\(\int \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx\) [61]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [F]
Sympy [F]
Maxima [F(-2)]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 67 \[ \int \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=-\frac {1}{4} b c \sqrt {\pi } x^2+\frac {1}{2} x \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))+\frac {\sqrt {\pi } (a+b \text {arcsinh}(c x))^2}{4 b c} \] Output:

-1/4*b*c*Pi^(1/2)*x^2+1/2*x*(Pi*c^2*x^2+Pi)^(1/2)*(a+b*arcsinh(c*x))+1/4*P 
i^(1/2)*(a+b*arcsinh(c*x))^2/b/c
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.03 \[ \int \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx=\frac {\sqrt {\pi } \left (4 a c x \sqrt {1+c^2 x^2}+2 b \text {arcsinh}(c x)^2-b \cosh (2 \text {arcsinh}(c x))+2 \text {arcsinh}(c x) (2 a+b \sinh (2 \text {arcsinh}(c x)))\right )}{8 c} \] Input:

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

Output:

(Sqrt[Pi]*(4*a*c*x*Sqrt[1 + c^2*x^2] + 2*b*ArcSinh[c*x]^2 - b*Cosh[2*ArcSi 
nh[c*x]] + 2*ArcSinh[c*x]*(2*a + b*Sinh[2*ArcSinh[c*x]])))/(8*c)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6200, 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.

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

\(\Big \downarrow \) 6200

\(\displaystyle \frac {1}{2} \sqrt {\pi } \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {c^2 x^2+1}}dx-\frac {1}{2} \sqrt {\pi } b c \int xdx+\frac {1}{2} x \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {1}{2} \sqrt {\pi } \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {c^2 x^2+1}}dx+\frac {1}{2} x \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))-\frac {1}{4} \sqrt {\pi } b c x^2\)

\(\Big \downarrow \) 6198

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

Input:

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

Output:

-1/4*(b*c*Sqrt[Pi]*x^2) + (x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/2 
 + (Sqrt[Pi]*(a + b*ArcSinh[c*x])^2)/(4*b*c)
 

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 6200
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_ 
Symbol] :> Simp[x*Sqrt[d + e*x^2]*((a + b*ArcSinh[c*x])^n/2), x] + (Simp[(1 
/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]]   Int[(a + b*ArcSinh[c*x])^n/Sq 
rt[1 + c^2*x^2], x], x] - Simp[b*c*(n/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2* 
x^2]]   Int[x*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e 
}, x] && EqQ[e, c^2*d] && GtQ[n, 0]
 
Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.49

method result size
default \(\frac {a x \sqrt {\pi \,c^{2} x^{2}+\pi }}{2}+\frac {a \pi \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{2 \sqrt {\pi \,c^{2}}}+\frac {b \sqrt {\pi }\, \left (2 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x c -c^{2} x^{2}+\operatorname {arcsinh}\left (x c \right )^{2}-1\right )}{4 c}\) \(100\)
parts \(\frac {a x \sqrt {\pi \,c^{2} x^{2}+\pi }}{2}+\frac {a \pi \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{2 \sqrt {\pi \,c^{2}}}+\frac {b \sqrt {\pi }\, \left (2 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x c -c^{2} x^{2}+\operatorname {arcsinh}\left (x c \right )^{2}-1\right )}{4 c}\) \(100\)

Input:

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

Output:

1/2*a*x*(Pi*c^2*x^2+Pi)^(1/2)+1/2*a*Pi*ln(Pi*c^2*x/(Pi*c^2)^(1/2)+(Pi*c^2* 
x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)+1/4*b*Pi^(1/2)*(2*arcsinh(x*c)*(c^2*x^2+1)^( 
1/2)*x*c-c^2*x^2+arcsinh(x*c)^2-1)/c
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

sqrt(pi)*(Integral(a*sqrt(c**2*x**2 + 1), x) + Integral(b*sqrt(c**2*x**2 + 
 1)*asinh(c*x), x))
 

Maxima [F(-2)]

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

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

Output:

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

Giac [F(-2)]

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

integrate((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
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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