3.512 \(\int x^2 \sqrt{d+c^2 d x^2} (a+b \sinh ^{-1}(c x))^n \, dx\)

Optimal. Leaf size=235 \[ \frac{2^{-2 (n+3)} e^{-\frac{4 a}{b}} \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac{a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c^3 \sqrt{c^2 x^2+1}}-\frac{2^{-2 (n+3)} e^{\frac{4 a}{b}} \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text{Gamma}\left (n+1,\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c^3 \sqrt{c^2 x^2+1}}-\frac{\sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^{n+1}}{8 b c^3 (n+1) \sqrt{c^2 x^2+1}} \]

[Out]

-(Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^(1 + n))/(8*b*c^3*(1 + n)*Sqrt[1 + c^2*x^2]) + (Sqrt[d + c^2*d*x^2]
*(a + b*ArcSinh[c*x])^n*Gamma[1 + n, (-4*(a + b*ArcSinh[c*x]))/b])/(2^(2*(3 + n))*c^3*E^((4*a)/b)*Sqrt[1 + c^2
*x^2]*(-((a + b*ArcSinh[c*x])/b))^n) - (E^((4*a)/b)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^n*Gamma[1 + n, (4
*(a + b*ArcSinh[c*x]))/b])/(2^(2*(3 + n))*c^3*Sqrt[1 + c^2*x^2]*((a + b*ArcSinh[c*x])/b)^n)

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Rubi [A]  time = 0.454066, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {5782, 5779, 5448, 3307, 2181} \[ \frac{2^{-2 (n+3)} e^{-\frac{4 a}{b}} \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac{a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c^3 \sqrt{c^2 x^2+1}}-\frac{2^{-2 (n+3)} e^{\frac{4 a}{b}} \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text{Gamma}\left (n+1,\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c^3 \sqrt{c^2 x^2+1}}-\frac{\sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^{n+1}}{8 b c^3 (n+1) \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^n,x]

[Out]

-(Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^(1 + n))/(8*b*c^3*(1 + n)*Sqrt[1 + c^2*x^2]) + (Sqrt[d + c^2*d*x^2]
*(a + b*ArcSinh[c*x])^n*Gamma[1 + n, (-4*(a + b*ArcSinh[c*x]))/b])/(2^(2*(3 + n))*c^3*E^((4*a)/b)*Sqrt[1 + c^2
*x^2]*(-((a + b*ArcSinh[c*x])/b))^n) - (E^((4*a)/b)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^n*Gamma[1 + n, (4
*(a + b*ArcSinh[c*x]))/b])/(2^(2*(3 + n))*c^3*Sqrt[1 + c^2*x^2]*((a + b*ArcSinh[c*x])/b)^n)

Rule 5782

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^IntPa
rt[p]*(d + e*x^2)^FracPart[p])/(1 + c^2*x^2)^FracPart[p], Int[x^m*(1 + c^2*x^2)^p*(a + b*ArcSinh[c*x])^n, x],
x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] &&  !(Integer
Q[p] || 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 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 3307

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rubi steps

\begin{align*} \int x^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \, dx &=\frac{\sqrt{d+c^2 d x^2} \int x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \, dx}{\sqrt{1+c^2 x^2}}\\ &=\frac{\sqrt{d+c^2 d x^2} \operatorname{Subst}\left (\int (a+b x)^n \cosh ^2(x) \sinh ^2(x) \, dx,x,\sinh ^{-1}(c x)\right )}{c^3 \sqrt{1+c^2 x^2}}\\ &=\frac{\sqrt{d+c^2 d x^2} \operatorname{Subst}\left (\int \left (-\frac{1}{8} (a+b x)^n+\frac{1}{8} (a+b x)^n \cosh (4 x)\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^3 \sqrt{1+c^2 x^2}}\\ &=-\frac{\sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^{1+n}}{8 b c^3 (1+n) \sqrt{1+c^2 x^2}}+\frac{\sqrt{d+c^2 d x^2} \operatorname{Subst}\left (\int (a+b x)^n \cosh (4 x) \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^3 \sqrt{1+c^2 x^2}}\\ &=-\frac{\sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^{1+n}}{8 b c^3 (1+n) \sqrt{1+c^2 x^2}}+\frac{\sqrt{d+c^2 d x^2} \operatorname{Subst}\left (\int e^{-4 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^3 \sqrt{1+c^2 x^2}}+\frac{\sqrt{d+c^2 d x^2} \operatorname{Subst}\left (\int e^{4 x} (a+b x)^n \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^3 \sqrt{1+c^2 x^2}}\\ &=-\frac{\sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^{1+n}}{8 b c^3 (1+n) \sqrt{1+c^2 x^2}}+\frac{4^{-3-n} e^{-\frac{4 a}{b}} \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac{a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,-\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c^3 \sqrt{1+c^2 x^2}}-\frac{4^{-3-n} e^{\frac{4 a}{b}} \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \Gamma \left (1+n,\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c^3 \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.945789, size = 170, normalized size = 0.72 \[ \frac{d \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^n \left (4^{-n} e^{-\frac{4 a}{b}} \left (-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{b^2}\right )^{-n} \left (\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )^n \text{Gamma}\left (n+1,-\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-e^{\frac{8 a}{b}} \left (-\frac{a+b \sinh ^{-1}(c x)}{b}\right )^n \text{Gamma}\left (n+1,\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )-\frac{8 \left (a+b \sinh ^{-1}(c x)\right )}{b n+b}\right )}{64 c^3 \sqrt{d \left (c^2 x^2+1\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^n,x]

[Out]

(d*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^n*((-8*(a + b*ArcSinh[c*x]))/(b + b*n) + ((a/b + ArcSinh[c*x])^n*Gam
ma[1 + n, (-4*(a + b*ArcSinh[c*x]))/b] - E^((8*a)/b)*(-((a + b*ArcSinh[c*x])/b))^n*Gamma[1 + n, (4*(a + b*ArcS
inh[c*x]))/b])/(4^n*E^((4*a)/b)*(-((a + b*ArcSinh[c*x])^2/b^2))^n)))/(64*c^3*Sqrt[d*(1 + c^2*x^2)])

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Maple [F]  time = 0.241, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{n}\sqrt{{c}^{2}d{x}^{2}+d}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arcsinh(c*x))^n*(c^2*d*x^2+d)^(1/2),x)

[Out]

int(x^2*(a+b*arcsinh(c*x))^n*(c^2*d*x^2+d)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c^{2} d x^{2} + d}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{n} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsinh(c*x))^n*(c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c^2*d*x^2 + d)*(b*arcsinh(c*x) + a)^n*x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c^2*d*x^2 + d)*(b*arcsinh(c*x) + a)^n*x^2, x)

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Sympy [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(x**2*(a+b*asinh(c*x))**n*(c**2*d*x**2+d)**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c^2*d*x^2 + d)*(b*arcsinh(c*x) + a)^n*x^2, x)