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

Optimal. Leaf size=135 \[ \frac {d^3 (c x+1)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}-\frac {d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac {b d^3 (c x+1)^4}{20 c^2}+\frac {b d^3 (c x+1)^3}{20 c^2}+\frac {3 b d^3 (c x+1)^2}{20 c^2}+\frac {6 b d^3 \log (1-c x)}{5 c^2}+\frac {3 b d^3 x}{5 c} \]

[Out]

3/5*b*d^3*x/c+3/20*b*d^3*(c*x+1)^2/c^2+1/20*b*d^3*(c*x+1)^3/c^2+1/20*b*d^3*(c*x+1)^4/c^2-1/4*d^3*(c*x+1)^4*(a+
b*arctanh(c*x))/c^2+1/5*d^3*(c*x+1)^5*(a+b*arctanh(c*x))/c^2+6/5*b*d^3*ln(-c*x+1)/c^2

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Rubi [A]  time = 0.10, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {43, 5936, 12, 77} \[ \frac {d^3 (c x+1)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}-\frac {d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac {b d^3 (c x+1)^4}{20 c^2}+\frac {b d^3 (c x+1)^3}{20 c^2}+\frac {3 b d^3 (c x+1)^2}{20 c^2}+\frac {6 b d^3 \log (1-c x)}{5 c^2}+\frac {3 b d^3 x}{5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*d^3*x)/(5*c) + (3*b*d^3*(1 + c*x)^2)/(20*c^2) + (b*d^3*(1 + c*x)^3)/(20*c^2) + (b*d^3*(1 + c*x)^4)/(20*c^
2) - (d^3*(1 + c*x)^4*(a + b*ArcTanh[c*x]))/(4*c^2) + (d^3*(1 + c*x)^5*(a + b*ArcTanh[c*x]))/(5*c^2) + (6*b*d^
3*Log[1 - c*x])/(5*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 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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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)^3 \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=-\frac {d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac {d^3 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}-(b c) \int \frac {(-1+4 c x) (d+c d x)^3}{20 c^2 (1-c x)} \, dx\\ &=-\frac {d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac {d^3 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}-\frac {b \int \frac {(-1+4 c x) (d+c d x)^3}{1-c x} \, dx}{20 c}\\ &=-\frac {d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac {d^3 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}-\frac {b \int \left (-12 d^3-\frac {24 d^3}{-1+c x}-6 d^2 (d+c d x)-3 d (d+c d x)^2-4 (d+c d x)^3\right ) \, dx}{20 c}\\ &=\frac {3 b d^3 x}{5 c}+\frac {3 b d^3 (1+c x)^2}{20 c^2}+\frac {b d^3 (1+c x)^3}{20 c^2}+\frac {b d^3 (1+c x)^4}{20 c^2}-\frac {d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac {d^3 (1+c x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 c^2}+\frac {6 b d^3 \log (1-c x)}{5 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 133, normalized size = 0.99 \[ \frac {d^3 \left (8 a c^5 x^5+30 a c^4 x^4+40 a c^3 x^3+20 a c^2 x^2+2 b c^4 x^4+10 b c^3 x^3+24 b c^2 x^2+2 b c^2 x^2 \left (4 c^3 x^3+15 c^2 x^2+20 c x+10\right ) \tanh ^{-1}(c x)+50 b c x+49 b \log (1-c x)-b \log (c x+1)\right )}{40 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*(50*b*c*x + 20*a*c^2*x^2 + 24*b*c^2*x^2 + 40*a*c^3*x^3 + 10*b*c^3*x^3 + 30*a*c^4*x^4 + 2*b*c^4*x^4 + 8*a*
c^5*x^5 + 2*b*c^2*x^2*(10 + 20*c*x + 15*c^2*x^2 + 4*c^3*x^3)*ArcTanh[c*x] + 49*b*Log[1 - c*x] - b*Log[1 + c*x]
))/(40*c^2)

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fricas [A]  time = 0.49, size = 165, normalized size = 1.22 \[ \frac {8 \, a c^{5} d^{3} x^{5} + 2 \, {\left (15 \, a + b\right )} c^{4} d^{3} x^{4} + 10 \, {\left (4 \, a + b\right )} c^{3} d^{3} x^{3} + 4 \, {\left (5 \, a + 6 \, b\right )} c^{2} d^{3} x^{2} + 50 \, b c d^{3} x - b d^{3} \log \left (c x + 1\right ) + 49 \, b d^{3} \log \left (c x - 1\right ) + {\left (4 \, b c^{5} d^{3} x^{5} + 15 \, b c^{4} d^{3} x^{4} + 20 \, b c^{3} d^{3} x^{3} + 10 \, b c^{2} d^{3} x^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{40 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(8*a*c^5*d^3*x^5 + 2*(15*a + b)*c^4*d^3*x^4 + 10*(4*a + b)*c^3*d^3*x^3 + 4*(5*a + 6*b)*c^2*d^3*x^2 + 50*b
*c*d^3*x - b*d^3*log(c*x + 1) + 49*b*d^3*log(c*x - 1) + (4*b*c^5*d^3*x^5 + 15*b*c^4*d^3*x^4 + 20*b*c^3*d^3*x^3
+ 10*b*c^2*d^3*x^2)*log(-(c*x + 1)/(c*x - 1)))/c^2

