\(\int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx\) [60]

Optimal result

Integrand size = 26, antiderivative size = 115 \[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=-\frac {b \sqrt {\pi } x^2}{16 c}-\frac {1}{16} b c \sqrt {\pi } x^4+\frac {x \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{8 c^2}+\frac {1}{4} x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))-\frac {\sqrt {\pi } (a+b \text {arcsinh}(c x))^2}{16 b c^3} \] Output:

-1/16*b*Pi^(1/2)*x^2/c-1/16*b*c*Pi^(1/2)*x^4+1/8*x*(Pi*c^2*x^2+Pi)^(1/2)*( 
a+b*arcsinh(c*x))/c^2+1/4*x^3*(Pi*c^2*x^2+Pi)^(1/2)*(a+b*arcsinh(c*x))-1/1 
6*Pi^(1/2)*(a+b*arcsinh(c*x))^2/b/c^3
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69 \[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\frac {\sqrt {\pi } \left (16 a c x \sqrt {1+c^2 x^2} \left (1+2 c^2 x^2\right )-8 b \text {arcsinh}(c x)^2-b \cosh (4 \text {arcsinh}(c x))+\text {arcsinh}(c x) (-16 a+4 b \sinh (4 \text {arcsinh}(c x)))\right )}{128 c^3} \] Input:

Integrate[x^2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]),x]
 

Output:

(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)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.99, 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

\(\displaystyle \frac {1}{4} x^3 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))+\frac {1}{4} \sqrt {\pi } \left (-\frac {(a+b \text {arcsinh}(c x))^2}{4 b c^3}+\frac {x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))}{2 c^2}-\frac {b x^2}{4 c}\right )-\frac {1}{16} \sqrt {\pi } b c x^4\)

Input:

Int[x^2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]),x]
 

Output:

-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
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 6198

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]
 

rule 6221

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])
 

rule 6227

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]
 

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.36

method	result	size
default	\(\frac {a x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{4 \pi \,c^{2}}-\frac {a x \sqrt {\pi \,c^{2} x^{2}+\pi }}{8 c^{2}}-\frac {a \pi \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{8 c^{2} \sqrt {\pi \,c^{2}}}-\frac {b \sqrt {\pi }\, \left (-4 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+c^{4} x^{4}-2 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x c +c^{2} x^{2}+\operatorname {arcsinh}\left (x c \right )^{2}\right )}{16 c^{3}}\)	\(156\)
parts	\(\frac {a x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{4 \pi \,c^{2}}-\frac {a x \sqrt {\pi \,c^{2} x^{2}+\pi }}{8 c^{2}}-\frac {a \pi \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{8 c^{2} \sqrt {\pi \,c^{2}}}-\frac {b \sqrt {\pi }\, \left (-4 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+c^{4} x^{4}-2 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x c +c^{2} x^{2}+\operatorname {arcsinh}\left (x c \right )^{2}\right )}{16 c^{3}}\)	\(156\)

Input:

int(x^2*(Pi*c^2*x^2+Pi)^(1/2)*(a+b*arcsinh(x*c)),x,method=_RETURNVERBOSE)
 

Output:

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(x*c)*(c^2*x^2+1)^(1/2)*x^3*c^3+c^4*x^4-2*arcs 
inh(x*c)*(c^2*x^2+1)^(1/2)*x*c+c^2*x^2+arcsinh(x*c)^2)/c^3
 

Fricas [F]

\[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\int { \sqrt {\pi + \pi c^{2} x^{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{2} \,d x } \] Input:

integrate(x^2*(pi*c^2*x^2+pi)^(1/2)*(a+b*arcsinh(c*x)),x, algorithm="frica 
s")
 

Output:

integral(sqrt(pi + pi*c^2*x^2)*(b*x^2*arcsinh(c*x) + a*x^2), x)
 

Sympy [F]

\[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\sqrt {\pi } \left (\int a x^{2} \sqrt {c^{2} x^{2} + 1}\, dx + \int b x^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}\, dx\right ) \] Input:

integrate(x**2*(pi*c**2*x**2+pi)**(1/2)*(a+b*asinh(c*x)),x)
 

Output:

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))
 

Maxima [F(-2)]

Exception generated. \[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(x^2*(pi*c^2*x^2+pi)^(1/2)*(a+b*arcsinh(c*x)),x, algorithm="maxim 
a")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [F]

\[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\int { \sqrt {\pi + \pi c^{2} x^{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{2} \,d x } \] Input:

integrate(x^2*(pi*c^2*x^2+pi)^(1/2)*(a+b*arcsinh(c*x)),x, algorithm="giac" 
)
 

Output:

integrate(sqrt(pi + pi*c^2*x^2)*(b*arcsinh(c*x) + a)*x^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\int x^2\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {\Pi \,c^2\,x^2+\Pi } \,d x \] Input:

int(x^2*(a + b*asinh(c*x))*(Pi + Pi*c^2*x^2)^(1/2),x)
 

Output:

int(x^2*(a + b*asinh(c*x))*(Pi + Pi*c^2*x^2)^(1/2), x)
 

Reduce [F]

\[ \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\frac {\sqrt {\pi }\, \left (2 \sqrt {c^{2} x^{2}+1}\, a \,c^{3} x^{3}+\sqrt {c^{2} x^{2}+1}\, a c x +8 \left (\int \sqrt {c^{2} x^{2}+1}\, \mathit {asinh} \left (c x \right ) x^{2}d x \right ) b \,c^{3}-\mathrm {log}\left (\sqrt {c^{2} x^{2}+1}+c x \right ) a \right )}{8 c^{3}} \] Input:

int(x^2*(Pi*c^2*x^2+Pi)^(1/2)*(a+b*asinh(c*x)),x)
 

Output:

(sqrt(pi)*(2*sqrt(c**2*x**2 + 1)*a*c**3*x**3 + sqrt(c**2*x**2 + 1)*a*c*x + 
 8*int(sqrt(c**2*x**2 + 1)*asinh(c*x)*x**2,x)*b*c**3 - log(sqrt(c**2*x**2 
+ 1) + c*x)*a))/(8*c**3)