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

Optimal. Leaf size=147 \[ d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{3} d e x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} e^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{b \sqrt{c^2 x^2+1} \left (15 c^4 d^2-10 c^2 d e+3 e^2\right )}{15 c^5}-\frac{2 b e \left (c^2 x^2+1\right )^{3/2} \left (5 c^2 d-3 e\right )}{45 c^5}-\frac{b e^2 \left (c^2 x^2+1\right )^{5/2}}{25 c^5} \]

[Out]

-(b*(15*c^4*d^2 - 10*c^2*d*e + 3*e^2)*Sqrt[1 + c^2*x^2])/(15*c^5) - (2*b*(5*c^2*d - 3*e)*e*(1 + c^2*x^2)^(3/2)
)/(45*c^5) - (b*e^2*(1 + c^2*x^2)^(5/2))/(25*c^5) + d^2*x*(a + b*ArcSinh[c*x]) + (2*d*e*x^3*(a + b*ArcSinh[c*x
]))/3 + (e^2*x^5*(a + b*ArcSinh[c*x]))/5

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Rubi [A]  time = 0.143094, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {194, 5704, 12, 1247, 698} \[ d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{3} d e x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} e^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{b \sqrt{c^2 x^2+1} \left (15 c^4 d^2-10 c^2 d e+3 e^2\right )}{15 c^5}-\frac{2 b e \left (c^2 x^2+1\right )^{3/2} \left (5 c^2 d-3 e\right )}{45 c^5}-\frac{b e^2 \left (c^2 x^2+1\right )^{5/2}}{25 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(15*c^4*d^2 - 10*c^2*d*e + 3*e^2)*Sqrt[1 + c^2*x^2])/(15*c^5) - (2*b*(5*c^2*d - 3*e)*e*(1 + c^2*x^2)^(3/2)
)/(45*c^5) - (b*e^2*(1 + c^2*x^2)^(5/2))/(25*c^5) + d^2*x*(a + b*ArcSinh[c*x]) + (2*d*e*x^3*(a + b*ArcSinh[c*x
]))/3 + (e^2*x^5*(a + b*ArcSinh[c*x]))/5

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 5704

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 + c^2*x^2], x], x], x]] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[e, c^2*d] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int \left (d+e x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{3} d e x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} e^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac{x \left (15 d^2+10 d e x^2+3 e^2 x^4\right )}{15 \sqrt{1+c^2 x^2}} \, dx\\ &=d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{3} d e x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} e^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{15} (b c) \int \frac{x \left (15 d^2+10 d e x^2+3 e^2 x^4\right )}{\sqrt{1+c^2 x^2}} \, dx\\ &=d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{3} d e x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} e^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{30} (b c) \operatorname{Subst}\left (\int \frac{15 d^2+10 d e x+3 e^2 x^2}{\sqrt{1+c^2 x}} \, dx,x,x^2\right )\\ &=d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{3} d e x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} e^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{30} (b c) \operatorname{Subst}\left (\int \left (\frac{15 c^4 d^2-10 c^2 d e+3 e^2}{c^4 \sqrt{1+c^2 x}}+\frac{2 \left (5 c^2 d-3 e\right ) e \sqrt{1+c^2 x}}{c^4}+\frac{3 e^2 \left (1+c^2 x\right )^{3/2}}{c^4}\right ) \, dx,x,x^2\right )\\ &=-\frac{b \left (15 c^4 d^2-10 c^2 d e+3 e^2\right ) \sqrt{1+c^2 x^2}}{15 c^5}-\frac{2 b \left (5 c^2 d-3 e\right ) e \left (1+c^2 x^2\right )^{3/2}}{45 c^5}-\frac{b e^2 \left (1+c^2 x^2\right )^{5/2}}{25 c^5}+d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{3} d e x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} e^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.163037, size = 125, normalized size = 0.85 \[ \frac{1}{225} \left (15 a x \left (15 d^2+10 d e x^2+3 e^2 x^4\right )-\frac{b \sqrt{c^2 x^2+1} \left (c^4 \left (225 d^2+50 d e x^2+9 e^2 x^4\right )-4 c^2 e \left (25 d+3 e x^2\right )+24 e^2\right )}{c^5}+15 b x \sinh ^{-1}(c x) \left (15 d^2+10 d e x^2+3 e^2 x^4\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a*x*(15*d^2 + 10*d*e*x^2 + 3*e^2*x^4) - (b*Sqrt[1 + c^2*x^2]*(24*e^2 - 4*c^2*e*(25*d + 3*e*x^2) + c^4*(225
*d^2 + 50*d*e*x^2 + 9*e^2*x^4)))/c^5 + 15*b*x*(15*d^2 + 10*d*e*x^2 + 3*e^2*x^4)*ArcSinh[c*x])/225

