3.45 \(\int \frac{\coth ^{-1}(a x)}{(c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=128 \[ -\frac{\left (3 a^2 c+2 d\right ) \tanh ^{-1}\left (\frac{a \sqrt{c+d x^2}}{\sqrt{a^2 c+d}}\right )}{3 c^2 \left (a^2 c+d\right )^{3/2}}+\frac{a}{3 c \left (a^2 c+d\right ) \sqrt{c+d x^2}}+\frac{2 x \coth ^{-1}(a x)}{3 c^2 \sqrt{c+d x^2}}+\frac{x \coth ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}} \]

[Out]

a/(3*c*(a^2*c + d)*Sqrt[c + d*x^2]) + (x*ArcCoth[a*x])/(3*c*(c + d*x^2)^(3/2)) + (2*x*ArcCoth[a*x])/(3*c^2*Sqr
t[c + d*x^2]) - ((3*a^2*c + 2*d)*ArcTanh[(a*Sqrt[c + d*x^2])/Sqrt[a^2*c + d]])/(3*c^2*(a^2*c + d)^(3/2))

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Rubi [A]  time = 0.339828, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 9, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.562, Rules used = {192, 191, 5977, 6688, 12, 571, 78, 63, 208} \[ -\frac{\left (3 a^2 c+2 d\right ) \tanh ^{-1}\left (\frac{a \sqrt{c+d x^2}}{\sqrt{a^2 c+d}}\right )}{3 c^2 \left (a^2 c+d\right )^{3/2}}+\frac{a}{3 c \left (a^2 c+d\right ) \sqrt{c+d x^2}}+\frac{2 x \coth ^{-1}(a x)}{3 c^2 \sqrt{c+d x^2}}+\frac{x \coth ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a*x]/(c + d*x^2)^(5/2),x]

[Out]

a/(3*c*(a^2*c + d)*Sqrt[c + d*x^2]) + (x*ArcCoth[a*x])/(3*c*(c + d*x^2)^(3/2)) + (2*x*ArcCoth[a*x])/(3*c^2*Sqr
t[c + d*x^2]) - ((3*a^2*c + 2*d)*ArcTanh[(a*Sqrt[c + d*x^2])/Sqrt[a^2*c + d]])/(3*c^2*(a^2*c + d)^(3/2))

Rule 192

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

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 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])

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 12

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

Rule 571

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 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]

