3.1.36 \(\int \frac {(d+c d x)^4 (a+b \tanh ^{-1}(c x))}{x^2} \, dx\) [36]

Optimal. Leaf size=178 \[ 6 a c^2 d^4 x+2 b c^2 d^4 x+\frac {1}{6} b c^3 d^4 x^2-2 b c d^4 \tanh ^{-1}(c x)+6 b c^2 d^4 x \tanh ^{-1}(c x)-\frac {d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+2 c^3 d^4 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{3} c^4 d^4 x^3 \left (a+b \tanh ^{-1}(c x)\right )+4 a c d^4 \log (x)+b c d^4 \log (x)+\frac {8}{3} b c d^4 \log \left (1-c^2 x^2\right )-2 b c d^4 \text {PolyLog}(2,-c x)+2 b c d^4 \text {PolyLog}(2,c x) \]

[Out]

6*a*c^2*d^4*x+2*b*c^2*d^4*x+1/6*b*c^3*d^4*x^2-2*b*c*d^4*arctanh(c*x)+6*b*c^2*d^4*x*arctanh(c*x)-d^4*(a+b*arcta
nh(c*x))/x+2*c^3*d^4*x^2*(a+b*arctanh(c*x))+1/3*c^4*d^4*x^3*(a+b*arctanh(c*x))+4*a*c*d^4*ln(x)+b*c*d^4*ln(x)+8
/3*b*c*d^4*ln(-c^2*x^2+1)-2*b*c*d^4*polylog(2,-c*x)+2*b*c*d^4*polylog(2,c*x)

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Rubi [A]
time = 0.20, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 12, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6087, 6021, 266, 6037, 272, 36, 29, 31, 6031, 327, 212, 45} \begin {gather*} \frac {1}{3} c^4 d^4 x^3 \left (a+b \tanh ^{-1}(c x)\right )+2 c^3 d^4 x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+6 a c^2 d^4 x+4 a c d^4 \log (x)+\frac {1}{6} b c^3 d^4 x^2+\frac {8}{3} b c d^4 \log \left (1-c^2 x^2\right )+2 b c^2 d^4 x+6 b c^2 d^4 x \tanh ^{-1}(c x)-2 b c d^4 \text {Li}_2(-c x)+2 b c d^4 \text {Li}_2(c x)+b c d^4 \log (x)-2 b c d^4 \tanh ^{-1}(c x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)^4*(a + b*ArcTanh[c*x]))/x^2,x]

[Out]

6*a*c^2*d^4*x + 2*b*c^2*d^4*x + (b*c^3*d^4*x^2)/6 - 2*b*c*d^4*ArcTanh[c*x] + 6*b*c^2*d^4*x*ArcTanh[c*x] - (d^4
*(a + b*ArcTanh[c*x]))/x + 2*c^3*d^4*x^2*(a + b*ArcTanh[c*x]) + (c^4*d^4*x^3*(a + b*ArcTanh[c*x]))/3 + 4*a*c*d
^4*Log[x] + b*c*d^4*Log[x] + (8*b*c*d^4*Log[1 - c^2*x^2])/3 - 2*b*c*d^4*PolyLog[2, -(c*x)] + 2*b*c*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 45

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

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Mathematica [A]
time = 0.08, size = 194, normalized size = 1.09 \begin {gather*} \frac {d^4 \left (-6 a+36 a c^2 x^2+12 b c^2 x^2+12 a c^3 x^3+b c^3 x^3+2 a c^4 x^4-6 b \tanh ^{-1}(c x)+36 b c^2 x^2 \tanh ^{-1}(c x)+12 b c^3 x^3 \tanh ^{-1}(c x)+2 b c^4 x^4 \tanh ^{-1}(c x)+24 a c x \log (x)+6 b c x \log (c x)+6 b c x \log (1-c x)-6 b c x \log (1+c x)+15 b c x \log \left (1-c^2 x^2\right )+b c x \log \left (-1+c^2 x^2\right )-12 b c x \text {PolyLog}(2,-c x)+12 b c x \text {PolyLog}(2,c x)\right )}{6 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + c*d*x)^4*(a + b*ArcTanh[c*x]))/x^2,x]

