3.5.7 \(\int \frac {(c+a^2 c x^2)^2}{\sinh ^{-1}(a x)^2} \, dx\) [407]

Optimal. Leaf size=77 \[ -\frac {c^2 \left (1+a^2 x^2\right )^{5/2}}{a \sinh ^{-1}(a x)}+\frac {5 c^2 \text {Shi}\left (\sinh ^{-1}(a x)\right )}{8 a}+\frac {15 c^2 \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{16 a}+\frac {5 c^2 \text {Shi}\left (5 \sinh ^{-1}(a x)\right )}{16 a} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {5790, 5819, 5556, 3379} \begin {gather*} -\frac {c^2 \left (a^2 x^2+1\right )^{5/2}}{a \sinh ^{-1}(a x)}+\frac {5 c^2 \text {Shi}\left (\sinh ^{-1}(a x)\right )}{8 a}+\frac {15 c^2 \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{16 a}+\frac {5 c^2 \text {Shi}\left (5 \sinh ^{-1}(a x)\right )}{16 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + a^2*c*x^2)^2/ArcSinh[a*x]^2,x]

[Out]

-((c^2*(1 + a^2*x^2)^(5/2))/(a*ArcSinh[a*x])) + (5*c^2*SinhIntegral[ArcSinh[a*x]])/(8*a) + (15*c^2*SinhIntegra
l[3*ArcSinh[a*x]])/(16*a) + (5*c^2*SinhIntegral[5*ArcSinh[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 5790

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

Rule 5819

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

Rubi steps

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

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Mathematica [A]
time = 0.17, size = 69, normalized size = 0.90 \begin {gather*} \frac {c^2 \left (-16 \left (1+a^2 x^2\right )^{5/2}+10 \sinh ^{-1}(a x) \text {Shi}\left (\sinh ^{-1}(a x)\right )+15 \sinh ^{-1}(a x) \text {Shi}\left (3 \sinh ^{-1}(a x)\right )+5 \sinh ^{-1}(a x) \text {Shi}\left (5 \sinh ^{-1}(a x)\right )\right )}{16 a \sinh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + a^2*c*x^2)^2/ArcSinh[a*x]^2,x]

[Out]

(c^2*(-16*(1 + a^2*x^2)^(5/2) + 10*ArcSinh[a*x]*SinhIntegral[ArcSinh[a*x]] + 15*ArcSinh[a*x]*SinhIntegral[3*Ar
cSinh[a*x]] + 5*ArcSinh[a*x]*SinhIntegral[5*ArcSinh[a*x]]))/(16*a*ArcSinh[a*x])

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Maple [A]
time = 3.43, size = 84, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {c^{2} \left (10 \hyperbolicSineIntegral \left (\arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+15 \hyperbolicSineIntegral \left (3 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+5 \hyperbolicSineIntegral \left (5 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )-5 \cosh \left (3 \arcsinh \left (a x \right )\right )-\cosh \left (5 \arcsinh \left (a x \right )\right )-10 \sqrt {a^{2} x^{2}+1}\right )}{16 a \arcsinh \left (a x \right )}\)	\(84\)
default	\(\frac {c^{2} \left (10 \hyperbolicSineIntegral \left (\arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+15 \hyperbolicSineIntegral \left (3 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+5 \hyperbolicSineIntegral \left (5 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )-5 \cosh \left (3 \arcsinh \left (a x \right )\right )-\cosh \left (5 \arcsinh \left (a x \right )\right )-10 \sqrt {a^{2} x^{2}+1}\right )}{16 a \arcsinh \left (a x \right )}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/a*c^2*(10*Shi(arcsinh(a*x))*arcsinh(a*x)+15*Shi(3*arcsinh(a*x))*arcsinh(a*x)+5*Shi(5*arcsinh(a*x))*arcsin
h(a*x)-5*cosh(3*arcsinh(a*x))-cosh(5*arcsinh(a*x))-10*(a^2*x^2+1)^(1/2))/arcsinh(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/arcsinh(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^2*x^2 + 1))/((a^3*x^2 + sqrt(a^2*x^2 + 1)*a^2*x + a)*log(a*x + sqrt(a^2*x^2 + 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^2*x^2 + 1) + c^2 + 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^2*x^2 + 1))/((a^4*
x^4 + (a^2*x^2 + 1)*a^2*x^2 + 2*a^2*x^2 + 2*(a^3*x^3 + a*x)*sqrt(a^2*x^2 + 1) + 1)*log(a*x + sqrt(a^2*x^2 + 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/arcsinh(a*x)^2,x, algorithm="fricas")

[Out]

integral((a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2)/arcsinh(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 \frac {2 a^{2} x^{2}}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx + \int \frac {a^{4} x^{4}}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx + \int \frac {1}{\operatorname {asinh}^{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/asinh(a*x)**2,x)

[Out]

c**2*(Integral(2*a**2*x**2/asinh(a*x)**2, x) + Integral(a**4*x**4/asinh(a*x)**2, x) + Integral(asinh(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/arcsinh(a*x)^2,x, algorithm="giac")

[Out]

integrate((a^2*c*x^2 + c)^2/arcsinh(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\,x^2+c\right )}^2}{{\mathrm {asinh}\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/asinh(a*x)^2,x)

[Out]

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

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