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

Optimal. Leaf size=108 \[ \frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {9 b d \log (1-c x)}{40 c^4}-\frac {b d \log (c x+1)}{40 c^4}+\frac {b d x}{4 c^3}+\frac {b d x^2}{10 c^2}+\frac {b d x^3}{12 c}+\frac {1}{20} b d x^4 \]

[Out]

1/4*b*d*x/c^3+1/10*b*d*x^2/c^2+1/12*b*d*x^3/c+1/20*b*d*x^4+1/4*d*x^4*(a+b*arctanh(c*x))+1/5*c*d*x^5*(a+b*arcta
nh(c*x))+9/40*b*d*ln(-c*x+1)/c^4-1/40*b*d*ln(c*x+1)/c^4

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Rubi [A]  time = 0.10, antiderivative size = 108, 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} \[ \frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {b d x^2}{10 c^2}+\frac {b d x}{4 c^3}+\frac {9 b d \log (1-c x)}{40 c^4}-\frac {b d \log (c x+1)}{40 c^4}+\frac {b d x^3}{12 c}+\frac {1}{20} b d x^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*x)/(4*c^3) + (b*d*x^2)/(10*c^2) + (b*d*x^3)/(12*c) + (b*d*x^4)/20 + (d*x^4*(a + b*ArcTanh[c*x]))/4 + (c*d
*x^5*(a + b*ArcTanh[c*x]))/5 + (9*b*d*Log[1 - c*x])/(40*c^4) - (b*d*Log[1 + c*x])/(40*c^4)

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^3 (d+c d x) \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )-(b c) \int \frac {d x^4 (5+4 c x)}{20 \left (1-c^2 x^2\right )} \, dx\\ &=\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{20} (b c d) \int \frac {x^4 (5+4 c x)}{1-c^2 x^2} \, dx\\ &=\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{20} (b c d) \int \left (-\frac {5}{c^4}-\frac {4 x}{c^3}-\frac {5 x^2}{c^2}-\frac {4 x^3}{c}+\frac {5+4 c x}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx\\ &=\frac {b d x}{4 c^3}+\frac {b d x^2}{10 c^2}+\frac {b d x^3}{12 c}+\frac {1}{20} b d x^4+\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )-\frac {(b d) \int \frac {5+4 c x}{1-c^2 x^2} \, dx}{20 c^3}\\ &=\frac {b d x}{4 c^3}+\frac {b d x^2}{10 c^2}+\frac {b d x^3}{12 c}+\frac {1}{20} b d x^4+\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac {(b d) \int \frac {1}{-c-c^2 x} \, dx}{40 c^2}-\frac {(9 b d) \int \frac {1}{c-c^2 x} \, dx}{40 c^2}\\ &=\frac {b d x}{4 c^3}+\frac {b d x^2}{10 c^2}+\frac {b d x^3}{12 c}+\frac {1}{20} b d x^4+\frac {1}{4} d x^4 \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{5} c d x^5 \left (a+b \tanh ^{-1}(c x)\right )+\frac {9 b d \log (1-c x)}{40 c^4}-\frac {b d \log (1+c x)}{40 c^4}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 97, normalized size = 0.90 \[ \frac {d \left (24 a c^5 x^5+30 a c^4 x^4+6 b c^4 x^4+6 b c^4 x^4 (4 c x+5) \tanh ^{-1}(c x)+10 b c^3 x^3+12 b c^2 x^2+30 b c x+27 b \log (1-c x)-3 b \log (c x+1)\right )}{120 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(30*b*c*x + 12*b*c^2*x^2 + 10*b*c^3*x^3 + 30*a*c^4*x^4 + 6*b*c^4*x^4 + 24*a*c^5*x^5 + 6*b*c^4*x^4*(5 + 4*c*
x)*ArcTanh[c*x] + 27*b*Log[1 - c*x] - 3*b*Log[1 + c*x]))/(120*c^4)

