3.4.51 \(\int \frac {(c+a^2 c x^2)^2}{\sinh ^{-1}(a x)} \, dx\) [351]

Optimal. Leaf size=50 \[ \frac {5 c^2 \text {Chi}\left (\sinh ^{-1}(a x)\right )}{8 a}+\frac {5 c^2 \text {Chi}\left (3 \sinh ^{-1}(a x)\right )}{16 a}+\frac {c^2 \text {Chi}\left (5 \sinh ^{-1}(a x)\right )}{16 a} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {5791, 3393, 3382} \begin {gather*} \frac {5 c^2 \text {Chi}\left (\sinh ^{-1}(a x)\right )}{8 a}+\frac {5 c^2 \text {Chi}\left (3 \sinh ^{-1}(a x)\right )}{16 a}+\frac {c^2 \text {Chi}\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],x]

[Out]

(5*c^2*CoshIntegral[ArcSinh[a*x]])/(8*a) + (5*c^2*CoshIntegral[3*ArcSinh[a*x]])/(16*a) + (c^2*CoshIntegral[5*A
rcSinh[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 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 5791

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

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 34, normalized size = 0.68 \begin {gather*} \frac {c^2 \left (10 \text {Chi}\left (\sinh ^{-1}(a x)\right )+5 \text {Chi}\left (3 \sinh ^{-1}(a x)\right )+\text {Chi}\left (5 \sinh ^{-1}(a x)\right )\right )}{16 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(10*CoshIntegral[ArcSinh[a*x]] + 5*CoshIntegral[3*ArcSinh[a*x]] + CoshIntegral[5*ArcSinh[a*x]]))/(16*a)

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Maple [A]
time = 1.78, size = 33, normalized size = 0.66

method	result	size
derivativedivides	\(\frac {c^{2} \left (10 \hyperbolicCosineIntegral \left (\arcsinh \left (a x \right )\right )+5 \hyperbolicCosineIntegral \left (3 \arcsinh \left (a x \right )\right )+\hyperbolicCosineIntegral \left (5 \arcsinh \left (a x \right )\right )\right )}{16 a}\)	\(33\)
default	\(\frac {c^{2} \left (10 \hyperbolicCosineIntegral \left (\arcsinh \left (a x \right )\right )+5 \hyperbolicCosineIntegral \left (3 \arcsinh \left (a x \right )\right )+\hyperbolicCosineIntegral \left (5 \arcsinh \left (a x \right )\right )\right )}{16 a}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/a*c^2*(10*Chi(arcsinh(a*x))+5*Chi(3*arcsinh(a*x))+Chi(5*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),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

c**2*(Integral(2*a**2*x**2/asinh(a*x), x) + Integral(a**4*x**4/asinh(a*x), x) + Integral(1/asinh(a*x), 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),x, algorithm="giac")

[Out]

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

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

[Out]

int((c + a^2*c*x^2)^2/asinh(a*x), x)

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