3.1.61 \(\int \frac {x (a+b \tanh ^{-1}(c x))}{(d+c d x)^3} \, dx\) [61]

Optimal. Leaf size=77 \[ \frac {b}{8 c^2 d^3 (1+c x)^2}-\frac {3 b}{8 c^2 d^3 (1+c x)}-\frac {b \tanh ^{-1}(c x)}{8 c^2 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2} \]

[Out]

1/8*b/c^2/d^3/(c*x+1)^2-3/8*b/c^2/d^3/(c*x+1)-1/8*b*arctanh(c*x)/c^2/d^3+1/2*x^2*(a+b*arctanh(c*x))/d^3/(c*x+1
)^2

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Rubi [A]
time = 0.06, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {37, 6083, 12, 90, 213} \begin {gather*} \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (c x+1)^2}-\frac {3 b}{8 c^2 d^3 (c x+1)}+\frac {b}{8 c^2 d^3 (c x+1)^2}-\frac {b \tanh ^{-1}(c x)}{8 c^2 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

b/(8*c^2*d^3*(1 + c*x)^2) - (3*b)/(8*c^2*d^3*(1 + c*x)) - (b*ArcTanh[c*x])/(8*c^2*d^3) + (x^2*(a + b*ArcTanh[c
*x]))/(2*d^3*(1 + c*x)^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 6083

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_))^(q_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x)^q, x]}, Dist[a + b*ArcTanh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 - c^2*
x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[q, -1] && IntegerQ[2*m] && ((IGtQ[m, 0] && IGtQ[q,
0]) || (ILtQ[m + q + 1, 0] && LtQ[m*q, 0]))

Rubi steps

\begin {align*} \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{(d+c d x)^3} \, dx &=\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}-(b c) \int \frac {x^2}{2 (1-c x) (d+c d x)^3} \, dx\\ &=\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}-\frac {1}{2} (b c) \int \frac {x^2}{(1-c x) (d+c d x)^3} \, dx\\ &=\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}-\frac {1}{2} (b c) \int \left (\frac {1}{2 c^2 d^3 (1+c x)^3}-\frac {3}{4 c^2 d^3 (1+c x)^2}-\frac {1}{4 c^2 d^3 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=\frac {b}{8 c^2 d^3 (1+c x)^2}-\frac {3 b}{8 c^2 d^3 (1+c x)}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}+\frac {b \int \frac {1}{-1+c^2 x^2} \, dx}{8 c d^3}\\ &=\frac {b}{8 c^2 d^3 (1+c x)^2}-\frac {3 b}{8 c^2 d^3 (1+c x)}-\frac {b \tanh ^{-1}(c x)}{8 c^2 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (1+c x)^2}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 99, normalized size = 1.29 \begin {gather*} -\frac {8 a+4 b+16 a c x+6 b c x+8 (b+2 b c x) \tanh ^{-1}(c x)+3 b (1+c x)^2 \log (1-c x)-3 b \log (1+c x)-6 b c x \log (1+c x)-3 b c^2 x^2 \log (1+c x)}{16 c^2 d^3 (1+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(8*a + 4*b + 16*a*c*x + 6*b*c*x + 8*(b + 2*b*c*x)*ArcTanh[c*x] + 3*b*(1 + c*x)^2*Log[1 - c*x] - 3*b*Log[
1 + c*x] - 6*b*c*x*Log[1 + c*x] - 3*b*c^2*x^2*Log[1 + c*x])/(c^2*d^3*(1 + c*x)^2)

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Maple [A]
time = 0.16, size = 114, normalized size = 1.48

method	result	size
derivativedivides	\(\frac {\frac {a \left (-\frac {1}{c x +1}+\frac {1}{2 \left (c x +1\right )^{2}}\right )}{d^{3}}-\frac {b \arctanh \left (c x \right )}{d^{3} \left (c x +1\right )}+\frac {b \arctanh \left (c x \right )}{2 d^{3} \left (c x +1\right )^{2}}-\frac {3 b \ln \left (c x -1\right )}{16 d^{3}}+\frac {b}{8 d^{3} \left (c x +1\right )^{2}}-\frac {3 b}{8 d^{3} \left (c x +1\right )}+\frac {3 b \ln \left (c x +1\right )}{16 d^{3}}}{c^{2}}\)	\(114\)
default	\(\frac {\frac {a \left (-\frac {1}{c x +1}+\frac {1}{2 \left (c x +1\right )^{2}}\right )}{d^{3}}-\frac {b \arctanh \left (c x \right )}{d^{3} \left (c x +1\right )}+\frac {b \arctanh \left (c x \right )}{2 d^{3} \left (c x +1\right )^{2}}-\frac {3 b \ln \left (c x -1\right )}{16 d^{3}}+\frac {b}{8 d^{3} \left (c x +1\right )^{2}}-\frac {3 b}{8 d^{3} \left (c x +1\right )}+\frac {3 b \ln \left (c x +1\right )}{16 d^{3}}}{c^{2}}\)	\(114\)
risch	\(-\frac {b \left (2 c x +1\right ) \ln \left (c x +1\right )}{4 c^{2} d^{3} \left (c x +1\right )^{2}}+\frac {3 b \,c^{2} \ln \left (-c x -1\right ) x^{2}-3 \ln \left (c x -1\right ) b \,c^{2} x^{2}+6 \ln \left (-c x -1\right ) b c x -6 \ln \left (c x -1\right ) b c x +8 b c x \ln \left (-c x +1\right )-16 c x a -6 b c x +3 b \ln \left (-c x -1\right )-3 b \ln \left (c x -1\right )+4 b \ln \left (-c x +1\right )-8 a -4 b}{16 c^{2} d^{3} \left (c x +1\right )^{2}}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(a/d^3*(-1/(c*x+1)+1/2/(c*x+1)^2)-b/d^3*arctanh(c*x)/(c*x+1)+1/2*b/d^3*arctanh(c*x)/(c*x+1)^2-3/16*b/d^3
*ln(c*x-1)+1/8*b/d^3/(c*x+1)^2-3/8*b/d^3/(c*x+1)+3/16*b/d^3*ln(c*x+1))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (69) = 138\).
time = 0.26, size = 152, normalized size = 1.97 \begin {gather*} -\frac {1}{16} \, {\left (c {\left (\frac {2 \, {\left (3 \, c x + 2\right )}}{c^{5} d^{3} x^{2} + 2 \, c^{4} d^{3} x + c^{3} d^{3}} - \frac {3 \, \log \left (c x + 1\right )}{c^{3} d^{3}} + \frac {3 \, \log \left (c x - 1\right )}{c^{3} d^{3}}\right )} + \frac {8 \, {\left (2 \, c x + 1\right )} \operatorname {artanh}\left (c x\right )}{c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}}\right )} b - \frac {{\left (2 \, c x + 1\right )} a}{2 \, {\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(c*(2*(3*c*x + 2)/(c^5*d^3*x^2 + 2*c^4*d^3*x + c^3*d^3) - 3*log(c*x + 1)/(c^3*d^3) + 3*log(c*x - 1)/(c^3
*d^3)) + 8*(2*c*x + 1)*arctanh(c*x)/(c^4*d^3*x^2 + 2*c^3*d^3*x + c^2*d^3))*b - 1/2*(2*c*x + 1)*a/(c^4*d^3*x^2
+ 2*c^3*d^3*x + c^2*d^3)

