3.316 \(\int \frac {(c-a^2 c x^2)^2}{\cosh ^{-1}(a x)^2} \, dx\)

Optimal. Leaf size=82 \[ \frac {5 c^2 \text {Chi}\left (\cosh ^{-1}(a x)\right )}{8 a}-\frac {15 c^2 \text {Chi}\left (3 \cosh ^{-1}(a x)\right )}{16 a}+\frac {5 c^2 \text {Chi}\left (5 \cosh ^{-1}(a x)\right )}{16 a}-\frac {c^2 (a x-1)^{5/2} (a x+1)^{5/2}}{a \cosh ^{-1}(a x)} \]

[Out]

-c^2*(a*x-1)^(5/2)*(a*x+1)^(5/2)/a/arccosh(a*x)+5/8*c^2*Chi(arccosh(a*x))/a-15/16*c^2*Chi(3*arccosh(a*x))/a+5/
16*c^2*Chi(5*arccosh(a*x))/a

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Rubi [A]  time = 0.31, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5695, 5781, 5448, 3301} \[ \frac {5 c^2 \text {Chi}\left (\cosh ^{-1}(a x)\right )}{8 a}-\frac {15 c^2 \text {Chi}\left (3 \cosh ^{-1}(a x)\right )}{16 a}+\frac {5 c^2 \text {Chi}\left (5 \cosh ^{-1}(a x)\right )}{16 a}-\frac {c^2 (a x-1)^{5/2} (a x+1)^{5/2}}{a \cosh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Int[(c - a^2*c*x^2)^2/ArcCosh[a*x]^2,x]

[Out]

-((c^2*(-1 + a*x)^(5/2)*(1 + a*x)^(5/2))/(a*ArcCosh[a*x])) + (5*c^2*CoshIntegral[ArcCosh[a*x]])/(8*a) - (15*c^
2*CoshIntegral[3*ArcCosh[a*x]])/(16*a) + (5*c^2*CoshIntegral[5*ArcCosh[a*x]])/(16*a)

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 5448

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 5695

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

Rule 5781

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d1_) + (e1_.)*(x_))^(p_.)*((d2_) + (e2_.)*(x_))^(p_
.), x_Symbol] :> Dist[(-(d1*d2))^p/c^(m + 1), Subst[Int[(a + b*x)^n*Cosh[x]^m*Sinh[x]^(2*p + 1), x], x, ArcCos
h[c*x]], x] /; FreeQ[{a, b, c, d1, e1, d2, e2, n}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && IntegerQ[p
+ 1/2] && GtQ[p, -1] && IGtQ[m, 0] && (GtQ[d1, 0] && LtQ[d2, 0])

Rubi steps

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

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Mathematica [B]  time = 0.32, size = 194, normalized size = 2.37 \[ -\frac {c^2 \left (16 a^5 x^5 \sqrt {\frac {a x-1}{a x+1}}+16 a^4 x^4 \sqrt {\frac {a x-1}{a x+1}}-32 a^3 x^3 \sqrt {\frac {a x-1}{a x+1}}-32 a^2 x^2 \sqrt {\frac {a x-1}{a x+1}}-10 \cosh ^{-1}(a x) \text {Chi}\left (\cosh ^{-1}(a x)\right )+15 \cosh ^{-1}(a x) \text {Chi}\left (3 \cosh ^{-1}(a x)\right )-5 \cosh ^{-1}(a x) \text {Chi}\left (5 \cosh ^{-1}(a x)\right )+16 a x \sqrt {\frac {a x-1}{a x+1}}+16 \sqrt {\frac {a x-1}{a x+1}}\right )}{16 a \cosh ^{-1}(a x)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - a^2*c*x^2)^2/ArcCosh[a*x]^2,x]

[Out]

-1/16*(c^2*(16*Sqrt[(-1 + a*x)/(1 + a*x)] + 16*a*x*Sqrt[(-1 + a*x)/(1 + a*x)] - 32*a^2*x^2*Sqrt[(-1 + a*x)/(1
+ a*x)] - 32*a^3*x^3*Sqrt[(-1 + a*x)/(1 + a*x)] + 16*a^4*x^4*Sqrt[(-1 + a*x)/(1 + a*x)] + 16*a^5*x^5*Sqrt[(-1
+ a*x)/(1 + a*x)] - 10*ArcCosh[a*x]*CoshIntegral[ArcCosh[a*x]] + 15*ArcCosh[a*x]*CoshIntegral[3*ArcCosh[a*x]]
- 5*ArcCosh[a*x]*CoshIntegral[5*ArcCosh[a*x]]))/(a*ArcCosh[a*x])

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{4} c^{2} x^{4} - 2 \, a^{2} c^{2} x^{2} + c^{2}}{\operatorname {arcosh}\left (a x\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^2/arccosh(a*x)^2,x, algorithm="fricas")

[Out]

integral((a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2)/arccosh(a*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^2/arccosh(a*x)^2,x, algorithm="giac")

[Out]

integrate((a^2*c*x^2 - c)^2/arccosh(a*x)^2, x)

