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

Optimal. Leaf size=98 \[ -\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {c d \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}+\frac {1}{3} b c^3 d \log (x)-\frac {5}{12} b c^3 d \log (1-c x)+\frac {1}{12} b c^3 d \log (c x+1)-\frac {b c^2 d}{2 x}-\frac {b c d}{6 x^2} \]

[Out]

-1/6*b*c*d/x^2-1/2*b*c^2*d/x-1/3*d*(a+b*arctanh(c*x))/x^3-1/2*c*d*(a+b*arctanh(c*x))/x^2+1/3*b*c^3*d*ln(x)-5/1
2*b*c^3*d*ln(-c*x+1)+1/12*b*c^3*d*ln(c*x+1)

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Rubi [A]  time = 0.09, antiderivative size = 98, 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, 801} \[ -\frac {c d \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {b c^2 d}{2 x}+\frac {1}{3} b c^3 d \log (x)-\frac {5}{12} b c^3 d \log (1-c x)+\frac {1}{12} b c^3 d \log (c x+1)-\frac {b c d}{6 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.07, size = 86, normalized size = 0.88 \[ -\frac {d \left (6 a c x+4 a-4 b c^3 x^3 \log (x)+5 b c^3 x^3 \log (1-c x)-b c^3 x^3 \log (c x+1)+6 b c^2 x^2+2 b c x+2 b (3 c x+2) \tanh ^{-1}(c x)\right )}{12 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 101, normalized size = 1.03 \[ \frac {b c^{3} d x^{3} \log \left (c x + 1\right ) - 5 \, b c^{3} d x^{3} \log \left (c x - 1\right ) + 4 \, b c^{3} d x^{3} \log \relax (x) - 6 \, b c^{2} d x^{2} - 2 \, {\left (3 \, a + b\right )} c d x - 4 \, a d - {\left (3 \, b c d x + 2 \, b d\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.29, size = 306, normalized size = 3.12 \[ \frac {1}{3} \, {\left (b c^{2} d \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - b c^{2} d \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {{\left (\frac {6 \, {\left (c x + 1\right )}^{2} b c^{2} d}{{\left (c x - 1\right )}^{2}} + \frac {3 \, {\left (c x + 1\right )} b c^{2} d}{c x - 1} + b c^{2} d\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {3 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {3 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {\frac {12 \, {\left (c x + 1\right )}^{2} a c^{2} d}{{\left (c x - 1\right )}^{2}} + \frac {6 \, {\left (c x + 1\right )} a c^{2} d}{c x - 1} + 2 \, a c^{2} d + \frac {5 \, {\left (c x + 1\right )}^{2} b c^{2} d}{{\left (c x - 1\right )}^{2}} + \frac {8 \, {\left (c x + 1\right )} b c^{2} d}{c x - 1} + 3 \, b c^{2} d}{\frac {{\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {3 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {3 \, {\left (c x + 1\right )}}{c x - 1} + 1}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*c^2*d*log(-(c*x + 1)/(c*x - 1) - 1) - b*c^2*d*log(-(c*x + 1)/(c*x - 1)) + (6*(c*x + 1)^2*b*c^2*d/(c*x -
1)^2 + 3*(c*x + 1)*b*c^2*d/(c*x - 1) + b*c^2*d)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^3/(c*x - 1)^3 + 3*(c*x +
1)^2/(c*x - 1)^2 + 3*(c*x + 1)/(c*x - 1) + 1) + (12*(c*x + 1)^2*a*c^2*d/(c*x - 1)^2 + 6*(c*x + 1)*a*c^2*d/(c*
x - 1) + 2*a*c^2*d + 5*(c*x + 1)^2*b*c^2*d/(c*x - 1)^2 + 8*(c*x + 1)*b*c^2*d/(c*x - 1) + 3*b*c^2*d)/((c*x + 1)
^3/(c*x - 1)^3 + 3*(c*x + 1)^2/(c*x - 1)^2 + 3*(c*x + 1)/(c*x - 1) + 1))*c

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maple [A]  time = 0.04, size = 95, normalized size = 0.97 \[ -\frac {d a}{3 x^{3}}-\frac {c d a}{2 x^{2}}-\frac {d b \arctanh \left (c x \right )}{3 x^{3}}-\frac {c d b \arctanh \left (c x \right )}{2 x^{2}}-\frac {b c d}{6 x^{2}}-\frac {b \,c^{2} d}{2 x}+\frac {c^{3} d b \ln \left (c x \right )}{3}-\frac {5 c^{3} d b \ln \left (c x -1\right )}{12}+\frac {b \,c^{3} d \ln \left (c x +1\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*d*a/x^3-1/2*c*d*a/x^2-1/3*d*b*arctanh(c*x)/x^3-1/2*c*d*b*arctanh(c*x)/x^2-1/6*b*c*d/x^2-1/2*b*c^2*d/x+1/3
*c^3*d*b*ln(c*x)-5/12*c^3*d*b*ln(c*x-1)+1/12*b*c^3*d*ln(c*x+1)

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maxima [A]  time = 0.31, size = 99, normalized size = 1.01 \[ \frac {1}{4} \, {\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{2}}\right )} b c d - \frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{3}}\right )} b d - \frac {a c d}{2 \, x^{2}} - \frac {a d}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.90, size = 110, normalized size = 1.12 \[ \frac {b\,c^3\,d\,\ln \relax (x)}{3}-\frac {a\,c\,d}{2\,x^2}-\frac {b\,c\,d}{6\,x^2}-\frac {b\,d\,\mathrm {atanh}\left (c\,x\right )}{3\,x^3}-\frac {b\,c^3\,d\,\ln \left (c^2\,x^2-1\right )}{6}-\frac {b\,c^2\,d}{2\,x}-\frac {a\,d}{3\,x^3}-\frac {b\,c^4\,d\,\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {-c^2}}\right )}{2\,\sqrt {-c^2}}-\frac {b\,c\,d\,\mathrm {atanh}\left (c\,x\right )}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atanh(c*x))*(d + c*d*x))/x^4,x)

[Out]

(b*c^3*d*log(x))/3 - (a*c*d)/(2*x^2) - (b*c*d)/(6*x^2) - (b*d*atanh(c*x))/(3*x^3) - (b*c^3*d*log(c^2*x^2 - 1))
/6 - (b*c^2*d)/(2*x) - (a*d)/(3*x^3) - (b*c^4*d*atan((c^2*x)/(-c^2)^(1/2)))/(2*(-c^2)^(1/2)) - (b*c*d*atanh(c*
x))/(2*x^2)

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sympy [A]  time = 1.42, size = 117, normalized size = 1.19 \[ \begin {cases} - \frac {a c d}{2 x^{2}} - \frac {a d}{3 x^{3}} + \frac {b c^{3} d \log {\relax (x )}}{3} - \frac {b c^{3} d \log {\left (x - \frac {1}{c} \right )}}{3} + \frac {b c^{3} d \operatorname {atanh}{\left (c x \right )}}{6} - \frac {b c^{2} d}{2 x} - \frac {b c d \operatorname {atanh}{\left (c x \right )}}{2 x^{2}} - \frac {b c d}{6 x^{2}} - \frac {b d \operatorname {atanh}{\left (c x \right )}}{3 x^{3}} & \text {for}\: c \neq 0 \\- \frac {a d}{3 x^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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