3.18 \(\int \frac {(d+c d x)^2 (a+b \tanh ^{-1}(c x))}{x^5} \, dx\)

Optimal. Leaf size=147 \[ -\frac {c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}+\frac {2}{3} b c^4 d^2 \log (x)-\frac {17}{24} b c^4 d^2 \log (1-c x)+\frac {1}{24} b c^4 d^2 \log (c x+1)-\frac {3 b c^3 d^2}{4 x}-\frac {b c^2 d^2}{3 x^2}-\frac {b c d^2}{12 x^3} \]

[Out]

-1/12*b*c*d^2/x^3-1/3*b*c^2*d^2/x^2-3/4*b*c^3*d^2/x-1/4*d^2*(a+b*arctanh(c*x))/x^4-2/3*c*d^2*(a+b*arctanh(c*x)
)/x^3-1/2*c^2*d^2*(a+b*arctanh(c*x))/x^2+2/3*b*c^4*d^2*ln(x)-17/24*b*c^4*d^2*ln(-c*x+1)+1/24*b*c^4*d^2*ln(c*x+
1)

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Rubi [A]  time = 0.15, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {43, 5936, 12, 1802} \[ -\frac {c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac {b c^2 d^2}{3 x^2}-\frac {3 b c^3 d^2}{4 x}+\frac {2}{3} b c^4 d^2 \log (x)-\frac {17}{24} b c^4 d^2 \log (1-c x)+\frac {1}{24} b c^4 d^2 \log (c x+1)-\frac {b c d^2}{12 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)^2*(a + b*ArcTanh[c*x]))/x^5,x]

[Out]

-(b*c*d^2)/(12*x^3) - (b*c^2*d^2)/(3*x^2) - (3*b*c^3*d^2)/(4*x) - (d^2*(a + b*ArcTanh[c*x]))/(4*x^4) - (2*c*d^
2*(a + b*ArcTanh[c*x]))/(3*x^3) - (c^2*d^2*(a + b*ArcTanh[c*x]))/(2*x^2) + (2*b*c^4*d^2*Log[x])/3 - (17*b*c^4*
d^2*Log[1 - c*x])/24 + (b*c^4*d^2*Log[1 + c*x])/24

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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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)^2 \left (a+b \tanh ^{-1}(c x)\right )}{x^5} \, dx &=-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-(b c) \int \frac {d^2 \left (-3-8 c x-6 c^2 x^2\right )}{12 x^4 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {1}{12} \left (b c d^2\right ) \int \frac {-3-8 c x-6 c^2 x^2}{x^4 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}-\frac {1}{12} \left (b c d^2\right ) \int \left (-\frac {3}{x^4}-\frac {8 c}{x^3}-\frac {9 c^2}{x^2}-\frac {8 c^3}{x}+\frac {17 c^4}{2 (-1+c x)}-\frac {c^4}{2 (1+c x)}\right ) \, dx\\ &=-\frac {b c d^2}{12 x^3}-\frac {b c^2 d^2}{3 x^2}-\frac {3 b c^3 d^2}{4 x}-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 x^2}+\frac {2}{3} b c^4 d^2 \log (x)-\frac {17}{24} b c^4 d^2 \log (1-c x)+\frac {1}{24} b c^4 d^2 \log (1+c x)\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 114, normalized size = 0.78 \[ -\frac {d^2 \left (12 a c^2 x^2+16 a c x+6 a-16 b c^4 x^4 \log (x)+17 b c^4 x^4 \log (1-c x)-b c^4 x^4 \log (c x+1)+18 b c^3 x^3+8 b c^2 x^2+2 b \left (6 c^2 x^2+8 c x+3\right ) \tanh ^{-1}(c x)+2 b c x\right )}{24 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + c*d*x)^2*(a + b*ArcTanh[c*x]))/x^5,x]

[Out]

-1/24*(d^2*(6*a + 16*a*c*x + 2*b*c*x + 12*a*c^2*x^2 + 8*b*c^2*x^2 + 18*b*c^3*x^3 + 2*b*(3 + 8*c*x + 6*c^2*x^2)
*ArcTanh[c*x] - 16*b*c^4*x^4*Log[x] + 17*b*c^4*x^4*Log[1 - c*x] - b*c^4*x^4*Log[1 + c*x]))/x^4

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fricas [A]  time = 0.70, size = 147, normalized size = 1.00 \[ \frac {b c^{4} d^{2} x^{4} \log \left (c x + 1\right ) - 17 \, b c^{4} d^{2} x^{4} \log \left (c x - 1\right ) + 16 \, b c^{4} d^{2} x^{4} \log \relax (x) - 18 \, b c^{3} d^{2} x^{3} - 4 \, {\left (3 \, a + 2 \, b\right )} c^{2} d^{2} x^{2} - 2 \, {\left (8 \, a + b\right )} c d^{2} x - 6 \, a d^{2} - {\left (6 \, b c^{2} d^{2} x^{2} + 8 \, b c d^{2} x + 3 \, b d^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{24 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^2*(a+b*arctanh(c*x))/x^5,x, algorithm="fricas")

