Integrand size = 26, antiderivative size = 109 \[ \int x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\frac {2 b \sqrt {\pi } x}{15 c^3}-\frac {b \sqrt {\pi } x^3}{45 c}-\frac {1}{25} b c \sqrt {\pi } x^5-\frac {\left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 c^4 \pi }+\frac {\left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{5 c^4 \pi ^2} \] Output:
2/15*b*Pi^(1/2)*x/c^3-1/45*b*Pi^(1/2)*x^3/c-1/25*b*c*Pi^(1/2)*x^5-1/3*(Pi* c^2*x^2+Pi)^(3/2)*(a+b*arcsinh(c*x))/c^4/Pi+1/5*(Pi*c^2*x^2+Pi)^(5/2)*(a+b *arcsinh(c*x))/c^4/Pi^2
Time = 0.17 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.97 \[ \int x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\frac {\sqrt {\pi } \left (15 a \sqrt {1+c^2 x^2} \left (-2+c^2 x^2+3 c^4 x^4\right )+b \left (30 c x-5 c^3 x^3-9 c^5 x^5\right )+15 b \sqrt {1+c^2 x^2} \left (-2+c^2 x^2+3 c^4 x^4\right ) \text {arcsinh}(c x)\right )}{225 c^4} \] Input:
Integrate[x^3*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]),x]
Output:
(Sqrt[Pi]*(15*a*Sqrt[1 + c^2*x^2]*(-2 + c^2*x^2 + 3*c^4*x^4) + b*(30*c*x - 5*c^3*x^3 - 9*c^5*x^5) + 15*b*Sqrt[1 + c^2*x^2]*(-2 + c^2*x^2 + 3*c^4*x^4 )*ArcSinh[c*x]))/(225*c^4)
Time = 0.38 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6219, 27, 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.
\(\displaystyle \int x^3 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x)) \, dx\) |
\(\Big \downarrow \) 6219 |
\(\displaystyle -\sqrt {\pi } b c \int -\frac {-3 c^4 x^4-c^2 x^2+2}{15 c^4}dx+\frac {\left (\pi c^2 x^2+\pi \right )^{5/2} (a+b \text {arcsinh}(c x))}{5 \pi ^2 c^4}-\frac {\left (\pi c^2 x^2+\pi \right )^{3/2} (a+b \text {arcsinh}(c x))}{3 \pi c^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {\pi } b \int \left (-3 c^4 x^4-c^2 x^2+2\right )dx}{15 c^3}+\frac {\left (\pi c^2 x^2+\pi \right )^{5/2} (a+b \text {arcsinh}(c x))}{5 \pi ^2 c^4}-\frac {\left (\pi c^2 x^2+\pi \right )^{3/2} (a+b \text {arcsinh}(c x))}{3 \pi c^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (\pi c^2 x^2+\pi \right )^{5/2} (a+b \text {arcsinh}(c x))}{5 \pi ^2 c^4}-\frac {\left (\pi c^2 x^2+\pi \right )^{3/2} (a+b \text {arcsinh}(c x))}{3 \pi c^4}+\frac {\sqrt {\pi } b \left (-\frac {3}{5} c^4 x^5-\frac {c^2 x^3}{3}+2 x\right )}{15 c^3}\) |
Input:
Int[x^3*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]),x]
Output:
(b*Sqrt[Pi]*(2*x - (c^2*x^3)/3 - (3*c^4*x^5)/5))/(15*c^3) - ((Pi + c^2*Pi* x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(3*c^4*Pi) + ((Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/(5*c^4*Pi^2)
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_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_ ), x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^2)^p, x]}, Simp[(a + b*ArcSi nh[c*x]) u, x] - Simp[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]] Int[S implifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x ] && EqQ[e, c^2*d] && IntegerQ[p - 1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1) /2, 0] || ILtQ[(m + 2*p + 3)/2, 0])
Time = 1.12 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.50
method | result | size |