[Out]

(d^4*(-6*a + 36*a*c^2*x^2 + 12*b*c^2*x^2 + 12*a*c^3*x^3 + b*c^3*x^3 + 2*a*c^4*x^4 - 6*b*ArcTanh[c*x] + 36*b*c^
2*x^2*ArcTanh[c*x] + 12*b*c^3*x^3*ArcTanh[c*x] + 2*b*c^4*x^4*ArcTanh[c*x] + 24*a*c*x*Log[x] + 6*b*c*x*Log[c*x]
+ 6*b*c*x*Log[1 - c*x] - 6*b*c*x*Log[1 + c*x] + 15*b*c*x*Log[1 - c^2*x^2] + b*c*x*Log[-1 + c^2*x^2] - 12*b*c*
x*PolyLog[2, -(c*x)] + 12*b*c*x*PolyLog[2, c*x]))/(6*x)

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Maple [A]
time = 0.22, size = 223, normalized size = 1.25

method	result	size
derivativedivides	\(c \left (\frac {d^{4} a \,c^{3} x^{3}}{3}+2 d^{4} a \,c^{2} x^{2}+6 a c \,d^{4} x -\frac {d^{4} a}{c x}+4 d^{4} a \ln \left (c x \right )+\frac {d^{4} b \arctanh \left (c x \right ) c^{3} x^{3}}{3}+2 d^{4} b \arctanh \left (c x \right ) c^{2} x^{2}+6 b c \,d^{4} x \arctanh \left (c x \right )-\frac {d^{4} b \arctanh \left (c x \right )}{c x}+4 d^{4} b \arctanh \left (c x \right ) \ln \left (c x \right )-2 d^{4} b \dilog \left (c x \right )-2 d^{4} b \dilog \left (c x +1\right )-2 d^{4} b \ln \left (c x \right ) \ln \left (c x +1\right )+\frac {b \,c^{2} d^{4} x^{2}}{6}+2 b c \,d^{4} x +d^{4} b \ln \left (c x \right )+\frac {5 d^{4} b \ln \left (c x +1\right )}{3}+\frac {11 d^{4} b \ln \left (c x -1\right )}{3}\right )\)	\(223\)
default	\(c \left (\frac {d^{4} a \,c^{3} x^{3}}{3}+2 d^{4} a \,c^{2} x^{2}+6 a c \,d^{4} x -\frac {d^{4} a}{c x}+4 d^{4} a \ln \left (c x \right )+\frac {d^{4} b \arctanh \left (c x \right ) c^{3} x^{3}}{3}+2 d^{4} b \arctanh \left (c x \right ) c^{2} x^{2}+6 b c \,d^{4} x \arctanh \left (c x \right )-\frac {d^{4} b \arctanh \left (c x \right )}{c x}+4 d^{4} b \arctanh \left (c x \right ) \ln \left (c x \right )-2 d^{4} b \dilog \left (c x \right )-2 d^{4} b \dilog \left (c x +1\right )-2 d^{4} b \ln \left (c x \right ) \ln \left (c x +1\right )+\frac {b \,c^{2} d^{4} x^{2}}{6}+2 b c \,d^{4} x +d^{4} b \ln \left (c x \right )+\frac {5 d^{4} b \ln \left (c x +1\right )}{3}+\frac {11 d^{4} b \ln \left (c x -1\right )}{3}\right )\)	\(223\)
risch	\(6 a \,c^{2} d^{4} x +2 b \,c^{2} d^{4} x +\frac {b \,c^{3} d^{4} x^{2}}{6}+\frac {c \,d^{4} b \ln \left (-c x \right )}{2}+\frac {11 c \,d^{4} b \ln \left (-c x +1\right )}{3}+2 c \,d^{4} \dilog \left (-c x +1\right ) b +4 c \,d^{4} a \ln \left (-c x \right )+\frac {c^{4} d^{4} x^{3} a}{3}+2 c^{3} d^{4} x^{2} a +\frac {d^{4} b \ln \left (-c x +1\right )}{2 x}-\frac {119 b c \,d^{4}}{18}-\frac {25 c \,d^{4} a}{3}+b \,c^{3} d^{4} \ln \left (c x +1\right ) x^{2}+3 b \,c^{2} d^{4} \ln \left (c x +1\right ) x -\frac {d^{4} a}{x}+\frac {b \,c^{4} d^{4} \ln \left (c x +1\right ) x^{3}}{6}-\frac {c^{4} d^{4} \ln \left (-c x +1\right ) x^{3} b}{6}-c^{3} d^{4} \ln \left (-c x +1\right ) x^{2} b -3 c^{2} d^{4} b \ln \left (-c x +1\right ) x +\frac {5 b c \,d^{4} \ln \left (c x +1\right )}{3}+\frac {b c \,d^{4} \ln \left (c x \right )}{2}-\frac {b \,d^{4} \ln \left (c x +1\right )}{2 x}-2 b c \,d^{4} \dilog \left (c x +1\right )\)	\(307\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^4*(a+b*arctanh(c*x))/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.35, size = 281, normalized size = 1.58 \begin {gather*} \frac {1}{3} \, a c^{4} d^{4} x^{3} + 2 \, a c^{3} d^{4} x^{2} + \frac {1}{6} \, b c^{3} d^{4} x^{2} + 6 \, a c^{2} d^{4} x + 2 \, b c^{2} d^{4} x + 3 \, {\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b c d^{4} - 2 \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} b c d^{4} + 2 \, {\left (\log \left (c x + 1\right ) \log \left (-c x\right ) + {\rm Li}_2\left (c x + 1\right )\right )} b c d^{4} - \frac {5}{6} \, b c d^{4} \log \left (c x + 1\right ) + \frac {7}{6} \, b c d^{4} \log \left (c x - 1\right ) + 4 \, a c d^{4} \log \left (x\right ) - \frac {1}{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 d^{4} - \frac {a d^{4}}{x} + \frac {1}{6} \, {\left (b c^{4} d^{4} x^{3} + 6 \, b c^{3} d^{4} x^{2}\right )} \log \left (c x + 1\right ) - \frac {1}{6} \, {\left (b c^{4} d^{4} x^{3} + 6 \, b c^{3} d^{4} x^{2}\right )} \log \left (-c x + 1\right ) \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^2,x, algorithm="maxima")

