3.5.8 \(\int \frac {c+a^2 c x^2}{\sinh ^{-1}(a x)^2} \, dx\) [408]

Optimal. Leaf size=54 \[ -\frac {c \left (1+a^2 x^2\right )^{3/2}}{a \sinh ^{-1}(a x)}+\frac {3 c \text {Shi}\left (\sinh ^{-1}(a x)\right )}{4 a}+\frac {3 c \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{4 a} \]

[Out]

-c*(a^2*x^2+1)^(3/2)/a/arcsinh(a*x)+3/4*c*Shi(arcsinh(a*x))/a+3/4*c*Shi(3*arcsinh(a*x))/a

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Rubi [A]
time = 0.09, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {5790, 5819, 5556, 3379} \begin {gather*} -\frac {c \left (a^2 x^2+1\right )^{3/2}}{a \sinh ^{-1}(a x)}+\frac {3 c \text {Shi}\left (\sinh ^{-1}(a x)\right )}{4 a}+\frac {3 c \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{4 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + a^2*c*x^2)/ArcSinh[a*x]^2,x]

[Out]

-((c*(1 + a^2*x^2)^(3/2))/(a*ArcSinh[a*x])) + (3*c*SinhIntegral[ArcSinh[a*x]])/(4*a) + (3*c*SinhIntegral[3*Arc
Sinh[a*x]])/(4*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 {c+a^2 c x^2}{\sinh ^{-1}(a x)^2} \, dx &=-\frac {c \left (1+a^2 x^2\right )^{3/2}}{a \sinh ^{-1}(a x)}+(3 a c) \int \frac {x \sqrt {1+a^2 x^2}}{\sinh ^{-1}(a x)} \, dx\\ &=-\frac {c \left (1+a^2 x^2\right )^{3/2}}{a \sinh ^{-1}(a x)}+\frac {(3 c) \text {Subst}\left (\int \frac {\cosh ^2(x) \sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{a}\\ &=-\frac {c \left (1+a^2 x^2\right )^{3/2}}{a \sinh ^{-1}(a x)}+\frac {(3 c) \text {Subst}\left (\int \left (\frac {\sinh (x)}{4 x}+\frac {\sinh (3 x)}{4 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a}\\ &=-\frac {c \left (1+a^2 x^2\right )^{3/2}}{a \sinh ^{-1}(a x)}+\frac {(3 c) \text {Subst}\left (\int \frac {\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{4 a}+\frac {(3 c) \text {Subst}\left (\int \frac {\sinh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{4 a}\\ &=-\frac {c \left (1+a^2 x^2\right )^{3/2}}{a \sinh ^{-1}(a x)}+\frac {3 c \text {Shi}\left (\sinh ^{-1}(a x)\right )}{4 a}+\frac {3 c \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{4 a}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 54, normalized size = 1.00 \begin {gather*} \frac {c \left (-4 \left (1+a^2 x^2\right )^{3/2}+3 \sinh ^{-1}(a x) \text {Shi}\left (\sinh ^{-1}(a x)\right )+3 \sinh ^{-1}(a x) \text {Shi}\left (3 \sinh ^{-1}(a x)\right )\right )}{4 a \sinh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(-4*(1 + a^2*x^2)^(3/2) + 3*ArcSinh[a*x]*SinhIntegral[ArcSinh[a*x]] + 3*ArcSinh[a*x]*SinhIntegral[3*ArcSinh
[a*x]]))/(4*a*ArcSinh[a*x])

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Maple [A]
time = 2.39, size = 60, normalized size = 1.11

method	result	size
derivativedivides	\(\frac {c \left (3 \hyperbolicSineIntegral \left (\arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+3 \hyperbolicSineIntegral \left (3 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )-\cosh \left (3 \arcsinh \left (a x \right )\right )-3 \sqrt {a^{2} x^{2}+1}\right )}{4 a \arcsinh \left (a x \right )}\)	\(60\)
default	\(\frac {c \left (3 \hyperbolicSineIntegral \left (\arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+3 \hyperbolicSineIntegral \left (3 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )-\cosh \left (3 \arcsinh \left (a x \right )\right )-3 \sqrt {a^{2} x^{2}+1}\right )}{4 a \arcsinh \left (a x \right )}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/a*c*(3*Shi(arcsinh(a*x))*arcsinh(a*x)+3*Shi(3*arcsinh(a*x))*arcsinh(a*x)-cosh(3*arcsinh(a*x))-3*(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)/arcsinh(a*x)^2,x, algorithm="maxima")

[Out]

-(a^5*c*x^5 + 2*a^3*c*x^3 + a*c*x + (a^4*c*x^4 + 2*a^2*c*x^2 + c)*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((3*a^6*c*x^6 + 7*a^4*c*x^4 + 5*a^2*c*x^2 + (3*a^4*c*
x^4 + 2*a^2*c*x^2 - c)*(a^2*x^2 + 1) + 3*(2*a^5*c*x^5 + 3*a^3*c*x^3 + a*c*x)*sqrt(a^2*x^2 + 1) + c)/((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)/arcsinh(a*x)^2,x, algorithm="fricas")

[Out]

integral((a^2*c*x^2 + c)/arcsinh(a*x)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} c \left (\int \frac {a^{2} x^{2}}{\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)/asinh(a*x)**2,x)

[Out]

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

[Out]

integrate((a^2*c*x^2 + c)/arcsinh(a*x)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {c\,a^2\,x^2+c}{{\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)/asinh(a*x)^2,x)

[Out]

int((c + a^2*c*x^2)/asinh(a*x)^2, x)

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