3.1 \(\int (d+e x)^4 (a+b \tanh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=149 \[ \frac {(d+e x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 e}-\frac {b (c d-e)^5 \log (c x+1)}{10 c^5 e}+\frac {b (c d+e)^5 \log (1-c x)}{10 c^5 e}+\frac {b e^2 x^2 \left (10 c^2 d^2+e^2\right )}{10 c^3}+\frac {b d e x \left (2 c^2 d^2+e^2\right )}{c^3}+\frac {b d e^3 x^3}{3 c}+\frac {b e^4 x^4}{20 c} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5926, 702, 633, 31} \[ \frac {(d+e x)^5 \left (a+b \tanh ^{-1}(c x)\right )}{5 e}+\frac {b e^2 x^2 \left (10 c^2 d^2+e^2\right )}{10 c^3}+\frac {b d e x \left (2 c^2 d^2+e^2\right )}{c^3}-\frac {b (c d-e)^5 \log (c x+1)}{10 c^5 e}+\frac {b (c d+e)^5 \log (1-c x)}{10 c^5 e}+\frac {b d e^3 x^3}{3 c}+\frac {b e^4 x^4}{20 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4*(a + b*ArcTanh[c*x]),x]

[Out]

(b*d*e*(2*c^2*d^2 + e^2)*x)/c^3 + (b*e^2*(10*c^2*d^2 + e^2)*x^2)/(10*c^3) + (b*d*e^3*x^3)/(3*c) + (b*e^4*x^4)/
(20*c) + ((d + e*x)^5*(a + b*ArcTanh[c*x]))/(5*e) + (b*(c*d + e)^5*Log[1 - c*x])/(10*c^5*e) - (b*(c*d - e)^5*L
og[1 + c*x])/(10*c^5*e)

Rule 31

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

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 702

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 5926

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b
*ArcTanh[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ
[{a, b, c, d, e, q}, x] && NeQ[q, -1]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 274, normalized size = 1.84 \[ \frac {3 c^4 e^3 x^4 (20 a c d+b e)+20 c^4 d e^2 x^3 (6 a c d+b e)+6 c^2 e x^2 \left (20 a c^3 d^3+b e \left (10 c^2 d^2+e^2\right )\right )+60 c^2 d x \left (a c^3 d^3+b e \left (2 c^2 d^2+e^2\right )\right )+12 a c^5 e^4 x^5+12 b c^5 x \tanh ^{-1}(c x) \left (5 d^4+10 d^3 e x+10 d^2 e^2 x^2+5 d e^3 x^3+e^4 x^4\right )+6 b \left (5 c^4 d^4+10 c^3 d^3 e+10 c^2 d^2 e^2+5 c d e^3+e^4\right ) \log (1-c x)+6 b \left (5 c^4 d^4-10 c^3 d^3 e+10 c^2 d^2 e^2-5 c d e^3+e^4\right ) \log (c x+1)}{60 c^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4*(a + b*ArcTanh[c*x]),x]

[Out]

(60*c^2*d*(a*c^3*d^3 + b*e*(2*c^2*d^2 + e^2))*x + 6*c^2*e*(20*a*c^3*d^3 + b*e*(10*c^2*d^2 + e^2))*x^2 + 20*c^4
*d*e^2*(6*a*c*d + b*e)*x^3 + 3*c^4*e^3*(20*a*c*d + b*e)*x^4 + 12*a*c^5*e^4*x^5 + 12*b*c^5*x*(5*d^4 + 10*d^3*e*
x + 10*d^2*e^2*x^2 + 5*d*e^3*x^3 + e^4*x^4)*ArcTanh[c*x] + 6*b*(5*c^4*d^4 + 10*c^3*d^3*e + 10*c^2*d^2*e^2 + 5*
c*d*e^3 + e^4)*Log[1 - c*x] + 6*b*(5*c^4*d^4 - 10*c^3*d^3*e + 10*c^2*d^2*e^2 - 5*c*d*e^3 + e^4)*Log[1 + c*x])/
(60*c^5)

