3.43 \(\int \frac {x^3 (a+b \tanh ^{-1}(c x))}{d+c d x} \, dx\)

Optimal. Leaf size=177 \[ \frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d}-\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^2 d}+\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 c d}+\frac {a x}{c^3 d}-\frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 c^4 d}+\frac {b \tanh ^{-1}(c x)}{2 c^4 d}-\frac {b x}{2 c^3 d}+\frac {b x \tanh ^{-1}(c x)}{c^3 d}+\frac {b x^2}{6 c^2 d}+\frac {2 b \log \left (1-c^2 x^2\right )}{3 c^4 d} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {5930, 5916, 266, 43, 321, 206, 5910, 260, 5918, 2402, 2315} \[ -\frac {b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 c^4 d}-\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^2 d}+\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d}+\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 c d}+\frac {a x}{c^3 d}+\frac {b x^2}{6 c^2 d}+\frac {2 b \log \left (1-c^2 x^2\right )}{3 c^4 d}-\frac {b x}{2 c^3 d}+\frac {b x \tanh ^{-1}(c x)}{c^3 d}+\frac {b \tanh ^{-1}(c x)}{2 c^4 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcTanh[c*x]))/(d + c*d*x),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, 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 5910

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

Rule 5916

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

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 5930

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

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{d+c d x} \, dx &=-\frac {\int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{d+c d x} \, dx}{c}+\frac {\int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c d}\\ &=\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 c d}+\frac {\int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{d+c d x} \, dx}{c^2}-\frac {b \int \frac {x^3}{1-c^2 x^2} \, dx}{3 d}-\frac {\int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c^2 d}\\ &=-\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^2 d}+\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 c d}-\frac {\int \frac {a+b \tanh ^{-1}(c x)}{d+c d x} \, dx}{c^3}-\frac {b \operatorname {Subst}\left (\int \frac {x}{1-c^2 x} \, dx,x,x^2\right )}{6 d}+\frac {\int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c^3 d}+\frac {b \int \frac {x^2}{1-c^2 x^2} \, dx}{2 c d}\\ &=\frac {a x}{c^3 d}-\frac {b x}{2 c^3 d}-\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^2 d}+\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 c d}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^4 d}-\frac {b \operatorname {Subst}\left (\int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{6 d}+\frac {b \int \frac {1}{1-c^2 x^2} \, dx}{2 c^3 d}+\frac {b \int \tanh ^{-1}(c x) \, dx}{c^3 d}-\frac {b \int \frac {\log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^3 d}\\ &=\frac {a x}{c^3 d}-\frac {b x}{2 c^3 d}+\frac {b x^2}{6 c^2 d}+\frac {b \tanh ^{-1}(c x)}{2 c^4 d}+\frac {b x \tanh ^{-1}(c x)}{c^3 d}-\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^2 d}+\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 c d}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^4 d}+\frac {b \log \left (1-c^2 x^2\right )}{6 c^4 d}-\frac {b \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c x}\right )}{c^4 d}-\frac {b \int \frac {x}{1-c^2 x^2} \, dx}{c^2 d}\\ &=\frac {a x}{c^3 d}-\frac {b x}{2 c^3 d}+\frac {b x^2}{6 c^2 d}+\frac {b \tanh ^{-1}(c x)}{2 c^4 d}+\frac {b x \tanh ^{-1}(c x)}{c^3 d}-\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^2 d}+\frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 c d}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^4 d}+\frac {2 b \log \left (1-c^2 x^2\right )}{3 c^4 d}-\frac {b \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 c^4 d}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 129, normalized size = 0.73 \[ \frac {2 a c^3 x^3-3 a c^2 x^2+6 a c x-6 a \log (c x+1)+b c^2 x^2+4 b \log \left (1-c^2 x^2\right )+b \tanh ^{-1}(c x) \left (2 c^3 x^3-3 c^2 x^2+6 c x+6 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+3\right )-3 b \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )-3 b c x-b}{6 c^4 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^3*(a + b*ArcTanh[c*x]))/(d + c*d*x),x]

[Out]

(-b + 6*a*c*x - 3*b*c*x - 3*a*c^2*x^2 + b*c^2*x^2 + 2*a*c^3*x^3 + b*ArcTanh[c*x]*(3 + 6*c*x - 3*c^2*x^2 + 2*c^
3*x^3 + 6*Log[1 + E^(-2*ArcTanh[c*x])]) - 6*a*Log[1 + c*x] + 4*b*Log[1 - c^2*x^2] - 3*b*PolyLog[2, -E^(-2*ArcT
anh[c*x])])/(6*c^4*d)

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{3} \operatorname {artanh}\left (c x\right ) + a x^{3}}{c d x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x))/(c*d*x+d),x, algorithm="fricas")

[Out]

integral((b*x^3*arctanh(c*x) + a*x^3)/(c*d*x + d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} x^{3}}{c d x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x))/(c*d*x+d),x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)*x^3/(c*d*x + d), x)