Rubi steps

\begin{align*} \int \frac{\coth ^{-1}(a x)}{\left (c+d x^2\right )^{5/2}} \, dx &=\frac{x \coth ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac{2 x \coth ^{-1}(a x)}{3 c^2 \sqrt{c+d x^2}}-a \int \frac{\frac{x}{3 c \left (c+d x^2\right )^{3/2}}+\frac{2 x}{3 c^2 \sqrt{c+d x^2}}}{1-a^2 x^2} \, dx\\ &=\frac{x \coth ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac{2 x \coth ^{-1}(a x)}{3 c^2 \sqrt{c+d x^2}}-a \int \frac{x \left (3 c+2 d x^2\right )}{3 c^2 \left (1-a^2 x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx\\ &=\frac{x \coth ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac{2 x \coth ^{-1}(a x)}{3 c^2 \sqrt{c+d x^2}}-\frac{a \int \frac{x \left (3 c+2 d x^2\right )}{\left (1-a^2 x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx}{3 c^2}\\ &=\frac{x \coth ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac{2 x \coth ^{-1}(a x)}{3 c^2 \sqrt{c+d x^2}}-\frac{a \operatorname{Subst}\left (\int \frac{3 c+2 d x}{\left (1-a^2 x\right ) (c+d x)^{3/2}} \, dx,x,x^2\right )}{6 c^2}\\ &=\frac{a}{3 c \left (a^2 c+d\right ) \sqrt{c+d x^2}}+\frac{x \coth ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac{2 x \coth ^{-1}(a x)}{3 c^2 \sqrt{c+d x^2}}-\frac{\left (a \left (3 a^2 c+2 d\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-a^2 x\right ) \sqrt{c+d x}} \, dx,x,x^2\right )}{6 c^2 \left (a^2 c+d\right )}\\ &=\frac{a}{3 c \left (a^2 c+d\right ) \sqrt{c+d x^2}}+\frac{x \coth ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac{2 x \coth ^{-1}(a x)}{3 c^2 \sqrt{c+d x^2}}-\frac{\left (a \left (3 a^2 c+2 d\right )\right ) \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 )}{3 c^2 d \left (a^2 c+d\right )}\\ &=\frac{a}{3 c \left (a^2 c+d\right ) \sqrt{c+d x^2}}+\frac{x \coth ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac{2 x \coth ^{-1}(a x)}{3 c^2 \sqrt{c+d x^2}}-\frac{\left (3 a^2 c+2 d\right ) \tanh ^{-1}\left (\frac{a \sqrt{c+d x^2}}{\sqrt{a^2 c+d}}\right )}{3 c^2 \left (a^2 c+d\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.321201, size = 226, normalized size = 1.77 \[ \frac{\frac{2 a c}{\left (a^2 c+d\right ) \sqrt{c+d x^2}}-\frac{\left (3 a^2 c+2 d\right ) \log \left (\sqrt{a^2 c+d} \sqrt{c+d x^2}+a c-d x\right )}{\left (a^2 c+d\right )^{3/2}}-\frac{\left (3 a^2 c+2 d\right ) \log \left (\sqrt{a^2 c+d} \sqrt{c+d x^2}+a c+d x\right )}{\left (a^2 c+d\right )^{3/2}}+\frac{\left (3 a^2 c+2 d\right ) \log (1-a x)}{\left (a^2 c+d\right )^{3/2}}+\frac{\left (3 a^2 c+2 d\right ) \log (a x+1)}{\left (a^2 c+d\right )^{3/2}}+\frac{2 x \coth ^{-1}(a x) \left (3 c+2 d x^2\right )}{\left (c+d x^2\right )^{3/2}}}{6 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*c)/((a^2*c + d)*Sqrt[c + d*x^2]) + (2*x*(3*c + 2*d*x^2)*ArcCoth[a*x])/(c + d*x^2)^(3/2) + ((3*a^2*c + 2*
d)*Log[1 - a*x])/(a^2*c + d)^(3/2) + ((3*a^2*c + 2*d)*Log[1 + a*x])/(a^2*c + d)^(3/2) - ((3*a^2*c + 2*d)*Log[a
*c - d*x + Sqrt[a^2*c + d]*Sqrt[c + d*x^2]])/(a^2*c + d)^(3/2) - ((3*a^2*c + 2*d)*Log[a*c + d*x + Sqrt[a^2*c +
d]*Sqrt[c + d*x^2]])/(a^2*c + d)^(3/2))/(6*c^2)