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giac [B]  time = 0.52, size = 527, normalized size = 3.90 \[ -\frac {1}{5} \, {\left (\frac {6 \, b d^{3} \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{3}} - \frac {6 \, b d^{3} \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{3}} - \frac {2 \, {\left (\frac {20 \, {\left (c x + 1\right )}^{4} b d^{3}}{{\left (c x - 1\right )}^{4}} - \frac {30 \, {\left (c x + 1\right )}^{3} b d^{3}}{{\left (c x - 1\right )}^{3}} + \frac {30 \, {\left (c x + 1\right )}^{2} b d^{3}}{{\left (c x - 1\right )}^{2}} - \frac {15 \, {\left (c x + 1\right )} b d^{3}}{c x - 1} + 3 \, b d^{3}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{5} c^{3}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{3}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{3}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{3}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{3}}{c x - 1} - c^{3}} - \frac {\frac {80 \, {\left (c x + 1\right )}^{4} a d^{3}}{{\left (c x - 1\right )}^{4}} - \frac {120 \, {\left (c x + 1\right )}^{3} a d^{3}}{{\left (c x - 1\right )}^{3}} + \frac {120 \, {\left (c x + 1\right )}^{2} a d^{3}}{{\left (c x - 1\right )}^{2}} - \frac {60 \, {\left (c x + 1\right )} a d^{3}}{c x - 1} + 12 \, a d^{3} + \frac {34 \, {\left (c x + 1\right )}^{4} b d^{3}}{{\left (c x - 1\right )}^{4}} - \frac {103 \, {\left (c x + 1\right )}^{3} b d^{3}}{{\left (c x - 1\right )}^{3}} + \frac {123 \, {\left (c x + 1\right )}^{2} b d^{3}}{{\left (c x - 1\right )}^{2}} - \frac {69 \, {\left (c x + 1\right )} b d^{3}}{c x - 1} + 15 \, b d^{3}}{\frac {{\left (c x + 1\right )}^{5} c^{3}}{{\left (c x - 1\right )}^{5}} - \frac {5 \, {\left (c x + 1\right )}^{4} c^{3}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3} c^{3}}{{\left (c x - 1\right )}^{3}} - \frac {10 \, {\left (c x + 1\right )}^{2} c^{3}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} c^{3}}{c x - 1} - c^{3}}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(6*b*d^3*log(-(c*x + 1)/(c*x - 1) + 1)/c^3 - 6*b*d^3*log(-(c*x + 1)/(c*x - 1))/c^3 - 2*(20*(c*x + 1)^4*b*
d^3/(c*x - 1)^4 - 30*(c*x + 1)^3*b*d^3/(c*x - 1)^3 + 30*(c*x + 1)^2*b*d^3/(c*x - 1)^2 - 15*(c*x + 1)*b*d^3/(c*
x - 1) + 3*b*d^3)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^5*c^3/(c*x - 1)^5 - 5*(c*x + 1)^4*c^3/(c*x - 1)^4 + 10*
(c*x + 1)^3*c^3/(c*x - 1)^3 - 10*(c*x + 1)^2*c^3/(c*x - 1)^2 + 5*(c*x + 1)*c^3/(c*x - 1) - c^3) - (80*(c*x + 1
)^4*a*d^3/(c*x - 1)^4 - 120*(c*x + 1)^3*a*d^3/(c*x - 1)^3 + 120*(c*x + 1)^2*a*d^3/(c*x - 1)^2 - 60*(c*x + 1)*a
*d^3/(c*x - 1) + 12*a*d^3 + 34*(c*x + 1)^4*b*d^3/(c*x - 1)^4 - 103*(c*x + 1)^3*b*d^3/(c*x - 1)^3 + 123*(c*x +
1)^2*b*d^3/(c*x - 1)^2 - 69*(c*x + 1)*b*d^3/(c*x - 1) + 15*b*d^3)/((c*x + 1)^5*c^3/(c*x - 1)^5 - 5*(c*x + 1)^4
*c^3/(c*x - 1)^4 + 10*(c*x + 1)^3*c^3/(c*x - 1)^3 - 10*(c*x + 1)^2*c^3/(c*x - 1)^2 + 5*(c*x + 1)*c^3/(c*x - 1)
- c^3))*c

