3.1.37 \(\int \frac {(d+c d x)^4 (a+b \tanh ^{-1}(c x))}{x^3} \, dx\) [37]

Optimal. Leaf size=156 \[ -\frac {b c d^4}{2 x}+4 a c^3 d^4 x+\frac {1}{2} b c^3 d^4 x+4 b c^3 d^4 x \tanh ^{-1}(c x)-\frac {d^4 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {1}{2} c^4 d^4 x^2 \left (a+b \tanh ^{-1}(c x)\right )+6 a c^2 d^4 \log (x)+4 b c^2 d^4 \log (x)-3 b c^2 d^4 \text {PolyLog}(2,-c x)+3 b c^2 d^4 \text {PolyLog}(2,c x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.14, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 12, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6087, 6021, 266, 6037, 331, 212, 272, 36, 29, 31, 6031, 327} \begin {gather*} \frac {1}{2} c^4 d^4 x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {d^4 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+4 a c^3 d^4 x+6 a c^2 d^4 \log (x)+\frac {1}{2} b c^3 d^4 x+4 b c^3 d^4 x \tanh ^{-1}(c x)-3 b c^2 d^4 \text {Li}_2(-c x)+3 b c^2 d^4 \text {Li}_2(c x)+4 b c^2 d^4 \log (x)-\frac {b c d^4}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 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 272

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 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 331

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

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 6031

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b/2)*PolyLog[2, (-c)*x]
, x] + Simp[(b/2)*PolyLog[2, c*x], x]) /; FreeQ[{a, b, c}, x]

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

________________________________________________________________________________________

Mathematica [A]
time = 0.08, size = 143, normalized size = 0.92 \begin {gather*} \frac {d^4 \left (-a-8 a c x-b c x+8 a c^3 x^3+b c^3 x^3+a c^4 x^4-b \tanh ^{-1}(c x)-8 b c x \tanh ^{-1}(c x)+8 b c^3 x^3 \tanh ^{-1}(c x)+b c^4 x^4 \tanh ^{-1}(c x)+12 a c^2 x^2 \log (x)+8 b c^2 x^2 \log (c x)-6 b c^2 x^2 \text {PolyLog}(2,-c x)+6 b c^2 x^2 \text {PolyLog}(2,c x)\right )}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^4*(-a - 8*a*c*x - b*c*x + 8*a*c^3*x^3 + b*c^3*x^3 + a*c^4*x^4 - b*ArcTanh[c*x] - 8*b*c*x*ArcTanh[c*x] + 8*b
*c^3*x^3*ArcTanh[c*x] + b*c^4*x^4*ArcTanh[c*x] + 12*a*c^2*x^2*Log[x] + 8*b*c^2*x^2*Log[c*x] - 6*b*c^2*x^2*Poly
Log[2, -(c*x)] + 6*b*c^2*x^2*PolyLog[2, c*x]))/(2*x^2)

