3.7.20 \(\int \frac {d+e x^2}{a+b \sinh ^{-1}(c x)} \, dx\) [620]

Optimal. Leaf size=180 \[ \frac {d \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b c}-\frac {e \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{4 b c^3}+\frac {e \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b c^3}-\frac {d \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b c}+\frac {e \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{4 b c^3}-\frac {e \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b c^3} \]

[Out]

d*Chi((a+b*arcsinh(c*x))/b)*cosh(a/b)/b/c-1/4*e*Chi((a+b*arcsinh(c*x))/b)*cosh(a/b)/b/c^3+1/4*e*Chi(3*(a+b*arc
sinh(c*x))/b)*cosh(3*a/b)/b/c^3-d*Shi((a+b*arcsinh(c*x))/b)*sinh(a/b)/b/c+1/4*e*Shi((a+b*arcsinh(c*x))/b)*sinh
(a/b)/b/c^3-1/4*e*Shi(3*(a+b*arcsinh(c*x))/b)*sinh(3*a/b)/b/c^3

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Rubi [A]
time = 0.25, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {5793, 5774, 3384, 3379, 3382, 5780, 5556} \begin {gather*} -\frac {e \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{4 b c^3}+\frac {e \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b c^3}+\frac {e \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{4 b c^3}-\frac {e \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b c^3}+\frac {d \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b c}-\frac {d \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Cosh[a/b]*CoshIntegral[(a + b*ArcSinh[c*x])/b])/(b*c) - (e*Cosh[a/b]*CoshIntegral[(a + b*ArcSinh[c*x])/b])/
(4*b*c^3) + (e*Cosh[(3*a)/b]*CoshIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(4*b*c^3) - (d*Sinh[a/b]*SinhIntegral[(
a + b*ArcSinh[c*x])/b])/(b*c) + (e*Sinh[a/b]*SinhIntegral[(a + b*ArcSinh[c*x])/b])/(4*b*c^3) - (e*Sinh[(3*a)/b
]*SinhIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(4*b*c^3)

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 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 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 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 5774

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[1/(b*c), Subst[Int[x^n*Cosh[-a/b + x/b], x], x
, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x]

Rule 5780

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

Rule 5793

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a
+ b*ArcSinh[c*x])^n, (d + e*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && NeQ[e, c^2*d] && IntegerQ[p] &&
(p > 0 || IGtQ[n, 0])

Rubi steps

\begin {align*} \int \frac {d+e x^2}{a+b \sinh ^{-1}(c x)} \, dx &=\int \left (\frac {d}{a+b \sinh ^{-1}(c x)}+\frac {e x^2}{a+b \sinh ^{-1}(c x)}\right ) \, dx\\ &=d \int \frac {1}{a+b \sinh ^{-1}(c x)} \, dx+e \int \frac {x^2}{a+b \sinh ^{-1}(c x)} \, dx\\ &=\frac {d \text {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \sinh ^{-1}(c x)\right )}{b c}+\frac {e \text {Subst}\left (\int \frac {\cosh (x) \sinh ^2(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^3}\\ &=\frac {e \text {Subst}\left (\int \left (-\frac {\cosh (x)}{4 (a+b x)}+\frac {\cosh (3 x)}{4 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^3}+\frac {\left (d \cosh \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {x}{b}\right )}{x} \, dx,x,a+b \sinh ^{-1}(c x)\right )}{b c}-\frac {\left (d \sinh \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {x}{b}\right )}{x} \, dx,x,a+b \sinh ^{-1}(c x)\right )}{b c}\\ &=\frac {d \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b c}-\frac {d \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b c}-\frac {e \text {Subst}\left (\int \frac {\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}+\frac {e \text {Subst}\left (\int \frac {\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}\\ &=\frac {d \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b c}-\frac {d \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b c}-\frac {\left (e \cosh \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}+\frac {\left (e \cosh \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}+\frac {\left (e \sinh \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}-\frac {\left (e \sinh \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}\\ &=-\frac {e \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^3}+\frac {e \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^3}+\frac {d \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b c}+\frac {e \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^3}-\frac {e \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^3}-\frac {d \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b c}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 126, normalized size = 0.70 \begin {gather*} \frac {\left (4 c^2 d-e\right ) \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )+e \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-4 c^2 d \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )+e \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )-e \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )}{4 b c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*c^2*d - e)*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]] + e*Cosh[(3*a)/b]*CoshIntegral[3*(a/b + ArcSinh[c*x]
)] - 4*c^2*d*Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]] + e*Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]] - e*Sin
h[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c*x])])/(4*b*c^3)

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Maple [A]
time = 6.94, size = 178, normalized size = 0.99

method	result	size
derivativedivides	\(\frac {-\frac {e \,{\mathrm e}^{-\frac {3 a}{b}} \expIntegral \left (1, -3 \arcsinh \left (c x \right )-\frac {3 a}{b}\right )}{8 c^{2} b}-\frac {e \,{\mathrm e}^{\frac {3 a}{b}} \expIntegral \left (1, 3 \arcsinh \left (c x \right )+\frac {3 a}{b}\right )}{8 c^{2} b}-\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right ) d}{2 b}+\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right ) e}{8 c^{2} b}-\frac {{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right ) d}{2 b}+\frac {{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right ) e}{8 c^{2} b}}{c}\)	\(178\)
default	\(\frac {-\frac {e \,{\mathrm e}^{-\frac {3 a}{b}} \expIntegral \left (1, -3 \arcsinh \left (c x \right )-\frac {3 a}{b}\right )}{8 c^{2} b}-\frac {e \,{\mathrm e}^{\frac {3 a}{b}} \expIntegral \left (1, 3 \arcsinh \left (c x \right )+\frac {3 a}{b}\right )}{8 c^{2} b}-\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right ) d}{2 b}+\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right ) e}{8 c^{2} b}-\frac {{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right ) d}{2 b}+\frac {{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right ) e}{8 c^{2} b}}{c}\)	\(178\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(-1/8/c^2*e/b*exp(-3*a/b)*Ei(1,-3*arcsinh(c*x)-3*a/b)-1/8/c^2*e/b*exp(3*a/b)*Ei(1,3*arcsinh(c*x)+3*a/b)-1/
2/b*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)*d+1/8/c^2/b*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)*e-1/2/b*exp(-a/b)*Ei(1,-arcsin
h(c*x)-a/b)*d+1/8/c^2/b*exp(-a/b)*Ei(1,-arcsinh(c*x)-a/b)*e)

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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((e*x^2+d)/(a+b*arcsinh(c*x)),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {d + e x^{2}}{a + b \operatorname {asinh}{\left (c x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)/(a+b*asinh(c*x)),x)

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x^2)/(a + b*asinh(c*x)),x)

[Out]

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

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