3.36 \(\int (c+d x^2)^3 \coth ^{-1}(a x) \, dx\)

Optimal. Leaf size=169 \[ \frac {d^2 x^4 \left (21 a^2 c+5 d\right )}{140 a^3}+\frac {d x^2 \left (35 a^4 c^2+21 a^2 c d+5 d^2\right )}{70 a^5}+\frac {\left (35 a^6 c^3+35 a^4 c^2 d+21 a^2 c d^2+5 d^3\right ) \log \left (1-a^2 x^2\right )}{70 a^7}+c^3 x \coth ^{-1}(a x)+c^2 d x^3 \coth ^{-1}(a x)+\frac {3}{5} c d^2 x^5 \coth ^{-1}(a x)+\frac {1}{7} d^3 x^7 \coth ^{-1}(a x)+\frac {d^3 x^6}{42 a} \]

[Out]

1/70*d*(35*a^4*c^2+21*a^2*c*d+5*d^2)*x^2/a^5+1/140*d^2*(21*a^2*c+5*d)*x^4/a^3+1/42*d^3*x^6/a+c^3*x*arccoth(a*x
)+c^2*d*x^3*arccoth(a*x)+3/5*c*d^2*x^5*arccoth(a*x)+1/7*d^3*x^7*arccoth(a*x)+1/70*(35*a^6*c^3+35*a^4*c^2*d+21*
a^2*c*d^2+5*d^3)*ln(-a^2*x^2+1)/a^7

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Rubi [A]  time = 0.13, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {194, 5977, 1810, 260} \[ \frac {d x^2 \left (35 a^4 c^2+21 a^2 c d+5 d^2\right )}{70 a^5}+\frac {\left (35 a^4 c^2 d+35 a^6 c^3+21 a^2 c d^2+5 d^3\right ) \log \left (1-a^2 x^2\right )}{70 a^7}+\frac {d^2 x^4 \left (21 a^2 c+5 d\right )}{140 a^3}+c^2 d x^3 \coth ^{-1}(a x)+c^3 x \coth ^{-1}(a x)+\frac {3}{5} c d^2 x^5 \coth ^{-1}(a x)+\frac {d^3 x^6}{42 a}+\frac {1}{7} d^3 x^7 \coth ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^3*ArcCoth[a*x],x]

[Out]

(d*(35*a^4*c^2 + 21*a^2*c*d + 5*d^2)*x^2)/(70*a^5) + (d^2*(21*a^2*c + 5*d)*x^4)/(140*a^3) + (d^3*x^6)/(42*a) +
c^3*x*ArcCoth[a*x] + c^2*d*x^3*ArcCoth[a*x] + (3*c*d^2*x^5*ArcCoth[a*x])/5 + (d^3*x^7*ArcCoth[a*x])/7 + ((35*
a^6*c^3 + 35*a^4*c^2*d + 21*a^2*c*d^2 + 5*d^3)*Log[1 - a^2*x^2])/(70*a^7)

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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 1810