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Maple [F]  time = 0.456, size = 0, normalized size = 0. \begin{align*} \int{{\rm arccoth} \left (ax\right ) \left ( d{x}^{2}+c \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.98491, size = 1496, normalized size = 11.69 \begin{align*} \left [\frac{{\left (3 \, a^{2} c^{3} +{\left (3 \, a^{2} c d^{2} + 2 \, d^{3}\right )} x^{4} + 2 \, c^{2} d + 2 \,{\left (3 \, a^{2} c^{2} d + 2 \, c d^{2}\right )} x^{2}\right )} \sqrt{a^{2} c + d} \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 ) + 2 \,{\left (2 \, a^{3} c^{3} + 2 \, a c^{2} d + 2 \,{\left (a^{3} c^{2} d + a c d^{2}\right )} x^{2} +{\left (2 \,{\left (a^{4} c^{2} d + 2 \, a^{2} c d^{2} + d^{3}\right )} x^{3} + 3 \,{\left (a^{4} c^{3} + 2 \, a^{2} c^{2} d + c d^{2}\right )} x\right )} \log \left (\frac{a x + 1}{a x - 1}\right )\right )} \sqrt{d x^{2} + c}}{12 \,{\left (a^{4} c^{6} + 2 \, a^{2} c^{5} d + c^{4} d^{2} +{\left (a^{4} c^{4} d^{2} + 2 \, a^{2} c^{3} d^{3} + c^{2} d^{4}\right )} x^{4} + 2 \,{\left (a^{4} c^{5} d + 2 \, a^{2} c^{4} d^{2} + c^{3} d^{3}\right )} x^{2}\right )}}, \frac{{\left (3 \, a^{2} c^{3} +{\left (3 \, a^{2} c d^{2} + 2 \, d^{3}\right )} x^{4} + 2 \, c^{2} d + 2 \,{\left (3 \, a^{2} c^{2} d + 2 \, c d^{2}\right )} x^{2}\right )} \sqrt{-a^{2} c - d} \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 ) +{\left (2 \, a^{3} c^{3} + 2 \, a c^{2} d + 2 \,{\left (a^{3} c^{2} d + a c d^{2}\right )} x^{2} +{\left (2 \,{\left (a^{4} c^{2} d + 2 \, a^{2} c d^{2} + d^{3}\right )} x^{3} + 3 \,{\left (a^{4} c^{3} + 2 \, a^{2} c^{2} d + c d^{2}\right )} x\right )} \log \left (\frac{a x + 1}{a x - 1}\right )\right )} \sqrt{d x^{2} + c}}{6 \,{\left (a^{4} c^{6} + 2 \, a^{2} c^{5} d + c^{4} d^{2} +{\left (a^{4} c^{4} d^{2} + 2 \, a^{2} c^{3} d^{3} + c^{2} d^{4}\right )} x^{4} + 2 \,{\left (a^{4} c^{5} d + 2 \, a^{2} c^{4} d^{2} + c^{3} d^{3}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*((3*a^2*c^3 + (3*a^2*c*d^2 + 2*d^3)*x^4 + 2*c^2*d + 2*(3*a^2*c^2*d + 2*c*d^2)*x^2)*sqrt(a^2*c + d)*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)) + 2*(2*a^3*c^3 + 2*a*c^2*d + 2*(a^3*c^2*d + a*c*d^2)*x
^2 + (2*(a^4*c^2*d + 2*a^2*c*d^2 + d^3)*x^3 + 3*(a^4*c^3 + 2*a^2*c^2*d + c*d^2)*x)*log((a*x + 1)/(a*x - 1)))*s
qrt(d*x^2 + c))/(a^4*c^6 + 2*a^2*c^5*d + c^4*d^2 + (a^4*c^4*d^2 + 2*a^2*c^3*d^3 + c^2*d^4)*x^4 + 2*(a^4*c^5*d
+ 2*a^2*c^4*d^2 + c^3*d^3)*x^2), 1/6*((3*a^2*c^3 + (3*a^2*c*d^2 + 2*d^3)*x^4 + 2*c^2*d + 2*(3*a^2*c^2*d + 2*c*
d^2)*x^2)*sqrt(-a^2*c - d)*arctan(1/2*(a^2*d*x^2 + 2*a^2*c + d)*sqrt(-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)) + (2*a^3*c^3 + 2*a*c^2*d + 2*(a^3*c^2*d + a*c*d^2)*x^2 + (2*(a^4*c^2*d + 2*a^2*c
*d^2 + d^3)*x^3 + 3*(a^4*c^3 + 2*a^2*c^2*d + c*d^2)*x)*log((a*x + 1)/(a*x - 1)))*sqrt(d*x^2 + c))/(a^4*c^6 + 2
*a^2*c^5*d + c^4*d^2 + (a^4*c^4*d^2 + 2*a^2*c^3*d^3 + c^2*d^4)*x^4 + 2*(a^4*c^5*d + 2*a^2*c^4*d^2 + c^3*d^3)*x
^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acoth}{\left (a x \right )}}{\left (c + d x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acoth(a*x)/(c + d*x**2)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (a x\right )}{{\left (d x^{2} + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(a*x)/(d*x^2 + c)^(5/2), x)