3.46 \(\int \frac{\coth ^{-1}(a x)}{(c+d x^2)^{7/2}} \, dx\)

Optimal. Leaf size=200 \[ -\frac{\left (15 a^4 c^2+20 a^2 c d+8 d^2\right ) \tanh ^{-1}\left (\frac{a \sqrt{c+d x^2}}{\sqrt{a^2 c+d}}\right )}{15 c^3 \left (a^2 c+d\right )^{5/2}}+\frac{a \left (7 a^2 c+4 d\right )}{15 c^2 \left (a^2 c+d\right )^2 \sqrt{c+d x^2}}+\frac{a}{15 c \left (a^2 c+d\right ) \left (c+d x^2\right )^{3/2}}+\frac{8 x \coth ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}+\frac{4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}} \]

[Out]

a/(15*c*(a^2*c + d)*(c + d*x^2)^(3/2)) + (a*(7*a^2*c + 4*d))/(15*c^2*(a^2*c + d)^2*Sqrt[c + d*x^2]) + (x*ArcCo
th[a*x])/(5*c*(c + d*x^2)^(5/2)) + (4*x*ArcCoth[a*x])/(15*c^2*(c + d*x^2)^(3/2)) + (8*x*ArcCoth[a*x])/(15*c^3*
Sqrt[c + d*x^2]) - ((15*a^4*c^2 + 20*a^2*c*d + 8*d^2)*ArcTanh[(a*Sqrt[c + d*x^2])/Sqrt[a^2*c + d]])/(15*c^3*(a
^2*c + d)^(5/2))

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Rubi [A]  time = 1.04748, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 8, 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, 6715, 897, 1261, 208} \[ -\frac{\left (15 a^4 c^2+20 a^2 c d+8 d^2\right ) \tanh ^{-1}\left (\frac{a \sqrt{c+d x^2}}{\sqrt{a^2 c+d}}\right )}{15 c^3 \left (a^2 c+d\right )^{5/2}}+\frac{a \left (7 a^2 c+4 d\right )}{15 c^2 \left (a^2 c+d\right )^2 \sqrt{c+d x^2}}+\frac{a}{15 c \left (a^2 c+d\right ) \left (c+d x^2\right )^{3/2}}+\frac{8 x \coth ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}+\frac{4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a*x]/(c + d*x^2)^(7/2),x]

[Out]

a/(15*c*(a^2*c + d)*(c + d*x^2)^(3/2)) + (a*(7*a^2*c + 4*d))/(15*c^2*(a^2*c + d)^2*Sqrt[c + d*x^2]) + (x*ArcCo
th[a*x])/(5*c*(c + d*x^2)^(5/2)) + (4*x*ArcCoth[a*x])/(15*c^2*(c + d*x^2)^(3/2)) + (8*x*ArcCoth[a*x])/(15*c^3*
Sqrt[c + d*x^2]) - ((15*a^4*c^2 + 20*a^2*c*d + 8*d^2)*ArcTanh[(a*Sqrt[c + d*x^2])/Sqrt[a^2*c + d]])/(15*c^3*(a
^2*c + d)^(5/2))