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

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 )^3 \coth ^{-1}(a x) \, dx &=c^3 x \coth ^{-1}(a x)+c^2 d x^3 \coth ^{-1}(a x)+\frac {3}{5} c d^2 x^5 \coth ^{-1}(a x)+\frac {1}{7} d^3 x^7 \coth ^{-1}(a x)-a \int \frac {c^3 x+c^2 d x^3+\frac {3}{5} c d^2 x^5+\frac {d^3 x^7}{7}}{1-a^2 x^2} \, dx\\ &=c^3 x \coth ^{-1}(a x)+c^2 d x^3 \coth ^{-1}(a x)+\frac {3}{5} c d^2 x^5 \coth ^{-1}(a x)+\frac {1}{7} d^3 x^7 \coth ^{-1}(a x)-a \int \left (-\frac {d \left (35 a^4 c^2+21 a^2 c d+5 d^2\right ) x}{35 a^6}-\frac {d^2 \left (21 a^2 c+5 d\right ) x^3}{35 a^4}-\frac {d^3 x^5}{7 a^2}+\frac {\left (35 a^6 c^3+35 a^4 c^2 d+21 a^2 c d^2+5 d^3\right ) x}{35 a^6 \left (1-a^2 x^2\right )}\right ) \, dx\\ &=\frac {d \left (35 a^4 c^2+21 a^2 c d+5 d^2\right ) x^2}{70 a^5}+\frac {d^2 \left (21 a^2 c+5 d\right ) x^4}{140 a^3}+\frac {d^3 x^6}{42 a}+c^3 x \coth ^{-1}(a x)+c^2 d x^3 \coth ^{-1}(a x)+\frac {3}{5} c d^2 x^5 \coth ^{-1}(a x)+\frac {1}{7} d^3 x^7 \coth ^{-1}(a x)-\frac {\left (35 a^6 c^3+35 a^4 c^2 d+21 a^2 c d^2+5 d^3\right ) \int \frac {x}{1-a^2 x^2} \, dx}{35 a^5}\\ &=\frac {d \left (35 a^4 c^2+21 a^2 c d+5 d^2\right ) x^2}{70 a^5}+\frac {d^2 \left (21 a^2 c+5 d\right ) x^4}{140 a^3}+\frac {d^3 x^6}{42 a}+c^3 x \coth ^{-1}(a x)+c^2 d x^3 \coth ^{-1}(a x)+\frac {3}{5} c d^2 x^5 \coth ^{-1}(a x)+\frac {1}{7} d^3 x^7 \coth ^{-1}(a x)+\frac {\left (35 a^6 c^3+35 a^4 c^2 d+21 a^2 c d^2+5 d^3\right ) \log \left (1-a^2 x^2\right )}{70 a^7}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 150, normalized size = 0.89 \[ \frac {12 a^7 x \coth ^{-1}(a x) \left (35 c^3+35 c^2 d x^2+21 c d^2 x^4+5 d^3 x^6\right )+a^2 d x^2 \left (a^4 \left (210 c^2+63 c d x^2+10 d^2 x^4\right )+3 a^2 d \left (42 c+5 d x^2\right )+30 d^2\right )+6 \left (35 a^6 c^3+35 a^4 c^2 d+21 a^2 c d^2+5 d^3\right ) \log \left (1-a^2 x^2\right )}{420 a^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^3*ArcCoth[a*x],x]

[Out]

(a^2*d*x^2*(30*d^2 + 3*a^2*d*(42*c + 5*d*x^2) + a^4*(210*c^2 + 63*c*d*x^2 + 10*d^2*x^4)) + 12*a^7*x*(35*c^3 +
35*c^2*d*x^2 + 21*c*d^2*x^4 + 5*d^3*x^6)*ArcCoth[a*x] + 6*(35*a^6*c^3 + 35*a^4*c^2*d + 21*a^2*c*d^2 + 5*d^3)*L
og[1 - a^2*x^2])/(420*a^7)

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fricas [A]  time = 0.60, size = 177, normalized size = 1.05 \[ \frac {10 \, a^{6} d^{3} x^{6} + 3 \, {\left (21 \, a^{6} c d^{2} + 5 \, a^{4} d^{3}\right )} x^{4} + 6 \, {\left (35 \, a^{6} c^{2} d + 21 \, a^{4} c d^{2} + 5 \, a^{2} d^{3}\right )} x^{2} + 6 \, {\left (35 \, a^{6} c^{3} + 35 \, a^{4} c^{2} d + 21 \, a^{2} c d^{2} + 5 \, d^{3}\right )} \log \left (a^{2} x^{2} - 1\right ) + 6 \, {\left (5 \, a^{7} d^{3} x^{7} + 21 \, a^{7} c d^{2} x^{5} + 35 \, a^{7} c^{2} d x^{3} + 35 \, a^{7} c^{3} x\right )} \log \left (\frac {a x + 1}{a x - 1}\right )}{420 \, a^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3*arccoth(a*x),x, algorithm="fricas")

[Out]

1/420*(10*a^6*d^3*x^6 + 3*(21*a^6*c*d^2 + 5*a^4*d^3)*x^4 + 6*(35*a^6*c^2*d + 21*a^4*c*d^2 + 5*a^2*d^3)*x^2 + 6
*(35*a^6*c^3 + 35*a^4*c^2*d + 21*a^2*c*d^2 + 5*d^3)*log(a^2*x^2 - 1) + 6*(5*a^7*d^3*x^7 + 21*a^7*c*d^2*x^5 + 3
5*a^7*c^2*d*x^3 + 35*a^7*c^3*x)*log((a*x + 1)/(a*x - 1)))/a^7

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x^{2} + c\right )}^{3} \operatorname {arcoth}\left (a x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3*arccoth(a*x),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)^3*arccoth(a*x), x)

