3.1.27 \(\int \frac {(d+c d x)^3 (a+b \tanh ^{-1}(c x))}{x^4} \, dx\) [27]

Optimal. Leaf size=176 \[ -\frac {b c d^3}{6 x^2}-\frac {3 b c^2 d^3}{2 x}+\frac {3}{2} b c^3 d^3 \tanh ^{-1}(c x)-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^3 d^3 \log (x)+\frac {10}{3} b c^3 d^3 \log (x)-\frac {5}{3} b c^3 d^3 \log \left (1-c^2 x^2\right )-\frac {1}{2} b c^3 d^3 \text {PolyLog}(2,-c x)+\frac {1}{2} b c^3 d^3 \text {PolyLog}(2,c x) \]

[Out]

-1/6*b*c*d^3/x^2-3/2*b*c^2*d^3/x+3/2*b*c^3*d^3*arctanh(c*x)-1/3*d^3*(a+b*arctanh(c*x))/x^3-3/2*c*d^3*(a+b*arct
anh(c*x))/x^2-3*c^2*d^3*(a+b*arctanh(c*x))/x+a*c^3*d^3*ln(x)+10/3*b*c^3*d^3*ln(x)-5/3*b*c^3*d^3*ln(-c^2*x^2+1)
-1/2*b*c^3*d^3*polylog(2,-c*x)+1/2*b*c^3*d^3*polylog(2,c*x)

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Rubi [A]
time = 0.14, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6087, 6037, 272, 46, 331, 212, 36, 29, 31, 6031} \begin {gather*} -\frac {3 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}+a c^3 d^3 \log (x)-\frac {1}{2} b c^3 d^3 \text {Li}_2(-c x)+\frac {1}{2} b c^3 d^3 \text {Li}_2(c x)+\frac {10}{3} b c^3 d^3 \log (x)+\frac {3}{2} b c^3 d^3 \tanh ^{-1}(c x)-\frac {3 b c^2 d^3}{2 x}-\frac {5}{3} b c^3 d^3 \log \left (1-c^2 x^2\right )-\frac {b c d^3}{6 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(b*c*d^3)/x^2 - (3*b*c^2*d^3)/(2*x) + (3*b*c^3*d^3*ArcTanh[c*x])/2 - (d^3*(a + b*ArcTanh[c*x]))/(3*x^3) -
(3*c*d^3*(a + b*ArcTanh[c*x]))/(2*x^2) - (3*c^2*d^3*(a + b*ArcTanh[c*x]))/x + a*c^3*d^3*Log[x] + (10*b*c^3*d^
3*Log[x])/3 - (5*b*c^3*d^3*Log[1 - c^2*x^2])/3 - (b*c^3*d^3*PolyLog[2, -(c*x)])/2 + (b*c^3*d^3*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)^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^4} \, dx &=\int \left (\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^4}+\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^3}+\frac {3 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}+\frac {c^3 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d^3 \int \frac {a+b \tanh ^{-1}(c x)}{x^4} \, dx+\left (3 c d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^3} \, dx+\left (3 c^2 d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^2} \, dx+\left (c^3 d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx\\ &=-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^3 d^3 \log (x)-\frac {1}{2} b c^3 d^3 \text {Li}_2(-c x)+\frac {1}{2} b c^3 d^3 \text {Li}_2(c x)+\frac {1}{3} \left (b c d^3\right ) \int \frac {1}{x^3 \left (1-c^2 x^2\right )} \, dx+\frac {1}{2} \left (3 b c^2 d^3\right ) \int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx+\left (3 b c^3 d^3\right ) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {3 b c^2 d^3}{2 x}-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^3 d^3 \log (x)-\frac {1}{2} b c^3 d^3 \text {Li}_2(-c x)+\frac {1}{2} b c^3 d^3 \text {Li}_2(c x)+\frac {1}{6} \left (b c d^3\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (1-c^2 x\right )} \, dx,x,x^2\right )+\frac {1}{2} \left (3 b c^3 d^3\right ) \text {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )+\frac {1}{2} \left (3 b c^4 d^3\right ) \int \frac {1}{1-c^2 x^2} \, dx\\ &=-\frac {3 b c^2 d^3}{2 x}+\frac {3}{2} b c^3 d^3 \tanh ^{-1}(c x)-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^3 d^3 \log (x)-\frac {1}{2} b c^3 d^3 \text {Li}_2(-c x)+\frac {1}{2} b c^3 d^3 \text {Li}_2(c x)+\frac {1}{6} \left (b c d^3\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 )+\frac {1}{2} \left (3 b c^3 d^3\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (3 b c^5 d^3\right ) \text {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b c d^3}{6 x^2}-\frac {3 b c^2 d^3}{2 x}+\frac {3}{2} b c^3 d^3 \tanh ^{-1}(c x)-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {3 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^3 d^3 \log (x)+\frac {10}{3} b c^3 d^3 \log (x)-\frac {5}{3} b c^3 d^3 \log \left (1-c^2 x^2\right )-\frac {1}{2} b c^3 d^3 \text {Li}_2(-c x)+\frac {1}{2} b c^3 d^3 \text {Li}_2(c x)\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 175, normalized size = 0.99 \begin {gather*} \frac {d^3 \left (-4 a-18 a c x-2 b c x-36 a c^2 x^2-18 b c^2 x^2-4 b \tanh ^{-1}(c x)-18 b c x \tanh ^{-1}(c x)-36 b c^2 x^2 \tanh ^{-1}(c x)+12 a c^3 x^3 \log (x)+40 b c^3 x^3 \log (c x)-9 b c^3 x^3 \log (1-c x)+9 b c^3 x^3 \log (1+c x)-20 b c^3 x^3 \log \left (1-c^2 x^2\right )-6 b c^3 x^3 \text {PolyLog}(2,-c x)+6 b c^3 x^3 \text {PolyLog}(2,c x)\right )}{12 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*(-4*a - 18*a*c*x - 2*b*c*x - 36*a*c^2*x^2 - 18*b*c^2*x^2 - 4*b*ArcTanh[c*x] - 18*b*c*x*ArcTanh[c*x] - 36*
b*c^2*x^2*ArcTanh[c*x] + 12*a*c^3*x^3*Log[x] + 40*b*c^3*x^3*Log[c*x] - 9*b*c^3*x^3*Log[1 - c*x] + 9*b*c^3*x^3*
Log[1 + c*x] - 20*b*c^3*x^3*Log[1 - c^2*x^2] - 6*b*c^3*x^3*PolyLog[2, -(c*x)] + 6*b*c^3*x^3*PolyLog[2, c*x]))/
(12*x^3)