[Out]

1/24*(b*c^4*d^2*x^4*log(c*x + 1) - 17*b*c^4*d^2*x^4*log(c*x - 1) + 16*b*c^4*d^2*x^4*log(x) - 18*b*c^3*d^2*x^3
- 4*(3*a + 2*b)*c^2*d^2*x^2 - 2*(8*a + b)*c*d^2*x - 6*a*d^2 - (6*b*c^2*d^2*x^2 + 8*b*c*d^2*x + 3*b*d^2)*log(-(
c*x + 1)/(c*x - 1)))/x^4

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giac [B]  time = 0.28, size = 431, normalized size = 2.93 \[ \frac {1}{3} \, {\left (2 \, b c^{3} d^{2} \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - 2 \, b c^{3} d^{2} \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {2 \, {\left (\frac {6 \, {\left (c x + 1\right )}^{3} b c^{3} d^{2}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} b c^{3} d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )} b c^{3} d^{2}}{c x - 1} + b c^{3} d^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {4 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {\frac {24 \, {\left (c x + 1\right )}^{3} a c^{3} d^{2}}{{\left (c x - 1\right )}^{3}} + \frac {24 \, {\left (c x + 1\right )}^{2} a c^{3} d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {16 \, {\left (c x + 1\right )} a c^{3} d^{2}}{c x - 1} + 4 \, a c^{3} d^{2} + \frac {10 \, {\left (c x + 1\right )}^{3} b c^{3} d^{2}}{{\left (c x - 1\right )}^{3}} + \frac {23 \, {\left (c x + 1\right )}^{2} b c^{3} d^{2}}{{\left (c x - 1\right )}^{2}} + \frac {18 \, {\left (c x + 1\right )} b c^{3} d^{2}}{c x - 1} + 5 \, b c^{3} d^{2}}{\frac {{\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {4 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )}}{c x - 1} + 1}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^2*(a+b*arctanh(c*x))/x^5,x, algorithm="giac")

[Out]

1/3*(2*b*c^3*d^2*log(-(c*x + 1)/(c*x - 1) - 1) - 2*b*c^3*d^2*log(-(c*x + 1)/(c*x - 1)) + 2*(6*(c*x + 1)^3*b*c^
3*d^2/(c*x - 1)^3 + 6*(c*x + 1)^2*b*c^3*d^2/(c*x - 1)^2 + 4*(c*x + 1)*b*c^3*d^2/(c*x - 1) + b*c^3*d^2)*log(-(c
*x + 1)/(c*x - 1))/((c*x + 1)^4/(c*x - 1)^4 + 4*(c*x + 1)^3/(c*x - 1)^3 + 6*(c*x + 1)^2/(c*x - 1)^2 + 4*(c*x +
1)/(c*x - 1) + 1) + (24*(c*x + 1)^3*a*c^3*d^2/(c*x - 1)^3 + 24*(c*x + 1)^2*a*c^3*d^2/(c*x - 1)^2 + 16*(c*x +
1)*a*c^3*d^2/(c*x - 1) + 4*a*c^3*d^2 + 10*(c*x + 1)^3*b*c^3*d^2/(c*x - 1)^3 + 23*(c*x + 1)^2*b*c^3*d^2/(c*x -
1)^2 + 18*(c*x + 1)*b*c^3*d^2/(c*x - 1) + 5*b*c^3*d^2)/((c*x + 1)^4/(c*x - 1)^4 + 4*(c*x + 1)^3/(c*x - 1)^3 +
6*(c*x + 1)^2/(c*x - 1)^2 + 4*(c*x + 1)/(c*x - 1) + 1))*c

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maple [A]  time = 0.04, size = 153, normalized size = 1.04 \[ -\frac {2 c \,d^{2} a}{3 x^{3}}-\frac {c^{2} d^{2} a}{2 x^{2}}-\frac {d^{2} a}{4 x^{4}}-\frac {2 c \,d^{2} b \arctanh \left (c x \right )}{3 x^{3}}-\frac {c^{2} d^{2} b \arctanh \left (c x \right )}{2 x^{2}}-\frac {d^{2} b \arctanh \left (c x \right )}{4 x^{4}}-\frac {b c \,d^{2}}{12 x^{3}}-\frac {b \,c^{2} d^{2}}{3 x^{2}}-\frac {3 b \,c^{3} d^{2}}{4 x}+\frac {2 c^{4} d^{2} b \ln \left (c x \right )}{3}-\frac {17 c^{4} d^{2} b \ln \left (c x -1\right )}{24}+\frac {b \,c^{4} d^{2} \ln \left (c x +1\right )}{24} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^2*(a+b*arctanh(c*x))/x^5,x)

