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

Optimal. Leaf size=87 \[ \frac{d \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2}-\frac{b d x \left (c^2 x^2+1\right )^{3/2}}{16 c}-\frac{3 b d x \sqrt{c^2 x^2+1}}{32 c}-\frac{3 b d \sinh ^{-1}(c x)}{32 c^2} \]

[Out]

(-3*b*d*x*Sqrt[1 + c^2*x^2])/(32*c) - (b*d*x*(1 + c^2*x^2)^(3/2))/(16*c) - (3*b*d*ArcSinh[c*x])/(32*c^2) + (d*
(1 + c^2*x^2)^2*(a + b*ArcSinh[c*x]))/(4*c^2)

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Rubi [A]  time = 0.0413919, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {5717, 195, 215} \[ \frac{d \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2}-\frac{b d x \left (c^2 x^2+1\right )^{3/2}}{16 c}-\frac{3 b d x \sqrt{c^2 x^2+1}}{32 c}-\frac{3 b d \sinh ^{-1}(c x)}{32 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*d*x*Sqrt[1 + c^2*x^2])/(32*c) - (b*d*x*(1 + c^2*x^2)^(3/2))/(16*c) - (3*b*d*ArcSinh[c*x])/(32*c^2) + (d*
(1 + c^2*x^2)^2*(a + b*ArcSinh[c*x]))/(4*c^2)

Rule 5717

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)
^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] - Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[p]), Int[(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] && GtQ[n, 0] && NeQ[p, -1]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int x \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{d \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2}-\frac{(b d) \int \left (1+c^2 x^2\right )^{3/2} \, dx}{4 c}\\ &=-\frac{b d x \left (1+c^2 x^2\right )^{3/2}}{16 c}+\frac{d \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2}-\frac{(3 b d) \int \sqrt{1+c^2 x^2} \, dx}{16 c}\\ &=-\frac{3 b d x \sqrt{1+c^2 x^2}}{32 c}-\frac{b d x \left (1+c^2 x^2\right )^{3/2}}{16 c}+\frac{d \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2}-\frac{(3 b d) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{32 c}\\ &=-\frac{3 b d x \sqrt{1+c^2 x^2}}{32 c}-\frac{b d x \left (1+c^2 x^2\right )^{3/2}}{16 c}-\frac{3 b d \sinh ^{-1}(c x)}{32 c^2}+\frac{d \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{4 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0520749, size = 77, normalized size = 0.89 \[ \frac{d \left (c x \left (8 a c x \left (c^2 x^2+2\right )-b \sqrt{c^2 x^2+1} \left (2 c^2 x^2+5\right )\right )+b \left (8 c^4 x^4+16 c^2 x^2+5\right ) \sinh ^{-1}(c x)\right )}{32 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(c*x*(8*a*c*x*(2 + c^2*x^2) - b*Sqrt[1 + c^2*x^2]*(5 + 2*c^2*x^2)) + b*(5 + 16*c^2*x^2 + 8*c^4*x^4)*ArcSinh
[c*x]))/(32*c^2)

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Maple [A]  time = 0.006, size = 94, normalized size = 1.1 \begin{align*}{\frac{1}{{c}^{2}} \left ( da \left ({\frac{{c}^{4}{x}^{4}}{4}}+{\frac{{c}^{2}{x}^{2}}{2}} \right ) +db \left ({\frac{{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{4}}{4}}+{\frac{{\it Arcsinh} \left ( cx \right ){c}^{2}{x}^{2}}{2}}-{\frac{{c}^{3}{x}^{3}}{16}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{5\,cx}{32}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{5\,{\it Arcsinh} \left ( cx \right ) }{32}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(d*a*(1/4*c^4*x^4+1/2*c^2*x^2)+d*b*(1/4*arcsinh(c*x)*c^4*x^4+1/2*arcsinh(c*x)*c^2*x^2-1/16*c^3*x^3*(c^2*
x^2+1)^(1/2)-5/32*c*x*(c^2*x^2+1)^(1/2)+5/32*arcsinh(c*x)))

