3.12 \(\int x (d+c d x)^2 (a+b \tanh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=129 \[ \frac {1}{4} c^2 d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{3} c d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{2} d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {17 b d^2 \log (1-c x)}{24 c^2}-\frac {b d^2 \log (c x+1)}{24 c^2}+\frac {1}{12} b c d^2 x^3+\frac {3 b d^2 x}{4 c}+\frac {1}{3} b d^2 x^2 \]

[Out]

3/4*b*d^2*x/c+1/3*b*d^2*x^2+1/12*b*c*d^2*x^3+1/2*d^2*x^2*(a+b*arctanh(c*x))+2/3*c*d^2*x^3*(a+b*arctanh(c*x))+1
/4*c^2*d^2*x^4*(a+b*arctanh(c*x))+17/24*b*d^2*ln(-c*x+1)/c^2-1/24*b*d^2*ln(c*x+1)/c^2

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {43, 5936, 12, 1802, 633, 31} \[ \frac {1}{4} c^2 d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{3} c d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{2} d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {17 b d^2 \log (1-c x)}{24 c^2}-\frac {b d^2 \log (c x+1)}{24 c^2}+\frac {1}{12} b c d^2 x^3+\frac {3 b d^2 x}{4 c}+\frac {1}{3} b d^2 x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*d^2*x)/(4*c) + (b*d^2*x^2)/3 + (b*c*d^2*x^3)/12 + (d^2*x^2*(a + b*ArcTanh[c*x]))/2 + (2*c*d^2*x^3*(a + b*
ArcTanh[c*x]))/3 + (c^2*d^2*x^4*(a + b*ArcTanh[c*x]))/4 + (17*b*d^2*Log[1 - c*x])/(24*c^2) - (b*d^2*Log[1 + c*
x])/(24*c^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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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 x (d+c d x)^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac {1}{2} d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{3} c d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} c^2 d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )-(b c) \int \frac {d^2 x^2 \left (6+8 c x+3 c^2 x^2\right )}{12 \left (1-c^2 x^2\right )} \, dx\\ &=\frac {1}{2} d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{3} c d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} c^2 d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{12} \left (b c d^2\right ) \int \frac {x^2 \left (6+8 c x+3 c^2 x^2\right )}{1-c^2 x^2} \, dx\\ &=\frac {1}{2} d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{3} c d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} c^2 d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{12} \left (b c d^2\right ) \int \left (-\frac {9}{c^2}-\frac {8 x}{c}-3 x^2+\frac {9+8 c x}{c^2 \left (1-c^2 x^2\right )}\right ) \, dx\\ &=\frac {3 b d^2 x}{4 c}+\frac {1}{3} b d^2 x^2+\frac {1}{12} b c d^2 x^3+\frac {1}{2} d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{3} c d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} c^2 d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {\left (b d^2\right ) \int \frac {9+8 c x}{1-c^2 x^2} \, dx}{12 c}\\ &=\frac {3 b d^2 x}{4 c}+\frac {1}{3} b d^2 x^2+\frac {1}{12} b c d^2 x^3+\frac {1}{2} d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{3} c d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} c^2 d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{24} \left (b d^2\right ) \int \frac {1}{-c-c^2 x} \, dx-\frac {1}{24} \left (17 b d^2\right ) \int \frac {1}{c-c^2 x} \, dx\\ &=\frac {3 b d^2 x}{4 c}+\frac {1}{3} b d^2 x^2+\frac {1}{12} b c d^2 x^3+\frac {1}{2} d^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {2}{3} c d^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} c^2 d^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {17 b d^2 \log (1-c x)}{24 c^2}-\frac {b d^2 \log (1+c x)}{24 c^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 107, normalized size = 0.83 \[ \frac {d^2 \left (6 a c^4 x^4+16 a c^3 x^3+12 a c^2 x^2+2 b c^3 x^3+8 b c^2 x^2+2 b c^2 x^2 \left (3 c^2 x^2+8 c x+6\right ) \tanh ^{-1}(c x)+18 b c x+17 b \log (1-c x)-b \log (c x+1)\right )}{24 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(18*b*c*x + 12*a*c^2*x^2 + 8*b*c^2*x^2 + 16*a*c^3*x^3 + 2*b*c^3*x^3 + 6*a*c^4*x^4 + 2*b*c^2*x^2*(6 + 8*c*
x + 3*c^2*x^2)*ArcTanh[c*x] + 17*b*Log[1 - c*x] - b*Log[1 + c*x]))/(24*c^2)

