3.91 \(\int \frac{1}{\sinh ^{-1}(a+b x)^3} \, dx\)
Optimal. Leaf size=63 \[ \frac{\text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{2 b}-\frac{a+b x}{2 b \sinh ^{-1}(a+b x)}-\frac{\sqrt{(a+b x)^2+1}}{2 b \sinh ^{-1}(a+b x)^2} \]
[Out]
-Sqrt[1 + (a + b*x)^2]/(2*b*ArcSinh[a + b*x]^2) - (a + b*x)/(2*b*ArcSinh[a + b*x]) + CoshIntegral[ArcSinh[a +
b*x]]/(2*b)
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Rubi [A] time = 0.0786598, antiderivative size = 63, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used =
{5863, 5655, 5774, 5657, 3301} \[ \frac{\text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{2 b}-\frac{a+b x}{2 b \sinh ^{-1}(a+b x)}-\frac{\sqrt{(a+b x)^2+1}}{2 b \sinh ^{-1}(a+b x)^2} \]
Antiderivative was successfully verified.
[In]
Int[ArcSinh[a + b*x]^(-3),x]
[Out]
-Sqrt[1 + (a + b*x)^2]/(2*b*ArcSinh[a + b*x]^2) - (a + b*x)/(2*b*ArcSinh[a + b*x]) + CoshIntegral[ArcSinh[a +
b*x]]/(2*b)
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 5657
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[1/(b*c), Subst[Int[x^n*Cosh[a/b - x/b], x], x,
a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x]
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}{\sinh ^{-1}(a+b x)^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\sinh ^{-1}(x)^3} \, dx,x,a+b x\right )}{b}\\ &=-\frac{\sqrt{1+(a+b x)^2}}{2 b \sinh ^{-1}(a+b x)^2}+\frac{\operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x^2} \sinh ^{-1}(x)^2} \, dx,x,a+b x\right )}{2 b}\\ &=-\frac{\sqrt{1+(a+b x)^2}}{2 b \sinh ^{-1}(a+b x)^2}-\frac{a+b x}{2 b \sinh ^{-1}(a+b x)}+\frac{\operatorname{Subst}\left (\int \frac{1}{\sinh ^{-1}(x)} \, dx,x,a+b x\right )}{2 b}\\ &=-\frac{\sqrt{1+(a+b x)^2}}{2 b \sinh ^{-1}(a+b x)^2}-\frac{a+b x}{2 b \sinh ^{-1}(a+b x)}+\frac{\operatorname{Subst}\left (\int \frac{\cosh (x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b}\\ &=-\frac{\sqrt{1+(a+b x)^2}}{2 b \sinh ^{-1}(a+b x)^2}-\frac{a+b x}{2 b \sinh ^{-1}(a+b x)}+\frac{\text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{2 b}\\ \end{align*}
Mathematica [A] time = 0.0699588, size = 53, normalized size = 0.84 \[ \frac{\text{Chi}\left (\sinh ^{-1}(a+b x)\right )-\frac{a+b x}{\sinh ^{-1}(a+b x)}-\frac{\sqrt{(a+b x)^2+1}}{\sinh ^{-1}(a+b x)^2}}{2 b} \]
Antiderivative was successfully verified.
