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

Optimal. Leaf size=103 \[ \frac{e \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{b^2 d}-\frac{e \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{b^2 d}-\frac{e \sqrt{(c+d x)^2+1} (c+d x)}{b d \left (a+b \sinh ^{-1}(c+d x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.147562, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5865, 12, 5665, 3303, 3298, 3301} \[ \frac{e \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{b^2 d}-\frac{e \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{b^2 d}-\frac{e \sqrt{(c+d x)^2+1} (c+d x)}{b d \left (a+b \sinh ^{-1}(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5865

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

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]

Rubi steps

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

Mathematica [A]  time = 0.295022, size = 97, normalized size = 0.94 \[ \frac{e \left (-\frac{b \sqrt{c^2+2 c d x+d^2 x^2+1} (c+d x)}{a+b \sinh ^{-1}(c+d x)}+\cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-\sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.049, size = 160, normalized size = 1.6 \begin{align*}{\frac{1}{d} \left ({\frac{e}{ \left ( 4\,a+4\,b{\it Arcsinh} \left ( dx+c \right ) \right ) b} \left ( 2\, \left ( dx+c \right ) ^{2}-2\, \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}}+1 \right ) }-{\frac{e}{2\,{b}^{2}}{{\rm e}^{2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,2\,{\it Arcsinh} \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) }-{\frac{e}{4\,b \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) } \left ( 2\, \left ( dx+c \right ) ^{2}+1+2\, \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }-{\frac{e}{2\,{b}^{2}}{{\rm e}^{-2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-2\,{\it Arcsinh} \left ( dx+c \right ) -2\,{\frac{a}{b}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d^4*e*x^4 + 4*c*d^3*e*x^3 + c^4*e + c^2*e + (6*c^2*d^2*e + d^2*e)*x^2 + 2*(2*c^3*d*e + c*d*e)*x + (d^3*e*x^3
+ 3*c*d^2*e*x^2 + c^3*e + c*e + (3*c^2*d*e + d*e)*x)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/(a*b*d^3*x^2 + 2*a*b*
c*d^2*x + (c^2*d + d)*a*b + (b^2*d^3*x^2 + 2*b^2*c*d^2*x + (c^2*d + d)*b^2 + (b^2*d^2*x + b^2*c*d)*sqrt(d^2*x^
2 + 2*c*d*x + c^2 + 1))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) + (a*b*d^2*x + a*b*c*d)*sqrt(d^2*x^2
+ 2*c*d*x + c^2 + 1)) + integrate((2*d^5*e*x^5 + 10*c*d^4*e*x^4 + 2*c^5*e + 4*c^3*e + 4*(5*c^2*d^3*e + d^3*e)*
x^3 + 4*(5*c^3*d^2*e + 3*c*d^2*e)*x^2 + 2*(d^3*e*x^3 + 3*c*d^2*e*x^2 + 3*c^2*d*e*x + c^3*e)*(d^2*x^2 + 2*c*d*x
+ c^2 + 1) + 2*c*e + 2*(5*c^4*d*e + 6*c^2*d*e + d*e)*x + (4*d^4*e*x^4 + 16*c*d^3*e*x^3 + 4*c^4*e + 4*c^2*e +
4*(6*c^2*d^2*e + d^2*e)*x^2 + 8*(2*c^3*d*e + c*d*e)*x + e)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/(a*b*d^4*x^4 + 4
*a*b*c*d^3*x^3 + 2*(3*c^2*d^2 + d^2)*a*b*x^2 + 4*(c^3*d + c*d)*a*b*x + (c^4 + 2*c^2 + 1)*a*b + (a*b*d^2*x^2 +
2*a*b*c*d*x + a*b*c^2)*(d^2*x^2 + 2*c*d*x + c^2 + 1) + (b^2*d^4*x^4 + 4*b^2*c*d^3*x^3 + 2*(3*c^2*d^2 + d^2)*b^
2*x^2 + 4*(c^3*d + c*d)*b^2*x + (c^4 + 2*c^2 + 1)*b^2 + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*(d^2*x^2 + 2*c*d
*x + c^2 + 1) + 2*(b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + (3*c^2*d + d)*b^2*x + (c^3 + c)*b^2)*sqrt(d^2*x^2 + 2*c*d*x
+ c^2 + 1))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) + 2*(a*b*d^3*x^3 + 3*a*b*c*d^2*x^2 + (3*c^2*d +
d)*a*b*x + (c^3 + c)*a*b)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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