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

Optimal. Leaf size=120 \[ \frac{d^2 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 c^2}-\frac{b d^2 x \left (c^2 x^2+1\right )^{5/2}}{36 c}-\frac{5 b d^2 x \left (c^2 x^2+1\right )^{3/2}}{144 c}-\frac{5 b d^2 x \sqrt{c^2 x^2+1}}{96 c}-\frac{5 b d^2 \sinh ^{-1}(c x)}{96 c^2} \]

[Out]

(-5*b*d^2*x*Sqrt[1 + c^2*x^2])/(96*c) - (5*b*d^2*x*(1 + c^2*x^2)^(3/2))/(144*c) - (b*d^2*x*(1 + c^2*x^2)^(5/2)
)/(36*c) - (5*b*d^2*ArcSinh[c*x])/(96*c^2) + (d^2*(1 + c^2*x^2)^3*(a + b*ArcSinh[c*x]))/(6*c^2)

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Rubi [A]  time = 0.0646363, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {5717, 195, 215} \[ \frac{d^2 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 c^2}-\frac{b d^2 x \left (c^2 x^2+1\right )^{5/2}}{36 c}-\frac{5 b d^2 x \left (c^2 x^2+1\right )^{3/2}}{144 c}-\frac{5 b d^2 x \sqrt{c^2 x^2+1}}{96 c}-\frac{5 b d^2 \sinh ^{-1}(c x)}{96 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b*d^2*x*Sqrt[1 + c^2*x^2])/(96*c) - (5*b*d^2*x*(1 + c^2*x^2)^(3/2))/(144*c) - (b*d^2*x*(1 + c^2*x^2)^(5/2)
)/(36*c) - (5*b*d^2*ArcSinh[c*x])/(96*c^2) + (d^2*(1 + c^2*x^2)^3*(a + b*ArcSinh[c*x]))/(6*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 )^2 \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{d^2 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 c^2}-\frac{\left (b d^2\right ) \int \left (1+c^2 x^2\right )^{5/2} \, dx}{6 c}\\ &=-\frac{b d^2 x \left (1+c^2 x^2\right )^{5/2}}{36 c}+\frac{d^2 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 c^2}-\frac{\left (5 b d^2\right ) \int \left (1+c^2 x^2\right )^{3/2} \, dx}{36 c}\\ &=-\frac{5 b d^2 x \left (1+c^2 x^2\right )^{3/2}}{144 c}-\frac{b d^2 x \left (1+c^2 x^2\right )^{5/2}}{36 c}+\frac{d^2 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 c^2}-\frac{\left (5 b d^2\right ) \int \sqrt{1+c^2 x^2} \, dx}{48 c}\\ &=-\frac{5 b d^2 x \sqrt{1+c^2 x^2}}{96 c}-\frac{5 b d^2 x \left (1+c^2 x^2\right )^{3/2}}{144 c}-\frac{b d^2 x \left (1+c^2 x^2\right )^{5/2}}{36 c}+\frac{d^2 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 c^2}-\frac{\left (5 b d^2\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{96 c}\\ &=-\frac{5 b d^2 x \sqrt{1+c^2 x^2}}{96 c}-\frac{5 b d^2 x \left (1+c^2 x^2\right )^{3/2}}{144 c}-\frac{b d^2 x \left (1+c^2 x^2\right )^{5/2}}{36 c}-\frac{5 b d^2 \sinh ^{-1}(c x)}{96 c^2}+\frac{d^2 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 c^2}\\ \end{align*}

Mathematica [A]  time = 0.132565, size = 104, normalized size = 0.87 \[ \frac{d^2 \left (c x \left (48 a c x \left (c^4 x^4+3 c^2 x^2+3\right )-b \sqrt{c^2 x^2+1} \left (8 c^4 x^4+26 c^2 x^2+33\right )\right )+3 b \left (16 c^6 x^6+48 c^4 x^4+48 c^2 x^2+11\right ) \sinh ^{-1}(c x)\right )}{288 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(c*x*(48*a*c*x*(3 + 3*c^2*x^2 + c^4*x^4) - b*Sqrt[1 + c^2*x^2]*(33 + 26*c^2*x^2 + 8*c^4*x^4)) + 3*b*(11 +
48*c^2*x^2 + 48*c^4*x^4 + 16*c^6*x^6)*ArcSinh[c*x]))/(288*c^2)

