3.4.53 \(\int x \tanh ^{-1}(a+b f^{c+d x}) \, dx\) [353]

Optimal. Leaf size=211 \[ -\frac {1}{4} x^2 \log \left (1-a-b f^{c+d x}\right )+\frac {1}{4} x^2 \log \left (1+a+b f^{c+d x}\right )+\frac {1}{4} x^2 \log \left (1-\frac {b f^{c+d x}}{1-a}\right )-\frac {1}{4} x^2 \log \left (1+\frac {b f^{c+d x}}{1+a}\right )+\frac {x \text {PolyLog}\left (2,\frac {b f^{c+d x}}{1-a}\right )}{2 d \log (f)}-\frac {x \text {PolyLog}\left (2,-\frac {b f^{c+d x}}{1+a}\right )}{2 d \log (f)}-\frac {\text {PolyLog}\left (3,\frac {b f^{c+d x}}{1-a}\right )}{2 d^2 \log ^2(f)}+\frac {\text {PolyLog}\left (3,-\frac {b f^{c+d x}}{1+a}\right )}{2 d^2 \log ^2(f)} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 211, normalized size of antiderivative = 1.00, 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 = {6348, 2612, 2611, 2320, 6724} \begin {gather*} -\frac {\text {Li}_3\left (\frac {b f^{c+d x}}{1-a}\right )}{2 d^2 \log ^2(f)}+\frac {\text {Li}_3\left (-\frac {b f^{c+d x}}{a+1}\right )}{2 d^2 \log ^2(f)}+\frac {x \text {Li}_2\left (\frac {b f^{c+d x}}{1-a}\right )}{2 d \log (f)}-\frac {x \text {Li}_2\left (-\frac {b f^{c+d x}}{a+1}\right )}{2 d \log (f)}-\frac {1}{4} x^2 \log \left (-a-b f^{c+d x}+1\right )+\frac {1}{4} x^2 \log \left (a+b f^{c+d x}+1\right )+\frac {1}{4} x^2 \log \left (1-\frac {b f^{c+d x}}{1-a}\right )-\frac {1}{4} x^2 \log \left (\frac {b f^{c+d x}}{a+1}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*ArcTanh[a + b*f^(c + d*x)],x]

[Out]

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

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 2612

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/d)*(F^(c*(a +
b*x)))^n], x] - Simp[(f + g*x)^(m + 1)*(Log[1 + (e/d)*(F^(c*(a + b*x)))^n]/(g*(m + 1))), x]) /; FreeQ[{F, a,
b, c, d, e, f, g, n}, x] && GtQ[m, 0] && NeQ[d, 1]

Rule 6348

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

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]

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 177, normalized size = 0.84 \begin {gather*} \frac {2 d^2 x^2 \tanh ^{-1}\left (a+b f^{c+d x}\right ) \log ^2(f)+d^2 x^2 \log ^2(f) \log \left (1+\frac {b f^{c+d x}}{-1+a}\right )-d^2 x^2 \log ^2(f) \log \left (1+\frac {b f^{c+d x}}{1+a}\right )+2 d x \log (f) \text {PolyLog}\left (2,-\frac {b f^{c+d x}}{-1+a}\right )-2 d x \log (f) \text {PolyLog}\left (2,-\frac {b f^{c+d x}}{1+a}\right )-2 \text {PolyLog}\left (3,-\frac {b f^{c+d x}}{-1+a}\right )+2 \text {PolyLog}\left (3,-\frac {b f^{c+d x}}{1+a}\right )}{4 d^2 \log ^2(f)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(595\) vs. \(2(195)=390\).
time = 0.06, size = 596, normalized size = 2.82