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 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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 )^{7/2}} \, dx &=\frac{x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \coth ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}-a \int \frac{\frac{x}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x}{15 c^3 \sqrt{c+d x^2}}}{1-a^2 x^2} \, dx\\ &=\frac{x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \coth ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}-a \int \frac{x \left (15 c^2+20 c d x^2+8 d^2 x^4\right )}{15 c^3 \left (1-a^2 x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx\\ &=\frac{x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \coth ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}-\frac{a \int \frac{x \left (15 c^2+20 c d x^2+8 d^2 x^4\right )}{\left (1-a^2 x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx}{15 c^3}\\ &=\frac{x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \coth ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}-\frac{a \operatorname{Subst}\left (\int \frac{15 c^2+20 c d x+8 d^2 x^2}{\left (1-a^2 x\right ) (c+d x)^{5/2}} \, dx,x,x^2\right )}{30 c^3}\\ &=\frac{x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \coth ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}-\frac{a \operatorname{Subst}\left (\int \frac{3 c^2+4 c x^2+8 x^4}{x^4 \left (\frac{a^2 c+d}{d}-\frac{a^2 x^2}{d}\right )} \, dx,x,\sqrt{c+d x^2}\right )}{15 c^3 d}\\ &=\frac{x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \coth ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}-\frac{a \operatorname{Subst}\left (\int \left (\frac{3 c^2 d}{\left (a^2 c+d\right ) x^4}+\frac{c d \left (7 a^2 c+4 d\right )}{\left (a^2 c+d\right )^2 x^2}+\frac{d \left (15 a^4 c^2+20 a^2 c d+8 d^2\right )}{\left (a^2 c+d\right )^2 \left (a^2 c+d-a^2 x^2\right )}\right ) \, dx,x,\sqrt{c+d x^2}\right )}{15 c^3 d}\\ &=\frac{a}{15 c \left (a^2 c+d\right ) \left (c+d x^2\right )^{3/2}}+\frac{a \left (7 a^2 c+4 d\right )}{15 c^2 \left (a^2 c+d\right )^2 \sqrt{c+d x^2}}+\frac{x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \coth ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}-\frac{\left (a \left (15 a^4 c^2+20 a^2 c d+8 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a^2 c+d-a^2 x^2} \, dx,x,\sqrt{c+d x^2}\right )}{15 c^3 \left (a^2 c+d\right )^2}\\ &=\frac{a}{15 c \left (a^2 c+d\right ) \left (c+d x^2\right )^{3/2}}+\frac{a \left (7 a^2 c+4 d\right )}{15 c^2 \left (a^2 c+d\right )^2 \sqrt{c+d x^2}}+\frac{x \coth ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \coth ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \coth ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}-\frac{\left (15 a^4 c^2+20 a^2 c d+8 d^2\right ) \tanh ^{-1}\left (\frac{a \sqrt{c+d x^2}}{\sqrt{a^2 c+d}}\right )}{15 c^3 \left (a^2 c+d\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.636018, size = 329, normalized size = 1.64 \[ \frac{\left (15 a^4 c^2+20 a^2 c d+8 d^2\right ) \log (1-a x) \left (c+d x^2\right )^{5/2}+\left (15 a^4 c^2+20 a^2 c d+8 d^2\right ) \log (a x+1) \left (c+d x^2\right )^{5/2}-\left (15 a^4 c^2+20 a^2 c d+8 d^2\right ) \left (c+d x^2\right )^{5/2} \log \left (\sqrt{a^2 c+d} \sqrt{c+d x^2}+a c-d x\right )-\left (15 a^4 c^2+20 a^2 c d+8 d^2\right ) \left (c+d x^2\right )^{5/2} \log \left (\sqrt{a^2 c+d} \sqrt{c+d x^2}+a c+d x\right )+2 x \left (a^2 c+d\right )^{5/2} \coth ^{-1}(a x) \left (15 c^2+20 c d x^2+8 d^2 x^4\right )+2 a c \sqrt{a^2 c+d} \left (c+d x^2\right ) \left (a^2 c \left (8 c+7 d x^2\right )+d \left (5 c+4 d x^2\right )\right )}{30 c^3 \left (a^2 c+d\right )^{5/2} \left (c+d x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[a*x]/(c + d*x^2)^(7/2),x]

[Out]

(2*a*c*Sqrt[a^2*c + d]*(c + d*x^2)*(d*(5*c + 4*d*x^2) + a^2*c*(8*c + 7*d*x^2)) + 2*(a^2*c + d)^(5/2)*x*(15*c^2
+ 20*c*d*x^2 + 8*d^2*x^4)*ArcCoth[a*x] + (15*a^4*c^2 + 20*a^2*c*d + 8*d^2)*(c + d*x^2)^(5/2)*Log[1 - a*x] + (
15*a^4*c^2 + 20*a^2*c*d + 8*d^2)*(c + d*x^2)^(5/2)*Log[1 + a*x] - (15*a^4*c^2 + 20*a^2*c*d + 8*d^2)*(c + d*x^2
)^(5/2)*Log[a*c - d*x + Sqrt[a^2*c + d]*Sqrt[c + d*x^2]] - (15*a^4*c^2 + 20*a^2*c*d + 8*d^2)*(c + d*x^2)^(5/2)
*Log[a*c + d*x + Sqrt[a^2*c + d]*Sqrt[c + d*x^2]])/(30*c^3*(a^2*c + d)^(5/2)*(c + d*x^2)^(5/2))

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Maple [F]  time = 0.455, size = 0, normalized size = 0. \begin{align*} \int{{\rm arccoth} \left (ax\right ) \left ( d{x}^{2}+c \right ) ^{-{\frac{7}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(a*x)/(d*x^2+c)^(7/2),x)

