3.5.89 \(\int \frac {1}{(1-a^2 x^2)^{7/2} \tanh ^{-1}(a x)^2} \, dx\) [489]

Optimal. Leaf size=66 \[ -\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac {5 \text {Shi}\left (\tanh ^{-1}(a x)\right )}{8 a}+\frac {15 \text {Shi}\left (3 \tanh ^{-1}(a x)\right )}{16 a}+\frac {5 \text {Shi}\left (5 \tanh ^{-1}(a x)\right )}{16 a} \]

[Out]

-1/a/(-a^2*x^2+1)^(5/2)/arctanh(a*x)+5/8*Shi(arctanh(a*x))/a+15/16*Shi(3*arctanh(a*x))/a+5/16*Shi(5*arctanh(a*
x))/a

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6113, 6181, 5556, 3379} \begin {gather*} -\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac {5 \text {Shi}\left (\tanh ^{-1}(a x)\right )}{8 a}+\frac {15 \text {Shi}\left (3 \tanh ^{-1}(a x)\right )}{16 a}+\frac {5 \text {Shi}\left (5 \tanh ^{-1}(a x)\right )}{16 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((1 - a^2*x^2)^(7/2)*ArcTanh[a*x]^2),x]

[Out]

-(1/(a*(1 - a^2*x^2)^(5/2)*ArcTanh[a*x])) + (5*SinhIntegral[ArcTanh[a*x]])/(8*a) + (15*SinhIntegral[3*ArcTanh[
a*x]])/(16*a) + (5*SinhIntegral[5*ArcTanh[a*x]])/(16*a)

Rule 3379

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[I*(SinhIntegral[c*f*(fz/
d) + f*fz*x]/d), x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 5556

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 6113

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

Rule 6181

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c^(
m + 1), Subst[Int[(a + b*x)^p*(Sinh[x]^m/Cosh[x]^(m + 2*(q + 1))), x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c,
d, e, p}, x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q] || GtQ[d, 0])

Rubi steps

\begin {align*} \int \frac {1}{\left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)^2} \, dx &=-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+(5 a) \int \frac {x}{\left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)} \, dx\\ &=-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac {5 \text {Subst}\left (\int \frac {\cosh ^4(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac {5 \text {Subst}\left (\int \left (\frac {\sinh (x)}{8 x}+\frac {3 \sinh (3 x)}{16 x}+\frac {\sinh (5 x)}{16 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac {5 \text {Subst}\left (\int \frac {\sinh (5 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}+\frac {5 \text {Subst}\left (\int \frac {\sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a}+\frac {15 \text {Subst}\left (\int \frac {\sinh (3 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}\\ &=-\frac {1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac {5 \text {Shi}\left (\tanh ^{-1}(a x)\right )}{8 a}+\frac {15 \text {Shi}\left (3 \tanh ^{-1}(a x)\right )}{16 a}+\frac {5 \text {Shi}\left (5 \tanh ^{-1}(a x)\right )}{16 a}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.12, size = 56, normalized size = 0.85 \begin {gather*} \frac {-\frac {16}{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+5 \left (2 \text {Shi}\left (\tanh ^{-1}(a x)\right )+3 \text {Shi}\left (3 \tanh ^{-1}(a x)\right )+\text {Shi}\left (5 \tanh ^{-1}(a x)\right )\right )}{16 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - a^2*x^2)^(7/2)*ArcTanh[a*x]^2),x]

[Out]

(-16/((1 - a^2*x^2)^(5/2)*ArcTanh[a*x]) + 5*(2*SinhIntegral[ArcTanh[a*x]] + 3*SinhIntegral[3*ArcTanh[a*x]] + S
inhIntegral[5*ArcTanh[a*x]]))/(16*a)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(175\) vs. \(2(58)=116\).
time = 2.97, size = 176, normalized size = 2.67

method	result	size
default	\(\frac {5 \arctanh \left (a x \right ) \hyperbolicSineIntegral \left (5 \arctanh \left (a x \right )\right ) a^{2} x^{2}+10 \arctanh \left (a x \right ) \hyperbolicSineIntegral \left (\arctanh \left (a x \right )\right ) a^{2} x^{2}+15 \arctanh \left (a x \right ) \hyperbolicSineIntegral \left (3 \arctanh \left (a x \right )\right ) a^{2} x^{2}-\cosh \left (5 \arctanh \left (a x \right )\right ) a^{2} x^{2}-5 \cosh \left (3 \arctanh \left (a x \right )\right ) a^{2} x^{2}-5 \hyperbolicSineIntegral \left (5 \arctanh \left (a x \right )\right ) \arctanh \left (a x \right )-10 \hyperbolicSineIntegral \left (\arctanh \left (a x \right )\right ) \arctanh \left (a x \right )-15 \hyperbolicSineIntegral \left (3 \arctanh \left (a x \right )\right ) \arctanh \left (a x \right )+\cosh \left (5 \arctanh \left (a x \right )\right )+10 \sqrt {-a^{2} x^{2}+1}+5 \cosh \left (3 \arctanh \left (a x \right )\right )}{16 a \arctanh \left (a x \right ) \left (a^{2} x^{2}-1\right )}\)	\(176\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-a^2*x^2+1)^(7/2)/arctanh(a*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/16/a*(5*arctanh(a*x)*Shi(5*arctanh(a*x))*a^2*x^2+10*arctanh(a*x)*Shi(arctanh(a*x))*a^2*x^2+15*arctanh(a*x)*S
hi(3*arctanh(a*x))*a^2*x^2-cosh(5*arctanh(a*x))*a^2*x^2-5*cosh(3*arctanh(a*x))*a^2*x^2-5*Shi(5*arctanh(a*x))*a
rctanh(a*x)-10*Shi(arctanh(a*x))*arctanh(a*x)-15*Shi(3*arctanh(a*x))*arctanh(a*x)+cosh(5*arctanh(a*x))+10*(-a^
2*x^2+1)^(1/2)+5*cosh(3*arctanh(a*x)))/arctanh(a*x)/(a^2*x^2-1)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^(7/2)/arctanh(a*x)^2,x, algorithm="maxima")

[Out]

integrate(1/((-a^2*x^2 + 1)^(7/2)*arctanh(a*x)^2), x)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^(7/2)/arctanh(a*x)^2,x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*x^2 + 1)/((a^8*x^8 - 4*a^6*x^6 + 6*a^4*x^4 - 4*a^2*x^2 + 1)*arctanh(a*x)^2), x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {7}{2}} \operatorname {atanh}^{2}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a**2*x**2+1)**(7/2)/atanh(a*x)**2,x)

[Out]

Integral(1/((-(a*x - 1)*(a*x + 1))**(7/2)*atanh(a*x)**2), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^(7/2)/arctanh(a*x)^2,x, algorithm="giac")

[Out]

integrate(1/((-a^2*x^2 + 1)^(7/2)*arctanh(a*x)^2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (1-a^2\,x^2\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(atanh(a*x)^2*(1 - a^2*x^2)^(7/2)),x)

[Out]

int(1/(atanh(a*x)^2*(1 - a^2*x^2)^(7/2)), x)

________________________________________________________________________________________