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maple [A]  time = 0.03, size = 173, normalized size = 1.28 \[ \frac {c^{3} d^{3} a \,x^{5}}{5}+\frac {3 c^{2} d^{3} a \,x^{4}}{4}+c \,d^{3} a \,x^{3}+\frac {d^{3} a \,x^{2}}{2}+\frac {c^{3} d^{3} b \arctanh \left (c x \right ) x^{5}}{5}+\frac {3 c^{2} d^{3} b \arctanh \left (c x \right ) x^{4}}{4}+c \,d^{3} b \arctanh \left (c x \right ) x^{3}+\frac {d^{3} b \arctanh \left (c x \right ) x^{2}}{2}+\frac {c^{2} d^{3} b \,x^{4}}{20}+\frac {c \,d^{3} b \,x^{3}}{4}+\frac {3 d^{3} b \,x^{2}}{5}+\frac {5 b \,d^{3} x}{4 c}+\frac {49 d^{3} b \ln \left (c x -1\right )}{40 c^{2}}-\frac {d^{3} b \ln \left (c x +1\right )}{40 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*c^3*d^3*a*x^5+3/4*c^2*d^3*a*x^4+c*d^3*a*x^3+1/2*d^3*a*x^2+1/5*c^3*d^3*b*arctanh(c*x)*x^5+3/4*c^2*d^3*b*arc
tanh(c*x)*x^4+c*d^3*b*arctanh(c*x)*x^3+1/2*d^3*b*arctanh(c*x)*x^2+1/20*c^2*d^3*b*x^4+1/4*c*d^3*b*x^3+3/5*d^3*b
*x^2+5/4*b*d^3*x/c+49/40/c^2*d^3*b*ln(c*x-1)-1/40/c^2*d^3*b*ln(c*x+1)

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maxima [B]  time = 0.32, size = 244, normalized size = 1.81 \[ \frac {1}{5} \, a c^{3} d^{3} x^{5} + \frac {3}{4} \, a c^{2} d^{3} x^{4} + \frac {1}{20} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b c^{3} d^{3} + a c d^{3} x^{3} + \frac {1}{8} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b c^{2} d^{3} + \frac {1}{2} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b c d^{3} + \frac {1}{2} \, a d^{3} x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b d^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.98, size = 153, normalized size = 1.13 \[ \frac {d^3\,\left (10\,a\,x^2+12\,b\,x^2+10\,b\,x^2\,\mathrm {atanh}\left (c\,x\right )\right )}{20}-\frac {\frac {d^3\,\left (25\,b\,\mathrm {atanh}\left (c\,x\right )-12\,b\,\ln \left (c^2\,x^2-1\right )\right )}{20}-\frac {5\,b\,c\,d^3\,x}{4}}{c^2}+\frac {c^3\,d^3\,\left (4\,a\,x^5+4\,b\,x^5\,\mathrm {atanh}\left (c\,x\right )\right )}{20}+\frac {c\,d^3\,\left (20\,a\,x^3+5\,b\,x^3+20\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{20}+\frac {c^2\,d^3\,\left (15\,a\,x^4+b\,x^4+15\,b\,x^4\,\mathrm {atanh}\left (c\,x\right )\right )}{20} \]

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]

(d^3*(10*a*x^2 + 12*b*x^2 + 10*b*x^2*atanh(c*x)))/20 - ((d^3*(25*b*atanh(c*x) - 12*b*log(c^2*x^2 - 1)))/20 - (
5*b*c*d^3*x)/4)/c^2 + (c^3*d^3*(4*a*x^5 + 4*b*x^5*atanh(c*x)))/20 + (c*d^3*(20*a*x^3 + 5*b*x^3 + 20*b*x^3*atan
h(c*x)))/20 + (c^2*d^3*(15*a*x^4 + b*x^4 + 15*b*x^4*atanh(c*x)))/20

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sympy [A]  time = 1.79, size = 211, normalized size = 1.56 \[ \begin {cases} \frac {a c^{3} d^{3} x^{5}}{5} + \frac {3 a c^{2} d^{3} x^{4}}{4} + a c d^{3} x^{3} + \frac {a d^{3} x^{2}}{2} + \frac {b c^{3} d^{3} x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + \frac {3 b c^{2} d^{3} x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + \frac {b c^{2} d^{3} x^{4}}{20} + b c d^{3} x^{3} \operatorname {atanh}{\left (c x \right )} + \frac {b c d^{3} x^{3}}{4} + \frac {b d^{3} x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {3 b d^{3} x^{2}}{5} + \frac {5 b d^{3} x}{4 c} + \frac {6 b d^{3} \log {\left (x - \frac {1}{c} \right )}}{5 c^{2}} - \frac {b d^{3} \operatorname {atanh}{\left (c x \right )}}{20 c^{2}} & \text {for}\: c \neq 0 \\\frac {a d^{3} x^{2}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**3*d**3*x**5/5 + 3*a*c**2*d**3*x**4/4 + a*c*d**3*x**3 + a*d**3*x**2/2 + b*c**3*d**3*x**5*atanh(
c*x)/5 + 3*b*c**2*d**3*x**4*atanh(c*x)/4 + b*c**2*d**3*x**4/20 + b*c*d**3*x**3*atanh(c*x) + b*c*d**3*x**3/4 +
b*d**3*x**2*atanh(c*x)/2 + 3*b*d**3*x**2/5 + 5*b*d**3*x/(4*c) + 6*b*d**3*log(x - 1/c)/(5*c**2) - b*d**3*atanh(
c*x)/(20*c**2), Ne(c, 0)), (a*d**3*x**2/2, True))

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