________________________________________________________________________________________

fricas [A]  time = 0.69, size = 137, normalized size = 1.06 \[ \frac {6 \, a c^{4} d^{2} x^{4} + 2 \, {\left (8 \, a + b\right )} c^{3} d^{2} x^{3} + 4 \, {\left (3 \, a + 2 \, b\right )} c^{2} d^{2} x^{2} + 18 \, b c d^{2} x - b d^{2} \log \left (c x + 1\right ) + 17 \, b d^{2} \log \left (c x - 1\right ) + {\left (3 \, b c^{4} d^{2} x^{4} + 8 \, b c^{3} d^{2} x^{3} + 6 \, b c^{2} d^{2} x^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{24 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*a*c^4*d^2*x^4 + 2*(8*a + b)*c^3*d^2*x^3 + 4*(3*a + 2*b)*c^2*d^2*x^2 + 18*b*c*d^2*x - b*d^2*log(c*x + 1
) + 17*b*d^2*log(c*x - 1) + (3*b*c^4*d^2*x^4 + 8*b*c^3*d^2*x^3 + 6*b*c^2*d^2*x^2)*log(-(c*x + 1)/(c*x - 1)))/c
^2

________________________________________________________________________________________

giac [B]  time = 0.41, size = 425, normalized size = 3.29 \[ -\frac {1}{3} \, c {\left (\frac {2 \, b d^{2} \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{3}} - \frac {2 \, {\left (\frac {6 \, {\left (c x + 1\right )}^{3} b d^{2}}{{\left (c x - 1\right )}^{3}} - \frac {6 \, {\left (c x + 1\right )}^{2} b d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )} b d^{2}}{c x - 1} - b d^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4} c^{3}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{3}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{3}}{c x - 1} + c^{3}} - \frac {2 \, b d^{2} \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{3}} - \frac {\frac {24 \, {\left (c x + 1\right )}^{3} a d^{2}}{{\left (c x - 1\right )}^{3}} - \frac {24 \, {\left (c x + 1\right )}^{2} a d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {16 \, {\left (c x + 1\right )} a d^{2}}{c x - 1} - 4 \, a d^{2} + \frac {10 \, {\left (c x + 1\right )}^{3} b d^{2}}{{\left (c x - 1\right )}^{3}} - \frac {23 \, {\left (c x + 1\right )}^{2} b d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {18 \, {\left (c x + 1\right )} b d^{2}}{c x - 1} - 5 \, b d^{2}}{\frac {{\left (c x + 1\right )}^{4} c^{3}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{3}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{3}}{c x - 1} + c^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*c*(2*b*d^2*log(-(c*x + 1)/(c*x - 1) + 1)/c^3 - 2*(6*(c*x + 1)^3*b*d^2/(c*x - 1)^3 - 6*(c*x + 1)^2*b*d^2/(
c*x - 1)^2 + 4*(c*x + 1)*b*d^2/(c*x - 1) - b*d^2)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^4*c^3/(c*x - 1)^4 - 4*(
c*x + 1)^3*c^3/(c*x - 1)^3 + 6*(c*x + 1)^2*c^3/(c*x - 1)^2 - 4*(c*x + 1)*c^3/(c*x - 1) + c^3) - 2*b*d^2*log(-(
c*x + 1)/(c*x - 1))/c^3 - (24*(c*x + 1)^3*a*d^2/(c*x - 1)^3 - 24*(c*x + 1)^2*a*d^2/(c*x - 1)^2 + 16*(c*x + 1)*
a*d^2/(c*x - 1) - 4*a*d^2 + 10*(c*x + 1)^3*b*d^2/(c*x - 1)^3 - 23*(c*x + 1)^2*b*d^2/(c*x - 1)^2 + 18*(c*x + 1)
*b*d^2/(c*x - 1) - 5*b*d^2)/((c*x + 1)^4*c^3/(c*x - 1)^4 - 4*(c*x + 1)^3*c^3/(c*x - 1)^3 + 6*(c*x + 1)^2*c^3/(
c*x - 1)^2 - 4*(c*x + 1)*c^3/(c*x - 1) + c^3))

