3.8 \(\int \frac{a+b \tanh ^{-1}(c x)}{(d+e x)^4} \, dx\)
Optimal. Leaf size=175 \[ -\frac{a+b \tanh ^{-1}(c x)}{3 e (d+e x)^3}+\frac{2 b c^3 d}{3 \left (c^2 d^2-e^2\right )^2 (d+e x)}+\frac{b c}{6 \left (c^2 d^2-e^2\right ) (d+e x)^2}-\frac{b c^3 \left (3 c^2 d^2+e^2\right ) \log (d+e x)}{3 (c d-e)^3 (c d+e)^3}-\frac{b c^3 \log (1-c x)}{6 e (c d+e)^3}+\frac{b c^3 \log (c x+1)}{6 e (c d-e)^3} \]
[Out]
(b*c)/(6*(c^2*d^2 - e^2)*(d + e*x)^2) + (2*b*c^3*d)/(3*(c^2*d^2 - e^2)^2*(d + e*x)) - (a + b*ArcTanh[c*x])/(3*
e*(d + e*x)^3) - (b*c^3*Log[1 - c*x])/(6*e*(c*d + e)^3) + (b*c^3*Log[1 + c*x])/(6*(c*d - e)^3*e) - (b*c^3*(3*c
^2*d^2 + e^2)*Log[d + e*x])/(3*(c*d - e)^3*(c*d + e)^3)
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Rubi [A] time = 0.188482, antiderivative size = 175, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used =
{5926, 710, 801} \[ -\frac{a+b \tanh ^{-1}(c x)}{3 e (d+e x)^3}+\frac{2 b c^3 d}{3 \left (c^2 d^2-e^2\right )^2 (d+e x)}+\frac{b c}{6 \left (c^2 d^2-e^2\right ) (d+e x)^2}-\frac{b c^3 \left (3 c^2 d^2+e^2\right ) \log (d+e x)}{3 (c d-e)^3 (c d+e)^3}-\frac{b c^3 \log (1-c x)}{6 e (c d+e)^3}+\frac{b c^3 \log (c x+1)}{6 e (c d-e)^3} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcTanh[c*x])/(d + e*x)^4,x]
[Out]
(b*c)/(6*(c^2*d^2 - e^2)*(d + e*x)^2) + (2*b*c^3*d)/(3*(c^2*d^2 - e^2)^2*(d + e*x)) - (a + b*ArcTanh[c*x])/(3*
e*(d + e*x)^3) - (b*c^3*Log[1 - c*x])/(6*e*(c*d + e)^3) + (b*c^3*Log[1 + c*x])/(6*(c*d - e)^3*e) - (b*c^3*(3*c
^2*d^2 + e^2)*Log[d + e*x])/(3*(c*d - e)^3*(c*d + e)^3)
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]
Rule 710
Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 +
a*e^2)), x] + Dist[c/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*(d - e*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d,
e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]
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]
Rubi steps
\begin{align*} \int \frac{a+b \tanh ^{-1}(c x)}{(d+e x)^4} \, dx &=-\frac{a+b \tanh ^{-1}(c x)}{3 e (d+e x)^3}+\frac{(b c) \int \frac{1}{(d+e x)^3 \left (1-c^2 x^2\right )} \, dx}{3 e}\\ &=\frac{b c}{6 \left (c^2 d^2-e^2\right ) (d+e x)^2}-\frac{a+b \tanh ^{-1}(c x)}{3 e (d+e x)^3}+\frac{\left (b c^3\right ) \int \frac{d-e x}{(d+e x)^2 \left (1-c^2 x^2\right )} \, dx}{3 e \left (c^2 d^2-e^2\right )}\\ &=\frac{b c}{6 \left (c^2 d^2-e^2\right ) (d+e x)^2}-\frac{a+b \tanh ^{-1}(c x)}{3 e (d+e x)^3}+\frac{\left (b c^3\right ) \int \left (-\frac{c (c d-e)}{2 (c d+e)^2 (-1+c x)}+\frac{c (c d+e)}{2 (c d-e)^2 (1+c x)}+\frac{2 d e^2}{(-c d+e) (c d+e) (d+e