Integrand size = 10, antiderivative size = 90 \[ \int x^2 \text {arcsinh}(a+b x) \, dx=-\frac {x^2 \sqrt {1+(a+b x)^2}}{9 b}+\frac {\left (4-11 a^2+5 a b x\right ) \sqrt {1+(a+b x)^2}}{18 b^3}-\frac {a \left (3-2 a^2\right ) \text {arcsinh}(a+b x)}{6 b^3}+\frac {1}{3} x^3 \text {arcsinh}(a+b x) \]
-1/6*a*(-2*a^2+3)*arcsinh(b*x+a)/b^3+1/3*x^3*arcsinh(b*x+a)-1/9*x^2*(1+(b* x+a)^2)^(1/2)/b+1/18*(5*a*b*x-11*a^2+4)*(1+(b*x+a)^2)^(1/2)/b^3
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.82 \[ \int x^2 \text {arcsinh}(a+b x) \, dx=\frac {\left (4-11 a^2+5 a b x-2 b^2 x^2\right ) \sqrt {1+a^2+2 a b x+b^2 x^2}+\left (-9 a+6 a^3+6 b^3 x^3\right ) \text {arcsinh}(a+b x)}{18 b^3} \]
((4 - 11*a^2 + 5*a*b*x - 2*b^2*x^2)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] + (- 9*a + 6*a^3 + 6*b^3*x^3)*ArcSinh[a + b*x])/(18*b^3)
Time = 0.33 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.30, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {6274, 27, 6243, 497, 25, 676, 222}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \text {arcsinh}(a+b x) \, dx\) |
\(\Big \downarrow \) 6274 |
\(\displaystyle \frac {\int x^2 \text {arcsinh}(a+b x)d(a+b x)}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int b^2 x^2 \text {arcsinh}(a+b x)d(a+b x)}{b^3}\) |
\(\Big \downarrow \) 6243 |
\(\displaystyle \frac {\frac {1}{3} \int -\frac {b^3 x^3}{\sqrt {(a+b x)^2+1}}d(a+b x)+\frac {1}{3} b^3 x^3 \text {arcsinh}(a+b x)}{b^3}\) |
\(\Big \downarrow \) 497 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{3} \int \frac {b x \left (-3 a^2+5 (a+b x) a+2\right )}{\sqrt {(a+b x)^2+1}}d(a+b x)-\frac {1}{3} b^2 x^2 \sqrt {(a+b x)^2+1}\right )+\frac {1}{3} b^3 x^3 \text {arcsinh}(a+b x)}{b^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {1}{3} \left (-\frac {1}{3} \int -\frac {b x \left (-3 a^2+5 (a+b x) a+2\right )}{\sqrt {(a+b x)^2+1}}d(a+b x)-\frac {1}{3} b^2 x^2 \sqrt {(a+b x)^2+1}\right )+\frac {1}{3} b^3 x^3 \text {arcsinh}(a+b x)}{b^3}\) |
\(\Big \downarrow \) 676 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{3} \left (-\frac {3}{2} a \left (3-2 a^2\right ) \int \frac {1}{\sqrt {(a+b x)^2+1}}d(a+b x)+2 \left (1-4 a^2\right ) \sqrt {(a+b x)^2+1}+\frac {5}{2} a (a+b x) \sqrt {(a+b x)^2+1}\right )-\frac {1}{3} b^2 x^2 \sqrt {(a+b x)^2+1}\right )+\frac {1}{3} b^3 x^3 \text {arcsinh}(a+b x)}{b^3}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{3} \left (-\frac {3}{2} a \left (3-2 a^2\right ) \text {arcsinh}(a+b x)+2 \left (1-4 a^2\right ) \sqrt {(a+b x)^2+1}+\frac {5}{2} a (a+b x) \sqrt {(a+b x)^2+1}\right )-\frac {1}{3} b^2 x^2 \sqrt {(a+b x)^2+1}\right )+\frac {1}{3} b^3 x^3 \text {arcsinh}(a+b x)}{b^3}\) |
((b^3*x^3*ArcSinh[a + b*x])/3 + (-1/3*(b^2*x^2*Sqrt[1 + (a + b*x)^2]) + (2 *(1 - 4*a^2)*Sqrt[1 + (a + b*x)^2] + (5*a*(a + b*x)*Sqrt[1 + (a + b*x)^2]) /2 - (3*a*(3 - 2*a^2)*ArcSinh[a + b*x])/2)/3)/3)/b^3