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maple [A]  time = 0.03, size = 233, normalized size = 1.38 \[ \frac {d^{3} x^{7} \mathrm {arccoth}\left (a x \right )}{7}+\frac {3 c \,d^{2} x^{5} \mathrm {arccoth}\left (a x \right )}{5}+c^{2} d \,x^{3} \mathrm {arccoth}\left (a x \right )+c^{3} x \,\mathrm {arccoth}\left (a x \right )+\frac {3 c \,d^{2} x^{2}}{10 a^{3}}+\frac {x^{2} c^{2} d}{2 a}+\frac {3 c \,d^{2} x^{4}}{20 a}+\frac {d^{3} x^{6}}{42 a}+\frac {x^{4} d^{3}}{28 a^{3}}+\frac {\ln \left (a x -1\right ) c^{3}}{2 a}+\frac {\ln \left (a x -1\right ) c^{2} d}{2 a^{3}}+\frac {3 \ln \left (a x -1\right ) c \,d^{2}}{10 a^{5}}+\frac {\ln \left (a x -1\right ) d^{3}}{14 a^{7}}+\frac {d^{3} x^{2}}{14 a^{5}}+\frac {c^{3} \ln \left (a x +1\right )}{2 a}+\frac {\ln \left (a x +1\right ) c^{2} d}{2 a^{3}}+\frac {3 \ln \left (a x +1\right ) c \,d^{2}}{10 a^{5}}+\frac {\ln \left (a x +1\right ) d^{3}}{14 a^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^3*arccoth(a*x),x)

[Out]

1/7*d^3*x^7*arccoth(a*x)+3/5*c*d^2*x^5*arccoth(a*x)+c^2*d*x^3*arccoth(a*x)+c^3*x*arccoth(a*x)+3/10/a^3*c*d^2*x
^2+1/2/a*x^2*c^2*d+3/20/a*c*d^2*x^4+1/42*d^3*x^6/a+1/28/a^3*x^4*d^3+1/2/a*ln(a*x-1)*c^3+1/2/a^3*ln(a*x-1)*c^2*
d+3/10/a^5*ln(a*x-1)*c*d^2+1/14/a^7*ln(a*x-1)*d^3+1/14/a^5*d^3*x^2+1/2*c^3*ln(a*x+1)/a+1/2/a^3*ln(a*x+1)*c^2*d
+3/10/a^5*ln(a*x+1)*c*d^2+1/14/a^7*ln(a*x+1)*d^3

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maxima [A]  time = 0.31, size = 198, normalized size = 1.17 \[ \frac {1}{420} \, a {\left (\frac {10 \, a^{4} d^{3} x^{6} + 3 \, {\left (21 \, a^{4} c d^{2} + 5 \, a^{2} d^{3}\right )} x^{4} + 6 \, {\left (35 \, a^{4} c^{2} d + 21 \, a^{2} c d^{2} + 5 \, d^{3}\right )} x^{2}}{a^{6}} + \frac {6 \, {\left (35 \, a^{6} c^{3} + 35 \, a^{4} c^{2} d + 21 \, a^{2} c d^{2} + 5 \, d^{3}\right )} \log \left (a x + 1\right )}{a^{8}} + \frac {6 \, {\left (35 \, a^{6} c^{3} + 35 \, a^{4} c^{2} d + 21 \, a^{2} c d^{2} + 5 \, d^{3}\right )} \log \left (a x - 1\right )}{a^{8}}\right )} + \frac {1}{35} \, {\left (5 \, d^{3} x^{7} + 21 \, c d^{2} x^{5} + 35 \, c^{2} d x^{3} + 35 \, c^{3} x\right )} \operatorname {arcoth}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3*arccoth(a*x),x, algorithm="maxima")

[Out]

1/420*a*((10*a^4*d^3*x^6 + 3*(21*a^4*c*d^2 + 5*a^2*d^3)*x^4 + 6*(35*a^4*c^2*d + 21*a^2*c*d^2 + 5*d^3)*x^2)/a^6
+ 6*(35*a^6*c^3 + 35*a^4*c^2*d + 21*a^2*c*d^2 + 5*d^3)*log(a*x + 1)/a^8 + 6*(35*a^6*c^3 + 35*a^4*c^2*d + 21*a
^2*c*d^2 + 5*d^3)*log(a*x - 1)/a^8) + 1/35*(5*d^3*x^7 + 21*c*d^2*x^5 + 35*c^2*d*x^3 + 35*c^3*x)*arccoth(a*x)

