3.62 \(\int \frac {(c e+d e x) (a+b \tanh ^{-1}(c+d x))}{1-(c+d x)^2} \, dx\)

Optimal. Leaf size=83 \[ -\frac {e \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 b d}+\frac {e \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d}+\frac {b e \text {Li}_2\left (-\frac {c+d x+1}{-c-d x+1}\right )}{2 d} \]

[Out]

-1/2*e*(a+b*arctanh(d*x+c))^2/b/d+e*(a+b*arctanh(d*x+c))*ln(2/(-d*x-c+1))/d+1/2*b*e*polylog(2,(-d*x-c-1)/(-d*x
-c+1))/d

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Rubi [A]  time = 0.26, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {12, 5984, 5918, 2402, 2315} \[ \frac {b e \text {PolyLog}\left (2,-\frac {c+d x+1}{-c-d x+1}\right )}{2 d}-\frac {e \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 b d}+\frac {e \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[((c*e + d*e*x)*(a + b*ArcTanh[c + d*x]))/(1 - (c + d*x)^2),x]

[Out]

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

Rule 12

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 - c^2
*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 5984

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 114, normalized size = 1.37 \[ -\frac {e \left (4 a \log (-c-d x+1)+4 a \log (c+d x+1)-2 b \text {Li}_2\left (\frac {1}{2} (-c-d x+1)\right )+2 b \text {Li}_2\left (\frac {1}{2} (c+d x+1)\right )-b \log ^2(-c-d x+1)+b \log ^2(c+d x+1)+b \log (4) \log (c+d x-1)-b \log (4) \log (c+d x+1)\right )}{8 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((c*e + d*e*x)*(a + b*ArcTanh[c + d*x]))/(1 - (c + d*x)^2),x]

[Out]

-1/8*(e*(4*a*Log[1 - c - d*x] - b*Log[1 - c - d*x]^2 + b*Log[4]*Log[-1 + c + d*x] + 4*a*Log[1 + c + d*x] - b*L
og[4]*Log[1 + c + d*x] + b*Log[1 + c + d*x]^2 - 2*b*PolyLog[2, (1 - c - d*x)/2] + 2*b*PolyLog[2, (1 + c + d*x)
/2]))/d

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {a d e x + a c e + {\left (b d e x + b c e\right )} \operatorname {artanh}\left (d x + c\right )}{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)*(a+b*arctanh(d*x+c))/(1-(d*x+c)^2),x, algorithm="fricas")

[Out]

integral(-(a*d*e*x + a*c*e + (b*d*e*x + b*c*e)*arctanh(d*x + c))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x)

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giac [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((d*e*x+c*e)*(a+b*arctanh(d*x+c))/(1-(d*x+c)^2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.06, size = 194, normalized size = 2.34 \[ -\frac {a e \ln \left (d x +c -1\right )}{2 d}-\frac {a e \ln \left (d x +c +1\right )}{2 d}-\frac {b e \arctanh \left (d x +c \right ) \ln \left (d x +c -1\right )}{2 d}-\frac {b e \arctanh \left (d x +c \right ) \ln \left (d x +c +1\right )}{2 d}-\frac {b e \ln \left (d x +c -1\right )^{2}}{8 d}+\frac {b e \dilog \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {b e \ln \left (d x +c -1\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {b e \ln \left (d x +c +1\right )^{2}}{8 d}-\frac {b e \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (d x +c +1\right )}{4 d}+\frac {b e \ln \left (-\frac {d x}{2}-\frac {c}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)*(a+b*arctanh(d*x+c))/(1-(d*x+c)^2),x)

[Out]

