3.1.15 \(\int \frac {(d+c d x)^2 (a+b \tanh ^{-1}(c x))}{x^2} \, dx\) [15]

Optimal. Leaf size=61 \[ \frac {d^2 \left (-1+c^2 x^2\right ) \left (a+b \tanh ^{-1}(c x)\right )}{x}+(2 a+b) c d^2 \log (x)-b c d^2 \text {PolyLog}(2,-c x)+b c d^2 \text {PolyLog}(2,c x) \]

[Out]

d^2*(c^2*x^2-1)*(a+b*arctanh(c*x))/x+(2*a+b)*c*d^2*ln(x)-b*c*d^2*polylog(2,-c*x)+b*c*d^2*polylog(2,c*x)

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Rubi [A]
time = 0.09, antiderivative size = 80, normalized size of antiderivative = 1.31, number of steps used = 11, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {6087, 6021, 266, 6037, 272, 36, 29, 31, 6031} \begin {gather*} -\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c^2 d^2 x+2 a c d^2 \log (x)+b c^2 d^2 x \tanh ^{-1}(c x)-b c d^2 \text {Li}_2(-c x)+b c d^2 \text {Li}_2(c x)+b c d^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.06, size = 73, normalized size = 1.20 \begin {gather*} \frac {d^2 \left (-a+a c^2 x^2-b \tanh ^{-1}(c x)+b c^2 x^2 \tanh ^{-1}(c x)+2 a c x \log (x)+b c x \log (c x)-b c x \text {PolyLog}(2,-c x)+b c x \text {PolyLog}(2,c x)\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(-a + a*c^2*x^2 - b*ArcTanh[c*x] + b*c^2*x^2*ArcTanh[c*x] + 2*a*c*x*Log[x] + b*c*x*Log[c*x] - b*c*x*PolyL
og[2, -(c*x)] + b*c*x*PolyLog[2, c*x]))/x

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Maple [A]
time = 0.21, size = 121, normalized size = 1.98

method	result	size
derivativedivides	\(c \left (d^{2} a c x -\frac {d^{2} a}{c x}+2 d^{2} a \ln \left (c x \right )+b c \,d^{2} x \arctanh \left (c x \right )-\frac {d^{2} b \arctanh \left (c x \right )}{c x}+2 d^{2} b \arctanh \left (c x \right ) \ln \left (c x \right )+d^{2} b \ln \left (c x \right )-d^{2} b \dilog \left (c x \right )-d^{2} b \dilog \left (c x +1\right )-d^{2} b \ln \left (c x \right ) \ln \left (c x +1\right )\right )\)	\(121\)
default	\(c \left (d^{2} a c x -\frac {d^{2} a}{c x}+2 d^{2} a \ln \left (c x \right )+b c \,d^{2} x \arctanh \left (c x \right )-\frac {d^{2} b \arctanh \left (c x \right )}{c x}+2 d^{2} b \arctanh \left (c x \right ) \ln \left (c x \right )+d^{2} b \ln \left (c x \right )-d^{2} b \dilog \left (c x \right )-d^{2} b \dilog \left (c x +1\right )-d^{2} b \ln \left (c x \right ) \ln \left (c x +1\right )\right )\)	\(121\)
risch	\(a \,c^{2} d^{2} x -c \,d^{2} a -\frac {d^{2} a}{x}+2 c \,d^{2} a \ln \left (-c x \right )-\frac {c^{2} d^{2} b \ln \left (-c x +1\right ) x}{2}-b c \,d^{2}+\frac {c \,d^{2} b \ln \left (-c x \right )}{2}+\frac {d^{2} b \ln \left (-c x +1\right )}{2 x}+c \,d^{2} \dilog \left (-c x +1\right ) b +\frac {b \,c^{2} d^{2} \ln \left (c x +1\right ) x}{2}+\frac {b c \,d^{2} \ln \left (c x \right )}{2}-\frac {b \,d^{2} \ln \left (c x +1\right )}{2 x}-b c \,d^{2} \dilog \left (c x +1\right )\)	\(159\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(d^2*a*c*x-d^2*a/c/x+2*d^2*a*ln(c*x)+b*c*d^2*x*arctanh(c*x)-d^2*b*arctanh(c*x)/c/x+2*d^2*b*arctanh(c*x)*ln(c
*x)+d^2*b*ln(c*x)-d^2*b*dilog(c*x)-d^2*b*dilog(c*x+1)-d^2*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)^2*(a+b*arctanh(c*x))/x^2,x, algorithm="maxima")