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mupad [B]  time = 1.52, size = 190, normalized size = 1.12 \[ c^3\,x\,\mathrm {acoth}\left (a\,x\right )+\frac {d^3\,x^7\,\mathrm {acoth}\left (a\,x\right )}{7}+\frac {c^3\,\ln \left (a^2\,x^2-1\right )}{2\,a}+\frac {d^3\,\ln \left (a^2\,x^2-1\right )}{14\,a^7}+\frac {d^3\,x^6}{42\,a}+\frac {d^3\,x^4}{28\,a^3}+\frac {d^3\,x^2}{14\,a^5}+\frac {c^2\,d\,\ln \left (a^2\,x^2-1\right )}{2\,a^3}+\frac {3\,c\,d^2\,\ln \left (a^2\,x^2-1\right )}{10\,a^5}+\frac {c^2\,d\,x^2}{2\,a}+\frac {3\,c\,d^2\,x^4}{20\,a}+\frac {3\,c\,d^2\,x^2}{10\,a^3}+c^2\,d\,x^3\,\mathrm {acoth}\left (a\,x\right )+\frac {3\,c\,d^2\,x^5\,\mathrm {acoth}\left (a\,x\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a*x)*(c + d*x^2)^3,x)

[Out]

c^3*x*acoth(a*x) + (d^3*x^7*acoth(a*x))/7 + (c^3*log(a^2*x^2 - 1))/(2*a) + (d^3*log(a^2*x^2 - 1))/(14*a^7) + (
d^3*x^6)/(42*a) + (d^3*x^4)/(28*a^3) + (d^3*x^2)/(14*a^5) + (c^2*d*log(a^2*x^2 - 1))/(2*a^3) + (3*c*d^2*log(a^
2*x^2 - 1))/(10*a^5) + (c^2*d*x^2)/(2*a) + (3*c*d^2*x^4)/(20*a) + (3*c*d^2*x^2)/(10*a^3) + c^2*d*x^3*acoth(a*x
) + (3*c*d^2*x^5*acoth(a*x))/5

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sympy [A]  time = 4.75, size = 282, normalized size = 1.67 \[ \begin {cases} c^{3} x \operatorname {acoth}{\left (a x \right )} + c^{2} d x^{3} \operatorname {acoth}{\left (a x \right )} + \frac {3 c d^{2} x^{5} \operatorname {acoth}{\left (a x \right )}}{5} + \frac {d^{3} x^{7} \operatorname {acoth}{\left (a x \right )}}{7} + \frac {c^{3} \log {\left (x - \frac {1}{a} \right )}}{a} + \frac {c^{3} \operatorname {acoth}{\left (a x \right )}}{a} + \frac {c^{2} d x^{2}}{2 a} + \frac {3 c d^{2} x^{4}}{20 a} + \frac {d^{3} x^{6}}{42 a} + \frac {c^{2} d \log {\left (x - \frac {1}{a} \right )}}{a^{3}} + \frac {c^{2} d \operatorname {acoth}{\left (a x \right )}}{a^{3}} + \frac {3 c d^{2} x^{2}}{10 a^{3}} + \frac {d^{3} x^{4}}{28 a^{3}} + \frac {3 c d^{2} \log {\left (x - \frac {1}{a} \right )}}{5 a^{5}} + \frac {3 c d^{2} \operatorname {acoth}{\left (a x \right )}}{5 a^{5}} + \frac {d^{3} x^{2}}{14 a^{5}} + \frac {d^{3} \log {\left (x - \frac {1}{a} \right )}}{7 a^{7}} + \frac {d^{3} \operatorname {acoth}{\left (a x \right )}}{7 a^{7}} & \text {for}\: a \neq 0 \\\frac {i \pi \left (c^{3} x + c^{2} d x^{3} + \frac {3 c d^{2} x^{5}}{5} + \frac {d^{3} x^{7}}{7}\right )}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**3*acoth(a*x),x)

[Out]

Piecewise((c**3*x*acoth(a*x) + c**2*d*x**3*acoth(a*x) + 3*c*d**2*x**5*acoth(a*x)/5 + d**3*x**7*acoth(a*x)/7 +
c**3*log(x - 1/a)/a + c**3*acoth(a*x)/a + c**2*d*x**2/(2*a) + 3*c*d**2*x**4/(20*a) + d**3*x**6/(42*a) + c**2*d
*log(x - 1/a)/a**3 + c**2*d*acoth(a*x)/a**3 + 3*c*d**2*x**2/(10*a**3) + d**3*x**4/(28*a**3) + 3*c*d**2*log(x -
1/a)/(5*a**5) + 3*c*d**2*acoth(a*x)/(5*a**5) + d**3*x**2/(14*a**5) + d**3*log(x - 1/a)/(7*a**7) + d**3*acoth(
a*x)/(7*a**7), Ne(a, 0)), (I*pi*(c**3*x + c**2*d*x**3 + 3*c*d**2*x**5/5 + d**3*x**7/7)/2, True))

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