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fricas [B]  time = 0.49, size = 322, normalized size = 2.16 \[ \frac {12 \, a c^{5} e^{4} x^{5} + 3 \, {\left (20 \, a c^{5} d e^{3} + b c^{4} e^{4}\right )} x^{4} + 20 \, {\left (6 \, a c^{5} d^{2} e^{2} + b c^{4} d e^{3}\right )} x^{3} + 6 \, {\left (20 \, a c^{5} d^{3} e + 10 \, b c^{4} d^{2} e^{2} + b c^{2} e^{4}\right )} x^{2} + 60 \, {\left (a c^{5} d^{4} + 2 \, b c^{4} d^{3} e + b c^{2} d e^{3}\right )} x + 6 \, {\left (5 \, b c^{4} d^{4} - 10 \, b c^{3} d^{3} e + 10 \, b c^{2} d^{2} e^{2} - 5 \, b c d e^{3} + b e^{4}\right )} \log \left (c x + 1\right ) + 6 \, {\left (5 \, b c^{4} d^{4} + 10 \, b c^{3} d^{3} e + 10 \, b c^{2} d^{2} e^{2} + 5 \, b c d e^{3} + b e^{4}\right )} \log \left (c x - 1\right ) + 6 \, {\left (b c^{5} e^{4} x^{5} + 5 \, b c^{5} d e^{3} x^{4} + 10 \, b c^{5} d^{2} e^{2} x^{3} + 10 \, b c^{5} d^{3} e x^{2} + 5 \, b c^{5} d^{4} x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{60 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="fricas")

[Out]

1/60*(12*a*c^5*e^4*x^5 + 3*(20*a*c^5*d*e^3 + b*c^4*e^4)*x^4 + 20*(6*a*c^5*d^2*e^2 + b*c^4*d*e^3)*x^3 + 6*(20*a
*c^5*d^3*e + 10*b*c^4*d^2*e^2 + b*c^2*e^4)*x^2 + 60*(a*c^5*d^4 + 2*b*c^4*d^3*e + b*c^2*d*e^3)*x + 6*(5*b*c^4*d
^4 - 10*b*c^3*d^3*e + 10*b*c^2*d^2*e^2 - 5*b*c*d*e^3 + b*e^4)*log(c*x + 1) + 6*(5*b*c^4*d^4 + 10*b*c^3*d^3*e +
10*b*c^2*d^2*e^2 + 5*b*c*d*e^3 + b*e^4)*log(c*x - 1) + 6*(b*c^5*e^4*x^5 + 5*b*c^5*d*e^3*x^4 + 10*b*c^5*d^2*e^
2*x^3 + 10*b*c^5*d^3*e*x^2 + 5*b*c^5*d^4*x)*log(-(c*x + 1)/(c*x - 1)))/c^5

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giac [B]  time = 0.23, size = 2351, normalized size = 15.78 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="giac")

[Out]