________________________________________________________________________________________

Maple [A]
time = 0.22, size = 202, normalized size = 1.29

method	result	size
derivativedivides	\(c^{2} \left (\frac {d^{4} a \,c^{2} x^{2}}{2}+4 a c \,d^{4} x +6 d^{4} a \ln \left (c x \right )-\frac {d^{4} a}{2 c^{2} x^{2}}-\frac {4 d^{4} a}{c x}+\frac {d^{4} b \arctanh \left (c x \right ) c^{2} x^{2}}{2}+4 b c \,d^{4} x \arctanh \left (c x \right )+6 d^{4} b \arctanh \left (c x \right ) \ln \left (c x \right )-\frac {d^{4} b \arctanh \left (c x \right )}{2 c^{2} x^{2}}-\frac {4 d^{4} b \arctanh \left (c x \right )}{c x}-3 d^{4} b \dilog \left (c x \right )-3 d^{4} b \dilog \left (c x +1\right )-3 d^{4} b \ln \left (c x \right ) \ln \left (c x +1\right )+\frac {b c \,d^{4} x}{2}+4 d^{4} b \ln \left (c x \right )-\frac {d^{4} b}{2 c x}\right )\)	\(202\)
default	\(c^{2} \left (\frac {d^{4} a \,c^{2} x^{2}}{2}+4 a c \,d^{4} x +6 d^{4} a \ln \left (c x \right )-\frac {d^{4} a}{2 c^{2} x^{2}}-\frac {4 d^{4} a}{c x}+\frac {d^{4} b \arctanh \left (c x \right ) c^{2} x^{2}}{2}+4 b c \,d^{4} x \arctanh \left (c x \right )+6 d^{4} b \arctanh \left (c x \right ) \ln \left (c x \right )-\frac {d^{4} b \arctanh \left (c x \right )}{2 c^{2} x^{2}}-\frac {4 d^{4} b \arctanh \left (c x \right )}{c x}-3 d^{4} b \dilog \left (c x \right )-3 d^{4} b \dilog \left (c x +1\right )-3 d^{4} b \ln \left (c x \right ) \ln \left (c x +1\right )+\frac {b c \,d^{4} x}{2}+4 d^{4} b \ln \left (c x \right )-\frac {d^{4} b}{2 c x}\right )\)	\(202\)
risch	\(-\frac {b c \,d^{4}}{2 x}+4 a \,c^{3} d^{4} x +\frac {b \,c^{3} d^{4} x}{2}-\frac {d^{4} a}{2 x^{2}}+\frac {b \,c^{4} d^{4} \ln \left (c x +1\right ) x^{2}}{4}+2 b \,c^{3} d^{4} \ln \left (c x +1\right ) x -\frac {2 b c \,d^{4} \ln \left (c x +1\right )}{x}-2 c^{3} d^{4} b \ln \left (-c x +1\right ) x +\frac {2 c \,d^{4} b \ln \left (-c x +1\right )}{x}-4 b \,c^{2} d^{4}-\frac {9 a \,c^{2} d^{4}}{2}-\frac {c^{4} d^{4} \ln \left (-c x +1\right ) x^{2} b}{4}+\frac {7 b \,c^{2} d^{4} \ln \left (c x \right )}{4}-\frac {b \,d^{4} \ln \left (c x +1\right )}{4 x^{2}}-3 b \,c^{2} d^{4} \dilog \left (c x +1\right )+\frac {c^{4} d^{4} x^{2} a}{2}-\frac {4 c \,d^{4} a}{x}+6 c^{2} d^{4} a \ln \left (-c x \right )+\frac {9 c^{2} d^{4} b \ln \left (-c x \right )}{4}+\frac {d^{4} b \ln \left (-c x +1\right )}{4 x^{2}}+3 c^{2} d^{4} \dilog \left (-c x +1\right ) b\)	\(287\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (146) = 292\).
time = 0.35, size = 293, normalized size = 1.88 \begin {gather*} \frac {1}{4} \, b c^{4} d^{4} x^{2} \log \left (c x + 1\right ) - \frac {1}{4} \, b c^{4} d^{4} x^{2} \log \left (-c x + 1\right ) + \frac {1}{2} \, a c^{4} d^{4} x^{2} + 4 \, a c^{3} d^{4} x + \frac {1}{2} \, b c^{3} d^{4} x + 2 \, {\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b c^{2} d^{4} - 3 \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} b c^{2} d^{4} + 3 \, {\left (\log \left (c x + 1\right ) \log \left (-c x\right ) + {\rm Li}_2\left (c x + 1\right )\right )} b c^{2} d^{4} - \frac {1}{4} \, b c^{2} d^{4} \log \left (c x + 1\right ) + \frac {1}{4} \, b c^{2} d^{4} \log \left (c x - 1\right ) + 6 \, a c^{2} d^{4} \log \left (x\right ) - 2 \, {\left (c {\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x}\right )} b c d^{4} + \frac {1}{4} \, {\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{2}}\right )} b d^{4} - \frac {4 \, a c d^{4}}{x} - \frac {a d^{4}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*c^4*d^4*x^2*log(c*x + 1) - 1/4*b*c^4*d^4*x^2*log(-c*x + 1) + 1/2*a*c^4*d^4*x^2 + 4*a*c^3*d^4*x + 1/2*b*c
^3*d^4*x + 2*(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*b*c^2*d^4 - 3*(log(c*x)*log(-c*x + 1) + dilog(-c*x + 1))
*b*c^2*d^4 + 3*(log(c*x + 1)*log(-c*x) + dilog(c*x + 1))*b*c^2*d^4 - 1/4*b*c^2*d^4*log(c*x + 1) + 1/4*b*c^2*d^
4*log(c*x - 1) + 6*a*c^2*d^4*log(x) - 2*(c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*b*c*d^4 + 1/4*((c
*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*b*d^4 - 4*a*c*d^4/x - 1/2*a*d^4/x^2

________________________________________________________________________________________

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

[Out]

integral((a*c^4*d^4*x^4 + 4*a*c^3*d^4*x^3 + 6*a*c^2*d^4*x^2 + 4*a*c*d^4*x + a*d^4 + (b*c^4*d^4*x^4 + 4*b*c^3*d
^4*x^3 + 6*b*c^2*d^4*x^2 + 4*b*c*d^4*x + b*d^4)*arctanh(c*x))/x^3, x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} d^{4} \left (\int 4 a c^{3}\, dx + \int \frac {a}{x^{3}}\, dx + \int \frac {4 a c}{x^{2}}\, dx + \int \frac {6 a c^{2}}{x}\, dx + \int a c^{4} x\, dx + \int 4 b c^{3} \operatorname {atanh}{\left (c x \right )}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{x^{3}}\, dx + \int \frac {4 b c \operatorname {atanh}{\left (c x \right )}}{x^{2}}\, dx + \int \frac {6 b c^{2} \operatorname {atanh}{\left (c x \right )}}{x}\, dx + \int b c^{4} x \operatorname {atanh}{\left (c x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**4*(Integral(4*a*c**3, x) + Integral(a/x**3, x) + Integral(4*a*c/x**2, x) + Integral(6*a*c**2/x, x) + Integr
al(a*c**4*x, x) + Integral(4*b*c**3*atanh(c*x), x) + Integral(b*atanh(c*x)/x**3, x) + Integral(4*b*c*atanh(c*x
)/x**2, x) + Integral(6*b*c**2*atanh(c*x)/x, x) + Integral(b*c**4*x*atanh(c*x), x))

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^4}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________