x)^2}-\frac{e^2 \left (3 c^2 d^2+e^2\right )}{(-c d+e)^2 (c d+e)^2 (d+e x)}\right ) \, dx}{3 e \left (c^2 d^2-e^2\right )}\\ &=\frac{b c}{6 \left (c^2 d^2-e^2\right ) (d+e x)^2}+\frac{2 b c^3 d}{3 \left (c^2 d^2-e^2\right )^2 (d+e x)}-\frac{a+b \tanh ^{-1}(c x)}{3 e (d+e x)^3}-\frac{b c^3 \log (1-c x)}{6 e (c d+e)^3}+\frac{b c^3 \log (1+c x)}{6 (c d-e)^3 e}-\frac{b c^3 \left (3 c^2 d^2+e^2\right ) \log (d+e x)}{3 (c d-e)^3 (c d+e)^3}\\ \end{align*}
Mathematica [A] time = 0.263068, size = 173, normalized size = 0.99 \[ \frac{1}{6} \left (-\frac{2 a}{e (d+e x)^3}+\frac{4 b c^3 d}{\left (e^2-c^2 d^2\right )^2 (d+e x)}+\frac{b c}{\left (c^2 d^2-e^2\right ) (d+e x)^2}-\frac{2 b c^3 \left (3 c^2 d^2+e^2\right ) \log (d+e x)}{\left (c^2 d^2-e^2\right )^3}-\frac{b c^3 \log (1-c x)}{e (c d+e)^3}+\frac{b c^3 \log (c x+1)}{e (c d-e)^3}-\frac{2 b \tanh ^{-1}(c x)}{e (d+e x)^3}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*ArcTanh[c*x])/(d + e*x)^4,x]
[Out]
((-2*a)/(e*(d + e*x)^3) + (b*c)/((c^2*d^2 - e^2)*(d + e*x)^2) + (4*b*c^3*d)/((-(c^2*d^2) + e^2)^2*(d + e*x)) -
(2*b*ArcTanh[c*x])/(e*(d + e*x)^3) - (b*c^3*Log[1 - c*x])/(e*(c*d + e)^3) + (b*c^3*Log[1 + c*x])/((c*d - e)^3
*e) - (2*b*c^3*(3*c^2*d^2 + e^2)*Log[d + e*x])/(c^2*d^2 - e^2)^3)/6
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Maple [A] time = 0.046, size = 223, normalized size = 1.3 \begin{align*} -{\frac{{c}^{3}a}{3\, \left ( cxe+cd \right ) ^{3}e}}-{\frac{{c}^{3}b{\it Artanh} \left ( cx \right ) }{3\, \left ( cxe+cd \right ) ^{3}e}}+{\frac{{c}^{3}b}{ \left ( 6\,cd+6\,e \right ) \left ( cd-e \right ) \left ( cxe+cd \right ) ^{2}}}-{\frac{{c}^{5}b\ln \left ( cxe+cd \right ){d}^{2}}{ \left ( cd+e \right ) ^{3} \left ( cd-e \right ) ^{3}}}-{\frac{{c}^{3}b{e}^{2}\ln \left ( cxe+cd \right ) }{3\, \left ( cd+e \right ) ^{3} \left ( cd-e \right ) ^{3}}}+{\frac{2\,b{c}^{4}d}{3\, \left ( cd+e \right ) ^{2} \left ( cd-e \right ) ^{2} \left ( cxe+cd \right ) }}-{\frac{{c}^{3}b\ln \left ( cx-1 \right ) }{6\,e \left ( cd+e \right ) ^{3}}}+{\frac{{c}^{3}b\ln \left ( cx+1 \right ) }{6\, \left ( cd-e \right ) ^{3}e}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arctanh(c*x))/(e*x+d)^4,x)
[Out]
-1/3*c^3*a/(c*e*x+c*d)^3/e-1/3*c^3*b/(c*e*x+c*d)^3/e*arctanh(c*x)+1/6*c^3*b/(c*d+e)/(c*d-e)/(c*e*x+c*d)^2-c^5*
b/(c*d+e)^3/(c*d-e)^3*ln(c*e*x+c*d)*d^2-1/3*c^3*b*e^2/(c*d+e)^3/(c*d-e)^3*ln(c*e*x+c*d)+2/3*c^4*b*d/(c*d+e)^2/
(c*d-e)^2/(c*e*x+c*d)-1/6*c^3*b/e/(c*d+e)^3*ln(c*x-1)+1/6*b*c^3*ln(c*x+1)/(c*d-e)^3/e