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maple [A]  time = 0.14, size = 87, normalized size = 1.06 \[ \frac {c^{2} \left (10 \Chi \left (\mathrm {arccosh}\left (a x \right )\right ) \mathrm {arccosh}\left (a x \right )-15 \Chi \left (3 \,\mathrm {arccosh}\left (a x \right )\right ) \mathrm {arccosh}\left (a x \right )+5 \Chi \left (5 \,\mathrm {arccosh}\left (a x \right )\right ) \mathrm {arccosh}\left (a x \right )-10 \sqrt {a x -1}\, \sqrt {a x +1}+5 \sinh \left (3 \,\mathrm {arccosh}\left (a x \right )\right )-\sinh \left (5 \,\mathrm {arccosh}\left (a x \right )\right )\right )}{16 a \,\mathrm {arccosh}\left (a x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*c*x^2+c)^2/arccosh(a*x)^2,x)

[Out]

1/16/a*c^2*(10*Chi(arccosh(a*x))*arccosh(a*x)-15*Chi(3*arccosh(a*x))*arccosh(a*x)+5*Chi(5*arccosh(a*x))*arccos
h(a*x)-10*(a*x-1)^(1/2)*(a*x+1)^(1/2)+5*sinh(3*arccosh(a*x))-sinh(5*arccosh(a*x)))/arccosh(a*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {a^{7} c^{2} x^{7} - 3 \, a^{5} c^{2} x^{5} + 3 \, a^{3} c^{2} x^{3} - a c^{2} x + {\left (a^{6} c^{2} x^{6} - 3 \, a^{4} c^{2} x^{4} + 3 \, a^{2} c^{2} x^{2} - c^{2}\right )} \sqrt {a x + 1} \sqrt {a x - 1}}{{\left (a^{3} x^{2} + \sqrt {a x + 1} \sqrt {a x - 1} a^{2} x - a\right )} \log \left (a x + \sqrt {a x + 1} \sqrt {a x - 1}\right )} + \int \frac {5 \, a^{8} c^{2} x^{8} - 16 \, a^{6} c^{2} x^{6} + 18 \, a^{4} c^{2} x^{4} - 8 \, a^{2} c^{2} x^{2} + {\left (5 \, a^{6} c^{2} x^{6} - 9 \, a^{4} c^{2} x^{4} + 3 \, a^{2} c^{2} x^{2} + c^{2}\right )} {\left (a x + 1\right )} {\left (a x - 1\right )} + 5 \, {\left (2 \, a^{7} c^{2} x^{7} - 5 \, a^{5} c^{2} x^{5} + 4 \, a^{3} c^{2} x^{3} - a c^{2} x\right )} \sqrt {a x + 1} \sqrt {a x - 1} + c^{2}}{{\left (a^{4} x^{4} + {\left (a x + 1\right )} {\left (a x - 1\right )} a^{2} x^{2} - 2 \, a^{2} x^{2} + 2 \, {\left (a^{3} x^{3} - a x\right )} \sqrt {a x + 1} \sqrt {a x - 1} + 1\right )} \log \left (a x + \sqrt {a x + 1} \sqrt {a x - 1}\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^2/arccosh(a*x)^2,x, algorithm="maxima")

[Out]

-(a^7*c^2*x^7 - 3*a^5*c^2*x^5 + 3*a^3*c^2*x^3 - a*c^2*x + (a^6*c^2*x^6 - 3*a^4*c^2*x^4 + 3*a^2*c^2*x^2 - c^2)*
sqrt(a*x + 1)*sqrt(a*x - 1))/((a^3*x^2 + sqrt(a*x + 1)*sqrt(a*x - 1)*a^2*x - a)*log(a*x + sqrt(a*x + 1)*sqrt(a
*x - 1))) + integrate((5*a^8*c^2*x^8 - 16*a^6*c^2*x^6 + 18*a^4*c^2*x^4 - 8*a^2*c^2*x^2 + (5*a^6*c^2*x^6 - 9*a^
4*c^2*x^4 + 3*a^2*c^2*x^2 + c^2)*(a*x + 1)*(a*x - 1) + 5*(2*a^7*c^2*x^7 - 5*a^5*c^2*x^5 + 4*a^3*c^2*x^3 - a*c^
2*x)*sqrt(a*x + 1)*sqrt(a*x - 1) + c^2)/((a^4*x^4 + (a*x + 1)*(a*x - 1)*a^2*x^2 - 2*a^2*x^2 + 2*(a^3*x^3 - a*x
)*sqrt(a*x + 1)*sqrt(a*x - 1) + 1)*log(a*x + sqrt(a*x + 1)*sqrt(a*x - 1))), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c-a^2\,c\,x^2\right )}^2}{{\mathrm {acosh}\left (a\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - a^2*c*x^2)^2/acosh(a*x)^2,x)

[Out]

int((c - a^2*c*x^2)^2/acosh(a*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c^{2} \left (\int \left (- \frac {2 a^{2} x^{2}}{\operatorname {acosh}^{2}{\left (a x \right )}}\right )\, dx + \int \frac {a^{4} x^{4}}{\operatorname {acosh}^{2}{\left (a x \right )}}\, dx + \int \frac {1}{\operatorname {acosh}^{2}{\left (a x \right )}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*c*x**2+c)**2/acosh(a*x)**2,x)

[Out]

c**2*(Integral(-2*a**2*x**2/acosh(a*x)**2, x) + Integral(a**4*x**4/acosh(a*x)**2, x) + Integral(acosh(a*x)**(-
2), x))

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