default | \(a \left (\frac {x^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{5 \pi \,c^{2}}-\frac {2 \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{15 \pi \,c^{4}}\right )+\frac {b \sqrt {\pi }\, \left (45 \,\operatorname {arcsinh}\left (x c \right ) x^{6} c^{6}+60 \,\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}-9 \sqrt {c^{2} x^{2}+1}\, x^{5} c^{5}-15 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}-5 \sqrt {c^{2} x^{2}+1}\, c^{3} x^{3}-30 \,\operatorname {arcsinh}\left (x c \right )+30 \sqrt {c^{2} x^{2}+1}\, x c \right )}{225 c^{4} \sqrt {c^{2} x^{2}+1}}\) | \(164\) |
parts | \(a \left (\frac {x^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{5 \pi \,c^{2}}-\frac {2 \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{15 \pi \,c^{4}}\right )+\frac {b \sqrt {\pi }\, \left (45 \,\operatorname {arcsinh}\left (x c \right ) x^{6} c^{6}+60 \,\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}-9 \sqrt {c^{2} x^{2}+1}\, x^{5} c^{5}-15 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}-5 \sqrt {c^{2} x^{2}+1}\, c^{3} x^{3}-30 \,\operatorname {arcsinh}\left (x c \right )+30 \sqrt {c^{2} x^{2}+1}\, x c \right )}{225 c^{4} \sqrt {c^{2} x^{2}+1}}\) | \(164\) |
orering | \(\frac {\left (81 c^{6} x^{6}+107 c^{4} x^{4}-120 c^{2} x^{2}-120\right ) \sqrt {\pi \,c^{2} x^{2}+\pi }\, \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}{225 c^{4} \left (c^{2} x^{2}+1\right )}-\frac {\left (9 c^{4} x^{4}+5 c^{2} x^{2}-30\right ) \left (3 x^{2} \sqrt {\pi \,c^{2} x^{2}+\pi }\, \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )+\frac {x^{4} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right ) \pi \,c^{2}}{\sqrt {\pi \,c^{2} x^{2}+\pi }}+\frac {x^{3} \sqrt {\pi \,c^{2} x^{2}+\pi }\, b c}{\sqrt {c^{2} x^{2}+1}}\right )}{225 x^{2} c^{4}}\) | \(173\) |
Input:
int(x^3*(Pi*c^2*x^2+Pi)^(1/2)*(a+b*arcsinh(x*c)),x,method=_RETURNVERBOSE)
Output:
a*(1/5*x^2*(Pi*c^2*x^2+Pi)^(3/2)/Pi/c^2-2/15/Pi/c^4*(Pi*c^2*x^2+Pi)^(3/2)) +1/225*b*Pi^(1/2)/c^4/(c^2*x^2+1)^(1/2)*(45*arcsinh(x*c)*x^6*c^6+60*arcsin h(x*c)*c^4*x^4-9*(c^2*x^2+1)^(1/2)*x^5*c^5-15*arcsinh(x*c)*c^2*x^2-5*(c^2* x^2+1)^(1/2)*c^3*x^3-30*arcsinh(x*c)+30*(c^2*x^2+1)^(1/2)*x*c)
Time = 0.10 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.45 \[ \int x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\frac {15 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (3 \, b c^{6} x^{6} + 4 \, b c^{4} x^{4} - b c^{2} x^{2} - 2 \, b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + \sqrt {\pi + \pi c^{2} x^{2}} {\left (45 \, a c^{6} x^{6} + 60 \, a c^{4} x^{4} - 15 \, a c^{2} x^{2} - {\left (9 \, b c^{5} x^{5} + 5 \, b c^{3} x^{3} - 30 \, b c x\right )} \sqrt {c^{2} x^{2} + 1} - 30 \, a\right )}}{225 \, {\left (c^{6} x^{2} + c^{4}\right )}} \] Input:
integrate(x^3*(pi*c^2*x^2+pi)^(1/2)*(a+b*arcsinh(c*x)),x, algorithm="frica s")
Output:
1/225*(15*sqrt(pi + pi*c^2*x^2)*(3*b*c^6*x^6 + 4*b*c^4*x^4 - b*c^2*x^2 - 2 *b)*log(c*x + sqrt(c^2*x^2 + 1)) + sqrt(pi + pi*c^2*x^2)*(45*a*c^6*x^6 + 6 0*a*c^4*x^4 - 15*a*c^2*x^2 - (9*b*c^5*x^5 + 5*b*c^3*x^3 - 30*b*c*x)*sqrt(c ^2*x^2 + 1) - 30*a))/(c^6*x^2 + c^4)
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (100) = 200\).
