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

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

[Out]

1/6*(-d*x-c)/b^2/d/(a+b*arcsinh(d*x+c))^2+1/6*cosh(a/b)*Shi((a+b*arcsinh(d*x+c))/b)/b^4/d-1/6*Chi((a+b*arcsinh
(d*x+c))/b)*sinh(a/b)/b^4/d-1/3*(1+(d*x+c)^2)^(1/2)/b/d/(a+b*arcsinh(d*x+c))^3-1/6*(1+(d*x+c)^2)^(1/2)/b^3/d/(
a+b*arcsinh(d*x+c))

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Rubi [A]  time = 0.27, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.44, 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]

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

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b^{4} \operatorname {arsinh}\left (d x + c\right )^{4} + 4 \, a b^{3} \operatorname {arsinh}\left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 4 \, a^{3} b \operatorname {arsinh}\left (d x + c\right ) + a^{4}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{4}}\,{d x} \]

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)

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maple [A]  time = 0.07, size = 272, normalized size = 1.70 \[ \frac {\frac {\left (-\sqrt {1+\left (d x +c \right )^{2}}+d x +c \right ) \left (b^{2} \arcsinh \left (d x +c \right )^{2}+2 a b \arcsinh \left (d x +c \right )-\arcsinh \left (d x +c \right ) b^{2}+a^{2}-a b +2 b^{2}\right )}{12 b^{3} \left (b^{3} \arcsinh \left (d x +c \right )^{3}+3 a \,b^{2} \arcsinh \left (d x +c \right )^{2}+3 a^{2} b \arcsinh \left (d x +c \right )+a^{3}\right )}+\frac {{\mathrm e}^{\frac {a}{b}} \Ei \left (1, \arcsinh \left (d x +c \right )+\frac {a}{b}\right )}{12 b^{4}}-\frac {d x +c +\sqrt {1+\left (d x +c \right )^{2}}}{6 b \left (a +b \arcsinh \left (d x +c \right )\right )^{3}}-\frac {d x +c +\sqrt {1+\left (d x +c \right )^{2}}}{12 b^{2} \left (a +b \arcsinh \left (d x +c \right )\right )^{2}}-\frac {d x +c +\sqrt {1+\left (d x +c \right )^{2}}}{12 b^{3} \left (a +b \arcsinh \left (d x +c \right )\right )}-\frac {{\mathrm e}^{-\frac {a}{b}} \Ei \left (1, -\arcsinh \left (d x +c \right )-\frac {a}{b}\right )}{12 b^{4}}}{d} \]

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*a*b^2*arcsinh(d*x+c)^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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asinh(c + d*x))^4,x)

[Out]

int(1/(a + b*asinh(c + d*x))^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{4}}\, dx \]

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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