3.1.59.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[1/(b *(n + 2*p + 1)) Int[(c + d*x)^(n - 2)*(a + b*x^2)^p*Simp[b*c^2*(n + 2*p + 1) - a*d^2*(n - 1) + 2*b*c*d*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, n , p}, x] && If[RationalQ[n], GtQ[n, 1], SumSimplerQ[n, -2]] && NeQ[n + 2*p + 1, 0] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x _Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p + 3))/(c*(2*p + 3)) Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g , p}, x] && !LeQ[p, -1]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x _Symbol] :> Simp[(d + e*x)^(m + 1)*((a + b*ArcSinh[c*x])^n/(e*(m + 1))), x] - Simp[b*c*(n/(e*(m + 1))) Int[(d + e*x)^(m + 1)*((a + b*ArcSinh[c*x])^( n - 1)/Sqrt[1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n , 0] && NeQ[m, -1]
Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^( m_.), x_Symbol] :> Simp[1/d Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b* ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.02 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.44
method | result | size |
derivativedivides | \(\frac {\operatorname {arcsinh}\left (b x +a \right ) a^{2} \left (b x +a \right )-\operatorname {arcsinh}\left (b x +a \right ) a \left (b x +a \right )^{2}+\frac {\operatorname {arcsinh}\left (b x +a \right ) \left (b x +a \right )^{3}}{3}-\frac {\left (b x +a \right )^{2} \sqrt {1+\left (b x +a \right )^{2}}}{9}+\frac {2 \sqrt {1+\left (b x +a \right )^{2}}}{9}-a^{2} \sqrt {1+\left (b x +a \right )^{2}}+a \left (\frac {\left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}}{2}-\frac {\operatorname {arcsinh}\left (b x +a \right )}{2}\right )}{b^{3}}\) | \(130\) |
default | \(\frac {\operatorname {arcsinh}\left (b x +a \right ) a^{2} \left (b x +a \right )-\operatorname {arcsinh}\left (b x +a \right ) a \left (b x +a \right )^{2}+\frac {\operatorname {arcsinh}\left (b x +a \right ) \left (b x +a \right )^{3}}{3}-\frac {\left (b x +a \right )^{2} \sqrt {1+\left (b x +a \right )^{2}}}{9}+\frac {2 \sqrt {1+\left (b x +a \right )^{2}}}{9}-a^{2} \sqrt {1+\left (b x +a \right )^{2}}+a \left (\frac {\left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}}{2}-\frac {\operatorname {arcsinh}\left (b x +a \right )}{2}\right )}{b^{3}}\) | \(130\) |
parts | \(\frac {x^{3} \operatorname {arcsinh}\left (b x +a \right )}{3}-\frac {b \left (\frac {x^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{3 b^{2}}-\frac {5 a \left (\frac {x \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {3 a \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{2 b}-\frac {\left (a^{2}+1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\right )}{3 b}-\frac {2 \left (a^{2}+1\right ) \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{3 b^{2}}\right )}{3}\) | \(285\) |
1/b^3*(arcsinh(b*x+a)*a^2*(b*x+a)-arcsinh(b*x+a)*a*(b*x+a)^2+1/3*arcsinh(b *x+a)*(b*x+a)^3-1/9*(b*x+a)^2*(1+(b*x+a)^2)^(1/2)+2/9*(1+(b*x+a)^2)^(1/2)- a^2*(1+(b*x+a)^2)^(1/2)+a*(1/2*(b*x+a)*(1+(b*x+a)^2)^(1/2)-1/2*arcsinh(b*x +a)))
Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01 \[ \int x^2 \text {arcsinh}(a+b x) \, dx=\frac {3 \, {\left (2 \, b^{3} x^{3} + 2 \, a^{3} - 3 \, a\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - {\left (2 \, b^{2} x^{2} - 5 \, a b x + 11 \, a^{2} - 4\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{18 \, b^{3}} \]
1/18*(3*(2*b^3*x^3 + 2*a^3 - 3*a)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a ^2 + 1)) - (2*b^2*x^2 - 5*a*b*x + 11*a^2 - 4)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/b^3
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (83) = 166\).
