \int x \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx &=\frac{1}{2} i \int x \log \left (1-\frac{i}{a+b f^{c+d x}}\right ) \, dx-\frac{1}{2} i \int x \log \left (1+\frac{i}{a+b f^{c+d x}}\right ) \, dx\\ &=\frac{1}{4} i x^2 \log \left (1-\frac{i}{a+b f^{c+d x}}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{i}{a+b f^{c+d x}}\right )+\frac{1}{4} \int \frac{b d f^{c+d x} x^2 \log (f)}{\left (i (1-i a)+b f^{c+d x}\right ) \left (a+b f^{c+d x}\right )} \, dx+\frac{1}{4} \int \frac{b d f^{c+d x} x^2 \log (f)}{\left (-i (1+i a)+b f^{c+d x}\right ) \left (a+b f^{c+d x}\right )} \, dx\\ &=\frac{1}{4} i x^2 \log \left (1-\frac{i}{a+b f^{c+d x}}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{i}{a+b f^{c+d x}}\right )+\frac{1}{4} (b d \log (f)) \int \frac{f^{c+d x} x^2}{\left (i (1-i a)+b f^{c+d x}\right ) \left (a+b f^{c+d x}\right )} \, dx+\frac{1}{4} (b d \log (f)) \int \frac{f^{c+d x} x^2}{\left (-i (1+i a)+b f^{c+d x}\right ) \left (a+b f^{c+d x}\right )} \, dx\\ &=\frac{1}{4} i x^2 \log \left (1-\frac{i}{a+b f^{c+d x}}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{i}{a+b f^{c+d x}}\right )+\frac{1}{4} (b d \log (f)) \int \left (\frac{i f^{c+d x} x^2}{a+b f^{c+d x}}-\frac{i f^{c+d x} x^2}{-i+a+b f^{c+d x}}\right ) \, dx+\frac{1}{4} (b d \log (f)) \int \left (-\frac{i f^{c+d x} x^2}{a+b f^{c+d x}}+\frac{i f^{c+d x} x^2}{i+a+b f^{c+d x}}\right ) \, dx\\ &=\frac{1}{4} i x^2 \log \left (1-\frac{i}{a+b f^{c+d x}}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{i}{a+b f^{c+d x}}\right )-\frac{1}{4} (i b d \log (f)) \int \frac{f^{c+d x} x^2}{-i+a+b f^{c+d x}} \, dx+\frac{1}{4} (i b d \log (f)) \int \frac{f^{c+d x} x^2}{i+a+b f^{c+d x}} \, dx\\ &=-\frac{1}{4} i x^2 \log \left (1-\frac{b f^{c+d x}}{i-a}\right )+\frac{1}{4} i x^2 \log \left (1+\frac{b f^{c+d x}}{i+a}\right )+\frac{1}{4} i x^2 \log \left (1-\frac{i}{a+b f^{c+d x}}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{i}{a+b f^{c+d x}}\right )+\frac{1}{2} i \int x \log \left (1+\frac{b f^{c+d x}}{-i+a}\right ) \, dx-\frac{1}{2} i \int x \log \left (1+\frac{b f^{c+d x}}{i+a}\right ) \, dx\\ &=-\frac{1}{4} i x^2 \log \left (1-\frac{b f^{c+d x}}{i-a}\right )+\frac{1}{4} i x^2 \log \left (1+\frac{b f^{c+d x}}{i+a}\right )+\frac{1}{4} i x^2 \log \left (1-\frac{i}{a+b f^{c+d x}}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{i}{a+b f^{c+d x}}\right )-\frac{i x \text{Li}_2\left (\frac{b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac{i x \text{Li}_2\left (-\frac{b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac{i \int \text{Li}_2\left (-\frac{b f^{c+d x}}{-i+a}\right ) \, dx}{2 d \log (f)}-\frac{i \int \text{Li}_2\left (-\frac{b f^{c+d x}}{i+a}\right ) \, dx}{2 d \log (f)}\\ &=-\frac{1}{4} i x^2 \log \left (1-\frac{b f^{c+d x}}{i-a}\right )+\frac{1}{4} i x^2 \log \left (1+\frac{b f^{c+d x}}{i+a}\right )+\frac{1}{4} i x^2 \log \left (1-\frac{i}{a+b f^{c+d x}}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{i}{a+b f^{c+d x}}\right )-\frac{i x \text{Li}_2\left (\frac{b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac{i x \text{Li}_2\left (-\frac{b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{b x}{i-a}\right )}{x} \, dx,x,f^{c+d x}\right )}{2 d^2 \log ^2(f)}-\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{b x}{i+a}\right )}{x} \, dx,x,f^{c+d x}\right )}{2 d^2 \log ^2(f)}\\ &=-\frac{1}{4} i x^2 \log \left (1-\frac{b f^{c+d x}}{i-a}\right )+\frac{1}{4} i x^2 \log \left (1+\frac{b f^{c+d x}}{i+a}\right )+\frac{1}{4} i x^2 \log \left (1-\frac{i}{a+b f^{c+d x}}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{i}{a+b f^{c+d x}}\right )-\frac{i x \text{Li}_2\left (\frac{b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac{i x \text{Li}_2\left (-\frac{b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac{i \text{Li}_3\left (\frac{b f^{c+d x}}{i-a}\right )}{2 d^2 \log ^2(f)}-\frac{i \text{Li}_3\left (-\frac{b f^{c+d x}}{i+a}\right )}{2 d^2 \log ^2(f)}\\ \end{align*}

