Integrand size = 26, antiderivative size = 75 \[ \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=-\frac {b x^2}{4 c \sqrt {\pi }}+\frac {x \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{2 c^2 \pi }-\frac {(a+b \text {arcsinh}(c x))^2}{4 b c^3 \sqrt {\pi }} \]
-1/4*b*x^2/c/Pi^(1/2)-1/4*(a+b*arcsinh(c*x))^2/b/c^3/Pi^(1/2)+1/2*x*(a+b*a rcsinh(c*x))*(Pi*c^2*x^2+Pi)^(1/2)/c^2/Pi
Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\frac {4 a c x \sqrt {1+c^2 x^2}-2 b \text {arcsinh}(c x)^2-b \cosh (2 \text {arcsinh}(c x))+\text {arcsinh}(c x) (-4 a+2 b \sinh (2 \text {arcsinh}(c x)))}{8 c^3 \sqrt {\pi }} \]
(4*a*c*x*Sqrt[1 + c^2*x^2] - 2*b*ArcSinh[c*x]^2 - b*Cosh[2*ArcSinh[c*x]] + ArcSinh[c*x]*(-4*a + 2*b*Sinh[2*ArcSinh[c*x]]))/(8*c^3*Sqrt[Pi])
Time = 0.38 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6227, 15, 6198}
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.
\(\displaystyle \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {\pi c^2 x^2+\pi }} \, dx\) |
\(\Big \downarrow \) 6227 |
\(\displaystyle -\frac {\int \frac {a+b \text {arcsinh}(c x)}{\sqrt {c^2 \pi x^2+\pi }}dx}{2 c^2}-\frac {b \int xdx}{2 \sqrt {\pi } c}+\frac {x \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{2 \pi c^2}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -\frac {\int \frac {a+b \text {arcsinh}(c x)}{\sqrt {c^2 \pi x^2+\pi }}dx}{2 c^2}+\frac {x \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{2 \pi c^2}-\frac {b x^2}{4 \sqrt {\pi } c}\) |
\(\Big \downarrow \) 6198 |
\(\displaystyle -\frac {(a+b \text {arcsinh}(c x))^2}{4 \sqrt {\pi } b c^3}+\frac {x \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{2 \pi c^2}-\frac {b x^2}{4 \sqrt {\pi } c}\) |
-1/4*(b*x^2)/(c*Sqrt[Pi]) + (x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x])) /(2*c^2*Pi) - (a + b*ArcSinh[c*x])^2/(4*b*c^3*Sqrt[Pi])
3.1.83.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_ Symbol] :> Simp[(1/(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*( a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c ^2*d] && NeQ[n, -1]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_ .)*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(e*(m + 2*p + 1))), x] + (-Simp[f^2*((m - 1)/(c^2*(m + 2*p + 1))) Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Simp[b*f*(n/(c*(m + 2*p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p] Int [(f*x)^(m - 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] ) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && IGtQ[ m, 1] && NeQ[m + 2*p + 1, 0]
Time = 0.16 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.43
method | result | size |
default | \(\frac {a x \sqrt {\pi \,c^{2} x^{2}+\pi }}{2 \pi \,c^{2}}-\frac {a \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{2 c^{2} \sqrt {\pi \,c^{2}}}-\frac {b \left (-2 \,\operatorname {arcsinh}\left (c x \right ) c x \sqrt {c^{2} x^{2}+1}+c^{2} x^{2}+\operatorname {arcsinh}\left (c x \right )^{2}+1\right )}{4 \sqrt {\pi }\, c^{3}}\) | \(107\) |
parts | \(\frac {a x \sqrt {\pi \,c^{2} x^{2}+\pi }}{2 \pi \,c^{2}}-\frac {a \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{2 c^{2} \sqrt {\pi \,c^{2}}}-\frac {b \left (-2 \,\operatorname {arcsinh}\left (c x \right ) c x \sqrt {c^{2} x^{2}+1}+c^{2} x^{2}+\operatorname {arcsinh}\left (c x \right )^{2}+1\right )}{4 \sqrt {\pi }\, c^{3}}\) | \(107\) |
1/2*a*x/Pi/c^2*(Pi*c^2*x^2+Pi)^(1/2)-1/2*a/c^2*ln(Pi*c^2*x/(Pi*c^2)^(1/2)+ (Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)-1/4*b/Pi^(1/2)*(-2*arcsinh(c*x)*c*x* (c^2*x^2+1)^(1/2)+c^2*x^2+arcsinh(c*x)^2+1)/c^3
\[ \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{2}}{\sqrt {\pi + \pi c^{2} x^{2}}} \,d x } \]
Time = 1.03 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.63 \[ \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\frac {a \left (\begin {cases} \frac {x \sqrt {c^{2} x^{2} + 1}}{2 c^{2}} - \frac {\log {\left (2 c^{2} x + 2 \sqrt {c^{2} x^{2} + 1} \sqrt {c^{2}} \right )}}{2 c^{2} \sqrt {c^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x^{3}}{3} & \text {otherwise} \end {cases}\right )}{\sqrt {\pi }} + \frac {b \left (\begin {cases} - \frac {x^{2}}{4 c} + \frac {x \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{2 c^{2}} - \frac {\operatorname {asinh}^{2}{\left (c x \right )}}{4 c^{3}} & \text {for}\: c \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{\sqrt {\pi }} \]
a*Piecewise((x*sqrt(c**2*x**2 + 1)/(2*c**2) - log(2*c**2*x + 2*sqrt(c**2*x **2 + 1)*sqrt(c**2))/(2*c**2*sqrt(c**2)), Ne(c**2, 0)), (x**3/3, True))/sq rt(pi) + b*Piecewise((-x**2/(4*c) + x*sqrt(c**2*x**2 + 1)*asinh(c*x)/(2*c* *2) - asinh(c*x)**2/(4*c**3), Ne(c, 0)), (0, True))/sqrt(pi)
Exception generated. \[ \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{2}}{\sqrt {\pi + \pi c^{2} x^{2}}} \,d x } \]
Timed out. \[ \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\int \frac {x^2\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{\sqrt {\Pi \,c^2\,x^2+\Pi }} \,d x \]