________________________________________________________________________________________

maple [A]  time = 0.03, size = 135, normalized size = 1.05 \[ \frac {c^{2} d^{2} a \,x^{4}}{4}+\frac {2 c \,d^{2} a \,x^{3}}{3}+\frac {d^{2} a \,x^{2}}{2}+\frac {c^{2} d^{2} b \arctanh \left (c x \right ) x^{4}}{4}+\frac {2 c \,d^{2} b \arctanh \left (c x \right ) x^{3}}{3}+\frac {d^{2} b \arctanh \left (c x \right ) x^{2}}{2}+\frac {b c \,d^{2} x^{3}}{12}+\frac {b \,d^{2} x^{2}}{3}+\frac {3 b \,d^{2} x}{4 c}+\frac {17 d^{2} b \ln \left (c x -1\right )}{24 c^{2}}-\frac {b \,d^{2} \ln \left (c x +1\right )}{24 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*c^2*d^2*a*x^4+2/3*c*d^2*a*x^3+1/2*d^2*a*x^2+1/4*c^2*d^2*b*arctanh(c*x)*x^4+2/3*c*d^2*b*arctanh(c*x)*x^3+1/
2*d^2*b*arctanh(c*x)*x^2+1/12*b*c*d^2*x^3+1/3*b*d^2*x^2+3/4*b*d^2*x/c+17/24/c^2*d^2*b*ln(c*x-1)-1/24*b*d^2*ln(
c*x+1)/c^2

________________________________________________________________________________________

maxima [A]  time = 0.31, size = 179, normalized size = 1.39 \[ \frac {1}{4} \, a c^{2} d^{2} x^{4} + \frac {2}{3} \, a c d^{2} x^{3} + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c^{2} d^{2} + \frac {1}{3} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b c d^{2} + \frac {1}{2} \, a d^{2} x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b d^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*c^2*d^2*x^4 + 2/3*a*c*d^2*x^3 + 1/24*(6*x^4*arctanh(c*x) + c*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5
+ 3*log(c*x - 1)/c^5))*b*c^2*d^2 + 1/3*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b*c*d^2 + 1/
2*a*d^2*x^2 + 1/4*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*b*d^2

________________________________________________________________________________________

mupad [B]  time = 0.97, size = 122, normalized size = 0.95 \[ \frac {d^2\,\left (6\,a\,x^2+4\,b\,x^2+6\,b\,x^2\,\mathrm {atanh}\left (c\,x\right )\right )}{12}-\frac {\frac {d^2\,\left (9\,b\,\mathrm {atanh}\left (c\,x\right )-4\,b\,\ln \left (c^2\,x^2-1\right )\right )}{12}-\frac {3\,b\,c\,d^2\,x}{4}}{c^2}+\frac {c^2\,d^2\,\left (3\,a\,x^4+3\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{12}+\frac {c\,d^2\,\left (8\,a\,x^3+b\,x^3+8\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d^2*(6*a*x^2 + 4*b*x^2 + 6*b*x^2*atanh(c*x)))/12 - ((d^2*(9*b*atanh(c*x) - 4*b*log(c^2*x^2 - 1)))/12 - (3*b*c
*d^2*x)/4)/c^2 + (c^2*d^2*(3*a*x^4 + 3*b*x^4*atanh(c*x)))/12 + (c*d^2*(8*a*x^3 + b*x^3 + 8*b*x^3*atanh(c*x)))/
12

________________________________________________________________________________________

sympy [A]  time = 1.33, size = 167, normalized size = 1.29 \[ \begin {cases} \frac {a c^{2} d^{2} x^{4}}{4} + \frac {2 a c d^{2} x^{3}}{3} + \frac {a d^{2} x^{2}}{2} + \frac {b c^{2} d^{2} x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + \frac {2 b c d^{2} x^{3} \operatorname {atanh}{\left (c x \right )}}{3} + \frac {b c d^{2} x^{3}}{12} + \frac {b d^{2} x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {b d^{2} x^{2}}{3} + \frac {3 b d^{2} x}{4 c} + \frac {2 b d^{2} \log {\left (x - \frac {1}{c} \right )}}{3 c^{2}} - \frac {b d^{2} \operatorname {atanh}{\left (c x \right )}}{12 c^{2}} & \text {for}\: c \neq 0 \\\frac {a d^{2} x^{2}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*d*x+d)**2*(a+b*atanh(c*x)),x)

[Out]

Piecewise((a*c**2*d**2*x**4/4 + 2*a*c*d**2*x**3/3 + a*d**2*x**2/2 + b*c**2*d**2*x**4*atanh(c*x)/4 + 2*b*c*d**2
*x**3*atanh(c*x)/3 + b*c*d**2*x**3/12 + b*d**2*x**2*atanh(c*x)/2 + b*d**2*x**2/3 + 3*b*d**2*x/(4*c) + 2*b*d**2
*log(x - 1/c)/(3*c**2) - b*d**2*atanh(c*x)/(12*c**2), Ne(c, 0)), (a*d**2*x**2/2, True))

________________________________________________________________________________________