\(\int \frac {x^5 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx\) [90]

Optimal result

Integrand size = 26, antiderivative size = 149 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=-\frac {8 b x}{15 c^5 \sqrt {\pi }}+\frac {4 b x^3}{45 c^3 \sqrt {\pi }}-\frac {b x^5}{25 c \sqrt {\pi }}+\frac {8 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{15 c^6 \pi }-\frac {4 x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{15 c^4 \pi }+\frac {x^4 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{5 c^2 \pi } \] Output:

-8/15*b*x/c^5/Pi^(1/2)+4/45*b*x^3/c^3/Pi^(1/2)-1/25*b*x^5/c/Pi^(1/2)+8/15* 
(Pi*c^2*x^2+Pi)^(1/2)*(a+b*arcsinh(c*x))/c^6/Pi-4/15*x^2*(Pi*c^2*x^2+Pi)^( 
1/2)*(a+b*arcsinh(c*x))/c^4/Pi+1/5*x^4*(Pi*c^2*x^2+Pi)^(1/2)*(a+b*arcsinh( 
c*x))/c^2/Pi
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.72 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\frac {15 a \sqrt {1+c^2 x^2} \left (8-4 c^2 x^2+3 c^4 x^4\right )+b \left (-120 c x+20 c^3 x^3-9 c^5 x^5\right )+15 b \sqrt {1+c^2 x^2} \left (8-4 c^2 x^2+3 c^4 x^4\right ) \text {arcsinh}(c x)}{225 c^6 \sqrt {\pi }} \] Input:

Integrate[(x^5*(a + b*ArcSinh[c*x]))/Sqrt[Pi + c^2*Pi*x^2],x]
 

Output:

(15*a*Sqrt[1 + c^2*x^2]*(8 - 4*c^2*x^2 + 3*c^4*x^4) + b*(-120*c*x + 20*c^3 
*x^3 - 9*c^5*x^5) + 15*b*Sqrt[1 + c^2*x^2]*(8 - 4*c^2*x^2 + 3*c^4*x^4)*Arc 
Sinh[c*x])/(225*c^6*Sqrt[Pi])
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {6227, 15, 6227, 15, 6213, 24}

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.

\(\displaystyle \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\sqrt {\pi c^2 x^2+\pi }} \, dx\)

\(\Big \downarrow \) 6227

\(\displaystyle -\frac {4 \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\sqrt {c^2 \pi x^2+\pi }}dx}{5 c^2}-\frac {b \int x^4dx}{5 \sqrt {\pi } c}+\frac {x^4 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{5 \pi c^2}\)

\(\Big \downarrow \) 15

\(\displaystyle -\frac {4 \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\sqrt {c^2 \pi x^2+\pi }}dx}{5 c^2}+\frac {x^4 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{5 \pi c^2}-\frac {b x^5}{25 \sqrt {\pi } c}\)

\(\Big \downarrow \) 6227

\(\displaystyle -\frac {4 \left (-\frac {2 \int \frac {x (a+b \text {arcsinh}(c x))}{\sqrt {c^2 \pi x^2+\pi }}dx}{3 c^2}-\frac {b \int x^2dx}{3 \sqrt {\pi } c}+\frac {x^2 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{3 \pi c^2}\right )}{5 c^2}+\frac {x^4 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{5 \pi c^2}-\frac {b x^5}{25 \sqrt {\pi } c}\)

\(\Big \downarrow \) 15

\(\displaystyle -\frac {4 \left (-\frac {2 \int \frac {x (a+b \text {arcsinh}(c x))}{\sqrt {c^2 \pi x^2+\pi }}dx}{3 c^2}+\frac {x^2 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{3 \pi c^2}-\frac {b x^3}{9 \sqrt {\pi } c}\right )}{5 c^2}+\frac {x^4 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{5 \pi c^2}-\frac {b x^5}{25 \sqrt {\pi } c}\)

