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

Optimal. Leaf size=62 \[ \frac {x \coth ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {\tanh ^{-1}\left (\frac {a \sqrt {c+d x^2}}{\sqrt {a^2 c+d}}\right )}{c \sqrt {a^2 c+d}} \]

[Out]

-arctanh(a*(d*x^2+c)^(1/2)/(a^2*c+d)^(1/2))/c/(a^2*c+d)^(1/2)+x*arccoth(a*x)/c/(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {191, 5977, 12, 444, 63, 208} \[ \frac {x \coth ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {\tanh ^{-1}\left (\frac {a \sqrt {c+d x^2}}{\sqrt {a^2 c+d}}\right )}{c \sqrt {a^2 c+d}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a*x]/(c + d*x^2)^(3/2),x]

[Out]

(x*ArcCoth[a*x])/(c*Sqrt[c + d*x^2]) - ArcTanh[(a*Sqrt[c + d*x^2])/Sqrt[a^2*c + d]]/(c*Sqrt[a^2*c + d])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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 \frac {\coth ^{-1}(a x)}{\left (c+d x^2\right )^{3/2}} \, dx &=\frac {x \coth ^{-1}(a x)}{c \sqrt {c+d x^2}}-a \int \frac {x}{c \left (1-a^2 x^2\right ) \sqrt {c+d x^2}} \, dx\\ &=\frac {x \coth ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {a \int \frac {x}{\left (1-a^2 x^2\right ) \sqrt {c+d x^2}} \, dx}{c}\\ &=\frac {x \coth ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\left (1-a^2 x\right ) \sqrt {c+d x}} \, dx,x,x^2\right )}{2 c}\\ &=\frac {x \coth ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {a \operatorname {Subst}\left (\int \frac {1}{1+\frac {a^2 c}{d}-\frac {a^2 x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{c d}\\ &=\frac {x \coth ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {\tanh ^{-1}\left (\frac {a \sqrt {c+d x^2}}{\sqrt {a^2 c+d}}\right )}{c \sqrt {a^2 c+d}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 119, normalized size = 1.92 \[ \frac {\frac {-\log \left (\sqrt {a^2 c+d} \sqrt {c+d x^2}+a c-d x\right )-\log \left (\sqrt {a^2 c+d} \sqrt {c+d x^2}+a c+d x\right )+\log (1-a x)+\log (a x+1)}{\sqrt {a^2 c+d}}+\frac {2 x \coth ^{-1}(a x)}{\sqrt {c+d x^2}}}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*x*ArcCoth[a*x])/Sqrt[c + d*x^2] + (Log[1 - a*x] + Log[1 + a*x] - Log[a*c - d*x + Sqrt[a^2*c + d]*Sqrt[c +
d*x^2]] - Log[a*c + d*x + Sqrt[a^2*c + d]*Sqrt[c + d*x^2]])/Sqrt[a^2*c + d])/(2*c)

