3.1.39 \(\int \frac {(d+c d x)^4 (a+b \tanh ^{-1}(c x))}{x^5} \, dx\) [39]

Optimal. Leaf size=209 \[ -\frac {b c d^4}{12 x^3}-\frac {2 b c^2 d^4}{3 x^2}-\frac {13 b c^3 d^4}{4 x}+\frac {13}{4} b c^4 d^4 \tanh ^{-1}(c x)-\frac {d^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac {4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac {4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^4 d^4 \log (x)+\frac {16}{3} b c^4 d^4 \log (x)-\frac {8}{3} b c^4 d^4 \log \left (1-c^2 x^2\right )-\frac {1}{2} b c^4 d^4 \text {PolyLog}(2,-c x)+\frac {1}{2} b c^4 d^4 \text {PolyLog}(2,c x) \]

[Out]

-1/12*b*c*d^4/x^3-2/3*b*c^2*d^4/x^2-13/4*b*c^3*d^4/x+13/4*b*c^4*d^4*arctanh(c*x)-1/4*d^4*(a+b*arctanh(c*x))/x^
4-4/3*c*d^4*(a+b*arctanh(c*x))/x^3-3*c^2*d^4*(a+b*arctanh(c*x))/x^2-4*c^3*d^4*(a+b*arctanh(c*x))/x+a*c^4*d^4*l
n(x)+16/3*b*c^4*d^4*ln(x)-8/3*b*c^4*d^4*ln(-c^2*x^2+1)-1/2*b*c^4*d^4*polylog(2,-c*x)+1/2*b*c^4*d^4*polylog(2,c
*x)

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Rubi [A]
time = 0.17, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6087, 6037, 331, 212, 272, 46, 36, 29, 31, 6031} \begin {gather*} -\frac {4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}-\frac {3 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac {d^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac {4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}+a c^4 d^4 \log (x)-\frac {1}{2} b c^4 d^4 \text {Li}_2(-c x)+\frac {1}{2} b c^4 d^4 \text {Li}_2(c x)+\frac {16}{3} b c^4 d^4 \log (x)+\frac {13}{4} b c^4 d^4 \tanh ^{-1}(c x)-\frac {13 b c^3 d^4}{4 x}-\frac {2 b c^2 d^4}{3 x^2}-\frac {8}{3} b c^4 d^4 \log \left (1-c^2 x^2\right )-\frac {b c d^4}{12 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)^4*(a + b*ArcTanh[c*x]))/x^5,x]

[Out]

-1/12*(b*c*d^4)/x^3 - (2*b*c^2*d^4)/(3*x^2) - (13*b*c^3*d^4)/(4*x) + (13*b*c^4*d^4*ArcTanh[c*x])/4 - (d^4*(a +
b*ArcTanh[c*x]))/(4*x^4) - (4*c*d^4*(a + b*ArcTanh[c*x]))/(3*x^3) - (3*c^2*d^4*(a + b*ArcTanh[c*x]))/x^2 - (4
*c^3*d^4*(a + b*ArcTanh[c*x]))/x + a*c^4*d^4*Log[x] + (16*b*c^4*d^4*Log[x])/3 - (8*b*c^4*d^4*Log[1 - c^2*x^2])
/3 - (b*c^4*d^4*PolyLog[2, -(c*x)])/2 + (b*c^4*d^4*PolyLog[2, c*x])/2