[In]
Integrate[ArcSinh[a + b*x]^(-3),x]
[Out]
(-(Sqrt[1 + (a + b*x)^2]/ArcSinh[a + b*x]^2) - (a + b*x)/ArcSinh[a + b*x] + CoshIntegral[ArcSinh[a + b*x]])/(2
*b)
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Maple [A] time = 0.024, size = 51, normalized size = 0.8 \begin{align*}{\frac{1}{b} \left ( -{\frac{1}{2\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}}\sqrt{1+ \left ( bx+a \right ) ^{2}}}-{\frac{bx+a}{2\,{\it Arcsinh} \left ( bx+a \right ) }}+{\frac{{\it Chi} \left ({\it Arcsinh} \left ( bx+a \right ) \right ) }{2}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/arcsinh(b*x+a)^3,x)
[Out]
1/b*(-1/2/arcsinh(b*x+a)^2*(1+(b*x+a)^2)^(1/2)-1/2/arcsinh(b*x+a)*(b*x+a)+1/2*Chi(arcsinh(b*x+a)))
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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(1/arcsinh(b*x+a)^3,x, algorithm="maxima")
[Out]
-1/2*(b^7*x^7 + 7*a*b^6*x^6 + a^7 + 3*(7*a^2*b^5 + b^5)*x^5 + 3*a^5 + 5*(7*a^3*b^4 + 3*a*b^4)*x^4 + (35*a^4*b^
3 + 30*a^2*b^3 + 3*b^3)*x^3 + 3*a^3 + 3*(7*a^5*b^2 + 10*a^3*b^2 + 3*a*b^2)*x^2 + (b^4*x^4 + 4*a*b^3*x^3 + a^4
+ (6*a^2*b^2 + b^2)*x^2 + a^2 + 2*(2*a^3*b + a*b)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + (3*b^5*x^5 + 15*a*b
^4*x^4 + 3*a^5 + 5*(6*a^2*b^3 + b^3)*x^3 + 5*a^3 + 15*(2*a^3*b^2 + a*b^2)*x^2 + (15*a^4*b + 15*a^2*b + 2*b)*x
+ 2*a)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + (7*a^6*b + 15*a^4*b + 9*a^2*b + b)*x + (b^7*x^7 + 7*a*b^6*x^6 + a^7 + 3
*(7*a^2*b^5 + b^5)*x^5 + 3*a^5 + 5*(7*a^3*b^4 + 3*a*b^4)*x^4 + (35*a^4*b^3 + 30*a^2*b^3 + 3*b^3)*x^3 + 3*a^3 +
3*(7*a^5*b^2 + 10*a^3*b^2 + 3*a*b^2)*x^2 + (b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4 - 1)*(b^2
*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + 3*(b^5*x^5 + 5*a*b^4*x^4 + a^5 + (10*a^2*b^3 + b^3)*x^3 + a^3 + (10*a^3*b^2
+ 3*a*b^2)*x^2 + (5*a^4*b + 3*a^2*b)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + (7*a^6*b + 15*a^4*b + 9*a^2*b + b)*x +
(3*b^6*x^6 + 18*a*b^5*x^5 + 3*a^6 + 3*(15*a^2*b^4 + 2*b^4)*x^4 + 6*a^4 + 12*(5*a^3*b^3 + 2*a*b^3)*x^3 + (45*a
^4*b^2 + 36*a^2*b^2 + 4*b^2)*x^2 + 4*a^2 + 2*(9*a^5*b + 12*a^3*b + 4*a*b)*x + 1)*sqrt(b^2*x^2 + 2*a*b*x + a^2
+ 1) + a)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + (3*b^6*x^6 + 18*a*b^5*x^5 + 3*a^6 + (45*a^2*b^4 +
7*b^4)*x^4 + 7*a^4 + 4*(15*a^3*b^3 + 7*a*b^3)*x^3 + (45*a^4*b^2 + 42*a^2*b^2 + 5*b^2)*x^2 + 5*a^2 + 2*(9*a^5*
b + 14*a^3*b + 5*a*b)*x + 1)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + a)/((b^7*x^6 + 6*a*b^6*x^5 + a^6*b + 3*a^4*b
+ 3*(5*a^2*b^5 + b^5)*x^4 + 4*(5*a^3*b^4 + 3*a*b^4)*x^3 + 3*a^2*b + 3*(5*a^4*b^3 + 6*a^2*b^3 + b^3)*x^2 + (b^4
*x^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a^3*b)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + 3*(b^5*x^4 + 4*a*b^4*x^3 + a^4
*b + a^2*b + (6*a^2*b^3 + b^3)*x^2 + 2*(2*a^3*b^2 + a*b^2)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + 6*(a^5*b^2 + 2*a
^3*b^2 + a*b^2)*x + 3*(b^6*x^5 + 5*a*b^5*x^4 + a^5*b + 2*a^3*b + 2*(5*a^2*b^4 + b^4)*x^3 + 2*(5*a^3*b^3 + 3*a*
