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

Optimal. Leaf size=84 \[ \frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 e}-\frac {b (c d-e)^2 \log (c x+1)}{4 c^2 e}+\frac {b (c d+e)^2 \log (1-c x)}{4 c^2 e}+\frac {b e x}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5926, 702, 633, 31} \[ \frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 e}-\frac {b (c d-e)^2 \log (c x+1)}{4 c^2 e}+\frac {b (c d+e)^2 \log (1-c x)}{4 c^2 e}+\frac {b e x}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + b*ArcTanh[c*x]),x]

[Out]

(b*e*x)/(2*c) + ((d + e*x)^2*(a + b*ArcTanh[c*x]))/(2*e) + (b*(c*d + e)^2*Log[1 - c*x])/(4*c^2*e) - (b*(c*d -
e)^2*Log[1 + c*x])/(4*c^2*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) \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 e}-\frac {(b c) \int \frac {(d+e x)^2}{1-c^2 x^2} \, dx}{2 e}\\ &=\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 e}-\frac {(b c) \int \left (-\frac {e^2}{c^2}+\frac {c^2 d^2+e^2+2 c^2 d e x}{c^2 \left (1-c^2 x^2\right )}\right ) \, dx}{2 e}\\ &=\frac {b e x}{2 c}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 e}-\frac {b \int \frac {c^2 d^2+e^2+2 c^2 d e x}{1-c^2 x^2} \, dx}{2 c e}\\ &=\frac {b e x}{2 c}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 e}+\frac {\left (b (c d-e)^2\right ) \int \frac {1}{-c-c^2 x} \, dx}{4 e}-\frac {\left (b (c d+e)^2\right ) \int \frac {1}{c-c^2 x} \, dx}{4 e}\\ &=\frac {b e x}{2 c}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 e}+\frac {b (c d+e)^2 \log (1-c x)}{4 c^2 e}-\frac {b (c d-e)^2 \log (1+c x)}{4 c^2 e}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 96, normalized size = 1.14 \[ a d x+\frac {1}{2} a e x^2+\frac {b d \log \left (1-c^2 x^2\right )}{2 c}+\frac {b e \log (1-c x)}{4 c^2}-\frac {b e \log (c x+1)}{4 c^2}+b d x \tanh ^{-1}(c x)+\frac {1}{2} b e x^2 \tanh ^{-1}(c x)+\frac {b e x}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*d*x + (b*e*x)/(2*c) + (a*e*x^2)/2 + b*d*x*ArcTanh[c*x] + (b*e*x^2*ArcTanh[c*x])/2 + (b*e*Log[1 - c*x])/(4*c^
2) - (b*e*Log[1 + c*x])/(4*c^2) + (b*d*Log[1 - c^2*x^2])/(2*c)

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fricas [A]  time = 0.60, size = 98, normalized size = 1.17 \[ \frac {2 \, a c^{2} e x^{2} + 2 \, {\left (2 \, a c^{2} d + b c e\right )} x + {\left (2 \, b c d - b e\right )} \log \left (c x + 1\right ) + {\left (2 \, b c d + b e\right )} \log \left (c x - 1\right ) + {\left (b c^{2} e x^{2} + 2 \, b c^{2} d x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.21, size = 292, normalized size = 3.48 \[ -\frac {{\left (\frac {{\left (c x + 1\right )}^{2} b c d \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{{\left (c x - 1\right )}^{2}} - \frac {2 \, {\left (c x + 1\right )} b c d \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c x - 1} + b c d \log \left (-\frac {c x + 1}{c x - 1} + 1\right ) - \frac {{\left (c x + 1\right )}^{2} b c d \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\left (c x - 1\right )}^{2}} + \frac {{\left (c x + 1\right )} b c d \log \left (-\frac {c x + 1}{c x - 1}\right )}{c x - 1} - \frac {2 \, {\left (c x + 1\right )} a c d}{c x - 1} + 2 \, a c d - \frac {{\left (c x + 1\right )} b e \log \left (-\frac {c x + 1}{c x - 1}\right )}{c x - 1} - \frac {2 \, {\left (c x + 1\right )} a e}{c x - 1} - \frac {{\left (c x + 1\right )} b e}{c x - 1} + b e\right )} c}{\frac {{\left (c x + 1\right )}^{2} c^{3}}{{\left (c x - 1\right )}^{2}} - \frac {2 \, {\left (c x + 1\right )} c^{3}}{c x - 1} + c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c*x + 1)^2*b*c*d*log(-(c*x + 1)/(c*x - 1) + 1)/(c*x - 1)^2 - 2*(c*x + 1)*b*c*d*log(-(c*x + 1)/(c*x - 1) + 1
)/(c*x - 1) + b*c*d*log(-(c*x + 1)/(c*x - 1) + 1) - (c*x + 1)^2*b*c*d*log(-(c*x + 1)/(c*x - 1))/(c*x - 1)^2 +
(c*x + 1)*b*c*d*log(-(c*x + 1)/(c*x - 1))/(c*x - 1) - 2*(c*x + 1)*a*c*d/(c*x - 1) + 2*a*c*d - (c*x + 1)*b*e*lo
g(-(c*x + 1)/(c*x - 1))/(c*x - 1) - 2*(c*x + 1)*a*e/(c*x - 1) - (c*x + 1)*b*e/(c*x - 1) + b*e)*c/((c*x + 1)^2*
c^3/(c*x - 1)^2 - 2*(c*x + 1)*c^3/(c*x - 1) + c^3)

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maple [A]  time = 0.03, size = 92, normalized size = 1.10 \[ \frac {a \,x^{2} e}{2}+a d x +\frac {b \arctanh \left (c x \right ) x^{2} e}{2}+b \arctanh \left (c x \right ) x d +\frac {b e x}{2 c}+\frac {b \ln \left (c x -1\right ) d}{2 c}+\frac {b \ln \left (c x -1\right ) e}{4 c^{2}}+\frac {b \ln \left (c x +1\right ) d}{2 c}-\frac {b \ln \left (c x +1\right ) e}{4 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 83, normalized size = 0.99 \[ \frac {1}{2} \, a e x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b e + a d x + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.83, size = 67, normalized size = 0.80 \[ a\,d\,x+\frac {a\,e\,x^2}{2}+b\,d\,x\,\mathrm {atanh}\left (c\,x\right )+\frac {b\,e\,x}{2\,c}-\frac {b\,e\,\mathrm {atanh}\left (c\,x\right )}{2\,c^2}+\frac {b\,e\,x^2\,\mathrm {atanh}\left (c\,x\right )}{2}+\frac {b\,d\,\ln \left (c^2\,x^2-1\right )}{2\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))*(d + e*x),x)

[Out]

a*d*x + (a*e*x^2)/2 + b*d*x*atanh(c*x) + (b*e*x)/(2*c) - (b*e*atanh(c*x))/(2*c^2) + (b*e*x^2*atanh(c*x))/2 + (
b*d*log(c^2*x^2 - 1))/(2*c)

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sympy [A]  time = 0.67, size = 92, normalized size = 1.10 \[ \begin {cases} a d x + \frac {a e x^{2}}{2} + b d x \operatorname {atanh}{\left (c x \right )} + \frac {b e x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {b d \log {\left (x - \frac {1}{c} \right )}}{c} + \frac {b d \operatorname {atanh}{\left (c x \right )}}{c} + \frac {b e x}{2 c} - \frac {b e \operatorname {atanh}{\left (c x \right )}}{2 c^{2}} & \text {for}\: c \neq 0 \\a \left (d x + \frac {e x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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