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Fricas [A]
time = 0.37, size = 84, normalized size = 1.09 \begin {gather*} -\frac {2 \, {\left (8 \, a + 3 \, b\right )} c x - {\left (3 \, b c^{2} x^{2} - 2 \, b c x - b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 8 \, a + 4 \, b}{16 \, {\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2*(8*a + 3*b)*c*x - (3*b*c^2*x^2 - 2*b*c*x - b)*log(-(c*x + 1)/(c*x - 1)) + 8*a + 4*b)/(c^4*d^3*x^2 + 2
*c^3*d^3*x + c^2*d^3)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (71) = 142\).
time = 0.84, size = 277, normalized size = 3.60 \begin {gather*} \begin {cases} - \frac {8 a c x}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {4 a}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} + \frac {3 b c^{2} x^{2} \operatorname {atanh}{\left (c x \right )}}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {2 b c x \operatorname {atanh}{\left (c x \right )}}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {3 b c x}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {b \operatorname {atanh}{\left (c x \right )}}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {2 b}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} & \text {for}\: c \neq 0 \\\frac {a x^{2}}{2 d^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-8*a*c*x/(8*c**4*d**3*x**2 + 16*c**3*d**3*x + 8*c**2*d**3) - 4*a/(8*c**4*d**3*x**2 + 16*c**3*d**3*x
+ 8*c**2*d**3) + 3*b*c**2*x**2*atanh(c*x)/(8*c**4*d**3*x**2 + 16*c**3*d**3*x + 8*c**2*d**3) - 2*b*c*x*atanh(c
*x)/(8*c**4*d**3*x**2 + 16*c**3*d**3*x + 8*c**2*d**3) - 3*b*c*x/(8*c**4*d**3*x**2 + 16*c**3*d**3*x + 8*c**2*d*
*3) - b*atanh(c*x)/(8*c**4*d**3*x**2 + 16*c**3*d**3*x + 8*c**2*d**3) - 2*b/(8*c**4*d**3*x**2 + 16*c**3*d**3*x
+ 8*c**2*d**3), Ne(c, 0)), (a*x**2/(2*d**3), True))

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Giac [A]
time = 0.42, size = 114, normalized size = 1.48 \begin {gather*} \frac {1}{32} \, c {\left (\frac {2 \, {\left (c x - 1\right )}^{2} {\left (\frac {2 \, {\left (c x + 1\right )} b}{c x - 1} + b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\left (c x + 1\right )}^{2} c^{3} d^{3}} + \frac {{\left (c x - 1\right )}^{2} {\left (\frac {8 \, {\left (c x + 1\right )} a}{c x - 1} + 4 \, a + \frac {4 \, {\left (c x + 1\right )} b}{c x - 1} + b\right )}}{{\left (c x + 1\right )}^{2} c^{3} d^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*c*(2*(c*x - 1)^2*(2*(c*x + 1)*b/(c*x - 1) + b)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^2*c^3*d^3) + (c*x - 1
)^2*(8*(c*x + 1)*a/(c*x - 1) + 4*a + 4*(c*x + 1)*b/(c*x - 1) + b)/((c*x + 1)^2*c^3*d^3))

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Mupad [B]
time = 1.27, size = 81, normalized size = 1.05 \begin {gather*} \frac {c\,\left (b\,x-2\,b\,x\,\mathrm {atanh}\left (c\,x\right )\right )-b\,\mathrm {atanh}\left (c\,x\right )+c^2\,\left (4\,a\,x^2+2\,b\,x^2+3\,b\,x^2\,\mathrm {atanh}\left (c\,x\right )\right )}{8\,c^4\,d^3\,x^2+16\,c^3\,d^3\,x+8\,c^2\,d^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*(b*x - 2*b*x*atanh(c*x)) - b*atanh(c*x) + c^2*(4*a*x^2 + 2*b*x^2 + 3*b*x^2*atanh(c*x)))/(8*c^2*d^3 + 16*c^3
*d^3*x + 8*c^4*d^3*x^2)

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