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maple [A]  time = 0.05, size = 253, normalized size = 1.43 \[ \frac {a \,x^{3}}{3 c d}-\frac {a \,x^{2}}{2 c^{2} d}+\frac {a x}{c^{3} d}-\frac {a \ln \left (c x +1\right )}{c^{4} d}+\frac {b \,x^{3} \arctanh \left (c x \right )}{3 c d}-\frac {b \arctanh \left (c x \right ) x^{2}}{2 c^{2} d}+\frac {b x \arctanh \left (c x \right )}{c^{3} d}-\frac {b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{c^{4} d}+\frac {b \ln \left (c x +1\right )^{2}}{4 c^{4} d}-\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 c^{4} d}+\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 c^{4} d}+\frac {b \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 c^{4} d}+\frac {b \,x^{2}}{6 c^{2} d}-\frac {b x}{2 c^{3} d}-\frac {2 b}{3 c^{4} d}+\frac {11 b \ln \left (c x +1\right )}{12 c^{4} d}+\frac {5 b \ln \left (c x -1\right )}{12 c^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctanh(c*x))/(c*d*x+d),x)

[Out]

1/3/c*a/d*x^3-1/2/c^2*a/d*x^2+a*x/c^3/d-1/c^4*a/d*ln(c*x+1)+1/3/c*b/d*x^3*arctanh(c*x)-1/2/c^2*b/d*arctanh(c*x
)*x^2+b*x*arctanh(c*x)/c^3/d-1/c^4*b/d*arctanh(c*x)*ln(c*x+1)+1/4/c^4*b/d*ln(c*x+1)^2-1/2/c^4*b/d*ln(-1/2*c*x+
1/2)*ln(c*x+1)+1/2/c^4*b/d*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+1/2/c^4*b/d*dilog(1/2+1/2*c*x)+1/6*b*x^2/c^2/d-1/2
*b*x/c^3/d-2/3/c^4*b/d+11/12/c^4*b/d*ln(c*x+1)+5/12/c^4*b/d*ln(c*x-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{72} \, {\left (2 \, c^{4} {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{7} d} - \frac {3 \, \log \left (c x + 1\right )}{c^{8} d} + \frac {3 \, \log \left (c x - 1\right )}{c^{8} d}\right )} + 216 \, c^{4} \int \frac {x^{4} \log \left (c x + 1\right )}{6 \, {\left (c^{5} d x^{2} - c^{3} d\right )}}\,{d x} - 3 \, c^{3} {\left (\frac {x^{2}}{c^{5} d} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{7} d}\right )} - 216 \, c^{3} \int \frac {x^{3} \log \left (c x + 1\right )}{6 \, {\left (c^{5} d x^{2} - c^{3} d\right )}}\,{d x} + 9 \, c^{2} {\left (\frac {2 \, x}{c^{5} d} - \frac {\log \left (c x + 1\right )}{c^{6} d} + \frac {\log \left (c x - 1\right )}{c^{6} d}\right )} - 216 \, c \int \frac {x \log \left (c x + 1\right )}{6 \, {\left (c^{5} d x^{2} - c^{3} d\right )}}\,{d x} - \frac {6 \, {\left (2 \, c^{3} x^{3} - 3 \, c^{2} x^{2} + 6 \, c x - 6 \, \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{c^{4} d} + \frac {18 \, \log \left (6 \, c^{5} d x^{2} - 6 \, c^{3} d\right )}{c^{4} d} - 216 \, \int \frac {\log \left (c x + 1\right )}{6 \, {\left (c^{5} d x^{2} - c^{3} d\right )}}\,{d x}\right )} b + \frac {1}{6} \, a {\left (\frac {2 \, c^{2} x^{3} - 3 \, c x^{2} + 6 \, x}{c^{3} d} - \frac {6 \, \log \left (c x + 1\right )}{c^{4} d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x))/(c*d*x+d),x, algorithm="maxima")

[Out]

1/72*(2*c^4*(2*(c^2*x^3 + 3*x)/(c^7*d) - 3*log(c*x + 1)/(c^8*d) + 3*log(c*x - 1)/(c^8*d)) + 216*c^4*integrate(
1/6*x^4*log(c*x + 1)/(c^5*d*x^2 - c^3*d), x) - 3*c^3*(x^2/(c^5*d) + log(c^2*x^2 - 1)/(c^7*d)) - 216*c^3*integr
ate(1/6*x^3*log(c*x + 1)/(c^5*d*x^2 - c^3*d), x) + 9*c^2*(2*x/(c^5*d) - log(c*x + 1)/(c^6*d) + log(c*x - 1)/(c
^6*d)) - 216*c*integrate(1/6*x*log(c*x + 1)/(c^5*d*x^2 - c^3*d), x) - 6*(2*c^3*x^3 - 3*c^2*x^2 + 6*c*x - 6*log
(c*x + 1))*log(-c*x + 1)/(c^4*d) + 18*log(6*c^5*d*x^2 - 6*c^3*d)/(c^4*d) - 216*integrate(1/6*log(c*x + 1)/(c^5
*d*x^2 - c^3*d), x))*b + 1/6*a*((2*c^2*x^3 - 3*c*x^2 + 6*x)/(c^3*d) - 6*log(c*x + 1)/(c^4*d))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}{d+c\,d\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*atanh(c*x)))/(d + c*d*x),x)

[Out]

int((x^3*(a + b*atanh(c*x)))/(d + c*d*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a x^{3}}{c x + 1}\, dx + \int \frac {b x^{3} \operatorname {atanh}{\left (c x \right )}}{c x + 1}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atanh(c*x))/(c*d*x+d),x)

[Out]

(Integral(a*x**3/(c*x + 1), x) + Integral(b*x**3*atanh(c*x)/(c*x + 1), x))/d

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