3.1.58 \(\int \frac {x^4 (a+b \tanh ^{-1}(c x))}{(d+c d x)^3} \, dx\) [58]

Optimal. Leaf size=227 \[ -\frac {3 a x}{c^4 d^3}+\frac {b x}{2 c^4 d^3}-\frac {b}{8 c^5 d^3 (1+c x)^2}+\frac {15 b}{8 c^5 d^3 (1+c x)}-\frac {19 b \tanh ^{-1}(c x)}{8 c^5 d^3}-\frac {3 b x \tanh ^{-1}(c x)}{c^4 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^3 d^3}-\frac {a+b \tanh ^{-1}(c x)}{2 c^5 d^3 (1+c x)^2}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3 (1+c x)}-\frac {6 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^5 d^3}-\frac {3 b \log \left (1-c^2 x^2\right )}{2 c^5 d^3}+\frac {3 b \text {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{c^5 d^3} \]

[Out]

-3*a*x/c^4/d^3+1/2*b*x/c^4/d^3-1/8*b/c^5/d^3/(c*x+1)^2+15/8*b/c^5/d^3/(c*x+1)-19/8*b*arctanh(c*x)/c^5/d^3-3*b*
x*arctanh(c*x)/c^4/d^3+1/2*x^2*(a+b*arctanh(c*x))/c^3/d^3+1/2*(-a-b*arctanh(c*x))/c^5/d^3/(c*x+1)^2+4*(a+b*arc
tanh(c*x))/c^5/d^3/(c*x+1)-6*(a+b*arctanh(c*x))*ln(2/(c*x+1))/c^5/d^3-3/2*b*ln(-c^2*x^2+1)/c^5/d^3+3*b*polylog
(2,1-2/(c*x+1))/c^5/d^3

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Rubi [A]
time = 0.21, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 13, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.650, Rules used = {6087, 6021, 266, 6037, 327, 212, 6063, 641, 46, 213, 6055, 2449, 2352} \begin {gather*} \frac {4 \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3 (c x+1)}-\frac {a+b \tanh ^{-1}(c x)}{2 c^5 d^3 (c x+1)^2}-\frac {6 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^3 d^3}-\frac {3 a x}{c^4 d^3}+\frac {3 b \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{c^5 d^3}+\frac {15 b}{8 c^5 d^3 (c x+1)}-\frac {b}{8 c^5 d^3 (c x+1)^2}-\frac {19 b \tanh ^{-1}(c x)}{8 c^5 d^3}+\frac {b x}{2 c^4 d^3}-\frac {3 b x \tanh ^{-1}(c x)}{c^4 d^3}-\frac {3 b \log \left (1-c^2 x^2\right )}{2 c^5 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcTanh[c*x]))/(d + c*d*x)^3,x]

[Out]

(-3*a*x)/(c^4*d^3) + (b*x)/(2*c^4*d^3) - b/(8*c^5*d^3*(1 + c*x)^2) + (15*b)/(8*c^5*d^3*(1 + c*x)) - (19*b*ArcT
anh[c*x])/(8*c^5*d^3) - (3*b*x*ArcTanh[c*x])/(c^4*d^3) + (x^2*(a + b*ArcTanh[c*x]))/(2*c^3*d^3) - (a + b*ArcTa
nh[c*x])/(2*c^5*d^3*(1 + c*x)^2) + (4*(a + b*ArcTanh[c*x]))/(c^5*d^3*(1 + c*x)) - (6*(a + b*ArcTanh[c*x])*Log[
2/(1 + c*x)])/(c^5*d^3) - (3*b*Log[1 - c^2*x^2])/(2*c^5*d^3) + (3*b*PolyLog[2, 1 - 2/(1 + c*x)])/(c^5*d^3)

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 212

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

Rule 213

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

Rule 266

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 327

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 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 2352