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fricas [B]  time = 0.50, size = 354, normalized size = 5.71 \[ \left [\frac {2 \, {\left (a^{2} c + d\right )} \sqrt {d x^{2} + c} x \log \left (\frac {a x + 1}{a x - 1}\right ) + \sqrt {a^{2} c + d} {\left (d x^{2} + c\right )} \log \left (\frac {a^{4} d^{2} x^{4} + 8 \, a^{4} c^{2} + 8 \, a^{2} c d + 2 \, {\left (4 \, a^{4} c d + 3 \, a^{2} d^{2}\right )} x^{2} - 4 \, {\left (a^{3} d x^{2} + 2 \, a^{3} c + a d\right )} \sqrt {a^{2} c + d} \sqrt {d x^{2} + c} + d^{2}}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1}\right )}{4 \, {\left (a^{2} c^{3} + c^{2} d + {\left (a^{2} c^{2} d + c d^{2}\right )} x^{2}\right )}}, \frac {{\left (a^{2} c + d\right )} \sqrt {d x^{2} + c} x \log \left (\frac {a x + 1}{a x - 1}\right ) + \sqrt {-a^{2} c - d} {\left (d x^{2} + c\right )} \arctan \left (\frac {{\left (a^{2} d x^{2} + 2 \, a^{2} c + d\right )} \sqrt {-a^{2} c - d} \sqrt {d x^{2} + c}}{2 \, {\left (a^{3} c^{2} + a c d + {\left (a^{3} c d + a d^{2}\right )} x^{2}\right )}}\right )}{2 \, {\left (a^{2} c^{3} + c^{2} d + {\left (a^{2} c^{2} d + c d^{2}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*(a^2*c + d)*sqrt(d*x^2 + c)*x*log((a*x + 1)/(a*x - 1)) + sqrt(a^2*c + d)*(d*x^2 + c)*log((a^4*d^2*x^4
+ 8*a^4*c^2 + 8*a^2*c*d + 2*(4*a^4*c*d + 3*a^2*d^2)*x^2 - 4*(a^3*d*x^2 + 2*a^3*c + a*d)*sqrt(a^2*c + d)*sqrt(d
*x^2 + c) + d^2)/(a^4*x^4 - 2*a^2*x^2 + 1)))/(a^2*c^3 + c^2*d + (a^2*c^2*d + c*d^2)*x^2), 1/2*((a^2*c + d)*sqr
t(d*x^2 + c)*x*log((a*x + 1)/(a*x - 1)) + sqrt(-a^2*c - d)*(d*x^2 + c)*arctan(1/2*(a^2*d*x^2 + 2*a^2*c + d)*sq
rt(-a^2*c - d)*sqrt(d*x^2 + c)/(a^3*c^2 + a*c*d + (a^3*c*d + a*d^2)*x^2)))/(a^2*c^3 + c^2*d + (a^2*c^2*d + c*d
^2)*x^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.94, size = 0, normalized size = 0.00 \[ \int \frac {\mathrm {arccoth}\left (a x \right )}{\left (d \,x^{2}+c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.34, size = 153, normalized size = 2.47 \[ \frac {a^{2} {\left (\frac {\operatorname {arsinh}\left (-\frac {2 \, a^{2} c}{\sqrt {c d} {\left | 2 \, a^{2} x + 2 \, a \right |}} + \frac {2 \, a d x}{\sqrt {c d} {\left | 2 \, a^{2} x + 2 \, a \right |}}\right )}{a^{3} \sqrt {c + \frac {d}{a^{2}}}} - \frac {\operatorname {arsinh}\left (\frac {2 \, a^{2} c}{\sqrt {c d} {\left | 2 \, a^{2} x - 2 \, a \right |}} + \frac {2 \, a d x}{\sqrt {c d} {\left | 2 \, a^{2} x - 2 \, a \right |}}\right )}{a^{3} \sqrt {c + \frac {d}{a^{2}}}}\right )}}{2 \, c} + \frac {x \operatorname {arcoth}\left (a x\right )}{\sqrt {d x^{2} + c} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*(arcsinh(-2*a^2*c/(sqrt(c*d)*abs(2*a^2*x + 2*a)) + 2*a*d*x/(sqrt(c*d)*abs(2*a^2*x + 2*a)))/(a^3*sqrt(c
+ d/a^2)) - arcsinh(2*a^2*c/(sqrt(c*d)*abs(2*a^2*x - 2*a)) + 2*a*d*x/(sqrt(c*d)*abs(2*a^2*x - 2*a)))/(a^3*sqr
t(c + d/a^2)))/c + x*arccoth(a*x)/(sqrt(d*x^2 + c)*c)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {acoth}\left (a\,x\right )}{{\left (d\,x^2+c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a*x)/(c + d*x^2)^(3/2),x)

[Out]

int(acoth(a*x)/(c + d*x^2)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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