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

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

[Out]

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

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Rubi [A]  time = 0.298012, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5702, 5699, 3312, 3307, 2181} \[ \frac{2^{-n-3} e^{-\frac{2 a}{b}} \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-\frac{a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c \sqrt{c^2 x^2+1}}-\frac{2^{-n-3} e^{\frac{2 a}{b}} \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^n \left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text{Gamma}\left (n+1,\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{c \sqrt{c^2 x^2+1}}+\frac{\sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^{n+1}}{2 b c (n+1) \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

Rule 5702

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(d^(p - 1/2)*Sqrt
[d + e*x^2])/Sqrt[1 + c^2*x^2], Int[(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] && IGtQ[2*p, 0] &&  !(IntegerQ[p] || GtQ[d, 0])

Rule 5699

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c, Subst[Int[
(a + b*x)^n*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IG
tQ[2*p, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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

Mathematica [A]  time = 0.581176, size = 160, normalized size = 0.68 \[ \frac{d \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^n \left (-2^{-n} e^{\frac{2 a}{b}} \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )^{-n} \text{Gamma}\left (n+1,\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+2^{-n} e^{-\frac{2 a}{b}} \left (-\frac{a+b \sinh ^{-1}(c x)}{b}\right )^{-n} \text{Gamma}\left (n+1,-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+\frac{4 a+4 b \sinh ^{-1}(c x)}{b n+b}\right )}{8 c \sqrt{d \left (c^2 x^2+1\right )}} \]

Antiderivative was successfully verified.

[In]

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

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Maple [F]  time = 0.154, size = 0, normalized size = 0. \begin{align*} \int \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((a+b*arcsinh(c*x))^n*(c^2*d*x^2+d)^(1/2),x)

[Out]

int((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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)

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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\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(d*(c**2*x**2 + 1))*(a + b*asinh(c*x))**n, x)

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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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)