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Maple [A]  time = 0.005, size = 204, normalized size = 1.4 \begin{align*}{\frac{1}{c} \left ({\frac{a}{{c}^{4}} \left ({\frac{{e}^{2}{c}^{5}{x}^{5}}{5}}+{\frac{2\,{c}^{5}de{x}^{3}}{3}}+x{c}^{5}{d}^{2} \right ) }+{\frac{b}{{c}^{4}} \left ({\frac{{\it Arcsinh} \left ( cx \right ){e}^{2}{c}^{5}{x}^{5}}{5}}+{\frac{2\,{\it Arcsinh} \left ( cx \right ){c}^{5}de{x}^{3}}{3}}+{\it Arcsinh} \left ( cx \right ){c}^{5}x{d}^{2}-{\frac{{e}^{2}}{5} \left ({\frac{{c}^{4}{x}^{4}}{5}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{4\,{c}^{2}{x}^{2}}{15}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{8}{15}\sqrt{{c}^{2}{x}^{2}+1}} \right ) }-{\frac{2\,{c}^{2}de}{3} \left ({\frac{{c}^{2}{x}^{2}}{3}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{2}{3}\sqrt{{c}^{2}{x}^{2}+1}} \right ) }-{d}^{2}{c}^{4}\sqrt{{c}^{2}{x}^{2}+1} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(a/c^4*(1/5*e^2*c^5*x^5+2/3*c^5*d*e*x^3+x*c^5*d^2)+b/c^4*(1/5*arcsinh(c*x)*e^2*c^5*x^5+2/3*arcsinh(c*x)*c^
5*d*e*x^3+arcsinh(c*x)*c^5*x*d^2-1/5*e^2*(1/5*c^4*x^4*(c^2*x^2+1)^(1/2)-4/15*c^2*x^2*(c^2*x^2+1)^(1/2)+8/15*(c
^2*x^2+1)^(1/2))-2/3*c^2*d*e*(1/3*c^2*x^2*(c^2*x^2+1)^(1/2)-2/3*(c^2*x^2+1)^(1/2))-d^2*c^4*(c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.17582, size = 243, normalized size = 1.65 \begin{align*} \frac{1}{5} \, a e^{2} x^{5} + \frac{2}{3} \, a d e x^{3} + \frac{2}{9} \,{\left (3 \, x^{3} \operatorname{arsinh}\left (c x\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac{2 \, \sqrt{c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b d e + \frac{1}{75} \,{\left (15 \, x^{5} \operatorname{arsinh}\left (c x\right ) -{\left (\frac{3 \, \sqrt{c^{2} x^{2} + 1} x^{4}}{c^{2}} - \frac{4 \, \sqrt{c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac{8 \, \sqrt{c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b e^{2} + a d^{2} x + \frac{{\left (c x \operatorname{arsinh}\left (c x\right ) - \sqrt{c^{2} x^{2} + 1}\right )} b d^{2}}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*e^2*x^5 + 2/3*a*d*e*x^3 + 2/9*(3*x^3*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c
^4))*b*d*e + 1/75*(15*x^5*arcsinh(c*x) - (3*sqrt(c^2*x^2 + 1)*x^4/c^2 - 4*sqrt(c^2*x^2 + 1)*x^2/c^4 + 8*sqrt(c
^2*x^2 + 1)/c^6)*c)*b*e^2 + a*d^2*x + (c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*b*d^2/c