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

Rule 2449

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 6021

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

Rule 6037

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

Rule 6055

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 6063

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

Rule 6087

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {x^4 \left (a+b \tanh ^{-1}(c x)\right )}{(d+c d x)^3} \, dx &=\int \left (-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^3}+\frac {a+b \tanh ^{-1}(c x)}{c^4 d^3 (1+c x)^3}-\frac {4 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (1+c x)^2}+\frac {6 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (1+c x)}\right ) \, dx\\ &=\frac {\int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{c^4 d^3}-\frac {3 \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c^4 d^3}-\frac {4 \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{c^4 d^3}+\frac {6 \int \frac {a+b \tanh ^{-1}(c x)}{1+c x} \, dx}{c^4 d^3}+\frac {\int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c^3 d^3}\\ &=-\frac {3 a x}{c^4 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^3 d^3}-\frac {a+b \tanh ^{-1}(c x)}{2 c^5 d^3 (1+c x)^2}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3 (1+c x)}-\frac {6 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {b \int \frac {1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{2 c^4 d^3}-\frac {(3 b) \int \tanh ^{-1}(c x) \, dx}{c^4 d^3}-\frac {(4 b) \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{c^4 d^3}+\frac {(6 b) \int \frac {\log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^4 d^3}-\frac {b \int \frac {x^2}{1-c^2 x^2} \, dx}{2 c^2 d^3}\\ &=-\frac {3 a x}{c^4 d^3}+\frac {b x}{2 c^4 d^3}-\frac {3 b x \tanh ^{-1}(c x)}{c^4 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^3 d^3}-\frac {a+b \tanh ^{-1}(c x)}{2 c^5 d^3 (1+c x)^2}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3 (1+c x)}-\frac {6 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {(6 b) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c x}\right )}{c^5 d^3}+\frac {b \int \frac {1}{(1-c x) (1+c x)^3} \, dx}{2 c^4 d^3}-\frac {b \int \frac {1}{1-c^2 x^2} \, dx}{2 c^4 d^3}-\frac {(4 b) \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{c^4 d^3}+\frac {(3 b) \int \frac {x}{1-c^2 x^2} \, dx}{c^3 d^3}\\ &=-\frac {3 a x}{c^4 d^3}+\frac {b x}{2 c^4 d^3}-\frac {b \tanh ^{-1}(c x)}{2 c^5 d^3}-\frac {3 b x \tanh ^{-1}(c x)}{c^4 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^3 d^3}-\frac {a+b \tanh ^{-1}(c x)}{2 c^5 d^3 (1+c x)^2}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3 (1+c x)}-\frac {6 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^5 d^3}-\frac {3 b \log \left (1-c^2 x^2\right )}{2 c^5 d^3}+\frac {3 b \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {b \int \left (\frac {1}{2 (1+c x)^3}+\frac {1}{4 (1+c x)^2}-\frac {1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{2 c^4 d^3}-\frac {(4 b) \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^4 d^3}\\ &=-\frac {3 a x}{c^4 d^3}+\frac {b x}{2 c^4 d^3}-\frac {b}{8 c^5 d^3 (1+c x)^2}+\frac {15 b}{8 c^5 d^3 (1+c x)}-\frac {b \tanh ^{-1}(c x)}{2 c^5 d^3}-\frac {3 b x \tanh ^{-1}(c x)}{c^4 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^3 d^3}-\frac {a+b \tanh ^{-1}(c x)}{2 c^5 d^3 (1+c x)^2}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3 (1+c x)}-\frac {6 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^5 d^3}-\frac {3 b \log \left (1-c^2 x^2\right )}{2 c^5 d^3}+\frac {3 b \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^5 d^3}-\frac {b \int \frac {1}{-1+c^2 x^2} \, dx}{8 c^4 d^3}+\frac {(2 b) \int \frac {1}{-1+c^2 x^2} \, dx}{c^4 d^3}\\ &=-\frac {3 a x}{c^4 d^3}+\frac {b x}{2 c^4 d^3}-\frac {b}{8 c^5 d^3 (1+c x)^2}+\frac {15 b}{8 c^5 d^3 (1+c x)}-\frac {19 b \tanh ^{-1}(c x)}{8 c^5 d^3}-\frac {3 b x \tanh ^{-1}(c x)}{c^4 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^3 d^3}-\frac {a+b \tanh ^{-1}(c x)}{2 c^5 d^3 (1+c x)^2}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3 (1+c x)}-\frac {6 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^5 d^3}-\frac {3 b \log \left (1-c^2 x^2\right )}{2 c^5 d^3}+\frac {3 b \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^5 d^3}\\ \end {align*}