\(\Big \downarrow \) 6213

\(\displaystyle -\frac {4 \left (-\frac {2 \left (\frac {\sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{\pi c^2}-\frac {b \int 1dx}{\sqrt {\pi } c}\right )}{3 c^2}+\frac {x^2 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{3 \pi c^2}-\frac {b x^3}{9 \sqrt {\pi } c}\right )}{5 c^2}+\frac {x^4 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{5 \pi c^2}-\frac {b x^5}{25 \sqrt {\pi } c}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {x^4 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{5 \pi c^2}-\frac {4 \left (\frac {x^2 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{3 \pi c^2}-\frac {2 \left (\frac {\sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{\pi c^2}-\frac {b x}{\sqrt {\pi } c}\right )}{3 c^2}-\frac {b x^3}{9 \sqrt {\pi } c}\right )}{5 c^2}-\frac {b x^5}{25 \sqrt {\pi } c}\)

Input:

Int[(x^5*(a + b*ArcSinh[c*x]))/Sqrt[Pi + c^2*Pi*x^2],x]
 

Output:

-1/25*(b*x^5)/(c*Sqrt[Pi]) + (x^4*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x 
]))/(5*c^2*Pi) - (4*(-1/9*(b*x^3)/(c*Sqrt[Pi]) + (x^2*Sqrt[Pi + c^2*Pi*x^2 
]*(a + b*ArcSinh[c*x]))/(3*c^2*Pi) - (2*(-((b*x)/(c*Sqrt[Pi])) + (Sqrt[Pi 
+ c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(c^2*Pi)))/(3*c^2)))/(5*c^2)
 

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 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 6213

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(2*e*(p 
+ 1))), x] - Simp[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p] 
 Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[ 
{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && NeQ[p, -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 = 1.33 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.15

method	result	size
orering	\(\frac {\left (81 c^{6} x^{6}-50 c^{4} x^{4}+440 c^{2} x^{2}+720\right ) \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}{225 c^{6} \sqrt {\pi \,c^{2} x^{2}+\pi }}-\frac {\left (9 c^{4} x^{4}-20 c^{2} x^{2}+120\right ) \left (c^{2} x^{2}+1\right ) \left (\frac {5 x^{4} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}{\sqrt {\pi \,c^{2} x^{2}+\pi }}+\frac {x^{5} b c}{\sqrt {c^{2} x^{2}+1}\, \sqrt {\pi \,c^{2} x^{2}+\pi }}-\frac {x^{6} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right ) \pi \,c^{2}}{\left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}\right )}{225 x^{4} c^{6}}\)	\(172\)
default	\(a \left (\frac {x^{4} \sqrt {\pi \,c^{2} x^{2}+\pi }}{5 \pi \,c^{2}}-\frac {4 \left (\frac {x^{2} \sqrt {\pi \,c^{2} x^{2}+\pi }}{3 \pi \,c^{2}}-\frac {2 \sqrt {\pi \,c^{2} x^{2}+\pi }}{3 \pi \,c^{4}}\right )}{5 c^{2}}\right )+\frac {b \left (45 \,\operatorname {arcsinh}\left (x c \right ) x^{6} c^{6}-15 \,\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}-9 \sqrt {c^{2} x^{2}+1}\, x^{5} c^{5}+60 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}+20 \sqrt {c^{2} x^{2}+1}\, c^{3} x^{3}+120 \,\operatorname {arcsinh}\left (x c \right )-120 \sqrt {c^{2} x^{2}+1}\, x c \right )}{225 \sqrt {\pi }\, c^{6} \sqrt {c^{2} x^{2}+1}}\)	\(193\)
parts	\(a \left (\frac {x^{4} \sqrt {\pi \,c^{2} x^{2}+\pi }}{5 \pi \,c^{2}}-\frac {4 \left (\frac {x^{2} \sqrt {\pi \,c^{2} x^{2}+\pi }}{3 \pi \,c^{2}}-\frac {2 \sqrt {\pi \,c^{2} x^{2}+\pi }}{3 \pi \,c^{4}}\right )}{5 c^{2}}\right )+\frac {b \left (45 \,\operatorname {arcsinh}\left (x c \right ) x^{6} c^{6}-15 \,\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}-9 \sqrt {c^{2} x^{2}+1}\, x^{5} c^{5}+60 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}+20 \sqrt {c^{2} x^{2}+1}\, c^{3} x^{3}+120 \,\operatorname {arcsinh}\left (x c \right )-120 \sqrt {c^{2} x^{2}+1}\, x c \right )}{225 \sqrt {\pi }\, c^{6} \sqrt {c^{2} x^{2}+1}}\)	\(193\)