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fricas [A]  time = 0.48, size = 114, normalized size = 1.06 \[ \frac {24 \, a c^{5} d x^{5} + 6 \, {\left (5 \, a + b\right )} c^{4} d x^{4} + 10 \, b c^{3} d x^{3} + 12 \, b c^{2} d x^{2} + 30 \, b c d x - 3 \, b d \log \left (c x + 1\right ) + 27 \, b d \log \left (c x - 1\right ) + 3 \, {\left (4 \, b c^{5} d x^{5} + 5 \, b c^{4} d x^{4}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{120 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(24*a*c^5*d*x^5 + 6*(5*a + b)*c^4*d*x^4 + 10*b*c^3*d*x^3 + 12*b*c^2*d*x^2 + 30*b*c*d*x - 3*b*d*log(c*x +
1) + 27*b*d*log(c*x - 1) + 3*(4*b*c^5*d*x^5 + 5*b*c^4*d*x^4)*log(-(c*x + 1)/(c*x - 1)))/c^4

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giac [B]  time = 0.25, size = 491, normalized size = 4.55 \[ \frac {1}{15} \, c {\left (\frac {3 \, {\left (\frac {10 \, {\left (c x + 1\right )}^{4} b d}{{\left (c x - 1\right )}^{4}} - \frac {5 \, {\left (c x + 1\right )}^{3} b d}{{\left (c x - 1\right )}^{3}} + \frac {15 \, {\left (c x + 1\right )}^{2} b d}{{\left (c x - 1\right )}^{2}} - \frac {5 \, {\left (c x + 1\right )} b d}{c x - 1} + b d\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{5} c^{5}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{5}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{5}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{5}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{5}}{c x - 1} - c^{5}} + \frac {\frac {60 \, {\left (c x + 1\right )}^{4} a d}{{\left (c x - 1\right )}^{4}} - \frac {30 \, {\left (c x + 1\right )}^{3} a d}{{\left (c x - 1\right )}^{3}} + \frac {90 \, {\left (c x + 1\right )}^{2} a d}{{\left (c x - 1\right )}^{2}} - \frac {30 \, {\left (c x + 1\right )} a d}{c x - 1} + 6 \, a d + \frac {27 \, {\left (c x + 1\right )}^{4} b d}{{\left (c x - 1\right )}^{4}} - \frac {69 \, {\left (c x + 1\right )}^{3} b d}{{\left (c x - 1\right )}^{3}} + \frac {79 \, {\left (c x + 1\right )}^{2} b d}{{\left (c x - 1\right )}^{2}} - \frac {47 \, {\left (c x + 1\right )} b d}{c x - 1} + 10 \, b d}{\frac {{\left (c x + 1\right )}^{5} c^{5}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{5}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{5}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{5}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{5}}{c x - 1} - c^{5}} - \frac {3 \, b d \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{5}} + \frac {3 \, b d \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{5}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*c*(3*(10*(c*x + 1)^4*b*d/(c*x - 1)^4 - 5*(c*x + 1)^3*b*d/(c*x - 1)^3 + 15*(c*x + 1)^2*b*d/(c*x - 1)^2 - 5
*(c*x + 1)*b*d/(c*x - 1) + b*d)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^5*c^5/(c*x - 1)^5 - 5*(c*x + 1)^4*c^5/(c*
x - 1)^4 + 10*(c*x + 1)^3*c^5/(c*x - 1)^3 - 10*(c*x + 1)^2*c^5/(c*x - 1)^2 + 5*(c*x + 1)*c^5/(c*x - 1) - c^5)
+ (60*(c*x + 1)^4*a*d/(c*x - 1)^4 - 30*(c*x + 1)^3*a*d/(c*x - 1)^3 + 90*(c*x + 1)^2*a*d/(c*x - 1)^2 - 30*(c*x
+ 1)*a*d/(c*x - 1) + 6*a*d + 27*(c*x + 1)^4*b*d/(c*x - 1)^4 - 69*(c*x + 1)^3*b*d/(c*x - 1)^3 + 79*(c*x + 1)^2*
b*d/(c*x - 1)^2 - 47*(c*x + 1)*b*d/(c*x - 1) + 10*b*d)/((c*x + 1)^5*c^5/(c*x - 1)^5 - 5*(c*x + 1)^4*c^5/(c*x -
1)^4 + 10*(c*x + 1)^3*c^5/(c*x - 1)^3 - 10*(c*x + 1)^2*c^5/(c*x - 1)^2 + 5*(c*x + 1)*c^5/(c*x - 1) - c^5) - 3
*b*d*log(-(c*x + 1)/(c*x - 1) + 1)/c^5 + 3*b*d*log(-(c*x + 1)/(c*x - 1))/c^5)