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Fricas [A]  time = 2.36344, size = 370, normalized size = 2.52 \begin{align*} \frac{45 \, a c^{5} e^{2} x^{5} + 150 \, a c^{5} d e x^{3} + 225 \, a c^{5} d^{2} x + 15 \,{\left (3 \, b c^{5} e^{2} x^{5} + 10 \, b c^{5} d e x^{3} + 15 \, b c^{5} d^{2} x\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (9 \, b c^{4} e^{2} x^{4} + 225 \, b c^{4} d^{2} - 100 \, b c^{2} d e + 24 \, b e^{2} + 2 \,{\left (25 \, b c^{4} d e - 6 \, b c^{2} e^{2}\right )} x^{2}\right )} \sqrt{c^{2} x^{2} + 1}}{225 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/225*(45*a*c^5*e^2*x^5 + 150*a*c^5*d*e*x^3 + 225*a*c^5*d^2*x + 15*(3*b*c^5*e^2*x^5 + 10*b*c^5*d*e*x^3 + 15*b*
c^5*d^2*x)*log(c*x + sqrt(c^2*x^2 + 1)) - (9*b*c^4*e^2*x^4 + 225*b*c^4*d^2 - 100*b*c^2*d*e + 24*b*e^2 + 2*(25*
b*c^4*d*e - 6*b*c^2*e^2)*x^2)*sqrt(c^2*x^2 + 1))/c^5

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Sympy [A]  time = 2.78751, size = 240, normalized size = 1.63 \begin{align*} \begin{cases} a d^{2} x + \frac{2 a d e x^{3}}{3} + \frac{a e^{2} x^{5}}{5} + b d^{2} x \operatorname{asinh}{\left (c x \right )} + \frac{2 b d e x^{3} \operatorname{asinh}{\left (c x \right )}}{3} + \frac{b e^{2} x^{5} \operatorname{asinh}{\left (c x \right )}}{5} - \frac{b d^{2} \sqrt{c^{2} x^{2} + 1}}{c} - \frac{2 b d e x^{2} \sqrt{c^{2} x^{2} + 1}}{9 c} - \frac{b e^{2} x^{4} \sqrt{c^{2} x^{2} + 1}}{25 c} + \frac{4 b d e \sqrt{c^{2} x^{2} + 1}}{9 c^{3}} + \frac{4 b e^{2} x^{2} \sqrt{c^{2} x^{2} + 1}}{75 c^{3}} - \frac{8 b e^{2} \sqrt{c^{2} x^{2} + 1}}{75 c^{5}} & \text{for}\: c \neq 0 \\a \left (d^{2} x + \frac{2 d e x^{3}}{3} + \frac{e^{2} x^{5}}{5}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d**2*x + 2*a*d*e*x**3/3 + a*e**2*x**5/5 + b*d**2*x*asinh(c*x) + 2*b*d*e*x**3*asinh(c*x)/3 + b*e**
2*x**5*asinh(c*x)/5 - b*d**2*sqrt(c**2*x**2 + 1)/c - 2*b*d*e*x**2*sqrt(c**2*x**2 + 1)/(9*c) - b*e**2*x**4*sqrt
(c**2*x**2 + 1)/(25*c) + 4*b*d*e*sqrt(c**2*x**2 + 1)/(9*c**3) + 4*b*e**2*x**2*sqrt(c**2*x**2 + 1)/(75*c**3) -
8*b*e**2*sqrt(c**2*x**2 + 1)/(75*c**5), Ne(c, 0)), (a*(d**2*x + 2*d*e*x**3/3 + e**2*x**5/5), True))

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Giac [A]  time = 1.62516, size = 262, normalized size = 1.78 \begin{align*}{\left (x \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{\sqrt{c^{2} x^{2} + 1}}{c}\right )} b d^{2} + a d^{2} x + \frac{1}{75} \,{\left (15 \, a x^{5} +{\left (15 \, x^{5} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{3 \,{\left (c^{2} x^{2} + 1\right )}^{\frac{5}{2}} - 10 \,{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} + 15 \, \sqrt{c^{2} x^{2} + 1}}{c^{5}}\right )} b\right )} e^{2} + \frac{2}{9} \,{\left (3 \, a d x^{3} +{\left (3 \, x^{3} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} - 3 \, \sqrt{c^{2} x^{2} + 1}}{c^{3}}\right )} b d\right )} e \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*log(c*x + sqrt(c^2*x^2 + 1)) - sqrt(c^2*x^2 + 1)/c)*b*d^2 + a*d^2*x + 1/75*(15*a*x^5 + (15*x^5*log(c*x + sq
rt(c^2*x^2 + 1)) - (3*(c^2*x^2 + 1)^(5/2) - 10*(c^2*x^2 + 1)^(3/2) + 15*sqrt(c^2*x^2 + 1))/c^5)*b)*e^2 + 2/9*(
3*a*d*x^3 + (3*x^3*log(c*x + sqrt(c^2*x^2 + 1)) - ((c^2*x^2 + 1)^(3/2) - 3*sqrt(c^2*x^2 + 1))/c^3)*b*d)*e