3.37 \(\int (c+d x^2)^2 \coth ^{-1}(a x) \, dx\)

Optimal. Leaf size=110 \[ \frac {d x^2 \left (10 a^2 c+3 d\right )}{30 a^3}+\frac {\left (15 a^4 c^2+10 a^2 c d+3 d^2\right ) \log \left (1-a^2 x^2\right )}{30 a^5}+c^2 x \coth ^{-1}(a x)+\frac {2}{3} c d x^3 \coth ^{-1}(a x)+\frac {1}{5} d^2 x^5 \coth ^{-1}(a x)+\frac {d^2 x^4}{20 a} \]

[Out]

1/30*d*(10*a^2*c+3*d)*x^2/a^3+1/20*d^2*x^4/a+c^2*x*arccoth(a*x)+2/3*c*d*x^3*arccoth(a*x)+1/5*d^2*x^5*arccoth(a
*x)+1/30*(15*a^4*c^2+10*a^2*c*d+3*d^2)*ln(-a^2*x^2+1)/a^5

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Rubi [A]  time = 0.13, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {194, 5977, 1594, 1247, 698} \[ \frac {\left (15 a^4 c^2+10 a^2 c d+3 d^2\right ) \log \left (1-a^2 x^2\right )}{30 a^5}+\frac {d x^2 \left (10 a^2 c+3 d\right )}{30 a^3}+c^2 x \coth ^{-1}(a x)+\frac {2}{3} c d x^3 \coth ^{-1}(a x)+\frac {d^2 x^4}{20 a}+\frac {1}{5} d^2 x^5 \coth ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^2*ArcCoth[a*x],x]

[Out]

(d*(10*a^2*c + 3*d)*x^2)/(30*a^3) + (d^2*x^4)/(20*a) + c^2*x*ArcCoth[a*x] + (2*c*d*x^3*ArcCoth[a*x])/3 + (d^2*
x^5*ArcCoth[a*x])/5 + ((15*a^4*c^2 + 10*a^2*c*d + 3*d^2)*Log[1 - a^2*x^2])/(30*a^5)

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 5977

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(d + e*x^
2)^q, x]}, Dist[a + b*ArcCoth[c*x], u, x] - Dist[b*c, Int[u/(1 - c^2*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x
] && (IntegerQ[q] || ILtQ[q + 1/2, 0])

Rubi steps

\begin {align*} \int \left (c+d x^2\right )^2 \coth ^{-1}(a x) \, dx &=c^2 x \coth ^{-1}(a x)+\frac {2}{3} c d x^3 \coth ^{-1}(a x)+\frac {1}{5} d^2 x^5 \coth ^{-1}(a x)-a \int \frac {c^2 x+\frac {2}{3} c d x^3+\frac {d^2 x^5}{5}}{1-a^2 x^2} \, dx\\ &=c^2 x \coth ^{-1}(a x)+\frac {2}{3} c d x^3 \coth ^{-1}(a x)+\frac {1}{5} d^2 x^5 \coth ^{-1}(a x)-a \int \frac {x \left (c^2+\frac {2}{3} c d x^2+\frac {d^2 x^4}{5}\right )}{1-a^2 x^2} \, dx\\ &=c^2 x \coth ^{-1}(a x)+\frac {2}{3} c d x^3 \coth ^{-1}(a x)+\frac {1}{5} d^2 x^5 \coth ^{-1}(a x)-\frac {1}{2} a \operatorname {Subst}\left (\int \frac {c^2+\frac {2 c d x}{3}+\frac {d^2 x^2}{5}}{1-a^2 x} \, dx,x,x^2\right )\\ &=c^2 x \coth ^{-1}(a x)+\frac {2}{3} c d x^3 \coth ^{-1}(a x)+\frac {1}{5} d^2 x^5 \coth ^{-1}(a x)-\frac {1}{2} a \operatorname {Subst}\left (\int \left (-\frac {d \left (10 a^2 c+3 d\right )}{15 a^4}-\frac {d^2 x}{5 a^2}+\frac {-15 a^4 c^2-10 a^2 c d-3 d^2}{15 a^4 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac {d \left (10 a^2 c+3 d\right ) x^2}{30 a^3}+\frac {d^2 x^4}{20 a}+c^2 x \coth ^{-1}(a x)+\frac {2}{3} c d x^3 \coth ^{-1}(a x)+\frac {1}{5} d^2 x^5 \coth ^{-1}(a x)+\frac {\left (15 a^4 c^2+10 a^2 c d+3 d^2\right ) \log \left (1-a^2 x^2\right )}{30 a^5}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 98, normalized size = 0.89 \[ \frac {4 a^5 x \coth ^{-1}(a x) \left (15 c^2+10 c d x^2+3 d^2 x^4\right )+a^2 d x^2 \left (a^2 \left (20 c+3 d x^2\right )+6 d\right )+\left (30 a^4 c^2+20 a^2 c d+6 d^2\right ) \log \left (1-a^2 x^2\right )}{60 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^2*ArcCoth[a*x],x]