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 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 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 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 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^5} \, dx &=\int \left (\frac {d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^5}+\frac {4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^4}+\frac {6 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^3}+\frac {4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}+\frac {c^4 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d^4 \int \frac {a+b \tanh ^{-1}(c x)}{x^5} \, dx+\left (4 c d^4\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^4} \, dx+\left (6 c^2 d^4\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^3} \, dx+\left (4 c^3 d^4\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^2} \, dx+\left (c^4 d^4\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx\\ &=-\frac {d^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac {4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac {4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^4 d^4 \log (x)-\frac {1}{2} b c^4 d^4 \text {Li}_2(-c x)+\frac {1}{2} b c^4 d^4 \text {Li}_2(c x)+\frac {1}{4} \left (b c d^4\right ) \int \frac {1}{x^4 \left (1-c^2 x^2\right )} \, dx+\frac {1}{3} \left (4 b c^2 d^4\right ) \int \frac {1}{x^3 \left (1-c^2 x^2\right )} \, dx+\left (3 b c^3 d^4\right ) \int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx+\left (4 b c^4 d^4\right ) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {b c d^4}{12 x^3}-\frac {3 b c^3 d^4}{x}-\frac {d^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac {4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac {4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^4 d^4 \log (x)-\frac {1}{2} b c^4 d^4 \text {Li}_2(-c x)+\frac {1}{2} b c^4 d^4 \text {Li}_2(c x)+\frac {1}{3} \left (2 b c^2 d^4\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (1-c^2 x\right )} \, dx,x,x^2\right )+\frac {1}{4} \left (b c^3 d^4\right ) \int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx+\left (2 b c^4 d^4\right ) \text {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )+\left (3 b c^5 d^4\right ) \int \frac {1}{1-c^2 x^2} \, dx\\ &=-\frac {b c d^4}{12 x^3}-\frac {13 b c^3 d^4}{4 x}+3 b c^4 d^4 \tanh ^{-1}(c x)-\frac {d^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac {4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac {4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^4 d^4 \log (x)-\frac {1}{2} b c^4 d^4 \text {Li}_2(-c x)+\frac {1}{2} b c^4 d^4 \text {Li}_2(c x)+\frac {1}{3} \left (2 b c^2 d^4\right ) \text {Subst}\left (\int \left (\frac {1}{x^2}+\frac {c^2}{x}-\frac {c^4}{-1+c^2 x}\right ) \, dx,x,x^2\right )+\left (2 b c^4 d^4\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{4} \left (b c^5 d^4\right ) \int \frac {1}{1-c^2 x^2} \, dx+\left (2 b c^6 d^4\right ) \text {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b c d^4}{12 x^3}-\frac {2 b c^2 d^4}{3 x^2}-\frac {13 b c^3 d^4}{4 x}+\frac {13}{4} b c^4 d^4 \tanh ^{-1}(c x)-\frac {d^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac {4 c d^4 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c^2 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac {4 c^3 d^4 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^4 d^4 \log (x)+\frac {16}{3} b c^4 d^4 \log (x)-\frac {8}{3} b c^4 d^4 \log \left (1-c^2 x^2\right )-\frac {1}{2} b c^4 d^4 \text {Li}_2(-c x)+\frac {1}{2} b c^4 d^4 \text {Li}_2(c x)\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 206, normalized size = 0.99 \begin {gather*} \frac {d^4 \left (-6 a-32 a c x-2 b c x-72 a c^2 x^2-16 b c^2 x^2-96 a c^3 x^3-78 b c^3 x^3-6 b \tanh ^{-1}(c x)-32 b c x \tanh ^{-1}(c x)-72 b c^2 x^2 \tanh ^{-1}(c x)-96 b c^3 x^3 \tanh ^{-1}(c x)+24 a c^4 x^4 \log (x)+128 b c^4 x^4 \log (c x)-39 b c^4 x^4 \log (1-c x)+39 b c^4 x^4 \log (1+c x)-64 b c^4 x^4 \log \left (1-c^2 x^2\right )-12 b c^4 x^4 \text {PolyLog}(2,-c x)+12 b c^4 x^4 \text {PolyLog}(2,c x)\right )}{24 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + c*d*x)^4*(a + b*ArcTanh[c*x]))/x^5,x]

[Out]

(d^4*(-6*a - 32*a*c*x - 2*b*c*x - 72*a*c^2*x^2 - 16*b*c^2*x^2 - 96*a*c^3*x^3 - 78*b*c^3*x^3 - 6*b*ArcTanh[c*x]
- 32*b*c*x*ArcTanh[c*x] - 72*b*c^2*x^2*ArcTanh[c*x] - 96*b*c^3*x^3*ArcTanh[c*x] + 24*a*c^4*x^4*Log[x] + 128*b
*c^4*x^4*Log[c*x] - 39*b*c^4*x^4*Log[1 - c*x] + 39*b*c^4*x^4*Log[1 + c*x] - 64*b*c^4*x^4*Log[1 - c^2*x^2] - 12
*b*c^4*x^4*PolyLog[2, -(c*x)] + 12*b*c^4*x^4*PolyLog[2, c*x]))/(24*x^4)