-1/15*(15*(c*x + 1)^5*b*c^4*d^4*log(-(c*x + 1)/(c*x - 1) + 1)/(c*x - 1)^5 - 75*(c*x + 1)^4*b*c^4*d^4*log(-(c*x
+ 1)/(c*x - 1) + 1)/(c*x - 1)^4 + 150*(c*x + 1)^3*b*c^4*d^4*log(-(c*x + 1)/(c*x - 1) + 1)/(c*x - 1)^3 - 150*(
c*x + 1)^2*b*c^4*d^4*log(-(c*x + 1)/(c*x - 1) + 1)/(c*x - 1)^2 + 75*(c*x + 1)*b*c^4*d^4*log(-(c*x + 1)/(c*x -
1) + 1)/(c*x - 1) - 15*b*c^4*d^4*log(-(c*x + 1)/(c*x - 1) + 1) - 15*(c*x + 1)^5*b*c^4*d^4*log(-(c*x + 1)/(c*x
- 1))/(c*x - 1)^5 + 60*(c*x + 1)^4*b*c^4*d^4*log(-(c*x + 1)/(c*x - 1))/(c*x - 1)^4 - 90*(c*x + 1)^3*b*c^4*d^4*
log(-(c*x + 1)/(c*x - 1))/(c*x - 1)^3 + 60*(c*x + 1)^2*b*c^4*d^4*log(-(c*x + 1)/(c*x - 1))/(c*x - 1)^2 - 15*(c
*x + 1)*b*c^4*d^4*log(-(c*x + 1)/(c*x - 1))/(c*x - 1) - 30*(c*x + 1)^4*a*c^4*d^4/(c*x - 1)^4 + 120*(c*x + 1)^3
*a*c^4*d^4/(c*x - 1)^3 - 180*(c*x + 1)^2*a*c^4*d^4/(c*x - 1)^2 + 120*(c*x + 1)*a*c^4*d^4/(c*x - 1) - 30*a*c^4*
d^4 - 60*(c*x + 1)^4*b*c^3*d^3*e*log(-(c*x + 1)/(c*x - 1))/(c*x - 1)^4 + 180*(c*x + 1)^3*b*c^3*d^3*e*log(-(c*x
+ 1)/(c*x - 1))/(c*x - 1)^3 - 180*(c*x + 1)^2*b*c^3*d^3*e*log(-(c*x + 1)/(c*x - 1))/(c*x - 1)^2 + 60*(c*x + 1
)*b*c^3*d^3*e*log(-(c*x + 1)/(c*x - 1))/(c*x - 1) - 120*(c*x + 1)^4*a*c^3*d^3*e/(c*x - 1)^4 + 360*(c*x + 1)^3*
a*c^3*d^3*e/(c*x - 1)^3 - 360*(c*x + 1)^2*a*c^3*d^3*e/(c*x - 1)^2 + 120*(c*x + 1)*a*c^3*d^3*e/(c*x - 1) - 60*(
c*x + 1)^4*b*c^3*d^3*e/(c*x - 1)^4 + 240*(c*x + 1)^3*b*c^3*d^3*e/(c*x - 1)^3 - 360*(c*x + 1)^2*b*c^3*d^3*e/(c*
x - 1)^2 + 240*(c*x + 1)*b*c^3*d^3*e/(c*x - 1) - 60*b*c^3*d^3*e + 30*(c*x + 1)^5*b*c^2*d^2*e^2*log(-(c*x + 1)/
(c*x - 1) + 1)/(c*x - 1)^5 - 150*(c*x + 1)^4*b*c^2*d^2*e^2*log(-(c*x + 1)/(c*x - 1) + 1)/(c*x - 1)^4 + 300*(c*
x + 1)^3*b*c^2*d^2*e^2*log(-(c*x + 1)/(c*x - 1) + 1)/(c*x - 1)^3 - 300*(c*x + 1)^2*b*c^2*d^2*e^2*log(-(c*x + 1
)/(c*x - 1) + 1)/(c*x - 1)^2 + 150*(c*x + 1)*b*c^2*d^2*e^2*log(-(c*x + 1)/(c*x - 1) + 1)/(c*x - 1) - 30*b*c^2*
d^2*e^2*log(-(c*x + 1)/(c*x - 1) + 1) - 30*(c*x + 1)^5*b*c^2*d^2*e^2*log(-(c*x + 1)/(c*x - 1))/(c*x - 1)^5 + 6
0*(c*x + 1)^4*b*c^2*d^2*e^2*log(-(c*x + 1)/(c*x - 1))/(c*x - 1)^4 - 120*(c*x + 1)^3*b*c^2*d^2*e^2*log(-(c*x +
1)/(c*x - 1))/(c*x - 1)^3 + 180*(c*x + 1)^2*b*c^2*d^2*e^2*log(-(c*x + 1)/(c*x - 1))/(c*x - 1)^2 - 90*(c*x + 1)
*b*c^2*d^2*e^2*log(-(c*x + 1)/(c*x - 1))/(c*x - 1) - 180*(c*x + 1)^4*a*c^2*d^2*e^2/(c*x - 1)^4 + 360*(c*x + 1)
^3*a*c^2*d^2*e^2/(c*x - 1)^3 - 240*(c*x + 1)^2*a*c^2*d^2*e^2/(c*x - 1)^2 + 120*(c*x + 1)*a*c^2*d^2*e^2/(c*x -
1) - 60*a*c^2*d^2*e^2 - 60*(c*x + 1)^4*b*c^2*d^2*e^2/(c*x - 1)^4 + 180*(c*x + 1)^3*b*c^2*d^2*e^2/(c*x - 1)^3 -
180*(c*x + 1)^2*b*c^2*d^2*e^2/(c*x - 1)^2 + 60*(c*x + 1)*b*c^2*d^2*e^2/(c*x - 1) - 60*(c*x + 1)^4*b*c*d*e^3*l
og(-(c*x + 1)/(c*x - 1))/(c*x - 1)^4 + 60*(c*x + 1)^3*b*c*d*e^3*log(-(c*x + 1)/(c*x - 1))/(c*x - 1)^3 - 60*(c*
x + 1)^2*b*c*d*e^3*log(-(c*x + 1)/(c*x - 1))/(c*x - 1)^2 + 60*(c*x + 1)*b*c*d*e^3*log(-(c*x + 1)/(c*x - 1))/(c
*x - 1) - 120*(c*x + 1)^4*a*c*d*e^3/(c*x - 1)^4 + 120*(c*x + 1)^3*a*c*d*e^3/(c*x - 1)^3 - 120*(c*x + 1)^2*a*c*
d*e^3/(c*x - 1)^2 + 120*(c*x + 1)*a*c*d*e^3/(c*x - 1) - 60*(c*x + 1)^4*b*c*d*e^3/(c*x - 1)^4 + 180*(c*x + 1)^3
*b*c*d*e^3/(c*x - 1)^3 - 220*(c*x + 1)^2*b*c*d*e^3/(c*x - 1)^2 + 140*(c*x + 1)*b*c*d*e^3/(c*x - 1) - 40*b*c*d*
e^3 + 3*(c*x + 1)^5*b*e^4*log(-(c*x + 1)/(c*x - 1) + 1)/(c*x - 1)^5 - 15*(c*x + 1)^4*b*e^4*log(-(c*x + 1)/(c*x
- 1) + 1)/(c*x - 1)^4 + 30*(c*x + 1)^3*b*e^4*log(-(c*x + 1)/(c*x - 1) + 1)/(c*x - 1)^3 - 30*(c*x + 1)^2*b*e^4
*log(-(c*x + 1)/(c*x - 1) + 1)/(c*x - 1)^2 + 15*(c*x + 1)*b*e^4*log(-(c*x + 1)/(c*x - 1) + 1)/(c*x - 1) - 3*b*
e^4*log(-(c*x + 1)/(c*x - 1) + 1) - 3*(c*x + 1)^5*b*e^4*log(-(c*x + 1)/(c*x - 1))/(c*x - 1)^5 - 30*(c*x + 1)^3
*b*e^4*log(-(c*x + 1)/(c*x - 1))/(c*x - 1)^3 - 15*(c*x + 1)*b*e^4*log(-(c*x + 1)/(c*x - 1))/(c*x - 1) - 30*(c*
x + 1)^4*a*e^4/(c*x - 1)^4 - 60*(c*x + 1)^2*a*e^4/(c*x - 1)^2 - 6*a*e^4 - 12*(c*x + 1)^4*b*e^4/(c*x - 1)^4 + 2
4*(c*x + 1)^3*b*e^4/(c*x - 1)^3 - 24*(c*x + 1)^2*b*e^4/(c*x - 1)^2 + 12*(c*x + 1)*b*e^4/(c*x - 1))*c/((c*x + 1
)^5*c^6/(c*x - 1)^5 - 5*(c*x + 1)^4*c^6/(c*x - 1)^4 + 10*(c*x + 1)^3*c^6/(c*x - 1)^3 - 10*(c*x + 1)^2*c^6/(c*x
- 1)^2 + 5*(c*x + 1)*c^6/(c*x - 1) - c^6)

