3.6.14 \(\int \frac {\tanh ^{-1}(a x)}{(c+d x^2)^{5/2}} \, dx\) [514]

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

[Out]

1/3*x*arctanh(a*x)/c/(d*x^2+c)^(3/2)-1/3*(3*a^2*c+2*d)*arctanh(a*(d*x^2+c)^(1/2)/(a^2*c+d)^(1/2))/c^2/(a^2*c+d
)^(3/2)+1/3*a/c/(a^2*c+d)/(d*x^2+c)^(1/2)+2/3*x*arctanh(a*x)/c^2/(d*x^2+c)^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 128, normalized size of antiderivative = 1.00, 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 = {198, 197, 6123, 6820, 12, 585, 79, 65, 214} \begin {gather*} -\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 \tanh ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}+\frac {x \tanh ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]/(c + d*x^2)^(5/2),x]

[Out]

a/(3*c*(a^2*c + d)*Sqrt[c + d*x^2]) + (x*ArcTanh[a*x])/(3*c*(c + d*x^2)^(3/2)) + (2*x*ArcTanh[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 12

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

Rule 65

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 79

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] || L
tQ[p, n]))))

Rule 197

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 198

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 214

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

Rule 585

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 6123

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(d + e*x^
2)^q, x]}, Dist[a + b*ArcTanh[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 6820

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

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 226, normalized size = 1.77 \begin {gather*} \frac {\frac {2 a c}{\left (a^2 c+d\right ) \sqrt {c+d x^2}}+\frac {2 x \left (3 c+2 d x^2\right ) \tanh ^{-1}(a x)}{\left (c+d x^2\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 (1+a x)}{\left (a^2 c+d\right )^{3/2}}-\frac {\left (3 a^2 c+2 d\right ) \log \left (a c-d x+\sqrt {a^2 c+d} \sqrt {c+d x^2}\right )}{\left (a^2 c+d\right )^{3/2}}-\frac {\left (3 a^2 c+2 d\right ) \log \left (a c+d x+\sqrt {a^2 c+d} \sqrt {c+d x^2}\right )}{\left (a^2 c+d\right )^{3/2}}}{6 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[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)*ArcTanh[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 = 3.20, size = 0, normalized size = 0.00 \[\int \frac {\arctanh \left (a x \right )}{\left (d \,x^{2}+c \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (108) = 216\).
time = 0.48, size = 223, normalized size = 1.74 \begin {gather*} \frac {1}{6} \, a {\left (\frac {\frac {a d \log \left (\frac {\sqrt {d x^{2} + c} a^{2} - \sqrt {a^{2} c + d} a}{\sqrt {d x^{2} + c} a^{2} + \sqrt {a^{2} c + d} a}\right )}{{\left (a^{2} c^{2} + c d\right )} \sqrt {a^{2} c + d}} + \frac {2 \, d}{{\left (a^{2} c^{2} + c d\right )} \sqrt {d x^{2} + c}}}{d} + \frac {2 \, \log \left (\frac {\sqrt {d x^{2} + c} a^{2} - \sqrt {a^{2} c + d} a}{\sqrt {d x^{2} + c} a^{2} + \sqrt {a^{2} c + d} a}\right )}{\sqrt {a^{2} c + d} a c^{2}}\right )} + \frac {1}{3} \, {\left (\frac {2 \, x}{\sqrt {d x^{2} + c} c^{2}} + \frac {x}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} c}\right )} \operatorname {artanh}\left (a x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*((a*d*log((sqrt(d*x^2 + c)*a^2 - sqrt(a^2*c + d)*a)/(sqrt(d*x^2 + c)*a^2 + sqrt(a^2*c + d)*a))/((a^2*c^2
+ c*d)*sqrt(a^2*c + d)) + 2*d/((a^2*c^2 + c*d)*sqrt(d*x^2 + c)))/d + 2*log((sqrt(d*x^2 + c)*a^2 - sqrt(a^2*c
+ d)*a)/(sqrt(d*x^2 + c)*a^2 + sqrt(a^2*c + d)*a))/(sqrt(a^2*c + d)*a*c^2)) + 1/3*(2*x/(sqrt(d*x^2 + c)*c^2) +
x/((d*x^2 + c)^(3/2)*c))*arctanh(a*x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (108) = 216\).
time = 0.38, size = 730, normalized size = 5.70 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(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)))*
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), 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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {atanh}{\left (a x \right )}}{\left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 135, normalized size = 1.05 \begin {gather*} \frac {1}{3} \, a {\left (\frac {{\left (3 \, a^{2} c + 2 \, d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} a}{\sqrt {-a^{2} c - d}}\right )}{{\left (a^{2} c^{3} + c^{2} d\right )} \sqrt {-a^{2} c - d} a} + \frac {1}{{\left (a^{2} c^{2} + c d\right )} \sqrt {d x^{2} + c}}\right )} + \frac {x {\left (\frac {2 \, d x^{2}}{c^{2}} + \frac {3}{c}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{6 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*((3*a^2*c + 2*d)*arctan(sqrt(d*x^2 + c)*a/sqrt(-a^2*c - d))/((a^2*c^3 + c^2*d)*sqrt(-a^2*c - d)*a) + 1/(
(a^2*c^2 + c*d)*sqrt(d*x^2 + c))) + 1/6*x*(2*d*x^2/c^2 + 3/c)*log(-(a*x + 1)/(a*x - 1))/(d*x^2 + c)^(3/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {atanh}\left (a\,x\right )}{{\left (d\,x^2+c\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)/(c + d*x^2)^(5/2),x)

[Out]

int(atanh(a*x)/(c + d*x^2)^(5/2), x)

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