[Out]

int(arccoth(a*x)/(d*x^2+c)^(7/2),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)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.24513, size = 2610, normalized size = 13.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)/(d*x^2+c)^(7/2),x, algorithm="fricas")

[Out]

[1/60*((15*a^4*c^5 + 20*a^2*c^4*d + (15*a^4*c^2*d^3 + 20*a^2*c*d^4 + 8*d^5)*x^6 + 8*c^3*d^2 + 3*(15*a^4*c^3*d^
2 + 20*a^2*c^2*d^3 + 8*c*d^4)*x^4 + 3*(15*a^4*c^4*d + 20*a^2*c^3*d^2 + 8*c^2*d^3)*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*(16*a^5*c^5 + 26*a^3*c^4*d + 10*a*c^3*d^2 + 2*(7*a^5
*c^3*d^2 + 11*a^3*c^2*d^3 + 4*a*c*d^4)*x^4 + 6*(5*a^5*c^4*d + 8*a^3*c^3*d^2 + 3*a*c^2*d^3)*x^2 + (8*(a^6*c^3*d
^2 + 3*a^4*c^2*d^3 + 3*a^2*c*d^4 + d^5)*x^5 + 20*(a^6*c^4*d + 3*a^4*c^3*d^2 + 3*a^2*c^2*d^3 + c*d^4)*x^3 + 15*
(a^6*c^5 + 3*a^4*c^4*d + 3*a^2*c^3*d^2 + c^2*d^3)*x)*log((a*x + 1)/(a*x - 1)))*sqrt(d*x^2 + c))/(a^6*c^9 + 3*a
^4*c^8*d + 3*a^2*c^7*d^2 + c^6*d^3 + (a^6*c^6*d^3 + 3*a^4*c^5*d^4 + 3*a^2*c^4*d^5 + c^3*d^6)*x^6 + 3*(a^6*c^7*
d^2 + 3*a^4*c^6*d^3 + 3*a^2*c^5*d^4 + c^4*d^5)*x^4 + 3*(a^6*c^8*d + 3*a^4*c^7*d^2 + 3*a^2*c^6*d^3 + c^5*d^4)*x
^2), 1/30*((15*a^4*c^5 + 20*a^2*c^4*d + (15*a^4*c^2*d^3 + 20*a^2*c*d^4 + 8*d^5)*x^6 + 8*c^3*d^2 + 3*(15*a^4*c^
3*d^2 + 20*a^2*c^2*d^3 + 8*c*d^4)*x^4 + 3*(15*a^4*c^4*d + 20*a^2*c^3*d^2 + 8*c^2*d^3)*x^2)*sqrt(-a^2*c - d)*ar
ctan(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))
+ (16*a^5*c^5 + 26*a^3*c^4*d + 10*a*c^3*d^2 + 2*(7*a^5*c^3*d^2 + 11*a^3*c^2*d^3 + 4*a*c*d^4)*x^4 + 6*(5*a^5*c
^4*d + 8*a^3*c^3*d^2 + 3*a*c^2*d^3)*x^2 + (8*(a^6*c^3*d^2 + 3*a^4*c^2*d^3 + 3*a^2*c*d^4 + d^5)*x^5 + 20*(a^6*c
^4*d + 3*a^4*c^3*d^2 + 3*a^2*c^2*d^3 + c*d^4)*x^3 + 15*(a^6*c^5 + 3*a^4*c^4*d + 3*a^2*c^3*d^2 + c^2*d^3)*x)*lo
g((a*x + 1)/(a*x - 1)))*sqrt(d*x^2 + c))/(a^6*c^9 + 3*a^4*c^8*d + 3*a^2*c^7*d^2 + c^6*d^3 + (a^6*c^6*d^3 + 3*a
^4*c^5*d^4 + 3*a^2*c^4*d^5 + c^3*d^6)*x^6 + 3*(a^6*c^7*d^2 + 3*a^4*c^6*d^3 + 3*a^2*c^5*d^4 + c^4*d^5)*x^4 + 3*
(a^6*c^8*d + 3*a^4*c^7*d^2 + 3*a^2*c^6*d^3 + c^5*d^4)*x^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(a*x)/(d*x**2+c)**(7/2),x)

[Out]

Timed out

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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{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)/(d*x^2+c)^(7/2),x, algorithm="giac")

[Out]

integrate(arccoth(a*x)/(d*x^2 + c)^(7/2), x)