Mathematica [A]  time = 0.284365, size = 250, normalized size = 1. \[ \frac{i \text{PolyLog}\left (3,\frac{b f^{c+d x}}{-a+i}\right )}{2 d^2 \log ^2(f)}-\frac{i \text{PolyLog}\left (3,-\frac{b f^{c+d x}}{a+i}\right )}{2 d^2 \log ^2(f)}-\frac{i x \text{PolyLog}\left (2,\frac{b f^{c+d x}}{-a+i}\right )}{2 d \log (f)}+\frac{i x \text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a+i}\right )}{2 d \log (f)}-\frac{1}{4} i x^2 \log \left (1-\frac{b f^{c+d x}}{-a+i}\right )+\frac{1}{4} i x^2 \log \left (1+\frac{b f^{c+d x}}{a+i}\right )+\frac{1}{4} i x^2 \log \left (1-\frac{i}{a+b f^{c+d x}}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{i}{a+b f^{c+d x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.414, size = 678, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I/d*c*ln((b*f^(d*x)*f^c+I+a)/(I+a))*x+1/4*Pi*x^2+1/4*I/d^2*ln(1-I*b/(1-I*a)*f^(d*x)*f^c)*c^2+1/2*I/d^2*c^
2*ln((b*f^(d*x)*f^c+a-I)/(-I+a))-1/2*I/d^2*c^2*ln((b*f^(d*x)*f^c+I+a)/(I+a))-1/2*I/d^2/ln(f)*c*dilog((b*f^(d*x
)*f^c+I+a)/(I+a))-1/4*I*x^2*ln(1-I*(a+b*f^(d*x+c)))-1/2*I/d*ln(1-I*b/(-I*a-1)*f^(d*x)*f^c)*x*c-1/4*I/d^2*c^2*l
n(I*f^(d*x)*f^c*b+I*a+1)+1/2*I/d^2/ln(f)*polylog(2,I*b/(1-I*a)*f^(d*x)*f^c)*c+1/2*I/d^2/ln(f)^2*polylog(3,I*b/
(-I*a-1)*f^(d*x)*f^c)+1/2*I/d^2/ln(f)*c*dilog((b*f^(d*x)*f^c+a-I)/(-I+a))-1/2*I/d/ln(f)*polylog(2,I*b/(-I*a-1)
*f^(d*x)*f^c)*x-1/2*I/d^2/ln(f)^2*polylog(3,I*b/(1-I*a)*f^(d*x)*f^c)+1/2*I/d*ln(1-I*b/(1-I*a)*f^(d*x)*f^c)*x*c
+1/4*I/d^2*c^2*ln(1-I*a-I*f^(d*x)*f^c*b)-1/4*I/d^2*ln(1-I*b/(-I*a-1)*f^(d*x)*f^c)*c^2+1/2*I/d/ln(f)*polylog(2,
I*b/(1-I*a)*f^(d*x)*f^c)*x-1/2*I/d^2/ln(f)*polylog(2,I*b/(-I*a-1)*f^(d*x)*f^c)*c+1/2*I/d*c*ln((b*f^(d*x)*f^c+a
-I)/(-I+a))*x+1/4*I*x^2*ln(1+I*(a+b*f^(d*x+c)))-1/4*I*ln(1-I*b/(-I*a-1)*f^(d*x)*f^c)*x^2+1/4*I*ln(1-I*b/(1-I*a
)*f^(d*x)*f^c)*x^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(x*arccot(a+b*f^(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.81829, size = 778, normalized size = 3.11 \begin{align*} \frac{2 \, d^{2} x^{2} \operatorname{arccot}\left (b f^{d x + c} + a\right ) \log \left (f\right )^{2} + i \, c^{2} \log \left (b f^{d x + c} + a + i\right ) \log \left (f\right )^{2} - i \, c^{2} \log \left (b f^{d x + c} + a - i\right ) \log \left (f\right )^{2} - 2 i \, d x{\rm Li}_2\left (-\frac{a^{2} +{\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \left (f\right ) + 2 i \, d x{\rm Li}_2\left (-\frac{a^{2} +{\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \left (f\right ) +{\left (-i \, d^{2} x^{2} + i \, c^{2}\right )} \log \left (f\right )^{2} \log \left (\frac{a^{2} +{\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) +{\left (i \, d^{2} x^{2} - i \, c^{2}\right )} \log \left (f\right )^{2} \log \left (\frac{a^{2} +{\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + 2 i \,{\rm polylog}\left (3, -\frac{{\left (a b + i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) - 2 i \,{\rm polylog}\left (3, -\frac{{\left (a b - i \, b\right )} f^{d x + c}}{a^{2} + 1}\right )}{4 \, d^{2} \log \left (f\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{arccot}\left (b f^{d x + c} + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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