3.4.16 \(\int \frac {(c-a^2 c x^2)^2}{\cosh ^{-1}(a x)^2} \, dx\) [316]

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

[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 = 0.20, 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 = {5904, 5953, 5556, 3382} \begin {gather*} \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)} \end {gather*}

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 3382

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 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 5904

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

Rule 5953

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d1_) + (e1_.)*(x_))^(p_.)*((d2_) + (e2_.)*(x_))^(p_
.), x_Symbol] :> Dist[(1/(b*c^(m + 1)))*Simp[(d1 + e1*x)^p/(1 + c*x)^p]*Simp[(d2 + e2*x)^p/(-1 + c*x)^p], Subs
t[Int[x^n*Cosh[-a/b + x/b]^m*Sinh[-a/b + x/b]^(2*p + 1), x], x, a + b*ArcCosh[c*x]], x] /; FreeQ[{a, b, c, d1,
e1, d2, e2, n}, x] && EqQ[e1, c*d1] && EqQ[e2, (-c)*d2] && IGtQ[p + 3/2, 0] && IGtQ[m, 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 ) \text {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 ) \text {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 ) \text {Subst}\left (\int \frac {\cosh (5 x)}{x} \, dx,x,\cosh ^{-1}(a x)\right )}{16 a}+\frac {\left (5 c^2\right ) \text {Subst}\left (\int \frac {\cosh (x)}{x} \, dx,x,\cosh ^{-1}(a x)\right )}{8 a}-\frac {\left (15 c^2\right ) \text {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] Leaf count is larger than twice the leaf count of optimal. \(194\) vs. \(2(82)=164\).
time = 0.21, size = 194, normalized size = 2.37 \begin {gather*} -\frac {c^2 \left (16 \sqrt {\frac {-1+a x}{1+a x}}+16 a x \sqrt {\frac {-1+a x}{1+a x}}-32 a^2 x^2 \sqrt {\frac {-1+a x}{1+a x}}-32 a^3 x^3 \sqrt {\frac {-1+a x}{1+a x}}+16 a^4 x^4 \sqrt {\frac {-1+a x}{1+a x}}+16 a^5 x^5 \sqrt {\frac {-1+a x}{1+a x}}-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 )\right )}{16 a \cosh ^{-1}(a x)} \end {gather*}

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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Maple [A]
time = 2.48, size = 85, normalized size = 1.04

method	result	size
derivativedivides	\(-\frac {c^{2} \left (10 \sqrt {a x -1}\, \sqrt {a x +1}+15 \hyperbolicCosineIntegral \left (3 \,\mathrm {arccosh}\left (a x \right )\right ) \mathrm {arccosh}\left (a x \right )-5 \hyperbolicCosineIntegral \left (5 \,\mathrm {arccosh}\left (a x \right )\right ) \mathrm {arccosh}\left (a x \right )-10 \hyperbolicCosineIntegral \left (\mathrm {arccosh}\left (a x \right )\right ) \mathrm {arccosh}\left (a x \right )-5 \sinh \left (3 \,\mathrm {arccosh}\left (a x \right )\right )+\sinh \left (5 \,\mathrm {arccosh}\left (a x \right )\right )\right )}{16 a \,\mathrm {arccosh}\left (a x \right )}\)	\(85\)
default	\(-\frac {c^{2} \left (10 \sqrt {a x -1}\, \sqrt {a x +1}+15 \hyperbolicCosineIntegral \left (3 \,\mathrm {arccosh}\left (a x \right )\right ) \mathrm {arccosh}\left (a x \right )-5 \hyperbolicCosineIntegral \left (5 \,\mathrm {arccosh}\left (a x \right )\right ) \mathrm {arccosh}\left (a x \right )-10 \hyperbolicCosineIntegral \left (\mathrm {arccosh}\left (a x \right )\right ) \mathrm {arccosh}\left (a x \right )-5 \sinh \left (3 \,\mathrm {arccosh}\left (a x \right )\right )+\sinh \left (5 \,\mathrm {arccosh}\left (a x \right )\right )\right )}{16 a \,\mathrm {arccosh}\left (a x \right )}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/a*c^2*(10*(a*x-1)^(1/2)*(a*x+1)^(1/2)+15*Chi(3*arccosh(a*x))*arccosh(a*x)-5*Chi(5*arccosh(a*x))*arccosh(
a*x)-10*Chi(arccosh(a*x))*arccosh(a*x)-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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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((-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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 ) \end {gather*}

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

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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