[Out]

(a^2*d*x^2*(6*d + a^2*(20*c + 3*d*x^2)) + 4*a^5*x*(15*c^2 + 10*c*d*x^2 + 3*d^2*x^4)*ArcCoth[a*x] + (30*a^4*c^2
+ 20*a^2*c*d + 6*d^2)*Log[1 - a^2*x^2])/(60*a^5)

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fricas [A]  time = 0.46, size = 118, normalized size = 1.07 \[ \frac {3 \, a^{4} d^{2} x^{4} + 2 \, {\left (10 \, a^{4} c d + 3 \, a^{2} d^{2}\right )} x^{2} + 2 \, {\left (15 \, a^{4} c^{2} + 10 \, a^{2} c d + 3 \, d^{2}\right )} \log \left (a^{2} x^{2} - 1\right ) + 2 \, {\left (3 \, a^{5} d^{2} x^{5} + 10 \, a^{5} c d x^{3} + 15 \, a^{5} c^{2} x\right )} \log \left (\frac {a x + 1}{a x - 1}\right )}{60 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^2*arccoth(a*x),x, algorithm="fricas")

[Out]

1/60*(3*a^4*d^2*x^4 + 2*(10*a^4*c*d + 3*a^2*d^2)*x^2 + 2*(15*a^4*c^2 + 10*a^2*c*d + 3*d^2)*log(a^2*x^2 - 1) +
2*(3*a^5*d^2*x^5 + 10*a^5*c*d*x^3 + 15*a^5*c^2*x)*log((a*x + 1)/(a*x - 1)))/a^5

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x^{2} + c\right )}^{2} \operatorname {arcoth}\left (a x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^2*arccoth(a*x),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)^2*arccoth(a*x), x)

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maple [A]  time = 0.04, size = 148, normalized size = 1.35 \[ \frac {d^{2} x^{5} \mathrm {arccoth}\left (a x \right )}{5}+\frac {2 c d \,x^{3} \mathrm {arccoth}\left (a x \right )}{3}+c^{2} x \,\mathrm {arccoth}\left (a x \right )+\frac {d^{2} x^{4}}{20 a}+\frac {c d \,x^{2}}{3 a}+\frac {x^{2} d^{2}}{10 a^{3}}+\frac {\ln \left (a x -1\right ) c^{2}}{2 a}+\frac {\ln \left (a x -1\right ) c d}{3 a^{3}}+\frac {\ln \left (a x -1\right ) d^{2}}{10 a^{5}}+\frac {c^{2} \ln \left (a x +1\right )}{2 a}+\frac {\ln \left (a x +1\right ) c d}{3 a^{3}}+\frac {\ln \left (a x +1\right ) d^{2}}{10 a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^2*arccoth(a*x),x)

[Out]

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

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maxima [A]  time = 0.30, size = 131, normalized size = 1.19 \[ \frac {1}{60} \, a {\left (\frac {3 \, a^{2} d^{2} x^{4} + 2 \, {\left (10 \, a^{2} c d + 3 \, d^{2}\right )} x^{2}}{a^{4}} + \frac {2 \, {\left (15 \, a^{4} c^{2} + 10 \, a^{2} c d + 3 \, d^{2}\right )} \log \left (a x + 1\right )}{a^{6}} + \frac {2 \, {\left (15 \, a^{4} c^{2} + 10 \, a^{2} c d + 3 \, d^{2}\right )} \log \left (a x - 1\right )}{a^{6}}\right )} + \frac {1}{15} \, {\left (3 \, d^{2} x^{5} + 10 \, c d x^{3} + 15 \, c^{2} x\right )} \operatorname {arcoth}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^2*arccoth(a*x),x, algorithm="maxima")

