Integrand size = 10, antiderivative size = 126 \[ \int x \text {arcsinh}(a+b x)^2 \, dx=-\frac {2 a x}{b}+\frac {(a+b x)^2}{4 b^2}+\frac {2 a \sqrt {1+(a+b x)^2} \text {arcsinh}(a+b x)}{b^2}-\frac {(a+b x) \sqrt {1+(a+b x)^2} \text {arcsinh}(a+b x)}{2 b^2}+\frac {\text {arcsinh}(a+b x)^2}{4 b^2}-\frac {a^2 \text {arcsinh}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \text {arcsinh}(a+b x)^2 \]
-2*a*x/b+1/4*(b*x+a)^2/b^2+1/4*arcsinh(b*x+a)^2/b^2-1/2*a^2*arcsinh(b*x+a) ^2/b^2+1/2*x^2*arcsinh(b*x+a)^2+2*a*arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)/b^2 -1/2*(b*x+a)*arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)/b^2
Time = 0.23 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.63 \[ \int x \text {arcsinh}(a+b x)^2 \, dx=\frac {b x (-6 a+b x)+2 (3 a-b x) \sqrt {1+a^2+2 a b x+b^2 x^2} \text {arcsinh}(a+b x)+\left (1-2 a^2+2 b^2 x^2\right ) \text {arcsinh}(a+b x)^2}{4 b^2} \]
(b*x*(-6*a + b*x) + 2*(3*a - b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*ArcSin h[a + b*x] + (1 - 2*a^2 + 2*b^2*x^2)*ArcSinh[a + b*x]^2)/(4*b^2)
Time = 0.50 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6274, 25, 27, 6243, 6253, 2009}
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.
\(\displaystyle \int x \text {arcsinh}(a+b x)^2 \, dx\) |
\(\Big \downarrow \) 6274 |
\(\displaystyle \frac {\int x \text {arcsinh}(a+b x)^2d(a+b x)}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int -x \text {arcsinh}(a+b x)^2d(a+b x)}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int -b x \text {arcsinh}(a+b x)^2d(a+b x)}{b^2}\) |
\(\Big \downarrow \) 6243 |
\(\displaystyle -\frac {\int \frac {b^2 x^2 \text {arcsinh}(a+b x)}{\sqrt {(a+b x)^2+1}}d(a+b x)-\frac {1}{2} b^2 x^2 \text {arcsinh}(a+b x)^2}{b^2}\) |
\(\Big \downarrow \) 6253 |
\(\displaystyle -\frac {\int \left (\frac {\text {arcsinh}(a+b x) a^2}{\sqrt {(a+b x)^2+1}}-\frac {2 (a+b x) \text {arcsinh}(a+b x) a}{\sqrt {(a+b x)^2+1}}+\frac {(a+b x)^2 \text {arcsinh}(a+b x)}{\sqrt {(a+b x)^2+1}}\right )d(a+b x)-\frac {1}{2} b^2 x^2 \text {arcsinh}(a+b x)^2}{b^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {1}{2} a^2 \text {arcsinh}(a+b x)^2-\frac {1}{2} b^2 x^2 \text {arcsinh}(a+b x)^2+\frac {1}{2} \sqrt {(a+b x)^2+1} (a+b x) \text {arcsinh}(a+b x)-\frac {1}{4} \text {arcsinh}(a+b x)^2-2 a \sqrt {(a+b x)^2+1} \text {arcsinh}(a+b x)-\frac {1}{4} (a+b x)^2+2 a (a+b x)}{b^2}\) |
-((2*a*(a + b*x) - (a + b*x)^2/4 - 2*a*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b *x] + ((a + b*x)*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/2 - ArcSinh[a + b *x]^2/4 + (a^2*ArcSinh[a + b*x]^2)/2 - (b^2*x^2*ArcSinh[a + b*x]^2)/2)/b^2 )
3.1.69.3.1 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[((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_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d _) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n , 0] && ((EqQ[n, 1] && GtQ[p, -1]) || GtQ[p, 0] || EqQ[m, 1] || (EqQ[m, 2] && LtQ[p, -2]))
