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

Optimal. Leaf size=180 \[ \frac{e \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{b^2 c^2}-\frac{e \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{b^2 c^2}-\frac{d \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{b^2 c}+\frac{d \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{b^2 c}-\frac{d \sqrt{c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac{e x \sqrt{c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

[Out]

-((d*Sqrt[1 + c^2*x^2])/(b*c*(a + b*ArcSinh[c*x]))) - (e*x*Sqrt[1 + c^2*x^2])/(b*c*(a + b*ArcSinh[c*x])) + (e*
Cosh[(2*a)/b]*CoshIntegral[(2*(a + b*ArcSinh[c*x]))/b])/(b^2*c^2) - (d*CoshIntegral[(a + b*ArcSinh[c*x])/b]*Si
nh[a/b])/(b^2*c) + (d*Cosh[a/b]*SinhIntegral[(a + b*ArcSinh[c*x])/b])/(b^2*c) - (e*Sinh[(2*a)/b]*SinhIntegral[
(2*(a + b*ArcSinh[c*x]))/b])/(b^2*c^2)

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Rubi [A]  time = 0.33673, antiderivative size = 176, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {5803, 5655, 5779, 3303, 3298, 3301, 5665} \[ \frac{e \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{b^2 c^2}-\frac{e \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{b^2 c^2}-\frac{d \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{b^2 c}+\frac{d \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{b^2 c}-\frac{d \sqrt{c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac{e x \sqrt{c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*Sqrt[1 + c^2*x^2])/(b*c*(a + b*ArcSinh[c*x]))) - (e*x*Sqrt[1 + c^2*x^2])/(b*c*(a + b*ArcSinh[c*x])) + (e*
Cosh[(2*a)/b]*CoshIntegral[(2*a)/b + 2*ArcSinh[c*x]])/(b^2*c^2) - (d*CoshIntegral[a/b + ArcSinh[c*x]]*Sinh[a/b
])/(b^2*c) + (d*Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]])/(b^2*c) - (e*Sinh[(2*a)/b]*SinhIntegral[(2*a)/b +
2*ArcSinh[c*x]])/(b^2*c^2)

Rule 5803

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*ArcSinh[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[m, 0] && LtQ[n, -1]

Rule 5655

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^(n + 1
))/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcSinh[c*x])^(n + 1))/Sqrt[1 + c^2*x^2], x], x] /; F
reeQ[{a, b, c}, x] && LtQ[n, -1]

Rule 5779

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

Rule 3303

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 3298

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 3301

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 5665

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 + c^2*x^2]*(a + b*ArcSi
nh[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0]
&& GeQ[n, -2] && LtQ[n, -1]

Rubi steps

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

Mathematica [A]  time = 0.734094, size = 150, normalized size = 0.83 \[ -\frac{\frac{b c d \sqrt{c^2 x^2+1}}{a+b \sinh ^{-1}(c x)}+\frac{b c e x \sqrt{c^2 x^2+1}}{a+b \sinh ^{-1}(c x)}+c d \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )-e \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-c d \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )+e \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )}{b^2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*c*d*Sqrt[1 + c^2*x^2])/(a + b*ArcSinh[c*x]) + (b*c*e*x*Sqrt[1 + c^2*x^2])/(a + b*ArcSinh[c*x]) - e*Cosh[
(2*a)/b]*CoshIntegral[2*(a/b + ArcSinh[c*x])] + c*d*CoshIntegral[a/b + ArcSinh[c*x]]*Sinh[a/b] - c*d*Cosh[a/b]
*SinhIntegral[a/b + ArcSinh[c*x]] + e*Sinh[(2*a)/b]*SinhIntegral[2*(a/b + ArcSinh[c*x])])/(b^2*c^2))

