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

Optimal. Leaf size=128 \[ \frac{1}{5} c^4 d^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{3} c^2 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+d^2 x \left (a+b \sinh ^{-1}(c x)\right )-\frac{b d^2 \left (c^2 x^2+1\right )^{5/2}}{25 c}-\frac{4 b d^2 \left (c^2 x^2+1\right )^{3/2}}{45 c}-\frac{8 b d^2 \sqrt{c^2 x^2+1}}{15 c} \]

[Out]

(-8*b*d^2*Sqrt[1 + c^2*x^2])/(15*c) - (4*b*d^2*(1 + c^2*x^2)^(3/2))/(45*c) - (b*d^2*(1 + c^2*x^2)^(5/2))/(25*c
) + d^2*x*(a + b*ArcSinh[c*x]) + (2*c^2*d^2*x^3*(a + b*ArcSinh[c*x]))/3 + (c^4*d^2*x^5*(a + b*ArcSinh[c*x]))/5

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Rubi [A]  time = 0.101974, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {194, 5679, 12, 1247, 698} \[ \frac{1}{5} c^4 d^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{3} c^2 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+d^2 x \left (a+b \sinh ^{-1}(c x)\right )-\frac{b d^2 \left (c^2 x^2+1\right )^{5/2}}{25 c}-\frac{4 b d^2 \left (c^2 x^2+1\right )^{3/2}}{45 c}-\frac{8 b d^2 \sqrt{c^2 x^2+1}}{15 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b*d^2*Sqrt[1 + c^2*x^2])/(15*c) - (4*b*d^2*(1 + c^2*x^2)^(3/2))/(45*c) - (b*d^2*(1 + c^2*x^2)^(5/2))/(25*c
) + d^2*x*(a + b*ArcSinh[c*x]) + (2*c^2*d^2*x^3*(a + b*ArcSinh[c*x]))/3 + (c^4*d^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 5679

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] && EqQ[e, c^2*d] && IGtQ[p, 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+c^2 d 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} c^2 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac{d^2 x \left (15+10 c^2 x^2+3 c^4 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} c^2 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{15} \left (b c d^2\right ) \int \frac{x \left (15+10 c^2 x^2+3 c^4 x^4\right )}{\sqrt{1+c^2 x^2}} \, dx\\ &=d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{3} c^2 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{30} \left (b c d^2\right ) \operatorname{Subst}\left (\int \frac{15+10 c^2 x+3 c^4 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} c^2 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{30} \left (b c d^2\right ) \operatorname{Subst}\left (\int \left (\frac{8}{\sqrt{1+c^2 x}}+4 \sqrt{1+c^2 x}+3 \left (1+c^2 x\right )^{3/2}\right ) \, dx,x,x^2\right )\\ &=-\frac{8 b d^2 \sqrt{1+c^2 x^2}}{15 c}-\frac{4 b d^2 \left (1+c^2 x^2\right )^{3/2}}{45 c}-\frac{b d^2 \left (1+c^2 x^2\right )^{5/2}}{25 c}+d^2 x \left (a+b \sinh ^{-1}(c x)\right )+\frac{2}{3} c^2 d^2 x^3 \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \sinh ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.120046, size = 95, normalized size = 0.74 \[ \frac{d^2 \left (15 a c x \left (3 c^4 x^4+10 c^2 x^2+15\right )-b \sqrt{c^2 x^2+1} \left (9 c^4 x^4+38 c^2 x^2+149\right )+15 b c x \left (3 c^4 x^4+10 c^2 x^2+15\right ) \sinh ^{-1}(c x)\right )}{225 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(15*a*c*x*(15 + 10*c^2*x^2 + 3*c^4*x^4) - b*Sqrt[1 + c^2*x^2]*(149 + 38*c^2*x^2 + 9*c^4*x^4) + 15*b*c*x*(
15 + 10*c^2*x^2 + 3*c^4*x^4)*ArcSinh[c*x]))/(225*c)

