3.270 \(\int \frac{(1+a^2+2 a b x+b^2 x^2)^{3/2}}{\sinh ^{-1}(a+b x)^2} \, dx\)

Optimal. Leaf size=54 \[ \frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b}+\frac{\text{Shi}\left (4 \sinh ^{-1}(a+b x)\right )}{2 b}-\frac{\left ((a+b x)^2+1\right )^2}{b \sinh ^{-1}(a+b x)} \]

[Out]

-((1 + (a + b*x)^2)^2/(b*ArcSinh[a + b*x])) + SinhIntegral[2*ArcSinh[a + b*x]]/b + SinhIntegral[4*ArcSinh[a +
b*x]]/(2*b)

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Rubi [A]  time = 0.156555, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {5867, 5696, 5779, 5448, 3298} \[ \frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b}+\frac{\text{Shi}\left (4 \sinh ^{-1}(a+b x)\right )}{2 b}-\frac{\left ((a+b x)^2+1\right )^2}{b \sinh ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2)/ArcSinh[a + b*x]^2,x]

[Out]

-((1 + (a + b*x)^2)^2/(b*ArcSinh[a + b*x])) + SinhIntegral[2*ArcSinh[a + b*x]]/b + SinhIntegral[4*ArcSinh[a +
b*x]]/(2*b)

Rule 5867

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> D
ist[1/d, Subst[Int[(C/d^2 + (C*x^2)/d^2)^p*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A,
B, C, n, p}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rule 5696

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(Sqrt[1 + c^2*x^2]
*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[(c*(2*p + 1)*d^IntPart[p]*(d + e*x^2)^Fr
acPart[p])/(b*(n + 1)*(1 + c^2*x^2)^FracPart[p]), Int[x*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n + 1),
x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && 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 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 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]

Rubi steps

\begin{align*} \int \frac{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{\sinh ^{-1}(a+b x)^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^{3/2}}{\sinh ^{-1}(x)^2} \, dx,x,a+b x\right )}{b}\\ &=-\frac{\left (1+(a+b x)^2\right )^2}{b \sinh ^{-1}(a+b x)}+\frac{4 \operatorname{Subst}\left (\int \frac{x \left (1+x^2\right )}{\sinh ^{-1}(x)} \, dx,x,a+b x\right )}{b}\\ &=-\frac{\left (1+(a+b x)^2\right )^2}{b \sinh ^{-1}(a+b x)}+\frac{4 \operatorname{Subst}\left (\int \frac{\cosh ^3(x) \sinh (x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=-\frac{\left (1+(a+b x)^2\right )^2}{b \sinh ^{-1}(a+b x)}+\frac{4 \operatorname{Subst}\left (\int \left (\frac{\sinh (2 x)}{4 x}+\frac{\sinh (4 x)}{8 x}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=-\frac{\left (1+(a+b x)^2\right )^2}{b \sinh ^{-1}(a+b x)}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=-\frac{\left (1+(a+b x)^2\right )^2}{b \sinh ^{-1}(a+b x)}+\frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b}+\frac{\text{Shi}\left (4 \sinh ^{-1}(a+b x)\right )}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.264959, size = 70, normalized size = 1.3 \[ \frac{-2 \left (a^2+2 a b x+b^2 x^2+1\right )^2+2 \sinh ^{-1}(a+b x) \text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )+\sinh ^{-1}(a+b x) \text{Shi}\left (4 \sinh ^{-1}(a+b x)\right )}{2 b \sinh ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2)/ArcSinh[a + b*x]^2,x]

[Out]

(-2*(1 + a^2 + 2*a*b*x + b^2*x^2)^2 + 2*ArcSinh[a + b*x]*SinhIntegral[2*ArcSinh[a + b*x]] + ArcSinh[a + b*x]*S
inhIntegral[4*ArcSinh[a + b*x]])/(2*b*ArcSinh[a + b*x])