b^3)*x^2 + a*b + (5*a^4*b^2 + 6*a^2*b^2 + b^2)*x)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + b)*log(b*x + a + sqrt(b^
2*x^2 + 2*a*b*x + a^2 + 1))^2) + integrate(1/2*(b^8*x^8 + 8*a*b^7*x^7 + a^8 + 4*(7*a^2*b^6 + b^6)*x^6 + 4*a^6
+ 8*(7*a^3*b^5 + 3*a*b^5)*x^5 + 2*(35*a^4*b^4 + 30*a^2*b^4 + 3*b^4)*x^4 + 6*a^4 + 8*(7*a^5*b^3 + 10*a^3*b^3 +
3*a*b^3)*x^3 + (b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4 + 3)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^2 +
4*(7*a^6*b^2 + 15*a^4*b^2 + 9*a^2*b^2 + b^2)*x^2 + (4*b^5*x^5 + 20*a*b^4*x^4 + 4*a^5 + 4*(10*a^2*b^3 + b^3)*x
^3 + 4*a^3 + 4*(10*a^3*b^2 + 3*a*b^2)*x^2 + (20*a^4*b + 12*a^2*b + 3*b)*x + 3*a)*(b^2*x^2 + 2*a*b*x + a^2 + 1)
^(3/2) + 3*(2*b^6*x^6 + 12*a*b^5*x^5 + 2*a^6 + 2*(15*a^2*b^4 + 2*b^4)*x^4 + 4*a^4 + 8*(5*a^3*b^3 + 2*a*b^3)*x^
3 + (30*a^4*b^2 + 24*a^2*b^2 + b^2)*x^2 + a^2 + 2*(6*a^5*b + 8*a^3*b + a*b)*x - 1)*(b^2*x^2 + 2*a*b*x + a^2 +
1) + 4*a^2 + 8*(a^7*b + 3*a^5*b + 3*a^3*b + a*b)*x + (4*b^7*x^7 + 28*a*b^6*x^6 + 4*a^7 + 12*(7*a^2*b^5 + b^5)*
x^5 + 12*a^5 + 20*(7*a^3*b^4 + 3*a*b^4)*x^4 + (140*a^4*b^3 + 120*a^2*b^3 + 9*b^3)*x^3 + 9*a^3 + 3*(28*a^5*b^2
+ 40*a^3*b^2 + 9*a*b^2)*x^2 + (28*a^6*b + 60*a^4*b + 27*a^2*b + b)*x + a)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) +
1)/((b^8*x^8 + 8*a*b^7*x^7 + a^8 + 4*(7*a^2*b^6 + b^6)*x^6 + 4*a^6 + 8*(7*a^3*b^5 + 3*a*b^5)*x^5 + 2*(35*a^4*b
^4 + 30*a^2*b^4 + 3*b^4)*x^4 + 6*a^4 + 8*(7*a^5*b^3 + 10*a^3*b^3 + 3*a*b^3)*x^3 + (b^4*x^4 + 4*a*b^3*x^3 + 6*a
^2*b^2*x^2 + 4*a^3*b*x + a^4)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^2 + 4*(7*a^6*b^2 + 15*a^4*b^2 + 9*a^2*b^2 + b^2)*x
^2 + 4*(b^5*x^5 + 5*a*b^4*x^4 + a^5 + (10*a^2*b^3 + b^3)*x^3 + a^3 + (10*a^3*b^2 + 3*a*b^2)*x^2 + (5*a^4*b + 3
*a^2*b)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + 6*(b^6*x^6 + 6*a*b^5*x^5 + a^6 + (15*a^2*b^4 + 2*b^4)*x^4 + 2
*a^4 + 4*(5*a^3*b^3 + 2*a*b^3)*x^3 + (15*a^4*b^2 + 12*a^2*b^2 + b^2)*x^2 + a^2 + 2*(3*a^5*b + 4*a^3*b + a*b)*x
)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + 4*a^2 + 8*(a^7*b + 3*a^5*b + 3*a^3*b + a*b)*x + 4*(b^7*x^7 + 7*a*b^6*x^6 + a
^7 + 3*(7*a^2*b^5 + b^5)*x^5 + 3*a^5 + 5*(7*a^3*b^4 + 3*a*b^4)*x^4 + (35*a^4*b^3 + 30*a^2*b^3 + 3*b^3)*x^3 + 3
*a^3 + 3*(7*a^5*b^2 + 10*a^3*b^2 + 3*a*b^2)*x^2 + (7*a^6*b + 15*a^4*b + 9*a^2*b + b)*x + a)*sqrt(b^2*x^2 + 2*a
*b*x + a^2 + 1) + 1)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\operatorname{arsinh}\left (b x + a\right )^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/arcsinh(b*x+a)^3,x, algorithm="fricas")
[Out]
integral(arcsinh(b*x + a)^(-3), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\operatorname{asinh}^{3}{\left (a + b x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/asinh(b*x+a)**3,x)
[Out]
Integral(asinh(a + b*x)**(-3), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\operatorname{arsinh}\left (b x + a\right )^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/arcsinh(b*x+a)^3,x, algorithm="giac")
[Out]
integrate(arcsinh(b*x + a)^(-3), x)