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Mathematica [A]
time = 0.51, size = 189, normalized size = 0.83 \begin {gather*} \frac {-96 a c x+16 a c^2 x^2-\frac {16 a}{(1+c x)^2}+\frac {128 a}{1+c x}+192 a \log (1+c x)+b \left (16 c x+28 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )-48 \log \left (1-c^2 x^2\right )+96 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )-28 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \left (-4-24 c x+4 c^2 x^2+14 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )-48 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )-14 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )\right )\right )}{32 c^5 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*ArcTanh[c*x]))/(d + c*d*x)^3,x]

[Out]

(-96*a*c*x + 16*a*c^2*x^2 - (16*a)/(1 + c*x)^2 + (128*a)/(1 + c*x) + 192*a*Log[1 + c*x] + b*(16*c*x + 28*Cosh[
2*ArcTanh[c*x]] - Cosh[4*ArcTanh[c*x]] - 48*Log[1 - c^2*x^2] + 96*PolyLog[2, -E^(-2*ArcTanh[c*x])] - 28*Sinh[2
*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]] + 4*ArcTanh[c*x]*(-4 - 24*c*x + 4*c^2*x^2 + 14*Cosh[2*ArcTanh[c*x]] - Co
sh[4*ArcTanh[c*x]] - 48*Log[1 + E^(-2*ArcTanh[c*x])] - 14*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]])))/(32*c
^5*d^3)

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Maple [A]
time = 0.35, size = 272, normalized size = 1.20

