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

Optimal. Leaf size=96 \[ \frac {1}{4} c d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{3} d x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {7 b d \log (1-c x)}{24 c^3}+\frac {b d \log (c x+1)}{24 c^3}+\frac {b d x}{4 c^2}+\frac {b d x^2}{6 c}+\frac {1}{12} b d x^3 \]

[Out]

1/4*b*d*x/c^2+1/6*b*d*x^2/c+1/12*b*d*x^3+1/3*d*x^3*(a+b*arctanh(c*x))+1/4*c*d*x^4*(a+b*arctanh(c*x))+7/24*b*d*
ln(-c*x+1)/c^3+1/24*b*d*ln(c*x+1)/c^3

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Rubi [A]  time = 0.10, antiderivative size = 96, 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, 801, 633, 31} \[ \frac {1}{4} c d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{3} d x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {b d x}{4 c^2}+\frac {7 b d \log (1-c x)}{24 c^3}+\frac {b d \log (c x+1)}{24 c^3}+\frac {b d x^2}{6 c}+\frac {1}{12} b d x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*x)/(4*c^2) + (b*d*x^2)/(6*c) + (b*d*x^3)/12 + (d*x^3*(a + b*ArcTanh[c*x]))/3 + (c*d*x^4*(a + b*ArcTanh[c*
x]))/4 + (7*b*d*Log[1 - c*x])/(24*c^3) + (b*d*Log[1 + c*x])/(24*c^3)

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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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^2 (d+c d x) \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac {1}{3} d x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} c d x^4 \left (a+b \tanh ^{-1}(c x)\right )-(b c) \int \frac {d x^3 (4+3 c x)}{12 \left (1-c^2 x^2\right )} \, dx\\ &=\frac {1}{3} d x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} c d x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{12} (b c d) \int \frac {x^3 (4+3 c x)}{1-c^2 x^2} \, dx\\ &=\frac {1}{3} d x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} c d x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{12} (b c d) \int \left (-\frac {3}{c^3}-\frac {4 x}{c^2}-\frac {3 x^2}{c}+\frac {3+4 c x}{c^3 \left (1-c^2 x^2\right )}\right ) \, dx\\ &=\frac {b d x}{4 c^2}+\frac {b d x^2}{6 c}+\frac {1}{12} b d x^3+\frac {1}{3} d x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} c d x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {(b d) \int \frac {3+4 c x}{1-c^2 x^2} \, dx}{12 c^2}\\ &=\frac {b d x}{4 c^2}+\frac {b d x^2}{6 c}+\frac {1}{12} b d x^3+\frac {1}{3} d x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} c d x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {(b d) \int \frac {1}{-c-c^2 x} \, dx}{24 c}-\frac {(7 b d) \int \frac {1}{c-c^2 x} \, dx}{24 c}\\ &=\frac {b d x}{4 c^2}+\frac {b d x^2}{6 c}+\frac {1}{12} b d x^3+\frac {1}{3} d x^3 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} c d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {7 b d \log (1-c x)}{24 c^3}+\frac {b d \log (1+c x)}{24 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 87, normalized size = 0.91 \[ \frac {d \left (6 a c^4 x^4+8 a c^3 x^3+2 b c^3 x^3+2 b c^3 x^3 (3 c x+4) \tanh ^{-1}(c x)+4 b c^2 x^2+6 b c x+7 b \log (1-c x)+b \log (c x+1)\right )}{24 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(6*b*c*x + 4*b*c^2*x^2 + 8*a*c^3*x^3 + 2*b*c^3*x^3 + 6*a*c^4*x^4 + 2*b*c^3*x^3*(4 + 3*c*x)*ArcTanh[c*x] + 7
*b*Log[1 - c*x] + b*Log[1 + c*x]))/(24*c^3)

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fricas [A]  time = 0.66, size = 102, normalized size = 1.06 \[ \frac {6 \, a c^{4} d x^{4} + 2 \, {\left (4 \, a + b\right )} c^{3} d x^{3} + 4 \, b c^{2} d x^{2} + 6 \, b c d x + b d \log \left (c x + 1\right ) + 7 \, b d \log \left (c x - 1\right ) + {\left (3 \, b c^{4} d x^{4} + 4 \, b c^{3} d x^{3}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{24 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*a*c^4*d*x^4 + 2*(4*a + b)*c^3*d*x^3 + 4*b*c^2*d*x^2 + 6*b*c*d*x + b*d*log(c*x + 1) + 7*b*d*log(c*x - 1
) + (3*b*c^4*d*x^4 + 4*b*c^3*d*x^3)*log(-(c*x + 1)/(c*x - 1)))/c^3