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maple [B]  time = 0.03, size = 395, normalized size = 2.65 \[ \frac {b \,e^{3} d x}{c^{3}}+\frac {b \,e^{4} \arctanh \left (c x \right ) x^{5}}{5}+b \arctanh \left (c x \right ) x \,d^{4}+\frac {b \arctanh \left (c x \right ) d^{5}}{5 e}-\frac {b \ln \left (c x +1\right ) d^{5}}{10 e}+\frac {b \,e^{4} x^{2}}{10 c^{3}}+2 a e \,x^{2} d^{3}+2 a \,e^{2} x^{3} d^{2}+a \,e^{3} x^{4} d +\frac {b \,e^{4} \ln \left (c x -1\right )}{10 c^{5}}+\frac {b \,e^{4} \ln \left (c x +1\right )}{10 c^{5}}+\frac {b \ln \left (c x -1\right ) d^{4}}{2 c}+\frac {b \ln \left (c x +1\right ) d^{4}}{2 c}+\frac {b \ln \left (c x -1\right ) d^{5}}{10 e}+\frac {a \,d^{5}}{5 e}+\frac {a \,e^{4} x^{5}}{5}+a x \,d^{4}+\frac {b \,e^{2} x^{2} d^{2}}{c}+b \,e^{3} \arctanh \left (c x \right ) x^{4} d +2 b \,e^{2} \arctanh \left (c x \right ) x^{3} d^{2}+2 b e \arctanh \left (c x \right ) x^{2} d^{3}+\frac {b e \ln \left (c x -1\right ) d^{3}}{c^{2}}+\frac {b \,e^{2} \ln \left (c x -1\right ) d^{2}}{c^{3}}+\frac {b \,e^{3} \ln \left (c x -1\right ) d}{2 c^{4}}-\frac {b \,e^{3} \ln \left (c x +1\right ) d}{2 c^{4}}-\frac {b e \ln \left (c x +1\right ) d^{3}}{c^{2}}+\frac {b \,e^{2} \ln \left (c x +1\right ) d^{2}}{c^{3}}+\frac {2 b e \,d^{3} x}{c}+\frac {b d \,e^{3} x^{3}}{3 c}+\frac {b \,e^{4} x^{4}}{20 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4*(a+b*arctanh(c*x)),x)