[Out]

1/3*a*c^4*d^4*x^3 + 2*a*c^3*d^4*x^2 + 1/6*b*c^3*d^4*x^2 + 6*a*c^2*d^4*x + 2*b*c^2*d^4*x + 3*(2*c*x*arctanh(c*x
) + log(-c^2*x^2 + 1))*b*c*d^4 - 2*(log(c*x)*log(-c*x + 1) + dilog(-c*x + 1))*b*c*d^4 + 2*(log(c*x + 1)*log(-c
*x) + dilog(c*x + 1))*b*c*d^4 - 5/6*b*c*d^4*log(c*x + 1) + 7/6*b*c*d^4*log(c*x - 1) + 4*a*c*d^4*log(x) - 1/2*(
c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*b*d^4 - a*d^4/x + 1/6*(b*c^4*d^4*x^3 + 6*b*c^3*d^4*x^2)*lo
g(c*x + 1) - 1/6*(b*c^4*d^4*x^3 + 6*b*c^3*d^4*x^2)*log(-c*x + 1)

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

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

[Out]

d**4*(Integral(6*a*c**2, x) + Integral(a/x**2, x) + Integral(4*a*c/x, x) + Integral(4*a*c**3*x, x) + Integral(
a*c**4*x**2, x) + Integral(6*b*c**2*atanh(c*x), x) + Integral(b*atanh(c*x)/x**2, x) + Integral(4*b*c*atanh(c*x
)/x, x) + Integral(4*b*c**3*x*atanh(c*x), x) + Integral(b*c**4*x**2*atanh(c*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((c*d*x+d)^4*(a+b*arctanh(c*x))/x^2,x, algorithm="giac")

[Out]

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

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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^2} \,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^2,x)

[Out]

int(((a + b*atanh(c*x))*(d + c*d*x)^4)/x^2, x)

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