method	result	size
derivativedivides	\(\frac {\frac {a \,c^{2} x^{2}}{2 d^{3}}-\frac {3 a c x}{d^{3}}-\frac {a}{2 d^{3} \left (c x +1\right )^{2}}+\frac {4 a}{d^{3} \left (c x +1\right )}+\frac {6 a \ln \left (c x +1\right )}{d^{3}}+\frac {b \arctanh \left (c x \right ) c^{2} x^{2}}{2 d^{3}}-\frac {3 b \arctanh \left (c x \right ) c x}{d^{3}}-\frac {b \arctanh \left (c x \right )}{2 d^{3} \left (c x +1\right )^{2}}+\frac {4 b \arctanh \left (c x \right )}{d^{3} \left (c x +1\right )}+\frac {6 b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{d^{3}}-\frac {3 b \ln \left (c x +1\right )^{2}}{2 d^{3}}+\frac {3 b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{d^{3}}-\frac {3 b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{d^{3}}-\frac {3 b \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{d^{3}}+\frac {b c x}{2 d^{3}}+\frac {b}{2 d^{3}}-\frac {b}{8 d^{3} \left (c x +1\right )^{2}}+\frac {15 b}{8 d^{3} \left (c x +1\right )}-\frac {43 b \ln \left (c x +1\right )}{16 d^{3}}-\frac {5 b \ln \left (c x -1\right )}{16 d^{3}}}{c^{5}}\)	\(272\)
default	\(\frac {\frac {a \,c^{2} x^{2}}{2 d^{3}}-\frac {3 a c x}{d^{3}}-\frac {a}{2 d^{3} \left (c x +1\right )^{2}}+\frac {4 a}{d^{3} \left (c x +1\right )}+\frac {6 a \ln \left (c x +1\right )}{d^{3}}+\frac {b \arctanh \left (c x \right ) c^{2} x^{2}}{2 d^{3}}-\frac {3 b \arctanh \left (c x \right ) c x}{d^{3}}-\frac {b \arctanh \left (c x \right )}{2 d^{3} \left (c x +1\right )^{2}}+\frac {4 b \arctanh \left (c x \right )}{d^{3} \left (c x +1\right )}+\frac {6 b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{d^{3}}-\frac {3 b \ln \left (c x +1\right )^{2}}{2 d^{3}}+\frac {3 b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{d^{3}}-\frac {3 b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{d^{3}}-\frac {3 b \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{d^{3}}+\frac {b c x}{2 d^{3}}+\frac {b}{2 d^{3}}-\frac {b}{8 d^{3} \left (c x +1\right )^{2}}+\frac {15 b}{8 d^{3} \left (c x +1\right )}-\frac {43 b \ln \left (c x +1\right )}{16 d^{3}}-\frac {5 b \ln \left (c x -1\right )}{16 d^{3}}}{c^{5}}\)	\(272\)
risch	\(\frac {3 b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{d^{3} c^{5}}-\frac {3 b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-c x +1\right )}{d^{3} c^{5}}+\frac {3 b \ln \left (-c x +1\right ) x}{2 d^{3} c^{4}}+\frac {b \ln \left (-c x +1\right )}{d^{3} c^{5} \left (-c x -1\right )}+\frac {3 b \ln \left (-c x +1\right )}{16 d^{3} c^{5} \left (-c x -1\right )^{2}}-\frac {\ln \left (-c x +1\right ) x^{2} b}{4 d^{3} c^{3}}-\frac {3 a x}{c^{4} d^{3}}+\frac {b x}{2 c^{4} d^{3}}-\frac {b}{8 c^{5} d^{3} \left (c x +1\right )^{2}}+\frac {2 b}{c^{5} d^{3} \left (c x +1\right )}-\frac {b \ln \left (-c x +1\right ) x}{d^{3} c^{4} \left (-c x -1\right )}-\frac {b \ln \left (-c x +1\right ) x^{2}}{16 d^{3} c^{3} \left (-c x -1\right )^{2}}-\frac {b \ln \left (-c x +1\right ) x}{8 d^{3} c^{4} \left (-c x -1\right )^{2}}-\frac {7 b \ln \left (c x +1\right )}{4 c^{5} d^{3}}+\frac {9 b}{8 c^{5} d^{3}}+\frac {3 b \dilog \left (-\frac {c x}{2}+\frac {1}{2}\right )}{d^{3} c^{5}}+\frac {5 a}{2 d^{3} c^{5}}+\left (\frac {b \left (\frac {1}{2} c \,x^{2}-3 x \right )}{2 c^{4} d^{3}}+\frac {4 b \,d^{3} x +\frac {7 d^{3} b}{2 c}}{2 c^{4} d^{6} \left (c x +1\right )^{2}}\right ) \ln \left (c x +1\right )+\frac {x^{2} a}{2 d^{3} c^{3}}-\frac {4 a}{d^{3} c^{5} \left (-c x -1\right )}+\frac {6 a \ln \left (-c x -1\right )}{d^{3} c^{5}}-\frac {15 b \ln \left (-c x -1\right )}{16 d^{3} c^{5}}+\frac {b}{8 d^{3} c^{5} \left (-c x -1\right )}-\frac {a}{2 d^{3} c^{5} \left (-c x -1\right )^{2}}+\frac {3 b \ln \left (c x +1\right )^{2}}{2 c^{5} d^{3}}-\frac {5 b \ln \left (-c x +1\right )}{4 d^{3} c^{5}}\)	\(496\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctanh(c*x))/(c*d*x+d)^3,x,method=_RETURNVERBOSE)

[Out]

1/c^5*(1/2*a/d^3*c^2*x^2-3*a/d^3*c*x-1/2*a/d^3/(c*x+1)^2+4*a/d^3/(c*x+1)+6*a/d^3*ln(c*x+1)+1/2*b/d^3*arctanh(c
*x)*c^2*x^2-3*b/d^3*arctanh(c*x)*c*x-1/2*b/d^3*arctanh(c*x)/(c*x+1)^2+4*b/d^3*arctanh(c*x)/(c*x+1)+6*b/d^3*arc
tanh(c*x)*ln(c*x+1)-3/2*b/d^3*ln(c*x+1)^2+3*b/d^3*ln(-1/2*c*x+1/2)*ln(c*x+1)-3*b/d^3*ln(-1/2*c*x+1/2)*ln(1/2*c
*x+1/2)-3*b/d^3*dilog(1/2*c*x+1/2)+1/2*b/d^3*c*x+1/2*b/d^3-1/8*b/d^3/(c*x+1)^2+15/8*b/d^3/(c*x+1)-43/16*b/d^3*
ln(c*x+1)-5/16*b/d^3*ln(c*x-1))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="maxima")

[Out]