Time = 0.25 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.89 \[ \int x^2 \text {arcsinh}(a+b x) \, dx=\begin {cases} \frac {a^{3} \operatorname {asinh}{\left (a + b x \right )}}{3 b^{3}} - \frac {11 a^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{18 b^{3}} + \frac {5 a x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{18 b^{2}} - \frac {a \operatorname {asinh}{\left (a + b x \right )}}{2 b^{3}} + \frac {x^{3} \operatorname {asinh}{\left (a + b x \right )}}{3} - \frac {x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{9 b} + \frac {2 \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{9 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \operatorname {asinh}{\left (a \right )}}{3} & \text {otherwise} \end {cases} \]
Piecewise((a**3*asinh(a + b*x)/(3*b**3) - 11*a**2*sqrt(a**2 + 2*a*b*x + b* *2*x**2 + 1)/(18*b**3) + 5*a*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(18*b* *2) - a*asinh(a + b*x)/(2*b**3) + x**3*asinh(a + b*x)/3 - x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(9*b) + 2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/( 9*b**3), Ne(b, 0)), (x**3*asinh(a)/3, True))
Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (78) = 156\).
Time = 0.18 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.33 \[ \int x^2 \text {arcsinh}(a+b x) \, dx=\frac {1}{3} \, x^{3} \operatorname {arsinh}\left (b x + a\right ) - \frac {1}{18} \, b {\left (\frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x^{2}}{b^{2}} - \frac {15 \, a^{3} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b^{4}} - \frac {5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x}{b^{3}} + \frac {9 \, {\left (a^{2} + 1\right )} a \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b^{4}} + \frac {15 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2}}{b^{4}} - \frac {4 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} + 1\right )}}{b^{4}}\right )} \]
1/3*x^3*arcsinh(b*x + a) - 1/18*b*(2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*x^2 /b^2 - 15*a^3*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/ b^4 - 5*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a*x/b^3 + 9*(a^2 + 1)*a*arcsinh( 2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^4 + 15*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^2/b^4 - 4*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(a^2 + 1)/b^4)
Time = 0.43 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.46 \[ \int x^2 \text {arcsinh}(a+b x) \, dx=\frac {1}{3} \, x^{3} \log \left (b x + a + \sqrt {{\left (b x + a\right )}^{2} + 1}\right ) - \frac {1}{18} \, {\left (\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (x {\left (\frac {2 \, x}{b^{2}} - \frac {5 \, a}{b^{3}}\right )} + \frac {11 \, a^{2} b - 4 \, b}{b^{5}}\right )} + \frac {3 \, {\left (2 \, a^{3} - 3 \, a\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} {\left | b \right |}\right )}{b^{3} {\left | b \right |}}\right )} b \]
1/3*x^3*log(b*x + a + sqrt((b*x + a)^2 + 1)) - 1/18*(sqrt(b^2*x^2 + 2*a*b* x + a^2 + 1)*(x*(2*x/b^2 - 5*a/b^3) + (11*a^2*b - 4*b)/b^5) + 3*(2*a^3 - 3 *a)*log(-a*b - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*abs(b))/(b^3 *abs(b)))*b
Timed out. \[ \int x^2 \text {arcsinh}(a+b x) \, dx=\int x^2\,\mathrm {asinh}\left (a+b\,x\right ) \,d x \]