-1/2/d*a*e*ln(d*x+c-1)-1/2/d*a*e*ln(d*x+c+1)-1/2/d*b*e*arctanh(d*x+c)*ln(d*x+c-1)-1/2/d*b*e*arctanh(d*x+c)*ln(
d*x+c+1)-1/8/d*b*e*ln(d*x+c-1)^2+1/2/d*b*e*dilog(1/2+1/2*d*x+1/2*c)+1/4/d*b*e*ln(d*x+c-1)*ln(1/2+1/2*d*x+1/2*c
)+1/8/d*b*e*ln(d*x+c+1)^2-1/4/d*b*e*ln(-1/2*d*x-1/2*c+1/2)*ln(d*x+c+1)+1/4/d*b*e*ln(-1/2*d*x-1/2*c+1/2)*ln(1/2
+1/2*d*x+1/2*c)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, b c e {\left (\frac {\log \left (d x + c + 1\right )}{d} - \frac {\log \left (d x + c - 1\right )}{d}\right )} \operatorname {artanh}\left (d x + c\right ) - \frac {1}{2} \, a d e {\left (\frac {{\left (c + 1\right )} \log \left (d x + c + 1\right )}{d^{2}} - \frac {{\left (c - 1\right )} \log \left (d x + c - 1\right )}{d^{2}}\right )} + \frac {1}{2} \, a c e {\left (\frac {\log \left (d x + c + 1\right )}{d} - \frac {\log \left (d x + c - 1\right )}{d}\right )} + \frac {1}{8} \, b d e {\left (\frac {2 \, {\left (c + 1\right )} \log \left (d x + c + 1\right ) \log \left (-d x - c + 1\right ) - {\left (c - 1\right )} \log \left (-d x - c + 1\right )^{2}}{d^{2}} - 4 \, \int \frac {{\left (c^{2} + {\left (c d + 3 \, d\right )} x + 2 \, c + 1\right )} \log \left (d x + c + 1\right )}{2 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d - d\right )}}\,{d x}\right )} - \frac {{\left (\log \left (d x + c + 1\right )^{2} - 2 \, \log \left (d x + c + 1\right ) \log \left (d x + c - 1\right ) + \log \left (d x + c - 1\right )^{2}\right )} b c e}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)*(a+b*arctanh(d*x+c))/(1-(d*x+c)^2),x, algorithm="maxima")

[Out]

1/2*b*c*e*(log(d*x + c + 1)/d - log(d*x + c - 1)/d)*arctanh(d*x + c) - 1/2*a*d*e*((c + 1)*log(d*x + c + 1)/d^2
- (c - 1)*log(d*x + c - 1)/d^2) + 1/2*a*c*e*(log(d*x + c + 1)/d - log(d*x + c - 1)/d) + 1/8*b*d*e*((2*(c + 1)
*log(d*x + c + 1)*log(-d*x - c + 1) - (c - 1)*log(-d*x - c + 1)^2)/d^2 - 4*integrate(1/2*(c^2 + (c*d + 3*d)*x
+ 2*c + 1)*log(d*x + c + 1)/(d^3*x^2 + 2*c*d^2*x + c^2*d - d), x)) - 1/8*(log(d*x + c + 1)^2 - 2*log(d*x + c +
1)*log(d*x + c - 1) + log(d*x + c - 1)^2)*b*c*e/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int -\frac {\left (c\,e+d\,e\,x\right )\,\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}{{\left (c+d\,x\right )}^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((c*e + d*e*x)*(a + b*atanh(c + d*x)))/((c + d*x)^2 - 1),x)

[Out]

int(-((c*e + d*e*x)*(a + b*atanh(c + d*x)))/((c + d*x)^2 - 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - e \left (\int \frac {a c}{c^{2} + 2 c d x + d^{2} x^{2} - 1}\, dx + \int \frac {a d x}{c^{2} + 2 c d x + d^{2} x^{2} - 1}\, dx + \int \frac {b c \operatorname {atanh}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2} - 1}\, dx + \int \frac {b d x \operatorname {atanh}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2} - 1}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)*(a+b*atanh(d*x+c))/(1-(d*x+c)**2),x)

[Out]

-e*(Integral(a*c/(c**2 + 2*c*d*x + d**2*x**2 - 1), x) + Integral(a*d*x/(c**2 + 2*c*d*x + d**2*x**2 - 1), x) +
Integral(b*c*atanh(c + d*x)/(c**2 + 2*c*d*x + d**2*x**2 - 1), x) + Integral(b*d*x*atanh(c + d*x)/(c**2 + 2*c*d
*x + d**2*x**2 - 1), x))

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