[Out]

-2/3*c*d^2*a/x^3-1/2*c^2*d^2*a/x^2-1/4*d^2*a/x^4-2/3*c*d^2*b*arctanh(c*x)/x^3-1/2*c^2*d^2*b*arctanh(c*x)/x^2-1
/4*d^2*b*arctanh(c*x)/x^4-1/12*b*c*d^2/x^3-1/3*b*c^2*d^2/x^2-3/4*b*c^3*d^2/x+2/3*c^4*d^2*b*ln(c*x)-17/24*c^4*d
^2*b*ln(c*x-1)+1/24*b*c^4*d^2*ln(c*x+1)

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maxima [A]  time = 0.32, size = 178, normalized size = 1.21 \[ \frac {1}{4} \, {\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{2}}\right )} b c^{2} d^{2} - \frac {1}{3} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{3}}\right )} b c d^{2} + \frac {1}{24} \, {\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x\right )}{x^{4}}\right )} b d^{2} - \frac {a c^{2} d^{2}}{2 \, x^{2}} - \frac {2 \, a c d^{2}}{3 \, x^{3}} - \frac {a d^{2}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^2*(a+b*arctanh(c*x))/x^5,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^2*d^2 - 1/3*((c^2*log(c^2*x^2 - 1) -
c^2*log(x^2) + 1/x^2)*c + 2*arctanh(c*x)/x^3)*b*c*d^2 + 1/24*((3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*
c^2*x^2 + 1)/x^3)*c - 6*arctanh(c*x)/x^4)*b*d^2 - 1/2*a*c^2*d^2/x^2 - 2/3*a*c*d^2/x^3 - 1/4*a*d^2/x^4

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mupad [B]  time = 1.01, size = 168, normalized size = 1.14 \[ \frac {2\,b\,c^4\,d^2\,\ln \relax (x)}{3}-\frac {b\,c^4\,d^2\,\ln \left (c^2\,x^2-1\right )}{3}-\frac {a\,c^2\,d^2}{2\,x^2}-\frac {b\,c^2\,d^2}{3\,x^2}-\frac {3\,b\,c^3\,d^2}{4\,x}-\frac {a\,d^2}{4\,x^4}-\frac {2\,a\,c\,d^2}{3\,x^3}-\frac {b\,c\,d^2}{12\,x^3}-\frac {b\,d^2\,\mathrm {atanh}\left (c\,x\right )}{4\,x^4}-\frac {3\,b\,c^5\,d^2\,\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {-c^2}}\right )}{4\,\sqrt {-c^2}}-\frac {2\,b\,c\,d^2\,\mathrm {atanh}\left (c\,x\right )}{3\,x^3}-\frac {b\,c^2\,d^2\,\mathrm {atanh}\left (c\,x\right )}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atanh(c*x))*(d + c*d*x)^2)/x^5,x)

[Out]

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

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sympy [A]  time = 2.02, size = 189, normalized size = 1.29 \[ \begin {cases} - \frac {a c^{2} d^{2}}{2 x^{2}} - \frac {2 a c d^{2}}{3 x^{3}} - \frac {a d^{2}}{4 x^{4}} + \frac {2 b c^{4} d^{2} \log {\relax (x )}}{3} - \frac {2 b c^{4} d^{2} \log {\left (x - \frac {1}{c} \right )}}{3} + \frac {b c^{4} d^{2} \operatorname {atanh}{\left (c x \right )}}{12} - \frac {3 b c^{3} d^{2}}{4 x} - \frac {b c^{2} d^{2} \operatorname {atanh}{\left (c x \right )}}{2 x^{2}} - \frac {b c^{2} d^{2}}{3 x^{2}} - \frac {2 b c d^{2} \operatorname {atanh}{\left (c x \right )}}{3 x^{3}} - \frac {b c d^{2}}{12 x^{3}} - \frac {b d^{2} \operatorname {atanh}{\left (c x \right )}}{4 x^{4}} & \text {for}\: c \neq 0 \\- \frac {a d^{2}}{4 x^{4}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)**2*(a+b*atanh(c*x))/x**5,x)

[Out]

Piecewise((-a*c**2*d**2/(2*x**2) - 2*a*c*d**2/(3*x**3) - a*d**2/(4*x**4) + 2*b*c**4*d**2*log(x)/3 - 2*b*c**4*d
**2*log(x - 1/c)/3 + b*c**4*d**2*atanh(c*x)/12 - 3*b*c**3*d**2/(4*x) - b*c**2*d**2*atanh(c*x)/(2*x**2) - b*c**
2*d**2/(3*x**2) - 2*b*c*d**2*atanh(c*x)/(3*x**3) - b*c*d**2/(12*x**3) - b*d**2*atanh(c*x)/(4*x**4), Ne(c, 0)),
(-a*d**2/(4*x**4), True))

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