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maple [A]  time = 0.03, size = 101, normalized size = 0.94 \[ \frac {c d a \,x^{5}}{5}+\frac {d a \,x^{4}}{4}+\frac {c d b \arctanh \left (c x \right ) x^{5}}{5}+\frac {d b \arctanh \left (c x \right ) x^{4}}{4}+\frac {b d \,x^{4}}{20}+\frac {b d \,x^{3}}{12 c}+\frac {b d \,x^{2}}{10 c^{2}}+\frac {b d x}{4 c^{3}}+\frac {9 d b \ln \left (c x -1\right )}{40 c^{4}}-\frac {b d \ln \left (c x +1\right )}{40 c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*c*d*a*x^5+1/4*d*a*x^4+1/5*c*d*b*arctanh(c*x)*x^5+1/4*d*b*arctanh(c*x)*x^4+1/20*b*d*x^4+1/12*b*d*x^3/c+1/10
*b*d*x^2/c^2+1/4*b*d*x/c^3+9/40/c^4*d*b*ln(c*x-1)-1/40*b*d*ln(c*x+1)/c^4

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maxima [A]  time = 0.31, size = 121, normalized size = 1.12 \[ \frac {1}{5} \, a c d x^{5} + \frac {1}{4} \, a d x^{4} + \frac {1}{20} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b c d + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*c*d*x^5 + 1/4*a*d*x^4 + 1/20*(4*x^5*arctanh(c*x) + c*((c^2*x^4 + 2*x^2)/c^4 + 2*log(c^2*x^2 - 1)/c^6))*b
*c*d + 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*d

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mupad [B]  time = 0.99, size = 103, normalized size = 0.95 \[ \frac {\frac {b\,c\,d\,x}{4}-\frac {d\,\left (15\,b\,\mathrm {atanh}\left (c\,x\right )-6\,b\,\ln \left (c^2\,x^2-1\right )\right )}{60}+\frac {b\,c^2\,d\,x^2}{10}+\frac {b\,c^3\,d\,x^3}{12}}{c^4}+\frac {d\,\left (15\,a\,x^4+3\,b\,x^4+15\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{60}+\frac {c\,d\,\left (12\,a\,x^5+12\,b\,x^5\,\mathrm {atanh}\left (c\,x\right )\right )}{60} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*c*d*x)/4 - (d*(15*b*atanh(c*x) - 6*b*log(c^2*x^2 - 1)))/60 + (b*c^2*d*x^2)/10 + (b*c^3*d*x^3)/12)/c^4 + (d
*(15*a*x^4 + 3*b*x^4 + 15*b*x^4*atanh(c*x)))/60 + (c*d*(12*a*x^5 + 12*b*x^5*atanh(c*x)))/60

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sympy [A]  time = 1.53, size = 124, normalized size = 1.15 \[ \begin {cases} \frac {a c d x^{5}}{5} + \frac {a d x^{4}}{4} + \frac {b c d x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + \frac {b d x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + \frac {b d x^{4}}{20} + \frac {b d x^{3}}{12 c} + \frac {b d x^{2}}{10 c^{2}} + \frac {b d x}{4 c^{3}} + \frac {b d \log {\left (x - \frac {1}{c} \right )}}{5 c^{4}} - \frac {b d \operatorname {atanh}{\left (c x \right )}}{20 c^{4}} & \text {for}\: c \neq 0 \\\frac {a d x^{4}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c*d*x**5/5 + a*d*x**4/4 + b*c*d*x**5*atanh(c*x)/5 + b*d*x**4*atanh(c*x)/4 + b*d*x**4/20 + b*d*x**
3/(12*c) + b*d*x**2/(10*c**2) + b*d*x/(4*c**3) + b*d*log(x - 1/c)/(5*c**4) - b*d*atanh(c*x)/(20*c**4), Ne(c, 0
)), (a*d*x**4/4, True))

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