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

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

[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

________________________________________________________________________________________

Rubi [A]  time = 0.08, 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, 5936, 12, 88, 207} \[ \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 d^3 (c x+1)^2}-\frac {3 b}{8 c^2 d^3 (c x+1)}+\frac {b}{8 c^2 d^3 (c x+1)^2}-\frac {b \tanh ^{-1}(c x)}{8 c^2 d^3} \]

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 88

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 207

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

Rule 5936

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*}

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 99, normalized size = 1.29 \[ -\frac {16 a c x+8 a-3 b c^2 x^2 \log (c x+1)+6 b c x-6 b c x \log (c x+1)+3 b (c x+1)^2 \log (1-c x)-3 b \log (c x+1)+8 (2 b c x+b) \tanh ^{-1}(c x)+4 b}{16 c^2 d^3 (c x+1)^2} \]

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)

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 84, normalized size = 1.09 \[ -\frac {2 \, {\left (8 \, a + 3 \, b\right )} c x - {\left (3 \, b c^{2} x^{2} - 2 \, b c x - b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 8 \, a + 4 \, b}{16 \, {\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.20, size = 114, normalized size = 1.48 \[ \frac {1}{32} \, c {\left (\frac {2 \, {\left (c x - 1\right )}^{2} {\left (\frac {2 \, {\left (c x + 1\right )} b}{c x - 1} + b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\left (c x + 1\right )}^{2} c^{3} d^{3}} + \frac {{\left (c x - 1\right )}^{2} {\left (\frac {8 \, {\left (c x + 1\right )} a}{c x - 1} + 4 \, a + \frac {4 \, {\left (c x + 1\right )} b}{c x - 1} + b\right )}}{{\left (c x + 1\right )}^{2} c^{3} d^{3}}\right )} \]

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))

________________________________________________________________________________________

maple [A]  time = 0.04, size = 136, normalized size = 1.77 \[ \frac {a}{2 c^{2} d^{3} \left (c x +1\right )^{2}}-\frac {a}{c^{2} d^{3} \left (c x +1\right )}+\frac {b \arctanh \left (c x \right )}{2 c^{2} d^{3} \left (c x +1\right )^{2}}-\frac {b \arctanh \left (c x \right )}{c^{2} d^{3} \left (c x +1\right )}-\frac {3 b \ln \left (c x -1\right )}{16 c^{2} d^{3}}+\frac {b}{8 c^{2} d^{3} \left (c x +1\right )^{2}}-\frac {3 b}{8 c^{2} d^{3} \left (c x +1\right )}+\frac {3 b \ln \left (c x +1\right )}{16 c^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.33, size = 152, normalized size = 1.97 \[ -\frac {1}{16} \, {\left (c {\left (\frac {2 \, {\left (3 \, c x + 2\right )}}{c^{5} d^{3} x^{2} + 2 \, c^{4} d^{3} x + c^{3} d^{3}} - \frac {3 \, \log \left (c x + 1\right )}{c^{3} d^{3}} + \frac {3 \, \log \left (c x - 1\right )}{c^{3} d^{3}}\right )} + \frac {8 \, {\left (2 \, c x + 1\right )} \operatorname {artanh}\left (c x\right )}{c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}}\right )} b - \frac {{\left (2 \, c x + 1\right )} a}{2 \, {\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} \]

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

sympy [A]  time = 2.26, size = 306, normalized size = 3.97 \[ \begin {cases} - \frac {8 a c x}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {4 a}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} + \frac {3 b c^{2} x^{2} \operatorname {atanh}{\left (c x \right )}}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {2 b c x \operatorname {atanh}{\left (c x \right )}}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {3 b c x}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {b \operatorname {atanh}{\left (c x \right )}}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {2 b}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} & \text {for}\: d \neq 0 \\\tilde {\infty } \left (\frac {a x^{2}}{2} + \frac {b x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {b x}{2 c} - \frac {b \operatorname {atanh}{\left (c x \right )}}{2 c^{2}}\right ) & \text {otherwise} \end {cases} \]

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(d, 0)), (zoo*(a*x**2/2 + b*x**2*atanh(c*x)/2 + b*x/(2*c) - b*atanh(c*x)/(2*c**2)), True))

________________________________________________________________________________________