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Maxima [B]  time = 1.15665, size = 204, normalized size = 2.34 \begin{align*} \frac{1}{4} \, a c^{2} d x^{4} + \frac{1}{32} \,{\left (8 \, x^{4} \operatorname{arsinh}\left (c x\right ) -{\left (\frac{2 \, \sqrt{c^{2} x^{2} + 1} x^{3}}{c^{2}} - \frac{3 \, \sqrt{c^{2} x^{2} + 1} x}{c^{4}} + \frac{3 \, \operatorname{arsinh}\left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{4}}\right )} c\right )} b c^{2} d + \frac{1}{2} \, a d x^{2} + \frac{1}{4} \,{\left (2 \, x^{2} \operatorname{arsinh}\left (c x\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x}{c^{2}} - \frac{\operatorname{arsinh}\left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{2}}\right )}\right )} b d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*c^2*d*x^4 + 1/32*(8*x^4*arcsinh(c*x) - (2*sqrt(c^2*x^2 + 1)*x^3/c^2 - 3*sqrt(c^2*x^2 + 1)*x/c^4 + 3*arcs
inh(c^2*x/sqrt(c^2))/(sqrt(c^2)*c^4))*c)*b*c^2*d + 1/2*a*d*x^2 + 1/4*(2*x^2*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1
)*x/c^2 - arcsinh(c^2*x/sqrt(c^2))/(sqrt(c^2)*c^2)))*b*d

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Fricas [A]  time = 2.66611, size = 220, normalized size = 2.53 \begin{align*} \frac{8 \, a c^{4} d x^{4} + 16 \, a c^{2} d x^{2} +{\left (8 \, b c^{4} d x^{4} + 16 \, b c^{2} d x^{2} + 5 \, b d\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (2 \, b c^{3} d x^{3} + 5 \, b c d x\right )} \sqrt{c^{2} x^{2} + 1}}{32 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(8*a*c^4*d*x^4 + 16*a*c^2*d*x^2 + (8*b*c^4*d*x^4 + 16*b*c^2*d*x^2 + 5*b*d)*log(c*x + sqrt(c^2*x^2 + 1)) -
(2*b*c^3*d*x^3 + 5*b*c*d*x)*sqrt(c^2*x^2 + 1))/c^2

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Sympy [A]  time = 1.45835, size = 117, normalized size = 1.34 \begin{align*} \begin{cases} \frac{a c^{2} d x^{4}}{4} + \frac{a d x^{2}}{2} + \frac{b c^{2} d x^{4} \operatorname{asinh}{\left (c x \right )}}{4} - \frac{b c d x^{3} \sqrt{c^{2} x^{2} + 1}}{16} + \frac{b d x^{2} \operatorname{asinh}{\left (c x \right )}}{2} - \frac{5 b d x \sqrt{c^{2} x^{2} + 1}}{32 c} + \frac{5 b d \operatorname{asinh}{\left (c x \right )}}{32 c^{2}} & \text{for}\: c \neq 0 \\\frac{a d x^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**2*d*x**4/4 + a*d*x**2/2 + b*c**2*d*x**4*asinh(c*x)/4 - b*c*d*x**3*sqrt(c**2*x**2 + 1)/16 + b*d
*x**2*asinh(c*x)/2 - 5*b*d*x*sqrt(c**2*x**2 + 1)/(32*c) + 5*b*d*asinh(c*x)/(32*c**2), Ne(c, 0)), (a*d*x**2/2,
True))

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Giac [B]  time = 1.69869, size = 242, normalized size = 2.78 \begin{align*} \frac{1}{4} \, a c^{2} d x^{4} + \frac{1}{32} \,{\left (8 \, x^{4} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (\sqrt{c^{2} x^{2} + 1} x{\left (\frac{2 \, x^{2}}{c^{2}} - \frac{3}{c^{4}}\right )} - \frac{3 \, \log \left ({\left | -x{\left | c \right |} + \sqrt{c^{2} x^{2} + 1} \right |}\right )}{c^{4}{\left | c \right |}}\right )} c\right )} b c^{2} d + \frac{1}{2} \, a d x^{2} + \frac{1}{4} \,{\left (2 \, x^{2} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x}{c^{2}} + \frac{\log \left ({\left | -x{\left | c \right |} + \sqrt{c^{2} x^{2} + 1} \right |}\right )}{c^{2}{\left | c \right |}}\right )}\right )} b d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*c^2*d*x^4 + 1/32*(8*x^4*log(c*x + sqrt(c^2*x^2 + 1)) - (sqrt(c^2*x^2 + 1)*x*(2*x^2/c^2 - 3/c^4) - 3*log(
abs(-x*abs(c) + sqrt(c^2*x^2 + 1)))/(c^4*abs(c)))*c)*b*c^2*d + 1/2*a*d*x^2 + 1/4*(2*x^2*log(c*x + sqrt(c^2*x^2
+ 1)) - c*(sqrt(c^2*x^2 + 1)*x/c^2 + log(abs(-x*abs(c) + sqrt(c^2*x^2 + 1)))/(c^2*abs(c))))*b*d