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giac [B]  time = 0.20, size = 394, normalized size = 4.10 \[ \frac {1}{3} \, c {\left (\frac {{\left (\frac {6 \, {\left (c x + 1\right )}^{3} b d}{{\left (c x - 1\right )}^{3}} - \frac {3 \, {\left (c x + 1\right )}^{2} b d}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )} b d}{c x - 1} - b d\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4} c^{4}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{4}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{4}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{4}}{c x - 1} + c^{4}} + \frac {\frac {12 \, {\left (c x + 1\right )}^{3} a d}{{\left (c x - 1\right )}^{3}} - \frac {6 \, {\left (c x + 1\right )}^{2} a d}{{\left (c x - 1\right )}^{2}} + \frac {8 \, {\left (c x + 1\right )} a d}{c x - 1} - 2 \, a d + \frac {5 \, {\left (c x + 1\right )}^{3} b d}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} b d}{{\left (c x - 1\right )}^{2}} + \frac {7 \, {\left (c x + 1\right )} b d}{c x - 1} - 2 \, b d}{\frac {{\left (c x + 1\right )}^{4} c^{4}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{4}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{4}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{4}}{c x - 1} + c^{4}} - \frac {b d \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{4}} + \frac {b d \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*c*((6*(c*x + 1)^3*b*d/(c*x - 1)^3 - 3*(c*x + 1)^2*b*d/(c*x - 1)^2 + 4*(c*x + 1)*b*d/(c*x - 1) - b*d)*log(-
(c*x + 1)/(c*x - 1))/((c*x + 1)^4*c^4/(c*x - 1)^4 - 4*(c*x + 1)^3*c^4/(c*x - 1)^3 + 6*(c*x + 1)^2*c^4/(c*x - 1
)^2 - 4*(c*x + 1)*c^4/(c*x - 1) + c^4) + (12*(c*x + 1)^3*a*d/(c*x - 1)^3 - 6*(c*x + 1)^2*a*d/(c*x - 1)^2 + 8*(
c*x + 1)*a*d/(c*x - 1) - 2*a*d + 5*(c*x + 1)^3*b*d/(c*x - 1)^3 - 10*(c*x + 1)^2*b*d/(c*x - 1)^2 + 7*(c*x + 1)*
b*d/(c*x - 1) - 2*b*d)/((c*x + 1)^4*c^4/(c*x - 1)^4 - 4*(c*x + 1)^3*c^4/(c*x - 1)^3 + 6*(c*x + 1)^2*c^4/(c*x -
1)^2 - 4*(c*x + 1)*c^4/(c*x - 1) + c^4) - b*d*log(-(c*x + 1)/(c*x - 1) + 1)/c^4 + b*d*log(-(c*x + 1)/(c*x - 1
))/c^4)

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maple [A]  time = 0.03, size = 91, normalized size = 0.95 \[ \frac {c d a \,x^{4}}{4}+\frac {d a \,x^{3}}{3}+\frac {c d b \arctanh \left (c x \right ) x^{4}}{4}+\frac {d b \arctanh \left (c x \right ) x^{3}}{3}+\frac {b d \,x^{3}}{12}+\frac {b d \,x^{2}}{6 c}+\frac {b d x}{4 c^{2}}+\frac {7 d b \ln \left (c x -1\right )}{24 c^{3}}+\frac {b d \ln \left (c x +1\right )}{24 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*c*d*a*x^4+1/3*d*a*x^3+1/4*c*d*b*arctanh(c*x)*x^4+1/3*d*b*arctanh(c*x)*x^3+1/12*b*d*x^3+1/6*b*d*x^2/c+1/4*b
*d*x/c^2+7/24/c^3*d*b*ln(c*x-1)+1/24*b*d*ln(c*x+1)/c^3

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maxima [A]  time = 0.32, size = 110, normalized size = 1.15 \[ \frac {1}{4} \, a c d x^{4} + \frac {1}{3} \, a d x^{3} + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c d + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.94, size = 92, normalized size = 0.96 \[ \frac {\frac {b\,c\,d\,x}{4}-\frac {d\,\left (3\,b\,\mathrm {atanh}\left (c\,x\right )-2\,b\,\ln \left (c^2\,x^2-1\right )\right )}{12}+\frac {b\,c^2\,d\,x^2}{6}}{c^3}+\frac {d\,\left (4\,a\,x^3+b\,x^3+4\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{12}+\frac {c\,d\,\left (3\,a\,x^4+3\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.20, size = 112, normalized size = 1.17 \[ \begin {cases} \frac {a c d x^{4}}{4} + \frac {a d x^{3}}{3} + \frac {b c d x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + \frac {b d x^{3} \operatorname {atanh}{\left (c x \right )}}{3} + \frac {b d x^{3}}{12} + \frac {b d x^{2}}{6 c} + \frac {b d x}{4 c^{2}} + \frac {b d \log {\left (x - \frac {1}{c} \right )}}{3 c^{3}} + \frac {b d \operatorname {atanh}{\left (c x \right )}}{12 c^{3}} & \text {for}\: c \neq 0 \\\frac {a d x^{3}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c*d*x**4/4 + a*d*x**3/3 + b*c*d*x**4*atanh(c*x)/4 + b*d*x**3*atanh(c*x)/3 + b*d*x**3/12 + b*d*x**
2/(6*c) + b*d*x/(4*c**2) + b*d*log(x - 1/c)/(3*c**3) + b*d*atanh(c*x)/(12*c**3), Ne(c, 0)), (a*d*x**3/3, True)
)

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