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Maple [A]  time = 0.005, size = 137, normalized size = 1.1 \begin{align*}{\frac{1}{{c}^{2}} \left ({d}^{2}a \left ({\frac{{c}^{6}{x}^{6}}{6}}+{\frac{{c}^{4}{x}^{4}}{2}}+{\frac{{c}^{2}{x}^{2}}{2}} \right ) +{d}^{2}b \left ({\frac{{\it Arcsinh} \left ( cx \right ){c}^{6}{x}^{6}}{6}}+{\frac{{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{4}}{2}}+{\frac{{\it Arcsinh} \left ( cx \right ){c}^{2}{x}^{2}}{2}}-{\frac{{c}^{5}{x}^{5}}{36}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{13\,{c}^{3}{x}^{3}}{144}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{11\,cx}{96}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{11\,{\it Arcsinh} \left ( cx \right ) }{96}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(d^2*a*(1/6*c^6*x^6+1/2*c^4*x^4+1/2*c^2*x^2)+d^2*b*(1/6*arcsinh(c*x)*c^6*x^6+1/2*arcsinh(c*x)*c^4*x^4+1/
2*arcsinh(c*x)*c^2*x^2-1/36*c^5*x^5*(c^2*x^2+1)^(1/2)-13/144*c^3*x^3*(c^2*x^2+1)^(1/2)-11/96*c*x*(c^2*x^2+1)^(
1/2)+11/96*arcsinh(c*x)))

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Maxima [B]  time = 1.05321, size = 365, normalized size = 3.04 \begin{align*} \frac{1}{6} \, a c^{4} d^{2} x^{6} + \frac{1}{2} \, a c^{2} d^{2} x^{4} + \frac{1}{288} \,{\left (48 \, x^{6} \operatorname{arsinh}\left (c x\right ) -{\left (\frac{8 \, \sqrt{c^{2} x^{2} + 1} x^{5}}{c^{2}} - \frac{10 \, \sqrt{c^{2} x^{2} + 1} x^{3}}{c^{4}} + \frac{15 \, \sqrt{c^{2} x^{2} + 1} x}{c^{6}} - \frac{15 \, \operatorname{arsinh}\left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{6}}\right )} c\right )} b c^{4} d^{2} + \frac{1}{16} \,{\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^{2} + \frac{1}{2} \, a d^{2} 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^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*c^4*d^2*x^6 + 1/2*a*c^2*d^2*x^4 + 1/288*(48*x^6*arcsinh(c*x) - (8*sqrt(c^2*x^2 + 1)*x^5/c^2 - 10*sqrt(c^
2*x^2 + 1)*x^3/c^4 + 15*sqrt(c^2*x^2 + 1)*x/c^6 - 15*arcsinh(c^2*x/sqrt(c^2))/(sqrt(c^2)*c^6))*c)*b*c^4*d^2 +
1/16*(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*arcsinh(c^2*x/sqrt(c^2
))/(sqrt(c^2)*c^4))*c)*b*c^2*d^2 + 1/2*a*d^2*x^2 + 1/4*(2*x^2*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x/c^2 - arcs
inh(c^2*x/sqrt(c^2))/(sqrt(c^2)*c^2)))*b*d^2