[Out]

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

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maxima [A]  time = 0.33, size = 273, normalized size = 1.83 \[ \frac {1}{5} \, a e^{4} x^{5} + a d e^{3} x^{4} + 2 \, a d^{2} e^{2} x^{3} + 2 \, a d^{3} e x^{2} + {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b d^{3} e + {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b d^{2} e^{2} + \frac {1}{6} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b d e^{3} + \frac {1}{20} \, {\left (4 \, x^{5} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b e^{4} + a d^{4} x + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d^{4}}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*(a+b*arctanh(c*x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.70, size = 272, normalized size = 1.83 \[ \frac {a\,e^4\,x^5}{5}+a\,d^4\,x+\frac {b\,d^4\,\ln \left (c^2\,x^2-1\right )}{2\,c}+\frac {b\,e^4\,\ln \left (c^2\,x^2-1\right )}{10\,c^5}+2\,a\,d^2\,e^2\,x^3+\frac {b\,e^4\,x^4}{20\,c}+\frac {b\,e^4\,x^2}{10\,c^3}+b\,d^4\,x\,\mathrm {atanh}\left (c\,x\right )+2\,a\,d^3\,e\,x^2+a\,d\,e^3\,x^4+\frac {b\,e^4\,x^5\,\mathrm {atanh}\left (c\,x\right )}{5}+\frac {2\,b\,d^3\,e\,x}{c}+\frac {b\,d\,e^3\,x}{c^3}-\frac {2\,b\,d^3\,e\,\mathrm {atanh}\left (c\,x\right )}{c^2}-\frac {b\,d\,e^3\,\mathrm {atanh}\left (c\,x\right )}{c^4}+2\,b\,d^3\,e\,x^2\,\mathrm {atanh}\left (c\,x\right )+b\,d\,e^3\,x^4\,\mathrm {atanh}\left (c\,x\right )+\frac {b\,d\,e^3\,x^3}{3\,c}+2\,b\,d^2\,e^2\,x^3\,\mathrm {atanh}\left (c\,x\right )+\frac {b\,d^2\,e^2\,\ln \left (c^2\,x^2-1\right )}{c^3}+\frac {b\,d^2\,e^2\,x^2}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))*(d + e*x)^4,x)

