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

Optimal. Leaf size=75 \[ \frac{1}{3} c^2 d x^3 \left (a+b \sinh ^{-1}(c x)\right )+d x \left (a+b \sinh ^{-1}(c x)\right )-\frac{b d \left (c^2 x^2+1\right )^{3/2}}{9 c}-\frac{2 b d \sqrt{c^2 x^2+1}}{3 c} \]

[Out]

(-2*b*d*Sqrt[1 + c^2*x^2])/(3*c) - (b*d*(1 + c^2*x^2)^(3/2))/(9*c) + d*x*(a + b*ArcSinh[c*x]) + (c^2*d*x^3*(a
+ b*ArcSinh[c*x]))/3

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Rubi [A]  time = 0.0599111, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {5679, 12, 444, 43} \[ \frac{1}{3} c^2 d x^3 \left (a+b \sinh ^{-1}(c x)\right )+d x \left (a+b \sinh ^{-1}(c x)\right )-\frac{b d \left (c^2 x^2+1\right )^{3/2}}{9 c}-\frac{2 b d \sqrt{c^2 x^2+1}}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*d*Sqrt[1 + c^2*x^2])/(3*c) - (b*d*(1 + c^2*x^2)^(3/2))/(9*c) + d*x*(a + b*ArcSinh[c*x]) + (c^2*d*x^3*(a
+ b*ArcSinh[c*x]))/3

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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=d x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^2 d x^3 \left (a+b \sinh ^{-1}(c x)\right )-(b c) \int \frac{d x \left (3+c^2 x^2\right )}{3 \sqrt{1+c^2 x^2}} \, dx\\ &=d x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^2 d x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{3} (b c d) \int \frac{x \left (3+c^2 x^2\right )}{\sqrt{1+c^2 x^2}} \, dx\\ &=d x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^2 d x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{6} (b c d) \operatorname{Subst}\left (\int \frac{3+c^2 x}{\sqrt{1+c^2 x}} \, dx,x,x^2\right )\\ &=d x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^2 d x^3 \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{6} (b c d) \operatorname{Subst}\left (\int \left (\frac{2}{\sqrt{1+c^2 x}}+\sqrt{1+c^2 x}\right ) \, dx,x,x^2\right )\\ &=-\frac{2 b d \sqrt{1+c^2 x^2}}{3 c}-\frac{b d \left (1+c^2 x^2\right )^{3/2}}{9 c}+d x \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{3} c^2 d x^3 \left (a+b \sinh ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0418292, size = 86, normalized size = 1.15 \[ \frac{1}{3} a c^2 d x^3+a d x-\frac{1}{9} b c d x^2 \sqrt{c^2 x^2+1}-\frac{7 b d \sqrt{c^2 x^2+1}}{9 c}+\frac{1}{3} b c^2 d x^3 \sinh ^{-1}(c x)+b d x \sinh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*d*x + (a*c^2*d*x^3)/3 - (7*b*d*Sqrt[1 + c^2*x^2])/(9*c) - (b*c*d*x^2*Sqrt[1 + c^2*x^2])/9 + b*d*x*ArcSinh[c*
x] + (b*c^2*d*x^3*ArcSinh[c*x])/3

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Maple [A]  time = 0.006, size = 76, normalized size = 1. \begin{align*}{\frac{1}{c} \left ( da \left ({\frac{{c}^{3}{x}^{3}}{3}}+cx \right ) +db \left ({\frac{{\it Arcsinh} \left ( cx \right ){c}^{3}{x}^{3}}{3}}+{\it Arcsinh} \left ( cx \right ) cx-{\frac{{c}^{2}{x}^{2}}{9}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{7}{9}\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)*(a+b*arcsinh(c*x)),x)

[Out]

1/c*(d*a*(1/3*c^3*x^3+c*x)+d*b*(1/3*arcsinh(c*x)*c^3*x^3+arcsinh(c*x)*c*x-1/9*c^2*x^2*(c^2*x^2+1)^(1/2)-7/9*(c
^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.11389, size = 131, normalized size = 1.75 \begin{align*} \frac{1}{3} \, a c^{2} d x^{3} + \frac{1}{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 + a d x + \frac{{\left (c x \operatorname{arsinh}\left (c x\right ) - \sqrt{c^{2} x^{2} + 1}\right )} b d}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*c^2*d*x^3 + 1/9*(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 +
a*d*x + (c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*b*d/c

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Fricas [A]  time = 2.60895, size = 184, normalized size = 2.45 \begin{align*} \frac{3 \, a c^{3} d x^{3} + 9 \, a c d x + 3 \,{\left (b c^{3} d x^{3} + 3 \, b c d x\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) -{\left (b c^{2} d x^{2} + 7 \, b d\right )} \sqrt{c^{2} x^{2} + 1}}{9 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(3*a*c^3*d*x^3 + 9*a*c*d*x + 3*(b*c^3*d*x^3 + 3*b*c*d*x)*log(c*x + sqrt(c^2*x^2 + 1)) - (b*c^2*d*x^2 + 7*b
*d)*sqrt(c^2*x^2 + 1))/c

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Sympy [A]  time = 0.729231, size = 90, normalized size = 1.2 \begin{align*} \begin{cases} \frac{a c^{2} d x^{3}}{3} + a d x + \frac{b c^{2} d x^{3} \operatorname{asinh}{\left (c x \right )}}{3} - \frac{b c d x^{2} \sqrt{c^{2} x^{2} + 1}}{9} + b d x \operatorname{asinh}{\left (c x \right )} - \frac{7 b d \sqrt{c^{2} x^{2} + 1}}{9 c} & \text{for}\: c \neq 0 \\a d 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)*(a+b*asinh(c*x)),x)

[Out]

Piecewise((a*c**2*d*x**3/3 + a*d*x + b*c**2*d*x**3*asinh(c*x)/3 - b*c*d*x**2*sqrt(c**2*x**2 + 1)/9 + b*d*x*asi
nh(c*x) - 7*b*d*sqrt(c**2*x**2 + 1)/(9*c), Ne(c, 0)), (a*d*x, True))

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Giac [A]  time = 1.36874, size = 151, normalized size = 2.01 \begin{align*} \frac{1}{3} \, a c^{2} d x^{3} + \frac{1}{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 +{\left (x \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{\sqrt{c^{2} x^{2} + 1}}{c}\right )} b d + a d x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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