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Maple [A]
time = 0.23, size = 248, normalized size = 1.19

method	result	size
derivativedivides	\(c^{4} \left (-\frac {3 d^{4} a}{c^{2} x^{2}}-\frac {d^{4} a}{4 c^{4} x^{4}}-\frac {4 d^{4} a}{c x}-\frac {4 d^{4} a}{3 c^{3} x^{3}}+d^{4} a \ln \left (c x \right )-\frac {3 d^{4} b \arctanh \left (c x \right )}{c^{2} x^{2}}-\frac {d^{4} b \arctanh \left (c x \right )}{4 c^{4} x^{4}}-\frac {4 d^{4} b \arctanh \left (c x \right )}{c x}-\frac {4 d^{4} b \arctanh \left (c x \right )}{3 c^{3} x^{3}}+d^{4} b \arctanh \left (c x \right ) \ln \left (c x \right )-\frac {103 d^{4} b \ln \left (c x -1\right )}{24}-\frac {25 d^{4} b \ln \left (c x +1\right )}{24}-\frac {d^{4} b}{12 c^{3} x^{3}}-\frac {2 d^{4} b}{3 c^{2} x^{2}}-\frac {13 d^{4} b}{4 c x}+\frac {16 d^{4} b \ln \left (c x \right )}{3}-\frac {d^{4} b \dilog \left (c x \right )}{2}-\frac {d^{4} b \dilog \left (c x +1\right )}{2}-\frac {d^{4} b \ln \left (c x \right ) \ln \left (c x +1\right )}{2}\right )\)	\(248\)
default	\(c^{4} \left (-\frac {3 d^{4} a}{c^{2} x^{2}}-\frac {d^{4} a}{4 c^{4} x^{4}}-\frac {4 d^{4} a}{c x}-\frac {4 d^{4} a}{3 c^{3} x^{3}}+d^{4} a \ln \left (c x \right )-\frac {3 d^{4} b \arctanh \left (c x \right )}{c^{2} x^{2}}-\frac {d^{4} b \arctanh \left (c x \right )}{4 c^{4} x^{4}}-\frac {4 d^{4} b \arctanh \left (c x \right )}{c x}-\frac {4 d^{4} b \arctanh \left (c x \right )}{3 c^{3} x^{3}}+d^{4} b \arctanh \left (c x \right ) \ln \left (c x \right )-\frac {103 d^{4} b \ln \left (c x -1\right )}{24}-\frac {25 d^{4} b \ln \left (c x +1\right )}{24}-\frac {d^{4} b}{12 c^{3} x^{3}}-\frac {2 d^{4} b}{3 c^{2} x^{2}}-\frac {13 d^{4} b}{4 c x}+\frac {16 d^{4} b \ln \left (c x \right )}{3}-\frac {d^{4} b \dilog \left (c x \right )}{2}-\frac {d^{4} b \dilog \left (c x +1\right )}{2}-\frac {d^{4} b \ln \left (c x \right ) \ln \left (c x +1\right )}{2}\right )\)	\(248\)
risch	\(\frac {2 c^{3} d^{4} b \ln \left (-c x +1\right )}{x}-\frac {b c \,d^{4}}{12 x^{3}}-\frac {2 b \,c^{2} d^{4}}{3 x^{2}}-\frac {13 b \,c^{3} d^{4}}{4 x}+\frac {3 c^{2} d^{4} b \ln \left (-c x +1\right )}{2 x^{2}}+\frac {2 c \,d^{4} b \ln \left (-c x +1\right )}{3 x^{3}}-\frac {2 b \,c^{3} d^{4} \ln \left (c x +1\right )}{x}-\frac {3 b \,c^{2} d^{4} \ln \left (c x +1\right )}{2 x^{2}}-\frac {2 b c \,d^{4} \ln \left (c x +1\right )}{3 x^{3}}-\frac {25 \ln \left (c x +1\right ) b \,c^{4} d^{4}}{24}-\frac {d^{4} a}{4 x^{4}}-\frac {103 \ln \left (-c x +1\right ) b \,c^{4} d^{4}}{24}+\frac {d^{4} b \ln \left (-c x +1\right )}{8 x^{4}}-\frac {4 c \,d^{4} a}{3 x^{3}}+c^{4} d^{4} a \ln \left (-c x \right )+\frac {103 c^{4} d^{4} b \ln \left (-c x \right )}{24}+\frac {c^{4} d^{4} \dilog \left (-c x +1\right ) b}{2}-\frac {4 c^{3} d^{4} a}{x}-\frac {3 c^{2} d^{4} a}{x^{2}}+\frac {25 b \,c^{4} d^{4} \ln \left (c x \right )}{24}-\frac {b \,d^{4} \ln \left (c x +1\right )}{8 x^{4}}-\frac {b \,c^{4} d^{4} \dilog \left (c x +1\right )}{2}\)	\(319\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^4*(a+b*arctanh(c*x))/x^5,x,method=_RETURNVERBOSE)