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Maple [A]  time = 0.125, size = 272, normalized size = 1.5 \begin{align*}{\frac{1}{c} \left ({\frac{d}{2\,b \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) } \left ( cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }+{\frac{d}{2\,{b}^{2}}{{\rm e}^{{\frac{a}{b}}}}{\it Ei} \left ( 1,{\it Arcsinh} \left ( cx \right ) +{\frac{a}{b}} \right ) }-{\frac{d}{2\,b \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) } \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }-{\frac{d}{2\,{b}^{2}}{{\rm e}^{-{\frac{a}{b}}}}{\it Ei} \left ( 1,-{\it Arcsinh} \left ( cx \right ) -{\frac{a}{b}} \right ) }+{\frac{e}{4\,c \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) b} \left ( 2\,{c}^{2}{x}^{2}-2\,cx\sqrt{{c}^{2}{x}^{2}+1}+1 \right ) }-{\frac{e}{2\,c{b}^{2}}{{\rm e}^{2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,2\,{\it Arcsinh} \left ( cx \right ) +2\,{\frac{a}{b}} \right ) }-{\frac{e}{4\,c \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) b} \left ( 2\,{c}^{2}{x}^{2}+1+2\,cx\sqrt{{c}^{2}{x}^{2}+1} \right ) }-{\frac{e}{2\,c{b}^{2}}{{\rm e}^{-2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-2\,{\it Arcsinh} \left ( cx \right ) -2\,{\frac{a}{b}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(1/2*(c*x-(c^2*x^2+1)^(1/2))*d/b/(a+b*arcsinh(c*x))+1/2*d/b^2*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)-1/2/b*d*(c*x
+(c^2*x^2+1)^(1/2))/(a+b*arcsinh(c*x))-1/2/b^2*d*exp(-a/b)*Ei(1,-arcsinh(c*x)-a/b)+1/4*(2*c^2*x^2-2*c*x*(c^2*x
^2+1)^(1/2)+1)*e/c/(a+b*arcsinh(c*x))/b-1/2/c*e/b^2*exp(2*a/b)*Ei(1,2*arcsinh(c*x)+2*a/b)-1/4/c*e/b*(2*c^2*x^2
+1+2*c*x*(c^2*x^2+1)^(1/2))/(a+b*arcsinh(c*x))-1/2/c*e/b^2*exp(-2*a/b)*Ei(1,-2*arcsinh(c*x)-2*a/b))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{c^{3} e x^{4} + c^{3} d x^{3} + c e x^{2} + c d x +{\left (c^{2} e x^{3} + c^{2} d x^{2} + e x + d\right )} \sqrt{c^{2} x^{2} + 1}}{a b c^{3} x^{2} + \sqrt{c^{2} x^{2} + 1} a b c^{2} x + a b c +{\left (b^{2} c^{3} x^{2} + \sqrt{c^{2} x^{2} + 1} b^{2} c^{2} x + b^{2} c\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )} + \int \frac{2 \, c^{5} e x^{5} + c^{5} d x^{4} + 4 \, c^{3} e x^{3} + 2 \, c^{3} d x^{2} + 2 \, c e x +{\left (2 \, c^{3} e x^{3} + c^{3} d x^{2} - c d\right )}{\left (c^{2} x^{2} + 1\right )} + c d +{\left (4 \, c^{4} e x^{4} + 2 \, c^{4} d x^{3} + 4 \, c^{2} e x^{2} + c^{2} d x + e\right )} \sqrt{c^{2} x^{2} + 1}}{a b c^{5} x^{4} +{\left (c^{2} x^{2} + 1\right )} a b c^{3} x^{2} + 2 \, a b c^{3} x^{2} + a b c +{\left (b^{2} c^{5} x^{4} +{\left (c^{2} x^{2} + 1\right )} b^{2} c^{3} x^{2} + 2 \, b^{2} c^{3} x^{2} + b^{2} c + 2 \,{\left (b^{2} c^{4} x^{3} + b^{2} c^{2} x\right )} \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 2 \,{\left (a b c^{4} x^{3} + a b c^{2} x\right )} \sqrt{c^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c^3*e*x^4 + c^3*d*x^3 + c*e*x^2 + c*d*x + (c^2*e*x^3 + c^2*d*x^2 + e*x + d)*sqrt(c^2*x^2 + 1))/(a*b*c^3*x^2
+ sqrt(c^2*x^2 + 1)*a*b*c^2*x + a*b*c + (b^2*c^3*x^2 + sqrt(c^2*x^2 + 1)*b^2*c^2*x + b^2*c)*log(c*x + sqrt(c^2
*x^2 + 1))) + integrate((2*c^5*e*x^5 + c^5*d*x^4 + 4*c^3*e*x^3 + 2*c^3*d*x^2 + 2*c*e*x + (2*c^3*e*x^3 + c^3*d*
x^2 - c*d)*(c^2*x^2 + 1) + c*d + (4*c^4*e*x^4 + 2*c^4*d*x^3 + 4*c^2*e*x^2 + c^2*d*x + e)*sqrt(c^2*x^2 + 1))/(a
*b*c^5*x^4 + (c^2*x^2 + 1)*a*b*c^3*x^2 + 2*a*b*c^3*x^2 + a*b*c + (b^2*c^5*x^4 + (c^2*x^2 + 1)*b^2*c^3*x^2 + 2*
b^2*c^3*x^2 + b^2*c + 2*(b^2*c^4*x^3 + b^2*c^2*x)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1)) + 2*(a*b*c^4
*x^3 + a*b*c^2*x)*sqrt(c^2*x^2 + 1)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{e x + d}{b^{2} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname{arsinh}\left (c x\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e x + d}{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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