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Maxima [B] time = 1.0115, size = 458, normalized size = 2.62 \begin{align*} \frac{1}{6} \,{\left ({\left (\frac{c^{2} \log \left (c x + 1\right )}{c^{3} d^{3} e - 3 \, c^{2} d^{2} e^{2} + 3 \, c d e^{3} - e^{4}} - \frac{c^{2} \log \left (c x - 1\right )}{c^{3} d^{3} e + 3 \, c^{2} d^{2} e^{2} + 3 \, c d e^{3} + e^{4}} - \frac{2 \,{\left (3 \, c^{4} d^{2} + c^{2} e^{2}\right )} \log \left (e x + d\right )}{c^{6} d^{6} - 3 \, c^{4} d^{4} e^{2} + 3 \, c^{2} d^{2} e^{4} - e^{6}} + \frac{4 \, c^{2} d e x + 5 \, c^{2} d^{2} - e^{2}}{c^{4} d^{6} - 2 \, c^{2} d^{4} e^{2} + d^{2} e^{4} +{\left (c^{4} d^{4} e^{2} - 2 \, c^{2} d^{2} e^{4} + e^{6}\right )} x^{2} + 2 \,{\left (c^{4} d^{5} e - 2 \, c^{2} d^{3} e^{3} + d e^{5}\right )} x}\right )} c - \frac{2 \, \operatorname{artanh}\left (c x\right )}{e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e}\right )} b - \frac{a}{3 \,{\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctanh(c*x))/(e*x+d)^4,x, algorithm="maxima")
[Out]
1/6*((c^2*log(c*x + 1)/(c^3*d^3*e - 3*c^2*d^2*e^2 + 3*c*d*e^3 - e^4) - c^2*log(c*x - 1)/(c^3*d^3*e + 3*c^2*d^2
*e^2 + 3*c*d*e^3 + e^4) - 2*(3*c^4*d^2 + c^2*e^2)*log(e*x + d)/(c^6*d^6 - 3*c^4*d^4*e^2 + 3*c^2*d^2*e^4 - e^6)
+ (4*c^2*d*e*x + 5*c^2*d^2 - e^2)/(c^4*d^6 - 2*c^2*d^4*e^2 + d^2*e^4 + (c^4*d^4*e^2 - 2*c^2*d^2*e^4 + e^6)*x^
2 + 2*(c^4*d^5*e - 2*c^2*d^3*e^3 + d*e^5)*x))*c - 2*arctanh(c*x)/(e^4*x^3 + 3*d*e^3*x^2 + 3*d^2*e^2*x + d^3*e)
)*b - 1/3*a/(e^4*x^3 + 3*d*e^3*x^2 + 3*d^2*e^2*x + d^3*e)
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Fricas [B] time = 4.12074, size = 1713, normalized size = 9.79 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctanh(c*x))/(e*x+d)^4,x, algorithm="fricas")
[Out]
-1/6*(2*a*c^6*d^6 - 5*b*c^5*d^5*e - 6*a*c^4*d^4*e^2 + 6*b*c^3*d^3*e^3 + 6*a*c^2*d^2*e^4 - b*c*d*e^5 - 2*a*e^6
- 4*(b*c^5*d^3*e^3 - b*c^3*d*e^5)*x^2 - (9*b*c^5*d^4*e^2 - 10*b*c^3*d^2*e^4 + b*c*e^6)*x - (b*c^6*d^6 + 3*b*c^
5*d^5*e + 3*b*c^4*d^4*e^2 + b*c^3*d^3*e^3 + (b*c^6*d^3*e^3 + 3*b*c^5*d^2*e^4 + 3*b*c^4*d*e^5 + b*c^3*e^6)*x^3
+ 3*(b*c^6*d^4*e^2 + 3*b*c^5*d^3*e^3 + 3*b*c^4*d^2*e^4 + b*c^3*d*e^5)*x^2 + 3*(b*c^6*d^5*e + 3*b*c^5*d^4*e^2 +
3*b*c^4*d^3*e^3 + b*c^3*d^2*e^4)*x)*log(c*x + 1) + (b*c^6*d^6 - 3*b*c^5*d^5*e + 3*b*c^4*d^4*e^2 - b*c^3*d^3*e
^3 + (b*c^6*d^3*e^3 - 3*b*c^5*d^2*e^4 + 3*b*c^4*d*e^5 - b*c^3*e^6)*x^3 + 3*(b*c^6*d^4*e^2 - 3*b*c^5*d^3*e^3 +
3*b*c^4*d^2*e^4 - b*c^3*d*e^5)*x^2 + 3*(b*c^6*d^5*e - 3*b*c^5*d^4*e^2 + 3*b*c^4*d^3*e^3 - b*c^3*d^2*e^4)*x)*lo
g(c*x - 1) + 2*(3*b*c^5*d^5*e + b*c^3*d^3*e^3 + (3*b*c^5*d^2*e^4 + b*c^3*e^6)*x^3 + 3*(3*b*c^5*d^3*e^3 + b*c^3