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 = 0.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {\frac {\operatorname {arcsinh}\left (b x +a \right )^{2} \left (1+\left (b x +a \right )^{2}\right )}{2}-\frac {\operatorname {arcsinh}\left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}\, \left (b x +a \right )}{2}-\frac {\operatorname {arcsinh}\left (b x +a \right )^{2}}{4}+\frac {\left (b x +a \right )^{2}}{4}+\frac {1}{4}-a \left (\operatorname {arcsinh}\left (b x +a \right )^{2} \left (b x +a \right )-2 \,\operatorname {arcsinh}\left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}+2 b x +2 a \right )}{b^{2}}\) | \(113\) |
default | \(\frac {\frac {\operatorname {arcsinh}\left (b x +a \right )^{2} \left (1+\left (b x +a \right )^{2}\right )}{2}-\frac {\operatorname {arcsinh}\left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}\, \left (b x +a \right )}{2}-\frac {\operatorname {arcsinh}\left (b x +a \right )^{2}}{4}+\frac {\left (b x +a \right )^{2}}{4}+\frac {1}{4}-a \left (\operatorname {arcsinh}\left (b x +a \right )^{2} \left (b x +a \right )-2 \,\operatorname {arcsinh}\left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}+2 b x +2 a \right )}{b^{2}}\) | \(113\) |
1/b^2*(1/2*arcsinh(b*x+a)^2*(1+(b*x+a)^2)-1/2*arcsinh(b*x+a)*(1+(b*x+a)^2) ^(1/2)*(b*x+a)-1/4*arcsinh(b*x+a)^2+1/4*(b*x+a)^2+1/4-a*(arcsinh(b*x+a)^2* (b*x+a)-2*arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)+2*b*x+2*a))
Time = 0.25 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.90 \[ \int x \text {arcsinh}(a+b x)^2 \, dx=\frac {b^{2} x^{2} - 6 \, a b x + {\left (2 \, b^{2} x^{2} - 2 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2} - 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b x - 3 \, a\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{4 \, b^{2}} \]
1/4*(b^2*x^2 - 6*a*b*x + (2*b^2*x^2 - 2*a^2 + 1)*log(b*x + a + sqrt(b^2*x^ 2 + 2*a*b*x + a^2 + 1))^2 - 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(b*x - 3*a )*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)))/b^2
Time = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.10 \[ \int x \text {arcsinh}(a+b x)^2 \, dx=\begin {cases} - \frac {a^{2} \operatorname {asinh}^{2}{\left (a + b x \right )}}{2 b^{2}} - \frac {3 a x}{2 b} + \frac {3 a \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{2 b^{2}} + \frac {x^{2} \operatorname {asinh}^{2}{\left (a + b x \right )}}{2} + \frac {x^{2}}{4} - \frac {x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{2 b} + \frac {\operatorname {asinh}^{2}{\left (a + b x \right )}}{4 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {asinh}^{2}{\left (a \right )}}{2} & \text {otherwise} \end {cases} \]
Piecewise((-a**2*asinh(a + b*x)**2/(2*b**2) - 3*a*x/(2*b) + 3*a*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x)/(2*b**2) + x**2*asinh(a + b*x)** 2/2 + x**2/4 - x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x)/(2*b) + asinh(a + b*x)**2/(4*b**2), Ne(b, 0)), (x**2*asinh(a)**2/2, True))
\[ \int x \text {arcsinh}(a+b x)^2 \, dx=\int { x \operatorname {arsinh}\left (b x + a\right )^{2} \,d x } \]
1/2*x^2*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2 - integrate((b^ 3*x^4 + 2*a*b^2*x^3 + (a^2*b + b)*x^2 + (b^2*x^3 + a*b*x^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/(b^3 *x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b + b)*x + (b^2*x^2 + 2*a*b*x + a^2 + 1) ^(3/2) + a), x)
\[ \int x \text {arcsinh}(a+b x)^2 \, dx=\int { x \operatorname {arsinh}\left (b x + a\right )^{2} \,d x } \]
Timed out. \[ \int x \text {arcsinh}(a+b x)^2 \, dx=\int x\,{\mathrm {asinh}\left (a+b\,x\right )}^2 \,d x \]