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Fricas [A]  time = 2.32668, size = 328, normalized size = 2.73 \begin{align*} \frac{48 \, a c^{6} d^{2} x^{6} + 144 \, a c^{4} d^{2} x^{4} + 144 \, a c^{2} d^{2} x^{2} + 3 \,{\left (16 \, b c^{6} d^{2} x^{6} + 48 \, b c^{4} d^{2} x^{4} + 48 \, b c^{2} d^{2} x^{2} + 11 \, b d^{2}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (8 \, b c^{5} d^{2} x^{5} + 26 \, b c^{3} d^{2} x^{3} + 33 \, b c d^{2} x\right )} \sqrt{c^{2} x^{2} + 1}}{288 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/288*(48*a*c^6*d^2*x^6 + 144*a*c^4*d^2*x^4 + 144*a*c^2*d^2*x^2 + 3*(16*b*c^6*d^2*x^6 + 48*b*c^4*d^2*x^4 + 48*
b*c^2*d^2*x^2 + 11*b*d^2)*log(c*x + sqrt(c^2*x^2 + 1)) - (8*b*c^5*d^2*x^5 + 26*b*c^3*d^2*x^3 + 33*b*c*d^2*x)*s
qrt(c^2*x^2 + 1))/c^2

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Sympy [A]  time = 5.81135, size = 190, normalized size = 1.58 \begin{align*} \begin{cases} \frac{a c^{4} d^{2} x^{6}}{6} + \frac{a c^{2} d^{2} x^{4}}{2} + \frac{a d^{2} x^{2}}{2} + \frac{b c^{4} d^{2} x^{6} \operatorname{asinh}{\left (c x \right )}}{6} - \frac{b c^{3} d^{2} x^{5} \sqrt{c^{2} x^{2} + 1}}{36} + \frac{b c^{2} d^{2} x^{4} \operatorname{asinh}{\left (c x \right )}}{2} - \frac{13 b c d^{2} x^{3} \sqrt{c^{2} x^{2} + 1}}{144} + \frac{b d^{2} x^{2} \operatorname{asinh}{\left (c x \right )}}{2} - \frac{11 b d^{2} x \sqrt{c^{2} x^{2} + 1}}{96 c} + \frac{11 b d^{2} \operatorname{asinh}{\left (c x \right )}}{96 c^{2}} & \text{for}\: c \neq 0 \\\frac{a d^{2} 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)**2*(a+b*asinh(c*x)),x)

[Out]

Piecewise((a*c**4*d**2*x**6/6 + a*c**2*d**2*x**4/2 + a*d**2*x**2/2 + b*c**4*d**2*x**6*asinh(c*x)/6 - b*c**3*d*
*2*x**5*sqrt(c**2*x**2 + 1)/36 + b*c**2*d**2*x**4*asinh(c*x)/2 - 13*b*c*d**2*x**3*sqrt(c**2*x**2 + 1)/144 + b*
d**2*x**2*asinh(c*x)/2 - 11*b*d**2*x*sqrt(c**2*x**2 + 1)/(96*c) + 11*b*d**2*asinh(c*x)/(96*c**2), Ne(c, 0)), (
a*d**2*x**2/2, True))

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Giac [B]  time = 1.77561, size = 405, normalized size = 3.38 \begin{align*} \frac{1}{6} \, a c^{4} d^{2} x^{6} + \frac{1}{2} \, a c^{2} d^{2} x^{4} + \frac{1}{288} \,{\left (48 \, x^{6} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (\sqrt{c^{2} x^{2} + 1}{\left (2 \, x^{2}{\left (\frac{4 \, x^{2}}{c^{2}} - \frac{5}{c^{4}}\right )} + \frac{15}{c^{6}}\right )} x + \frac{15 \, \log \left ({\left | -x{\left | c \right |} + \sqrt{c^{2} x^{2} + 1} \right |}\right )}{c^{6}{\left | c \right |}}\right )} c\right )} b c^{4} d^{2} + \frac{1}{16} \,{\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^{2} + \frac{1}{2} \, a d^{2} 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^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*c^4*d^2*x^6 + 1/2*a*c^2*d^2*x^4 + 1/288*(48*x^6*log(c*x + sqrt(c^2*x^2 + 1)) - (sqrt(c^2*x^2 + 1)*(2*x^2
*(4*x^2/c^2 - 5/c^4) + 15/c^6)*x + 15*log(abs(-x*abs(c) + sqrt(c^2*x^2 + 1)))/(c^6*abs(c)))*c)*b*c^4*d^2 + 1/1
6*(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^2 + 1/2*a*d^2*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^2