*d*e^5)*x^2 + 3*(3*b*c^5*d^4*e^2 + b*c^3*d^2*e^4)*x)*log(e*x + d) + (b*c^6*d^6 - 3*b*c^4*d^4*e^2 + 3*b*c^2*d^2
*e^4 - b*e^6)*log(-(c*x + 1)/(c*x - 1)))/(c^6*d^9*e - 3*c^4*d^7*e^3 + 3*c^2*d^5*e^5 - d^3*e^7 + (c^6*d^6*e^4 -
3*c^4*d^4*e^6 + 3*c^2*d^2*e^8 - e^10)*x^3 + 3*(c^6*d^7*e^3 - 3*c^4*d^5*e^5 + 3*c^2*d^3*e^7 - d*e^9)*x^2 + 3*(
c^6*d^8*e^2 - 3*c^4*d^6*e^4 + 3*c^2*d^4*e^6 - d^2*e^8)*x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*atanh(c*x))/(e*x+d)**4,x)
[Out]
Timed out
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Giac [B] time = 3.09371, size = 1511, normalized size = 8.63 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctanh(c*x))/(e*x+d)^4,x, algorithm="giac")
[Out]
1/6*(b*c^6*d^3*x^3*e^3*log(c*x + 1) + 3*b*c^6*d^4*x^2*e^2*log(c*x + 1) + 3*b*c^6*d^5*x*e*log(c*x + 1) - b*c^6*
d^3*x^3*e^3*log(c*x - 1) - 3*b*c^6*d^4*x^2*e^2*log(c*x - 1) - 3*b*c^6*d^5*x*e*log(c*x - 1) + b*c^6*d^6*log(c*x
+ 1) - b*c^6*d^6*log(c*x - 1) - b*c^6*d^6*log(-(c*x + 1)/(c*x - 1)) - 2*a*c^6*d^6 + 3*b*c^5*d^2*x^3*e^4*log(c
*x + 1) + 9*b*c^5*d^3*x^2*e^3*log(c*x + 1) + 9*b*c^5*d^4*x*e^2*log(c*x + 1) + 3*b*c^5*d^5*e*log(c*x + 1) + 3*b
*c^5*d^2*x^3*e^4*log(c*x - 1) + 9*b*c^5*d^3*x^2*e^3*log(c*x - 1) + 9*b*c^5*d^4*x*e^2*log(c*x - 1) + 3*b*c^5*d^
5*e*log(c*x - 1) - 6*b*c^5*d^2*x^3*e^4*log(x*e + d) - 18*b*c^5*d^3*x^2*e^3*log(x*e + d) - 18*b*c^5*d^4*x*e^2*l
og(x*e + d) - 6*b*c^5*d^5*e*log(x*e + d) + 4*b*c^5*d^3*x^2*e^3 + 9*b*c^5*d^4*x*e^2 + 5*b*c^5*d^5*e + 3*b*c^4*d
*x^3*e^5*log(c*x + 1) + 9*b*c^4*d^2*x^2*e^4*log(c*x + 1) + 9*b*c^4*d^3*x*e^3*log(c*x + 1) + 3*b*c^4*d^4*e^2*lo
g(c*x + 1) - 3*b*c^4*d*x^3*e^5*log(c*x - 1) - 9*b*c^4*d^2*x^2*e^4*log(c*x - 1) - 9*b*c^4*d^3*x*e^3*log(c*x - 1
) - 3*b*c^4*d^4*e^2*log(c*x - 1) + 3*b*c^4*d^4*e^2*log(-(c*x + 1)/(c*x - 1)) + 6*a*c^4*d^4*e^2 + b*c^3*x^3*e^6
*log(c*x + 1) + 3*b*c^3*d*x^2*e^5*log(c*x + 1) + 3*b*c^3*d^2*x*e^4*log(c*x + 1) + b*c^3*d^3*e^3*log(c*x + 1) +
b*c^3*x^3*e^6*log(c*x - 1) + 3*b*c^3*d*x^2*e^5*log(c*x - 1) + 3*b*c^3*d^2*x*e^4*log(c*x - 1) + b*c^3*d^3*e^3*
log(c*x - 1) - 2*b*c^3*x^3*e^6*log(x*e + d) - 6*b*c^3*d*x^2*e^5*log(x*e + d) - 6*b*c^3*d^2*x*e^4*log(x*e + d)
- 2*b*c^3*d^3*e^3*log(x*e + d) - 4*b*c^3*d*x^2*e^5 - 10*b*c^3*d^2*x*e^4 - 6*b*c^3*d^3*e^3 - 3*b*c^2*d^2*e^4*lo
g(-(c*x + 1)/(c*x - 1)) - 6*a*c^2*d^2*e^4 + b*c*x*e^6 + b*c*d*e^5 + b*e^6*log(-(c*x + 1)/(c*x - 1)) + 2*a*e^6)
/(c^6*d^6*x^3*e^4 + 3*c^6*d^7*x^2*e^3 + 3*c^6*d^8*x*e^2 + c^6*d^9*e - 3*c^4*d^4*x^3*e^6 - 9*c^4*d^5*x^2*e^5 -
9*c^4*d^6*x*e^4 - 3*c^4*d^7*e^3 + 3*c^2*d^2*x^3*e^8 + 9*c^2*d^3*x^2*e^7 + 9*c^2*d^4*x*e^6 + 3*c^2*d^5*e^5 - x^
3*e^10 - 3*d*x^2*e^9 - 3*d^2*x*e^8 - d^3*e^7)