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

Optimal. Leaf size=232 \[ \frac {i \text {Li}_3\left (\frac {i b f^{c+d x}}{1-i a}\right )}{2 d^2 \log ^2(f)}-\frac {i \text {Li}_3\left (-\frac {i b f^{c+d x}}{i a+1}\right )}{2 d^2 \log ^2(f)}-\frac {i x \text {Li}_2\left (\frac {i b f^{c+d x}}{1-i a}\right )}{2 d \log (f)}+\frac {i x \text {Li}_2\left (-\frac {i b f^{c+d x}}{i a+1}\right )}{2 d \log (f)}-\frac {1}{4} i x^2 \log \left (1-\frac {i b f^{c+d x}}{1-i a}\right )+\frac {1}{4} i x^2 \log \left (1+\frac {i b f^{c+d x}}{1+i a}\right )+\frac {1}{2} x^2 \tan ^{-1}\left (a+b f^{c+d x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 250, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5143, 2532, 2531, 2282, 6589} \[ -\frac {i \text {PolyLog}\left (3,\frac {b f^{c+d x}}{-a+i}\right )}{2 d^2 \log ^2(f)}+\frac {i \text {PolyLog}\left (3,-\frac {b f^{c+d x}}{a+i}\right )}{2 d^2 \log ^2(f)}+\frac {i x \text {PolyLog}\left (2,\frac {b f^{c+d x}}{-a+i}\right )}{2 d \log (f)}-\frac {i x \text {PolyLog}\left (2,-\frac {b f^{c+d x}}{a+i}\right )}{2 d \log (f)}+\frac {1}{4} i x^2 \log \left (-i a-i b f^{c+d x}+1\right )-\frac {1}{4} i x^2 \log \left (i a+i b f^{c+d x}+1\right )+\frac {1}{4} i x^2 \log \left (1-\frac {b f^{c+d x}}{-a+i}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {b f^{c+d x}}{a+i}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Int[x*ArcTan[a + b*f^(c + d*x)],x]

[Out]

(I/4)*x^2*Log[1 - I*a - I*b*f^(c + d*x)] - (I/4)*x^2*Log[1 + I*a + I*b*f^(c + d*x)] + (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/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 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 2532

Int[Log[(d_) + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[
((f + g*x)^(m + 1)*Log[d + e*(F^(c*(a + b*x)))^n])/(g*(m + 1)), x] + (Int[(f + g*x)^m*Log[1 + (e*(F^(c*(a + b*
x)))^n)/d], x] - Simp[((f + g*x)^(m + 1)*Log[1 + (e*(F^(c*(a + b*x)))^n)/d])/(g*(m + 1)), x]) /; FreeQ[{F, a,
b, c, d, e, f, g, n}, x] && GtQ[m, 0] && NeQ[d, 1]

Rule 5143

Int[ArcTan[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Dist[I/2, Int[x^m*Log[1 - I*a - I
*b*f^(c + d*x)], x], x] - Dist[I/2, Int[x^m*Log[1 + I*a + I*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]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 236, normalized size = 1.02 \[ \frac {i \left (d^2 x^2 \log ^2(f) \log \left (-i a-i b f^{c+d x}+1\right )-d^2 x^2 \log ^2(f) \log \left (i a+i b f^{c+d x}+1\right )-d^2 x^2 \log ^2(f) \log \left (\frac {a+b f^{c+d x}+i}{a+i}\right )+d^2 x^2 \log ^2(f) \log \left (1+\frac {b f^{c+d x}}{a-i}\right )-2 \text {Li}_3\left (\frac {b f^{c+d x}}{i-a}\right )+2 \text {Li}_3\left (-\frac {b f^{c+d x}}{a+i}\right )+2 d x \log (f) \text {Li}_2\left (\frac {b f^{c+d x}}{i-a}\right )-2 d x \log (f) \text {Li}_2\left (-\frac {b f^{c+d x}}{a+i}\right )\right )}{4 d^2 \log ^2(f)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [C]  time = 0.45, size = 304, normalized size = 1.31 \[ \frac {2 \, d^{2} x^{2} \arctan \left (b f^{d x + c} + a\right ) \log \relax (f)^{2} - i \, c^{2} \log \left (b f^{d x + c} + a + i\right ) \log \relax (f)^{2} + i \, c^{2} \log \left (b f^{d x + c} + a - i\right ) \log \relax (f)^{2} + 2 i \, d x {\rm Li}_2\left (-\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \relax (f) - 2 i \, d x {\rm Li}_2\left (-\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \relax (f) + {\left (i \, d^{2} x^{2} - i \, c^{2}\right )} \log \relax (f)^{2} \log \left (\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + {\left (-i \, d^{2} x^{2} + i \, c^{2}\right )} \log \relax (f)^{2} \log \left (\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) - 2 i \, {\rm polylog}\left (3, -\frac {{\left (a b + i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) + 2 i \, {\rm polylog}\left (3, -\frac {{\left (a b - i \, b\right )} f^{d x + c}}{a^{2} + 1}\right )}{4 \, d^{2} \log \relax (f)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*d^2*x^2*arctan(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \arctan \left (b f^{d x + c} + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.32, size = 672, normalized size = 2.90 \[ -\frac {i x^{2} \ln \left (1+i \left (a +b \,f^{d x +c}\right )\right )}{4}+\frac {i c^{2} \ln \left (\frac {f^{d x} f^{c} b +i+a}{i+a}\right )}{2 d^{2}}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x^{2}}{4}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) c^{2}}{4 d^{2}}+\frac {i c \dilog \left (\frac {f^{d x} f^{c} b +i+a}{i+a}\right )}{2 d^{2} \ln \relax (f )}+\frac {i \polylog \left (3, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right )}{2 d^{2} \ln \relax (f )^{2}}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x^{2}}{4}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) c^{2}}{4 d^{2}}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x c}{2 d}-\frac {i \polylog \left (2, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x}{2 d \ln \relax (f )}+\frac {i \polylog \left (2, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x}{2 d \ln \relax (f )}-\frac {i c \dilog \left (\frac {f^{d x} f^{c} b +a -i}{-i+a}\right )}{2 d^{2} \ln \relax (f )}-\frac {i c \ln \left (\frac {f^{d x} f^{c} b +a -i}{-i+a}\right ) x}{2 d}-\frac {i c^{2} \ln \left (1-i a -i f^{d x} f^{c} b \right )}{4 d^{2}}+\frac {i c^{2} \ln \left (i f^{d x} f^{c} b +i a +1\right )}{4 d^{2}}-\frac {i c^{2} \ln \left (\frac {f^{d x} f^{c} b +a -i}{-i+a}\right )}{2 d^{2}}+\frac {i \polylog \left (2, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) c}{2 d^{2} \ln \relax (f )}-\frac {i \polylog \left (3, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right )}{2 d^{2} \ln \relax (f )^{2}}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x c}{2 d}-\frac {i \polylog \left (2, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) c}{2 d^{2} \ln \relax (f )}+\frac {i c \ln \left (\frac {f^{d x} f^{c} b +i+a}{i+a}\right ) x}{2 d}+\frac {i x^{2} \ln \left (1-i \left (a +b \,f^{d x +c}\right )\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atan(a + b*f^(c + d*x)),x)

[Out]

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

[Out]

Timed out

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