method	result	size
risch	\(\frac {x^{2} \ln \left (1+a +b \,f^{d x +c}\right )}{4}-\frac {x^{2} \ln \left (1-a -b \,f^{d x +c}\right )}{4}+\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{1-a}\right ) x^{2}}{4}+\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{1-a}\right ) x c}{2 d}+\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{1-a}\right ) c^{2}}{4 d^{2}}+\frac {\polylog \left (2, \frac {b \,f^{d x} f^{c}}{1-a}\right ) x}{2 \ln \left (f \right ) d}+\frac {\polylog \left (2, \frac {b \,f^{d x} f^{c}}{1-a}\right ) c}{2 \ln \left (f \right ) d^{2}}-\frac {\polylog \left (3, \frac {b \,f^{d x} f^{c}}{1-a}\right )}{2 \ln \left (f \right )^{2} d^{2}}+\frac {c^{2} \ln \left (1-a -b \,f^{d x} f^{c}\right )}{4 d^{2}}-\frac {c \dilog \left (\frac {b \,f^{d x} f^{c}+a -1}{-1+a}\right )}{2 \ln \left (f \right ) d^{2}}-\frac {c \ln \left (\frac {b \,f^{d x} f^{c}+a -1}{-1+a}\right ) x}{2 d}-\frac {c^{2} \ln \left (\frac {b \,f^{d x} f^{c}+a -1}{-1+a}\right )}{2 d^{2}}-\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{-a -1}\right ) x^{2}}{4}-\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{-a -1}\right ) x c}{2 d}-\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{-a -1}\right ) c^{2}}{4 d^{2}}-\frac {\polylog \left (2, \frac {b \,f^{d x} f^{c}}{-a -1}\right ) x}{2 \ln \left (f \right ) d}-\frac {\polylog \left (2, \frac {b \,f^{d x} f^{c}}{-a -1}\right ) c}{2 \ln \left (f \right ) d^{2}}+\frac {\polylog \left (3, \frac {b \,f^{d x} f^{c}}{-a -1}\right )}{2 \ln \left (f \right )^{2} d^{2}}-\frac {c^{2} \ln \left (1+a +b \,f^{d x} f^{c}\right )}{4 d^{2}}+\frac {c \dilog \left (\frac {1+a +b \,f^{d x} f^{c}}{1+a}\right )}{2 \ln \left (f \right ) d^{2}}+\frac {c \ln \left (\frac {1+a +b \,f^{d x} f^{c}}{1+a}\right ) x}{2 d}+\frac {c^{2} \ln \left (\frac {1+a +b \,f^{d x} f^{c}}{1+a}\right )}{2 d^{2}}\)	\(596\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctanh(a+b*f^(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 194, normalized size = 0.92 \begin {gather*} -\frac {1}{4} \, b d {\left (\frac {d^{2} x^{2} \log \left (\frac {b f^{d x} f^{c}}{a + 1} + 1\right ) \log \left (f\right )^{2} + 2 \, d x {\rm Li}_2\left (-\frac {b f^{d x} f^{c}}{a + 1}\right ) \log \left (f\right ) - 2 \, {\rm Li}_{3}(-\frac {b f^{d x} f^{c}}{a + 1})}{b d^{3} \log \left (f\right )^{3}} - \frac {d^{2} x^{2} \log \left (\frac {b f^{d x} f^{c}}{a - 1} + 1\right ) \log \left (f\right )^{2} + 2 \, d x {\rm Li}_2\left (-\frac {b f^{d x} f^{c}}{a - 1}\right ) \log \left (f\right ) - 2 \, {\rm Li}_{3}(-\frac {b f^{d x} f^{c}}{a - 1})}{b d^{3} \log \left (f\right )^{3}}\right )} \log \left (f\right ) + \frac {1}{2} \, x^{2} \operatorname {artanh}\left (b f^{d x + c} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 396 vs. \(2 (188) = 376\).
time = 0.40, size = 396, normalized size = 1.88 \begin {gather*} \frac {d^{2} x^{2} \log \left (f\right )^{2} \log \left (-\frac {b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a + 1}{b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a - 1}\right ) - c^{2} \log \left (b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a + 1\right ) \log \left (f\right )^{2} + c^{2} \log \left (b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a - 1\right ) \log \left (f\right )^{2} - 2 \, d x {\rm Li}_2\left (-\frac {b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a + 1}{a + 1} + 1\right ) \log \left (f\right ) + 2 \, d x {\rm Li}_2\left (-\frac {b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a - 1}{a - 1} + 1\right ) \log \left (f\right ) - {\left (d^{2} x^{2} - c^{2}\right )} \log \left (f\right )^{2} \log \left (\frac {b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a + 1}{a + 1}\right ) + {\left (d^{2} x^{2} - c^{2}\right )} \log \left (f\right )^{2} \log \left (\frac {b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + a - 1}{a - 1}\right ) + 2 \, {\rm polylog}\left (3, -\frac {b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right )}{a + 1}\right ) - 2 \, {\rm polylog}\left (3, -\frac {b \cosh \left ({\left (d x + c\right )} \log \left (f\right )\right ) + b \sinh \left ({\left (d x + c\right )} \log \left (f\right )\right )}{a - 1}\right )}{4 \, d^{2} \log \left (f\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \operatorname {atanh}{\left (a + b f^{c} f^{d x} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*atanh(a + b*f**c*f**(d*x)), x)

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x\,\mathrm {atanh}\left (a+b\,f^{c+d\,x}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh(a + b*f^(c + d*x)),x)

[Out]

int(x*atanh(a + b*f^(c + d*x)), x)

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