1/32*(c^5*(2*(9*c*x + 8)/(c^12*d^3*x^2 + 2*c^11*d^3*x + c^10*d^3) + 4*(c*x^2 - 4*x)/(c^9*d^3) + 31*log(c*x + 1
)/(c^10*d^3) + log(c*x - 1)/(c^10*d^3)) + 32*c^5*integrate(1/2*x^5*log(c*x + 1)/(c^8*d^3*x^4 + 2*c^7*d^3*x^3 -
2*c^5*d^3*x - c^4*d^3), x) + 3*c^4*(2*(7*c*x + 6)/(c^11*d^3*x^2 + 2*c^10*d^3*x + c^9*d^3) - 8*x/(c^8*d^3) + 1
7*log(c*x + 1)/(c^9*d^3) - log(c*x - 1)/(c^9*d^3)) - 32*c^4*integrate(1/2*x^4*log(c*x + 1)/(c^8*d^3*x^4 + 2*c^
7*d^3*x^3 - 2*c^5*d^3*x - c^4*d^3), x) - 15*c^3*(2*(5*c*x + 4)/(c^10*d^3*x^2 + 2*c^9*d^3*x + c^8*d^3) + 7*log(
c*x + 1)/(c^8*d^3) + log(c*x - 1)/(c^8*d^3)) + 192*c^3*integrate(1/2*x^3*log(c*x + 1)/(c^8*d^3*x^4 + 2*c^7*d^3
*x^3 - 2*c^5*d^3*x - c^4*d^3), x) + 9*c^2*(2*(3*c*x + 2)/(c^9*d^3*x^2 + 2*c^8*d^3*x + c^7*d^3) + log(c*x + 1)/
(c^7*d^3) - log(c*x - 1)/(c^7*d^3)) + 576*c^2*integrate(1/2*x^2*log(c*x + 1)/(c^8*d^3*x^4 + 2*c^7*d^3*x^3 - 2*
c^5*d^3*x - c^4*d^3), x) + 9*c*(2*x/(c^7*d^3*x^2 + 2*c^6*d^3*x + c^5*d^3) - log(c*x + 1)/(c^6*d^3) + log(c*x -
1)/(c^6*d^3)) + 576*c*integrate(1/2*x*log(c*x + 1)/(c^8*d^3*x^4 + 2*c^7*d^3*x^3 - 2*c^5*d^3*x - c^4*d^3), x)
- 8*(c^4*x^4 - 4*c^3*x^3 - 11*c^2*x^2 + 2*c*x + 12*(c^2*x^2 + 2*c*x + 1)*log(c*x + 1) + 7)*log(-c*x + 1)/(c^7*
d^3*x^2 + 2*c^6*d^3*x + c^5*d^3) + 14*(c*x + 2)/(c^7*d^3*x^2 + 2*c^6*d^3*x + c^5*d^3) - 7*log(c*x + 1)/(c^5*d^
3) + 7*log(c*x - 1)/(c^5*d^3) + 192*integrate(1/2*log(c*x + 1)/(c^8*d^3*x^4 + 2*c^7*d^3*x^3 - 2*c^5*d^3*x - c^
4*d^3), x))*b + 1/2*a*((8*c*x + 7)/(c^7*d^3*x^2 + 2*c^6*d^3*x + c^5*d^3) + (c*x^2 - 6*x)/(c^4*d^3) + 12*log(c*
x + 1)/(c^5*d^3))

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Fricas [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^4*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="fricas")

[Out]

integral((b*x^4*arctanh(c*x) + a*x^4)/(c^3*d^3*x^3 + 3*c^2*d^3*x^2 + 3*c*d^3*x + d^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a x^{4}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {b x^{4} \operatorname {atanh}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx}{d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*atanh(c*x))/(c*d*x+d)**3,x)

[Out]

(Integral(a*x**4/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integral(b*x**4*atanh(c*x)/(c**3*x**3 + 3*c**2*x*
*2 + 3*c*x + 1), x))/d**3

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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^4*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)*x^4/(c*d*x + d)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4\,\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}{{\left (d+c\,d\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*atanh(c*x)))/(d + c*d*x)^3,x)

[Out]

int((x^4*(a + b*atanh(c*x)))/(d + c*d*x)^3, x)

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