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]

e*Chi(2*(a+b*arcsinh(c*x))/b)*cosh(2*a/b)/b^2/c^2+d*cosh(a/b)*Shi((a+b*arcsinh(c*x))/b)/b^2/c-d*Chi((a+b*arcsi
nh(c*x))/b)*sinh(a/b)/b^2/c-e*Shi(2*(a+b*arcsinh(c*x))/b)*sinh(2*a/b)/b^2/c^2-d*(c^2*x^2+1)^(1/2)/b/c/(a+b*arc
sinh(c*x))-e*x*(c^2*x^2+1)^(1/2)/b/c/(a+b*arcsinh(c*x))

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Rubi [A]  time = 0.34, 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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.75, 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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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e x + d}{b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e x + d}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}\,{d x} \]

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)

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maple [A]  time = 0.27, size = 272, normalized size = 1.51 \[ \frac {\frac {\left (c x -\sqrt {c^{2} x^{2}+1}\right ) d}{2 b \left (a +b \arcsinh \left (c x \right )\right )}+\frac {d \,{\mathrm e}^{\frac {a}{b}} \Ei \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right )}{2 b^{2}}-\frac {d \left (c x +\sqrt {c^{2} x^{2}+1}\right )}{2 b \left (a +b \arcsinh \left (c x \right )\right )}-\frac {d \,{\mathrm e}^{-\frac {a}{b}} \Ei \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right )}{2 b^{2}}+\frac {\left (2 c^{2} x^{2}-2 c x \sqrt {c^{2} x^{2}+1}+1\right ) e}{4 c \left (a +b \arcsinh \left (c x \right )\right ) b}-\frac {e \,{\mathrm e}^{\frac {2 a}{b}} \Ei \left (1, 2 \arcsinh \left (c x \right )+\frac {2 a}{b}\right )}{2 c \,b^{2}}-\frac {e \left (2 c^{2} x^{2}+1+2 c x \sqrt {c^{2} x^{2}+1}\right )}{4 c b \left (a +b \arcsinh \left (c x \right )\right )}-\frac {e \,{\mathrm e}^{-\frac {2 a}{b}} \Ei \left (1, -2 \arcsinh \left (c x \right )-\frac {2 a}{b}\right )}{2 c \,b^{2}}}{c} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ -\frac {c^{3} e x^{4} + c^{3} d x^{3} + c e x^{2} + c d x + {\left (c^{2} e x^{3} + c^{2} d x^{2} + e x + d\right )} \sqrt {c^{2} x^{2} + 1}}{a b c^{3} x^{2} + \sqrt {c^{2} x^{2} + 1} a b c^{2} x + a b c + {\left (b^{2} c^{3} x^{2} + \sqrt {c^{2} x^{2} + 1} b^{2} c^{2} x + b^{2} c\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )} + \int \frac {2 \, c^{5} e x^{5} + c^{5} d x^{4} + 4 \, c^{3} e x^{3} + 2 \, c^{3} d x^{2} + 2 \, c e x + {\left (2 \, c^{3} e x^{3} + c^{3} d x^{2} - c d\right )} {\left (c^{2} x^{2} + 1\right )} + c d + {\left (4 \, c^{4} e x^{4} + 2 \, c^{4} d x^{3} + 4 \, c^{2} e x^{2} + c^{2} d x + e\right )} \sqrt {c^{2} x^{2} + 1}}{a b c^{5} x^{4} + {\left (c^{2} x^{2} + 1\right )} a b c^{3} x^{2} + 2 \, a b c^{3} x^{2} + a b c + {\left (b^{2} c^{5} x^{4} + {\left (c^{2} x^{2} + 1\right )} b^{2} c^{3} x^{2} + 2 \, b^{2} c^{3} x^{2} + b^{2} c + 2 \, {\left (b^{2} c^{4} x^{3} + b^{2} c^{2} x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (a b c^{4} x^{3} + a b c^{2} x\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)/(a + b*asinh(c*x))^2,x)

[Out]

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

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

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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