[Out]

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

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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)^2*(a+b*arctanh(c*x))/x^2,x, algorithm="fricas")

[Out]

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

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

[Out]

d**2*(Integral(a*c**2, x) + Integral(a/x**2, x) + Integral(2*a*c/x, x) + Integral(b*c**2*atanh(c*x), x) + Inte
gral(b*atanh(c*x)/x**2, x) + Integral(2*b*c*atanh(c*x)/x, x))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 410 vs. \(2 (59) = 118\).
time = 1.15, size = 410, normalized size = 6.72 \begin {gather*} \frac {1}{6} \, {\left (\frac {6 \, a d^{2}}{\frac {{\left (c x + 1\right )} c^{2}}{c x - 1} + c^{2}} + \frac {5 \, b d^{2} \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{2}} + \frac {3 \, b d^{2} \log \left (-\frac {c x + 1}{c x - 1} - 1\right )}{c^{2}} + {\left (\frac {3 \, b d^{2}}{\frac {{\left (c x + 1\right )} c^{2}}{c x - 1} + c^{2}} - \frac {\frac {3 \, {\left (c x + 1\right )}^{2} b d^{2}}{{\left (c x - 1\right )}^{2}} - \frac {12 \, {\left (c x + 1\right )} b d^{2}}{c x - 1} + 5 \, b d^{2}}{\frac {{\left (c x + 1\right )}^{3} c^{2}}{{\left (c x - 1\right )}^{3}} - \frac {3 \, {\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} + \frac {3 \, {\left (c x + 1\right )} c^{2}}{c x - 1} - c^{2}}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right ) - \frac {8 \, b d^{2} \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{2}} - \frac {2 \, {\left (\frac {3 \, {\left (c x + 1\right )}^{2} a d^{2}}{{\left (c x - 1\right )}^{2}} - \frac {12 \, {\left (c x + 1\right )} a d^{2}}{c x - 1} + 5 \, a d^{2} - \frac {{\left (c x + 1\right )}^{2} b d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {{\left (c x + 1\right )} b d^{2}}{c x - 1}\right )}}{\frac {{\left (c x + 1\right )}^{3} c^{2}}{{\left (c x - 1\right )}^{3}} - \frac {3 \, {\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} + \frac {3 \, {\left (c x + 1\right )} c^{2}}{c x - 1} - c^{2}}\right )} c^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^2*(a+b*arctanh(c*x))/x^2,x, algorithm="giac")

[Out]

1/6*(6*a*d^2/((c*x + 1)*c^2/(c*x - 1) + c^2) + 5*b*d^2*log(-(c*x + 1)/(c*x - 1) + 1)/c^2 + 3*b*d^2*log(-(c*x +
1)/(c*x - 1) - 1)/c^2 + (3*b*d^2/((c*x + 1)*c^2/(c*x - 1) + c^2) - (3*(c*x + 1)^2*b*d^2/(c*x - 1)^2 - 12*(c*x
+ 1)*b*d^2/(c*x - 1) + 5*b*d^2)/((c*x + 1)^3*c^2/(c*x - 1)^3 - 3*(c*x + 1)^2*c^2/(c*x - 1)^2 + 3*(c*x + 1)*c^
2/(c*x - 1) - c^2))*log(-(c*x + 1)/(c*x - 1)) - 8*b*d^2*log(-(c*x + 1)/(c*x - 1))/c^2 - 2*(3*(c*x + 1)^2*a*d^2
/(c*x - 1)^2 - 12*(c*x + 1)*a*d^2/(c*x - 1) + 5*a*d^2 - (c*x + 1)^2*b*d^2/(c*x - 1)^2 + (c*x + 1)*b*d^2/(c*x -
1))/((c*x + 1)^3*c^2/(c*x - 1)^3 - 3*(c*x + 1)^2*c^2/(c*x - 1)^2 + 3*(c*x + 1)*c^2/(c*x - 1) - c^2))*c^2

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^2}{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)^2)/x^2,x)

[Out]

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

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