[Out]

1/60*a*((3*a^2*d^2*x^4 + 2*(10*a^2*c*d + 3*d^2)*x^2)/a^4 + 2*(15*a^4*c^2 + 10*a^2*c*d + 3*d^2)*log(a*x + 1)/a^
6 + 2*(15*a^4*c^2 + 10*a^2*c*d + 3*d^2)*log(a*x - 1)/a^6) + 1/15*(3*d^2*x^5 + 10*c*d*x^3 + 15*c^2*x)*arccoth(a
*x)

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mupad [B]  time = 1.37, size = 115, normalized size = 1.05 \[ \frac {a^4\,\left (\frac {c^2\,\ln \left (a^2\,x^2-1\right )}{2}+\frac {d^2\,x^4}{20}+\frac {c\,d\,x^2}{3}\right )+a^2\,\left (\frac {d^2\,x^2}{10}+\frac {c\,d\,\ln \left (a^2\,x^2-1\right )}{3}\right )+\frac {d^2\,\ln \left (a^2\,x^2-1\right )}{10}}{a^5}+c^2\,x\,\mathrm {acoth}\left (a\,x\right )+\frac {d^2\,x^5\,\mathrm {acoth}\left (a\,x\right )}{5}+\frac {2\,c\,d\,x^3\,\mathrm {acoth}\left (a\,x\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a*x)*(c + d*x^2)^2,x)

[Out]

(a^4*((c^2*log(a^2*x^2 - 1))/2 + (d^2*x^4)/20 + (c*d*x^2)/3) + a^2*((d^2*x^2)/10 + (c*d*log(a^2*x^2 - 1))/3) +
(d^2*log(a^2*x^2 - 1))/10)/a^5 + c^2*x*acoth(a*x) + (d^2*x^5*acoth(a*x))/5 + (2*c*d*x^3*acoth(a*x))/3

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sympy [A]  time = 2.13, size = 182, normalized size = 1.65 \[ \begin {cases} c^{2} x \operatorname {acoth}{\left (a x \right )} + \frac {2 c d x^{3} \operatorname {acoth}{\left (a x \right )}}{3} + \frac {d^{2} x^{5} \operatorname {acoth}{\left (a x \right )}}{5} + \frac {c^{2} \log {\left (x - \frac {1}{a} \right )}}{a} + \frac {c^{2} \operatorname {acoth}{\left (a x \right )}}{a} + \frac {c d x^{2}}{3 a} + \frac {d^{2} x^{4}}{20 a} + \frac {2 c d \log {\left (x - \frac {1}{a} \right )}}{3 a^{3}} + \frac {2 c d \operatorname {acoth}{\left (a x \right )}}{3 a^{3}} + \frac {d^{2} x^{2}}{10 a^{3}} + \frac {d^{2} \log {\left (x - \frac {1}{a} \right )}}{5 a^{5}} + \frac {d^{2} \operatorname {acoth}{\left (a x \right )}}{5 a^{5}} & \text {for}\: a \neq 0 \\\frac {i \pi \left (c^{2} x + \frac {2 c d x^{3}}{3} + \frac {d^{2} x^{5}}{5}\right )}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**2*acoth(a*x),x)

[Out]

Piecewise((c**2*x*acoth(a*x) + 2*c*d*x**3*acoth(a*x)/3 + d**2*x**5*acoth(a*x)/5 + c**2*log(x - 1/a)/a + c**2*a
coth(a*x)/a + c*d*x**2/(3*a) + d**2*x**4/(20*a) + 2*c*d*log(x - 1/a)/(3*a**3) + 2*c*d*acoth(a*x)/(3*a**3) + d*
*2*x**2/(10*a**3) + d**2*log(x - 1/a)/(5*a**5) + d**2*acoth(a*x)/(5*a**5), Ne(a, 0)), (I*pi*(c**2*x + 2*c*d*x*
*3/3 + d**2*x**5/5)/2, True))

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