3.1.91 \(\int \frac {1}{\sinh ^{-1}(a+b x)^3} \, dx\) [91]

Optimal. Leaf size=63 \[ -\frac {\sqrt {1+(a+b x)^2}}{2 b \sinh ^{-1}(a+b x)^2}-\frac {a+b x}{2 b \sinh ^{-1}(a+b x)}+\frac {\text {Chi}\left (\sinh ^{-1}(a+b x)\right )}{2 b} \]

[Out]

1/2*(-b*x-a)/b/arcsinh(b*x+a)+1/2*Chi(arcsinh(b*x+a))/b-1/2*(1+(b*x+a)^2)^(1/2)/b/arcsinh(b*x+a)^2

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 63, normalized size of antiderivative = 1.00, 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 = {5858, 5773, 5818, 5774, 3382} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]^(-3),x]

[Out]

-1/2*Sqrt[1 + (a + b*x)^2]/(b*ArcSinh[a + b*x]^2) - (a + b*x)/(2*b*ArcSinh[a + b*x]) + CoshIntegral[ArcSinh[a
+ b*x]]/(2*b)

Rule 3382

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 5773

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_), 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 5818

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

Rule 5858

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}{\sinh ^{-1}(a+b x)^3} \, dx &=\frac {\text {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 {\text {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 {\text {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 {\text {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 = 0.06, size = 53, normalized size = 0.84 \begin {gather*} \frac {-\frac {\sqrt {1+(a+b x)^2}}{\sinh ^{-1}(a+b x)^2}-\frac {a+b x}{\sinh ^{-1}(a+b x)}+\text {Chi}\left (\sinh ^{-1}(a+b x)\right )}{2 b} \end {gather*}

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)

________________________________________________________________________________________

Maple [A]
time = 1.93, size = 51, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {-\frac {\sqrt {1+\left (b x +a \right )^{2}}}{2 \arcsinh \left (b x +a \right )^{2}}-\frac {b x +a}{2 \arcsinh \left (b x +a \right )}+\frac {\hyperbolicCosineIntegral \left (\arcsinh \left (b x +a \right )\right )}{2}}{b}\)	\(51\)
default	\(\frac {-\frac {\sqrt {1+\left (b x +a \right )^{2}}}{2 \arcsinh \left (b x +a \right )^{2}}-\frac {b x +a}{2 \arcsinh \left (b x +a \right )}+\frac {\hyperbolicCosineIntegral \left (\arcsinh \left (b x +a \right )\right )}{2}}{b}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arcsinh(b*x+a)^3,x,method=_RETURNVERBOSE)

[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)))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\operatorname {asinh}^{3}{\left (a + b x \right )}}\, dx \end {gather*}

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)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\mathrm {asinh}\left (a+b\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/asinh(a + b*x)^3,x)

[Out]

int(1/asinh(a + b*x)^3, x)

________________________________________________________________________________________