3.178 \(\int \frac{1}{(a+b \sinh ^{-1}(c+d x))^4} \, dx\)

Optimal. Leaf size=160 \[ -\frac{\sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{6 b^4 d}+\frac{\cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{6 b^4 d}-\frac{c+d x}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{\sqrt{(c+d x)^2+1}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{\sqrt{(c+d x)^2+1}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3} \]

[Out]

-Sqrt[1 + (c + d*x)^2]/(3*b*d*(a + b*ArcSinh[c + d*x])^3) - (c + d*x)/(6*b^2*d*(a + b*ArcSinh[c + d*x])^2) - S
qrt[1 + (c + d*x)^2]/(6*b^3*d*(a + b*ArcSinh[c + d*x])) - (CoshIntegral[(a + b*ArcSinh[c + d*x])/b]*Sinh[a/b])
/(6*b^4*d) + (Cosh[a/b]*SinhIntegral[(a + b*ArcSinh[c + d*x])/b])/(6*b^4*d)

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Rubi [A]  time = 0.267411, antiderivative size = 156, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {5863, 5655, 5774, 5779, 3303, 3298, 3301} \[ -\frac{\sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{6 b^4 d}+\frac{\cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{6 b^4 d}-\frac{c+d x}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{\sqrt{(c+d x)^2+1}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{\sqrt{(c+d x)^2+1}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])^(-4),x]

[Out]

-Sqrt[1 + (c + d*x)^2]/(3*b*d*(a + b*ArcSinh[c + d*x])^3) - (c + d*x)/(6*b^2*d*(a + b*ArcSinh[c + d*x])^2) - S
qrt[1 + (c + d*x)^2]/(6*b^3*d*(a + b*ArcSinh[c + d*x])) - (CoshIntegral[a/b + ArcSinh[c + d*x]]*Sinh[a/b])/(6*
b^4*d) + (Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c + d*x]])/(6*b^4*d)

Rule 5863

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

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 5774

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x
)^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && LtQ[n, -
1] && GtQ[d, 0]

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]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \sinh ^{-1}(c+d x)\right )^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (a+b \sinh ^{-1}(x)\right )^4} \, dx,x,c+d x\right )}{d}\\ &=-\frac{\sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}+\frac{\operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{3 b d}\\ &=-\frac{\sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{c+d x}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{1}{\left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{6 b^2 d}\\ &=-\frac{\sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{c+d x}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{\sqrt{1+(c+d x)^2}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )} \, dx,x,c+d x\right )}{6 b^3 d}\\ &=-\frac{\sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{c+d x}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{\sqrt{1+(c+d x)^2}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{6 b^3 d}\\ &=-\frac{\sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{c+d x}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{\sqrt{1+(c+d x)^2}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\cosh \left (\frac{a}{b}\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{6 b^3 d}-\frac{\sinh \left (\frac{a}{b}\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{6 b^3 d}\\ &=-\frac{\sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^3}-\frac{c+d x}{6 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{\sqrt{1+(c+d x)^2}}{6 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{\text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right ) \sinh \left (\frac{a}{b}\right )}{6 b^4 d}+\frac{\cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{6 b^4 d}\\ \end{align*}

Mathematica [A]  time = 0.436369, size = 130, normalized size = 0.81 \[ -\frac{\frac{2 b^3 \sqrt{(c+d x)^2+1}}{\left (a+b \sinh ^{-1}(c+d x)\right )^3}+\frac{b^2 (c+d x)}{\left (a+b \sinh ^{-1}(c+d x)\right )^2}+\sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )-\cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )+\frac{b \sqrt{(c+d x)^2+1}}{a+b \sinh ^{-1}(c+d x)}}{6 b^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c + d*x])^(-4),x]

[Out]

-((2*b^3*Sqrt[1 + (c + d*x)^2])/(a + b*ArcSinh[c + d*x])^3 + (b^2*(c + d*x))/(a + b*ArcSinh[c + d*x])^2 + (b*S
qrt[1 + (c + d*x)^2])/(a + b*ArcSinh[c + d*x]) + CoshIntegral[a/b + ArcSinh[c + d*x]]*Sinh[a/b] - Cosh[a/b]*Si
nhIntegral[a/b + ArcSinh[c + d*x]])/(6*b^4*d)

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Maple [A]  time = 0.062, size = 272, normalized size = 1.7 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{2} \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}+2\,ab{\it Arcsinh} \left ( dx+c \right ) -{\it Arcsinh} \left ( dx+c \right ){b}^{2}+{a}^{2}-ab+2\,{b}^{2}}{12\,{b}^{3} \left ({b}^{3} \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{3}+3\, \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}a{b}^{2}+3\,{a}^{2}b{\it Arcsinh} \left ( dx+c \right ) +{a}^{3} \right ) } \left ( -\sqrt{1+ \left ( dx+c \right ) ^{2}}+dx+c \right ) }+{\frac{1}{12\,{b}^{4}}{{\rm e}^{{\frac{a}{b}}}}{\it Ei} \left ( 1,{\it Arcsinh} \left ( dx+c \right ) +{\frac{a}{b}} \right ) }-{\frac{1}{6\,b \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{3}} \left ( dx+c+\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }-{\frac{1}{12\,{b}^{2} \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}} \left ( dx+c+\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }-{\frac{1}{12\,{b}^{3} \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) } \left ( dx+c+\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }-{\frac{1}{12\,{b}^{4}}{{\rm e}^{-{\frac{a}{b}}}}{\it Ei} \left ( 1,-{\it Arcsinh} \left ( dx+c \right ) -{\frac{a}{b}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsinh(d*x+c))^4,x)

[Out]

1/d*(1/12*(-(1+(d*x+c)^2)^(1/2)+d*x+c)*(b^2*arcsinh(d*x+c)^2+2*a*b*arcsinh(d*x+c)-arcsinh(d*x+c)*b^2+a^2-a*b+2
*b^2)/b^3/(b^3*arcsinh(d*x+c)^3+3*arcsinh(d*x+c)^2*a*b^2+3*a^2*b*arcsinh(d*x+c)+a^3)+1/12/b^4*exp(a/b)*Ei(1,ar
csinh(d*x+c)+a/b)-1/6/b*(d*x+c+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))^3-1/12/b^2*(d*x+c+(1+(d*x+c)^2)^(1/2)
)/(a+b*arcsinh(d*x+c))^2-1/12/b^3*(d*x+c+(1+(d*x+c)^2)^(1/2))/(a+b*arcsinh(d*x+c))-1/12/b^4*exp(-a/b)*Ei(1,-ar
csinh(d*x+c)-a/b))

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsinh(d*x+c))^4,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{b^{4} \operatorname{arsinh}\left (d x + c\right )^{4} + 4 \, a b^{3} \operatorname{arsinh}\left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \operatorname{arsinh}\left (d x + c\right )^{2} + 4 \, a^{3} b \operatorname{arsinh}\left (d x + c\right ) + a^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsinh(d*x+c))^4,x, algorithm="fricas")

[Out]

integral(1/(b^4*arcsinh(d*x + c)^4 + 4*a*b^3*arcsinh(d*x + c)^3 + 6*a^2*b^2*arcsinh(d*x + c)^2 + 4*a^3*b*arcsi
nh(d*x + c) + a^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \operatorname{asinh}{\left (c + d x \right )}\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*asinh(d*x+c))**4,x)

[Out]

Integral((a + b*asinh(c + d*x))**(-4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsinh(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)^(-4), x)