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Maple [A]
time = 0.22, size = 208, normalized size = 1.18

method	result	size
derivativedivides	\(c^{3} \left (-\frac {3 d^{3} a}{c x}-\frac {d^{3} a}{3 c^{3} x^{3}}-\frac {3 d^{3} a}{2 c^{2} x^{2}}+d^{3} a \ln \left (c x \right )-\frac {3 d^{3} b \arctanh \left (c x \right )}{c x}-\frac {d^{3} b \arctanh \left (c x \right )}{3 c^{3} x^{3}}-\frac {3 d^{3} b \arctanh \left (c x \right )}{2 c^{2} x^{2}}+d^{3} b \arctanh \left (c x \right ) \ln \left (c x \right )-\frac {d^{3} b}{6 c^{2} x^{2}}-\frac {3 d^{3} b}{2 c x}+\frac {10 d^{3} b \ln \left (c x \right )}{3}-\frac {29 d^{3} b \ln \left (c x -1\right )}{12}-\frac {11 d^{3} b \ln \left (c x +1\right )}{12}-\frac {d^{3} b \dilog \left (c x \right )}{2}-\frac {d^{3} b \dilog \left (c x +1\right )}{2}-\frac {d^{3} b \ln \left (c x \right ) \ln \left (c x +1\right )}{2}\right )\)	\(208\)
default	\(c^{3} \left (-\frac {3 d^{3} a}{c x}-\frac {d^{3} a}{3 c^{3} x^{3}}-\frac {3 d^{3} a}{2 c^{2} x^{2}}+d^{3} a \ln \left (c x \right )-\frac {3 d^{3} b \arctanh \left (c x \right )}{c x}-\frac {d^{3} b \arctanh \left (c x \right )}{3 c^{3} x^{3}}-\frac {3 d^{3} b \arctanh \left (c x \right )}{2 c^{2} x^{2}}+d^{3} b \arctanh \left (c x \right ) \ln \left (c x \right )-\frac {d^{3} b}{6 c^{2} x^{2}}-\frac {3 d^{3} b}{2 c x}+\frac {10 d^{3} b \ln \left (c x \right )}{3}-\frac {29 d^{3} b \ln \left (c x -1\right )}{12}-\frac {11 d^{3} b \ln \left (c x +1\right )}{12}-\frac {d^{3} b \dilog \left (c x \right )}{2}-\frac {d^{3} b \dilog \left (c x +1\right )}{2}-\frac {d^{3} b \ln \left (c x \right ) \ln \left (c x +1\right )}{2}\right )\)	\(208\)
risch	\(-\frac {3 b \,c^{2} d^{3}}{2 x}+\frac {29 c^{3} d^{3} b \ln \left (-c x \right )}{12}-\frac {29 \ln \left (-c x +1\right ) b \,c^{3} d^{3}}{12}+\frac {3 c^{2} d^{3} b \ln \left (-c x +1\right )}{2 x}+\frac {3 c \,d^{3} b \ln \left (-c x +1\right )}{4 x^{2}}-\frac {b c \,d^{3}}{6 x^{2}}+\frac {d^{3} b \ln \left (-c x +1\right )}{6 x^{3}}+\frac {c^{3} d^{3} \dilog \left (-c x +1\right ) b}{2}-\frac {3 c \,d^{3} a}{2 x^{2}}-\frac {d^{3} a}{3 x^{3}}-\frac {3 c^{2} d^{3} a}{x}+c^{3} d^{3} a \ln \left (-c x \right )+\frac {11 b \,c^{3} d^{3} \ln \left (c x \right )}{12}-\frac {11 \ln \left (c x +1\right ) b \,c^{3} d^{3}}{12}-\frac {3 b c \,d^{3} \ln \left (c x +1\right )}{4 x^{2}}-\frac {b \,d^{3} \ln \left (c x +1\right )}{6 x^{3}}-\frac {3 b \,c^{2} d^{3} \ln \left (c x +1\right )}{2 x}-\frac {b \,c^{3} d^{3} \dilog \left (c x +1\right )}{2}\)	\(258\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*(-3*d^3*a/c/x-1/3*d^3*a/c^3/x^3-3/2*d^3*a/c^2/x^2+d^3*a*ln(c*x)-3*d^3*b*arctanh(c*x)/c/x-1/3*d^3*b*arctanh
(c*x)/c^3/x^3-3/2*d^3*b*arctanh(c*x)/c^2/x^2+d^3*b*arctanh(c*x)*ln(c*x)-1/6*d^3*b/c^2/x^2-3/2*d^3*b/c/x+10/3*d
^3*b*ln(c*x)-29/12*d^3*b*ln(c*x-1)-11/12*d^3*b*ln(c*x+1)-1/2*d^3*b*dilog(c*x)-1/2*d^3*b*dilog(c*x+1)-1/2*d^3*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)^3*(a+b*arctanh(c*x))/x^4,x, algorithm="maxima")

[Out]

1/2*b*c^3*d^3*integrate((log(c*x + 1) - log(-c*x + 1))/x, x) + a*c^3*d^3*log(x) - 3/2*(c*(log(c^2*x^2 - 1) - l
og(x^2)) + 2*arctanh(c*x)/x)*b*c^2*d^3 + 3/4*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*
b*c*d^3 - 1/6*((c^2*log(c^2*x^2 - 1) - c^2*log(x^2) + 1/x^2)*c + 2*arctanh(c*x)/x^3)*b*d^3 - 3*a*c^2*d^3/x - 3
/2*a*c*d^3/x^2 - 1/3*a*d^3/x^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((c*d*x+d)^3*(a+b*arctanh(c*x))/x^4,x, algorithm="fricas")

[Out]

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

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

[Out]

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

[Out]

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

[Out]

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

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