Time = 1.00 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.03 \[ \int x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\begin {cases} \frac {\sqrt {\pi } a x^{4} \sqrt {c^{2} x^{2} + 1}}{5} + \frac {\sqrt {\pi } a x^{2} \sqrt {c^{2} x^{2} + 1}}{15 c^{2}} - \frac {2 \sqrt {\pi } a \sqrt {c^{2} x^{2} + 1}}{15 c^{4}} - \frac {\sqrt {\pi } b c x^{5}}{25} + \frac {\sqrt {\pi } b x^{4} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{5} - \frac {\sqrt {\pi } b x^{3}}{45 c} + \frac {\sqrt {\pi } b x^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{15 c^{2}} + \frac {2 \sqrt {\pi } b x}{15 c^{3}} - \frac {2 \sqrt {\pi } b \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{15 c^{4}} & \text {for}\: c \neq 0 \\\frac {\sqrt {\pi } a x^{4}}{4} & \text {otherwise} \end {cases} \] Input:
integrate(x**3*(pi*c**2*x**2+pi)**(1/2)*(a+b*asinh(c*x)),x)
Output:
Piecewise((sqrt(pi)*a*x**4*sqrt(c**2*x**2 + 1)/5 + sqrt(pi)*a*x**2*sqrt(c* *2*x**2 + 1)/(15*c**2) - 2*sqrt(pi)*a*sqrt(c**2*x**2 + 1)/(15*c**4) - sqrt (pi)*b*c*x**5/25 + sqrt(pi)*b*x**4*sqrt(c**2*x**2 + 1)*asinh(c*x)/5 - sqrt (pi)*b*x**3/(45*c) + sqrt(pi)*b*x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/(15*c* *2) + 2*sqrt(pi)*b*x/(15*c**3) - 2*sqrt(pi)*b*sqrt(c**2*x**2 + 1)*asinh(c* x)/(15*c**4), Ne(c, 0)), (sqrt(pi)*a*x**4/4, True))
Time = 0.04 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.23 \[ \int x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\frac {1}{15} \, b {\left (\frac {3 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x^{2}}{\pi c^{2}} - \frac {2 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}}{\pi c^{4}}\right )} \operatorname {arsinh}\left (c x\right ) + \frac {1}{15} \, a {\left (\frac {3 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x^{2}}{\pi c^{2}} - \frac {2 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}}{\pi c^{4}}\right )} - \frac {{\left (9 \, \sqrt {\pi } c^{4} x^{5} + 5 \, \sqrt {\pi } c^{2} x^{3} - 30 \, \sqrt {\pi } x\right )} b}{225 \, c^{3}} \] Input:
integrate(x^3*(pi*c^2*x^2+pi)^(1/2)*(a+b*arcsinh(c*x)),x, algorithm="maxim a")
Output:
1/15*b*(3*(pi + pi*c^2*x^2)^(3/2)*x^2/(pi*c^2) - 2*(pi + pi*c^2*x^2)^(3/2) /(pi*c^4))*arcsinh(c*x) + 1/15*a*(3*(pi + pi*c^2*x^2)^(3/2)*x^2/(pi*c^2) - 2*(pi + pi*c^2*x^2)^(3/2)/(pi*c^4)) - 1/225*(9*sqrt(pi)*c^4*x^5 + 5*sqrt( pi)*c^2*x^3 - 30*sqrt(pi)*x)*b/c^3
Exception generated. \[ \int x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x^3*(pi*c^2*x^2+pi)^(1/2)*(a+b*arcsinh(c*x)),x, algorithm="giac" )
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\int x^3\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {\Pi \,c^2\,x^2+\Pi } \,d x \] Input:
int(x^3*(a + b*asinh(c*x))*(Pi + Pi*c^2*x^2)^(1/2),x)
Output:
int(x^3*(a + b*asinh(c*x))*(Pi + Pi*c^2*x^2)^(1/2), x)
\[ \int x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\frac {\sqrt {\pi }\, \left (3 \sqrt {c^{2} x^{2}+1}\, a \,c^{4} x^{4}+\sqrt {c^{2} x^{2}+1}\, a \,c^{2} x^{2}-2 \sqrt {c^{2} x^{2}+1}\, a +15 \left (\int \sqrt {c^{2} x^{2}+1}\, \mathit {asinh} \left (c x \right ) x^{3}d x \right ) b \,c^{4}\right )}{15 c^{4}} \] Input:
int(x^3*(Pi*c^2*x^2+Pi)^(1/2)*(a+b*asinh(c*x)),x)
Output:
(sqrt(pi)*(3*sqrt(c**2*x**2 + 1)*a*c**4*x**4 + sqrt(c**2*x**2 + 1)*a*c**2* x**2 - 2*sqrt(c**2*x**2 + 1)*a + 15*int(sqrt(c**2*x**2 + 1)*asinh(c*x)*x** 3,x)*b*c**4))/(15*c**4)