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Maple [A]  time = 0.05, size = 72, normalized size = 1.3 \begin{align*}{\frac{8\,{\it Shi} \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ){\it Arcsinh} \left ( bx+a \right ) +4\,{\it Shi} \left ( 4\,{\it Arcsinh} \left ( bx+a \right ) \right ){\it Arcsinh} \left ( bx+a \right ) -4\,\cosh \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ) -\cosh \left ( 4\,{\it Arcsinh} \left ( bx+a \right ) \right ) -3}{8\,b{\it Arcsinh} \left ( bx+a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2+1)^(3/2)/arcsinh(b*x+a)^2,x)

[Out]

1/8/b*(8*Shi(2*arcsinh(b*x+a))*arcsinh(b*x+a)+4*Shi(4*arcsinh(b*x+a))*arcsinh(b*x+a)-4*cosh(2*arcsinh(b*x+a))-
cosh(4*arcsinh(b*x+a))-3)/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((b^2*x^2+2*a*b*x+a^2+1)^(3/2)/arcsinh(b*x+a)^2,x, algorithm="maxima")

[Out]

-((b^4*x^4 + 4*a*b^3*x^3 + a^4 + 2*(3*a^2*b^2 + b^2)*x^2 + 2*a^2 + 4*(a^3*b + a*b)*x + 1)*(b^2*x^2 + 2*a*b*x +
a^2 + 1) + (b^5*x^5 + 5*a*b^4*x^4 + a^5 + 2*(5*a^2*b^3 + b^3)*x^3 + 2*a^3 + 2*(5*a^3*b^2 + 3*a*b^2)*x^2 + (5*
a^4*b + 6*a^2*b + b)*x + a)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/((b^3*x^2 + 2*a*b^2*x + a^2*b + sqrt(b^2*x^2 +
2*a*b*x + a^2 + 1)*(b^2*x + a*b) + b)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))) + integrate(((4*b^4*x^
4 + 16*a*b^3*x^3 + 4*a^4 + 3*(8*a^2*b^2 + b^2)*x^2 + 3*a^2 + 2*(8*a^3*b + 3*a*b)*x - 1)*(b^2*x^2 + 2*a*b*x + a
^2 + 1)^(3/2) + 4*(2*b^5*x^5 + 10*a*b^4*x^4 + 2*a^5 + (20*a^2*b^3 + 3*b^3)*x^3 + 3*a^3 + (20*a^3*b^2 + 9*a*b^2
)*x^2 + (10*a^4*b + 9*a^2*b + b)*x + a)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + (4*b^6*x^6 + 24*a*b^5*x^5 + 4*a^6 + 3*
(20*a^2*b^4 + 3*b^4)*x^4 + 9*a^4 + 4*(20*a^3*b^3 + 9*a*b^3)*x^3 + 6*(10*a^4*b^2 + 9*a^2*b^2 + b^2)*x^2 + 6*a^2
+ 12*(2*a^5*b + 3*a^3*b + a*b)*x + 1)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/((b^4*x^4 + 4*a*b^3*x^3 + a^4 + 2*(3
*a^2*b^2 + b^2)*x^2 + (b^2*x^2 + 2*a*b*x + a^2 + 1)*(b^2*x^2 + 2*a*b*x + a^2) + 2*a^2 + 4*(a^3*b + a*b)*x + 2*
(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*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{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{3}{2}}}{\operatorname{arsinh}\left (b x + a\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2+1)^(3/2)/arcsinh(b*x+a)^2,x, algorithm="fricas")

[Out]

integral((b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/arcsinh(b*x + a)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )^{\frac{3}{2}}}{\operatorname{asinh}^{2}{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2+1)**(3/2)/asinh(b*x+a)**2,x)

[Out]

Integral((a**2 + 2*a*b*x + b**2*x**2 + 1)**(3/2)/asinh(a + b*x)**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2+1)^(3/2)/arcsinh(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/arcsinh(b*x + a)^2, x)