Integrand size = 26, antiderivative size = 119 \[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=-\frac {b \sqrt {\pi } x^2}{16 c}-\frac {1}{16} b c \sqrt {\pi } x^4+\frac {\sqrt {\pi } x \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))}{8 c^2}+\frac {1}{4} x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))-\frac {\sqrt {\pi } (a+b \text {arcsinh}(c x))^2}{16 b c^3} \]
-1/16*b*x^2*Pi^(1/2)/c-1/16*b*c*x^4*Pi^(1/2)-1/16*(a+b*arcsinh(c*x))^2*Pi^ (1/2)/b/c^3+1/8*x*(a+b*arcsinh(c*x))*Pi^(1/2)*(c^2*x^2+1)^(1/2)/c^2+1/4*x^ 3*(a+b*arcsinh(c*x))*(Pi*c^2*x^2+Pi)^(1/2)
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.66 \[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\frac {\sqrt {\pi } \left (16 a c x \sqrt {1+c^2 x^2} \left (1+2 c^2 x^2\right )-8 b \text {arcsinh}(c x)^2-b \cosh (4 \text {arcsinh}(c x))+\text {arcsinh}(c x) (-16 a+4 b \sinh (4 \text {arcsinh}(c x)))\right )}{128 c^3} \]
(Sqrt[Pi]*(16*a*c*x*Sqrt[1 + c^2*x^2]*(1 + 2*c^2*x^2) - 8*b*ArcSinh[c*x]^2 - b*Cosh[4*ArcSinh[c*x]] + ArcSinh[c*x]*(-16*a + 4*b*Sinh[4*ArcSinh[c*x]] )))/(128*c^3)
Time = 0.51 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {6221, 15, 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.
\(\displaystyle \int x^2 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x)) \, dx\) |
\(\Big \downarrow \) 6221 |
\(\displaystyle \frac {1}{4} \sqrt {\pi } \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}dx-\frac {1}{4} \sqrt {\pi } b c \int x^3dx+\frac {1}{4} x^3 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {1}{4} \sqrt {\pi } \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}dx+\frac {1}{4} x^3 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))-\frac {1}{16} \sqrt {\pi } b c x^4\) |
\(\Big \downarrow \) 6227 |
\(\displaystyle \frac {1}{4} \sqrt {\pi } \left (-\frac {\int \frac {a+b \text {arcsinh}(c x)}{\sqrt {c^2 x^2+1}}dx}{2 c^2}-\frac {b \int xdx}{2 c}+\frac {x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))}{2 c^2}\right )+\frac {1}{4} x^3 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))-\frac {1}{16} \sqrt {\pi } b c x^4\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {1}{4} \sqrt {\pi } \left (-\frac {\int \frac {a+b \text {arcsinh}(c x)}{\sqrt {c^2 x^2+1}}dx}{2 c^2}+\frac {x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))}{2 c^2}-\frac {b x^2}{4 c}\right )+\frac {1}{4} x^3 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))-\frac {1}{16} \sqrt {\pi } b c x^4\) |
\(\Big \downarrow \) 6198 |
\(\displaystyle \frac {1}{4} x^3 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))+\frac {1}{4} \sqrt {\pi } \left (-\frac {(a+b \text {arcsinh}(c x))^2}{4 b c^3}+\frac {x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))}{2 c^2}-\frac {b x^2}{4 c}\right )-\frac {1}{16} \sqrt {\pi } b c x^4\) |
-1/16*(b*c*Sqrt[Pi]*x^4) + (x^3*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]) )/4 + (Sqrt[Pi]*(-1/4*(b*x^2)/c + (x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x] ))/(2*c^2) - (a + b*ArcSinh[c*x])^2/(4*b*c^3)))/4
3.1.56.3.1 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_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*Arc Sinh[c*x])^n/(f*(m + 2))), x] + (Simp[(1/(m + 2))*Simp[Sqrt[d + e*x^2]/Sqrt [1 + c^2*x^2]] Int[(f*x)^m*((a + b*ArcSinh[c*x])^n/Sqrt[1 + c^2*x^2]), x] , x] - Simp[b*c*(n/(f*(m + 2)))*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]] I nt[(f*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d , e, f, m}, x] && EqQ[e, c^2*d] && IGtQ[n, 0] && (IGtQ[m, -2] || EqQ[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 = 0.21 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.31
method | result | size |
default | \(\frac {a x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{4 \pi \,c^{2}}-\frac {a x \sqrt {\pi \,c^{2} x^{2}+\pi }}{8 c^{2}}-\frac {a \pi \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{8 c^{2} \sqrt {\pi \,c^{2}}}-\frac {b \sqrt {\pi }\, \left (-4 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+c^{4} x^{4}-2 \,\operatorname {arcsinh}\left (c x \right ) c x \sqrt {c^{2} x^{2}+1}+c^{2} x^{2}+\operatorname {arcsinh}\left (c x \right )^{2}\right )}{16 c^{3}}\) | \(156\) |
parts | \(\frac {a x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{4 \pi \,c^{2}}-\frac {a x \sqrt {\pi \,c^{2} x^{2}+\pi }}{8 c^{2}}-\frac {a \pi \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{8 c^{2} \sqrt {\pi \,c^{2}}}-\frac {b \sqrt {\pi }\, \left (-4 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+c^{4} x^{4}-2 \,\operatorname {arcsinh}\left (c x \right ) c x \sqrt {c^{2} x^{2}+1}+c^{2} x^{2}+\operatorname {arcsinh}\left (c x \right )^{2}\right )}{16 c^{3}}\) | \(156\) |
1/4*a*x*(Pi*c^2*x^2+Pi)^(3/2)/Pi/c^2-1/8*a/c^2*x*(Pi*c^2*x^2+Pi)^(1/2)-1/8 *a/c^2*Pi*ln(Pi*c^2*x/(Pi*c^2)^(1/2)+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2) -1/16*b*Pi^(1/2)*(-4*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^3*c^3+c^4*x^4-2*arcs inh(c*x)*c*x*(c^2*x^2+1)^(1/2)+c^2*x^2+arcsinh(c*x)^2)/c^3
\[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\int { \sqrt {\pi + \pi c^{2} x^{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{2} \,d x } \]
\[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\sqrt {\pi } \left (\int a x^{2} \sqrt {c^{2} x^{2} + 1}\, dx + \int b x^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}\, dx\right ) \]
sqrt(pi)*(Integral(a*x**2*sqrt(c**2*x**2 + 1), x) + Integral(b*x**2*sqrt(c **2*x**2 + 1)*asinh(c*x), x))
Exception generated. \[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: RuntimeError} \]
\[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\int { \sqrt {\pi + \pi c^{2} x^{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{2} \,d x } \]
Timed out. \[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\int x^2\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {\Pi \,c^2\,x^2+\Pi } \,d x \]