Input:

int(x^5*(a+b*arcsinh(x*c))/(Pi*c^2*x^2+Pi)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/225*(81*c^6*x^6-50*c^4*x^4+440*c^2*x^2+720)/c^6*(a+b*arcsinh(x*c))/(Pi*c 
^2*x^2+Pi)^(1/2)-1/225/x^4*(9*c^4*x^4-20*c^2*x^2+120)/c^6*(c^2*x^2+1)*(5*x 
^4*(a+b*arcsinh(x*c))/(Pi*c^2*x^2+Pi)^(1/2)+x^5*b*c/(c^2*x^2+1)^(1/2)/(Pi* 
c^2*x^2+Pi)^(1/2)-x^6*(a+b*arcsinh(x*c))/(Pi*c^2*x^2+Pi)^(3/2)*Pi*c^2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.08 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\frac {15 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (3 \, b c^{6} x^{6} - b c^{4} x^{4} + 4 \, b c^{2} x^{2} + 8 \, b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + \sqrt {\pi + \pi c^{2} x^{2}} {\left (45 \, a c^{6} x^{6} - 15 \, a c^{4} x^{4} + 60 \, a c^{2} x^{2} - {\left (9 \, b c^{5} x^{5} - 20 \, b c^{3} x^{3} + 120 \, b c x\right )} \sqrt {c^{2} x^{2} + 1} + 120 \, a\right )}}{225 \, {\left (\pi c^{8} x^{2} + \pi c^{6}\right )}} \] Input:

integrate(x^5*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="frica 
s")
 

Output:

1/225*(15*sqrt(pi + pi*c^2*x^2)*(3*b*c^6*x^6 - b*c^4*x^4 + 4*b*c^2*x^2 + 8 
*b)*log(c*x + sqrt(c^2*x^2 + 1)) + sqrt(pi + pi*c^2*x^2)*(45*a*c^6*x^6 - 1 
5*a*c^4*x^4 + 60*a*c^2*x^2 - (9*b*c^5*x^5 - 20*b*c^3*x^3 + 120*b*c*x)*sqrt 
(c^2*x^2 + 1) + 120*a))/(pi*c^8*x^2 + pi*c^6)
 

Sympy [A] (verification not implemented)

Time = 10.68 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.23 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\frac {a \left (\begin {cases} \frac {x^{4} \sqrt {c^{2} x^{2} + 1}}{5 c^{2}} - \frac {4 x^{2} \sqrt {c^{2} x^{2} + 1}}{15 c^{4}} + \frac {8 \sqrt {c^{2} x^{2} + 1}}{15 c^{6}} & \text {for}\: c^{2} \neq 0 \\\frac {x^{6}}{6} & \text {otherwise} \end {cases}\right )}{\sqrt {\pi }} + \frac {b \left (\begin {cases} - \frac {x^{5}}{25 c} + \frac {x^{4} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{5 c^{2}} + \frac {4 x^{3}}{45 c^{3}} - \frac {4 x^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{15 c^{4}} - \frac {8 x}{15 c^{5}} + \frac {8 \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{15 c^{6}} & \text {for}\: c \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{\sqrt {\pi }} \] Input:

integrate(x**5*(a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(1/2),x)
 