[Out]

(a*e^4*x^5)/5 + a*d^4*x + (b*d^4*log(c^2*x^2 - 1))/(2*c) + (b*e^4*log(c^2*x^2 - 1))/(10*c^5) + 2*a*d^2*e^2*x^3
+ (b*e^4*x^4)/(20*c) + (b*e^4*x^2)/(10*c^3) + b*d^4*x*atanh(c*x) + 2*a*d^3*e*x^2 + a*d*e^3*x^4 + (b*e^4*x^5*a
tanh(c*x))/5 + (2*b*d^3*e*x)/c + (b*d*e^3*x)/c^3 - (2*b*d^3*e*atanh(c*x))/c^2 - (b*d*e^3*atanh(c*x))/c^4 + 2*b
*d^3*e*x^2*atanh(c*x) + b*d*e^3*x^4*atanh(c*x) + (b*d*e^3*x^3)/(3*c) + 2*b*d^2*e^2*x^3*atanh(c*x) + (b*d^2*e^2
*log(c^2*x^2 - 1))/c^3 + (b*d^2*e^2*x^2)/c

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sympy [A]  time = 2.96, size = 381, normalized size = 2.56 \[ \begin {cases} a d^{4} x + 2 a d^{3} e x^{2} + 2 a d^{2} e^{2} x^{3} + a d e^{3} x^{4} + \frac {a e^{4} x^{5}}{5} + b d^{4} x \operatorname {atanh}{\left (c x \right )} + 2 b d^{3} e x^{2} \operatorname {atanh}{\left (c x \right )} + 2 b d^{2} e^{2} x^{3} \operatorname {atanh}{\left (c x \right )} + b d e^{3} x^{4} \operatorname {atanh}{\left (c x \right )} + \frac {b e^{4} x^{5} \operatorname {atanh}{\left (c x \right )}}{5} + \frac {b d^{4} \log {\left (x - \frac {1}{c} \right )}}{c} + \frac {b d^{4} \operatorname {atanh}{\left (c x \right )}}{c} + \frac {2 b d^{3} e x}{c} + \frac {b d^{2} e^{2} x^{2}}{c} + \frac {b d e^{3} x^{3}}{3 c} + \frac {b e^{4} x^{4}}{20 c} - \frac {2 b d^{3} e \operatorname {atanh}{\left (c x \right )}}{c^{2}} + \frac {2 b d^{2} e^{2} \log {\left (x - \frac {1}{c} \right )}}{c^{3}} + \frac {2 b d^{2} e^{2} \operatorname {atanh}{\left (c x \right )}}{c^{3}} + \frac {b d e^{3} x}{c^{3}} + \frac {b e^{4} x^{2}}{10 c^{3}} - \frac {b d e^{3} \operatorname {atanh}{\left (c x \right )}}{c^{4}} + \frac {b e^{4} \log {\left (x - \frac {1}{c} \right )}}{5 c^{5}} + \frac {b e^{4} \operatorname {atanh}{\left (c x \right )}}{5 c^{5}} & \text {for}\: c \neq 0 \\a \left (d^{4} x + 2 d^{3} e x^{2} + 2 d^{2} e^{2} x^{3} + d e^{3} x^{4} + \frac {e^{4} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4*(a+b*atanh(c*x)),x)

[Out]

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

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