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Maple [A]  time = 0.005, size = 119, normalized size = 0.9 \begin{align*}{\frac{1}{c} \left ({d}^{2}a \left ({\frac{{c}^{5}{x}^{5}}{5}}+{\frac{2\,{c}^{3}{x}^{3}}{3}}+cx \right ) +{d}^{2}b \left ({\frac{{\it Arcsinh} \left ( cx \right ){c}^{5}{x}^{5}}{5}}+{\frac{2\,{\it Arcsinh} \left ( cx \right ){c}^{3}{x}^{3}}{3}}+{\it Arcsinh} \left ( cx \right ) cx-{\frac{{c}^{4}{x}^{4}}{25}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{38\,{c}^{2}{x}^{2}}{225}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{149}{225}\sqrt{{c}^{2}{x}^{2}+1}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(d^2*a*(1/5*c^5*x^5+2/3*c^3*x^3+c*x)+d^2*b*(1/5*arcsinh(c*x)*c^5*x^5+2/3*arcsinh(c*x)*c^3*x^3+arcsinh(c*x)
*c*x-1/25*c^4*x^4*(c^2*x^2+1)^(1/2)-38/225*c^2*x^2*(c^2*x^2+1)^(1/2)-149/225*(c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.19368, size = 262, normalized size = 2.05 \begin{align*} \frac{1}{5} \, a c^{4} d^{2} x^{5} + \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 c^{4} d^{2} + \frac{2}{3} \, a c^{2} d^{2} 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 c^{2} d^{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((c^2*d*x^2+d)^2*(a+b*arcsinh(c*x)),x, algorithm="maxima")

[Out]

1/5*a*c^4*d^2*x^5 + 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*c^4*d^2 + 2/3*a*c^2*d^2*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*c^2*d^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 = 2.43673, size = 296, normalized size = 2.31 \begin{align*} \frac{45 \, a c^{5} d^{2} x^{5} + 150 \, a c^{3} d^{2} x^{3} + 225 \, a c d^{2} x + 15 \,{\left (3 \, b c^{5} d^{2} x^{5} + 10 \, b c^{3} d^{2} x^{3} + 15 \, b c d^{2} x\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (9 \, b c^{4} d^{2} x^{4} + 38 \, b c^{2} d^{2} x^{2} + 149 \, b d^{2}\right )} \sqrt{c^{2} x^{2} + 1}}{225 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/225*(45*a*c^5*d^2*x^5 + 150*a*c^3*d^2*x^3 + 225*a*c*d^2*x + 15*(3*b*c^5*d^2*x^5 + 10*b*c^3*d^2*x^3 + 15*b*c*
d^2*x)*log(c*x + sqrt(c^2*x^2 + 1)) - (9*b*c^4*d^2*x^4 + 38*b*c^2*d^2*x^2 + 149*b*d^2)*sqrt(c^2*x^2 + 1))/c

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Sympy [A]  time = 2.86785, size = 165, normalized size = 1.29 \begin{align*} \begin{cases} \frac{a c^{4} d^{2} x^{5}}{5} + \frac{2 a c^{2} d^{2} x^{3}}{3} + a d^{2} x + \frac{b c^{4} d^{2} x^{5} \operatorname{asinh}{\left (c x \right )}}{5} - \frac{b c^{3} d^{2} x^{4} \sqrt{c^{2} x^{2} + 1}}{25} + \frac{2 b c^{2} d^{2} x^{3} \operatorname{asinh}{\left (c x \right )}}{3} - \frac{38 b c d^{2} x^{2} \sqrt{c^{2} x^{2} + 1}}{225} + b d^{2} x \operatorname{asinh}{\left (c x \right )} - \frac{149 b d^{2} \sqrt{c^{2} x^{2} + 1}}{225 c} & \text{for}\: c \neq 0 \\a d^{2} x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**4*d**2*x**5/5 + 2*a*c**2*d**2*x**3/3 + a*d**2*x + b*c**4*d**2*x**5*asinh(c*x)/5 - b*c**3*d**2*
x**4*sqrt(c**2*x**2 + 1)/25 + 2*b*c**2*d**2*x**3*asinh(c*x)/3 - 38*b*c*d**2*x**2*sqrt(c**2*x**2 + 1)/225 + b*d
**2*x*asinh(c*x) - 149*b*d**2*sqrt(c**2*x**2 + 1)/(225*c), Ne(c, 0)), (a*d**2*x, True))

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Giac [A]  time = 1.56878, size = 281, normalized size = 2.2 \begin{align*} \frac{1}{5} \, a c^{4} d^{2} x^{5} + \frac{1}{75} \,{\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 c^{4} d^{2} + \frac{2}{3} \, a c^{2} d^{2} x^{3} + \frac{2}{9} \,{\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 c^{2} d^{2} +{\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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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