[Out]

c^4*(-3*d^4*a/c^2/x^2-1/4*d^4*a/c^4/x^4-4*d^4*a/c/x-4/3*d^4*a/c^3/x^3+d^4*a*ln(c*x)-3*d^4*b*arctanh(c*x)/c^2/x
^2-1/4*d^4*b*arctanh(c*x)/c^4/x^4-4*d^4*b*arctanh(c*x)/c/x-4/3*d^4*b*arctanh(c*x)/c^3/x^3+d^4*b*arctanh(c*x)*l
n(c*x)-103/24*d^4*b*ln(c*x-1)-25/24*d^4*b*ln(c*x+1)-1/12*d^4*b/c^3/x^3-2/3*d^4*b/c^2/x^2-13/4*d^4*b/c/x+16/3*d
^4*b*ln(c*x)-1/2*d^4*b*dilog(c*x)-1/2*d^4*b*dilog(c*x+1)-1/2*d^4*b*ln(c*x)*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((c*d*x+d)^4*(a+b*arctanh(c*x))/x^5,x, algorithm="maxima")

[Out]

1/2*b*c^4*d^4*integrate((log(c*x + 1) - log(-c*x + 1))/x, x) + a*c^4*d^4*log(x) - 2*(c*(log(c^2*x^2 - 1) - log
(x^2)) + 2*arctanh(c*x)/x)*b*c^3*d^4 + 3/2*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*b*
c^2*d^4 - 2/3*((c^2*log(c^2*x^2 - 1) - c^2*log(x^2) + 1/x^2)*c + 2*arctanh(c*x)/x^3)*b*c*d^4 - 4*a*c^3*d^4/x +
1/24*((3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^2*x^2 + 1)/x^3)*c - 6*arctanh(c*x)/x^4)*b*d^4 - 3*a*c
^2*d^4/x^2 - 4/3*a*c*d^4/x^3 - 1/4*a*d^4/x^4

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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^5,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^5, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} d^{4} \left (\int \frac {a}{x^{5}}\, dx + \int \frac {4 a c}{x^{4}}\, dx + \int \frac {6 a c^{2}}{x^{3}}\, dx + \int \frac {4 a c^{3}}{x^{2}}\, dx + \int \frac {a c^{4}}{x}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{x^{5}}\, dx + \int \frac {4 b c \operatorname {atanh}{\left (c x \right )}}{x^{4}}\, dx + \int \frac {6 b c^{2} \operatorname {atanh}{\left (c x \right )}}{x^{3}}\, dx + \int \frac {4 b c^{3} \operatorname {atanh}{\left (c x \right )}}{x^{2}}\, dx + \int \frac {b c^{4} \operatorname {atanh}{\left (c x \right )}}{x}\, 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**5,x)

[Out]

d**4*(Integral(a/x**5, x) + Integral(4*a*c/x**4, x) + Integral(6*a*c**2/x**3, x) + Integral(4*a*c**3/x**2, x)
+ Integral(a*c**4/x, x) + Integral(b*atanh(c*x)/x**5, x) + Integral(4*b*c*atanh(c*x)/x**4, x) + Integral(6*b*c
**2*atanh(c*x)/x**3, x) + Integral(4*b*c**3*atanh(c*x)/x**2, x) + Integral(b*c**4*atanh(c*x)/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^5,x, algorithm="giac")

[Out]

integrate((c*d*x + d)^4*(b*arctanh(c*x) + a)/x^5, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^4}{x^5} \,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^5,x)

[Out]

int(((a + b*atanh(c*x))*(d + c*d*x)^4)/x^5, x)

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