Output:

a*Piecewise((x**4*sqrt(c**2*x**2 + 1)/(5*c**2) - 4*x**2*sqrt(c**2*x**2 + 1 
)/(15*c**4) + 8*sqrt(c**2*x**2 + 1)/(15*c**6), Ne(c**2, 0)), (x**6/6, True 
))/sqrt(pi) + b*Piecewise((-x**5/(25*c) + x**4*sqrt(c**2*x**2 + 1)*asinh(c 
*x)/(5*c**2) + 4*x**3/(45*c**3) - 4*x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/(1 
5*c**4) - 8*x/(15*c**5) + 8*sqrt(c**2*x**2 + 1)*asinh(c*x)/(15*c**6), Ne(c 
, 0)), (0, True))/sqrt(pi)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.17 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\frac {1}{15} \, {\left (\frac {3 \, \sqrt {\pi + \pi c^{2} x^{2}} x^{4}}{\pi c^{2}} - \frac {4 \, \sqrt {\pi + \pi c^{2} x^{2}} x^{2}}{\pi c^{4}} + \frac {8 \, \sqrt {\pi + \pi c^{2} x^{2}}}{\pi c^{6}}\right )} b \operatorname {arsinh}\left (c x\right ) + \frac {1}{15} \, {\left (\frac {3 \, \sqrt {\pi + \pi c^{2} x^{2}} x^{4}}{\pi c^{2}} - \frac {4 \, \sqrt {\pi + \pi c^{2} x^{2}} x^{2}}{\pi c^{4}} + \frac {8 \, \sqrt {\pi + \pi c^{2} x^{2}}}{\pi c^{6}}\right )} a - \frac {{\left (9 \, c^{4} x^{5} - 20 \, c^{2} x^{3} + 120 \, x\right )} b}{225 \, \sqrt {\pi } c^{5}} \] Input:

integrate(x^5*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="maxim 
a")
 

Output:

1/15*(3*sqrt(pi + pi*c^2*x^2)*x^4/(pi*c^2) - 4*sqrt(pi + pi*c^2*x^2)*x^2/( 
pi*c^4) + 8*sqrt(pi + pi*c^2*x^2)/(pi*c^6))*b*arcsinh(c*x) + 1/15*(3*sqrt( 
pi + pi*c^2*x^2)*x^4/(pi*c^2) - 4*sqrt(pi + pi*c^2*x^2)*x^2/(pi*c^4) + 8*s 
qrt(pi + pi*c^2*x^2)/(pi*c^6))*a - 1/225*(9*c^4*x^5 - 20*c^2*x^3 + 120*x)* 
b/(sqrt(pi)*c^5)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^5*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),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. \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\int \frac {x^5\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{\sqrt {\Pi \,c^2\,x^2+\Pi }} \,d x \] Input:

int((x^5*(a + b*asinh(c*x)))/(Pi + Pi*c^2*x^2)^(1/2),x)
 

Output:

int((x^5*(a + b*asinh(c*x)))/(Pi + Pi*c^2*x^2)^(1/2), x)
 

Reduce [F]

\[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\frac {3 \sqrt {c^{2} x^{2}+1}\, a \,c^{4} x^{4}-4 \sqrt {c^{2} x^{2}+1}\, a \,c^{2} x^{2}+8 \sqrt {c^{2} x^{2}+1}\, a +15 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{5}}{\sqrt {c^{2} x^{2}+1}}d x \right ) b \,c^{6}}{15 \sqrt {\pi }\, c^{6}} \] Input:

int(x^5*(a+b*asinh(c*x))/(Pi*c^2*x^2+Pi)^(1/2),x)
 

Output:

(3*sqrt(c**2*x**2 + 1)*a*c**4*x**4 - 4*sqrt(c**2*x**2 + 1)*a*c**2*x**2 + 8 
*sqrt(c**2*x**2 + 1)*a + 15*int((asinh(c*x)*x**5)/sqrt(c**2*x**2 + 1),x)*b 
*c**6)/(15*sqrt(pi)*c**6)