\(\int \frac {x^5 (a+b \text {arcsinh}(c x))}{(\pi +c^2 \pi x^2)^{5/2}} \, dx\) [111]

Optimal result

Integrand size = 26, antiderivative size = 146 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=-\frac {b x}{c^5 \pi ^{5/2}}+\frac {b x}{6 c^5 \pi ^{5/2} \left (1+c^2 x^2\right )}-\frac {a+b \text {arcsinh}(c x)}{3 c^6 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {2 (a+b \text {arcsinh}(c x))}{c^6 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {\sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{c^6 \pi ^3}-\frac {11 b \arctan (c x)}{6 c^6 \pi ^{5/2}} \] Output:

-b*x/c^5/Pi^(5/2)+1/6*b*x/c^5/Pi^(5/2)/(c^2*x^2+1)-1/3*(a+b*arcsinh(c*x))/ 
c^6/Pi/(Pi*c^2*x^2+Pi)^(3/2)+2*(a+b*arcsinh(c*x))/c^6/Pi^2/(Pi*c^2*x^2+Pi) 
^(1/2)+(Pi*c^2*x^2+Pi)^(1/2)*(a+b*arcsinh(c*x))/c^6/Pi^3-11/6*b*arctan(c*x 
)/c^6/Pi^(5/2)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.90 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\frac {16 a+24 a c^2 x^2+6 a c^4 x^4-5 b c x \sqrt {1+c^2 x^2}-6 b c^3 x^3 \sqrt {1+c^2 x^2}+2 b \left (8+12 c^2 x^2+3 c^4 x^4\right ) \text {arcsinh}(c x)-11 b \left (1+c^2 x^2\right )^{3/2} \arctan (c x)}{6 c^6 \pi ^{5/2} \left (1+c^2 x^2\right )^{3/2}} \] Input:

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

Output:

(16*a + 24*a*c^2*x^2 + 6*a*c^4*x^4 - 5*b*c*x*Sqrt[1 + c^2*x^2] - 6*b*c^3*x 
^3*Sqrt[1 + c^2*x^2] + 2*b*(8 + 12*c^2*x^2 + 3*c^4*x^4)*ArcSinh[c*x] - 11* 
b*(1 + c^2*x^2)^(3/2)*ArcTan[c*x])/(6*c^6*Pi^(5/2)*(1 + c^2*x^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {6219, 27, 1471, 25, 299, 216}

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.

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

\(\Big \downarrow \) 6219

\(\displaystyle -\sqrt {\pi } b c \int \frac {3 c^4 x^4+12 c^2 x^2+8}{3 c^6 \pi ^3 \left (c^2 x^2+1\right )^2}dx+\frac {\sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{\pi ^3 c^6}+\frac {2 (a+b \text {arcsinh}(c x))}{\pi ^2 c^6 \sqrt {\pi c^2 x^2+\pi }}-\frac {a+b \text {arcsinh}(c x)}{3 \pi c^6 \left (\pi c^2 x^2+\pi \right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {b \int \frac {3 c^4 x^4+12 c^2 x^2+8}{\left (c^2 x^2+1\right )^2}dx}{3 \pi ^{5/2} c^5}+\frac {\sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{\pi ^3 c^6}+\frac {2 (a+b \text {arcsinh}(c x))}{\pi ^2 c^6 \sqrt {\pi c^2 x^2+\pi }}-\frac {a+b \text {arcsinh}(c x)}{3 \pi c^6 \left (\pi c^2 x^2+\pi \right )^{3/2}}\)

\(\Big \downarrow \) 1471

\(\displaystyle -\frac {b \left (-\frac {1}{2} \int -\frac {6 c^2 x^2+17}{c^2 x^2+1}dx-\frac {x}{2 \left (c^2 x^2+1\right )}\right )}{3 \pi ^{5/2} c^5}+\frac {\sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{\pi ^3 c^6}+\frac {2 (a+b \text {arcsinh}(c x))}{\pi ^2 c^6 \sqrt {\pi c^2 x^2+\pi }}-\frac {a+b \text {arcsinh}(c x)}{3 \pi c^6 \left (\pi c^2 x^2+\pi \right )^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {b \left (\frac {1}{2} \int \frac {6 c^2 x^2+17}{c^2 x^2+1}dx-\frac {x}{2 \left (c^2 x^2+1\right )}\right )}{3 \pi ^{5/2} c^5}+\frac {\sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{\pi ^3 c^6}+\frac {2 (a+b \text {arcsinh}(c x))}{\pi ^2 c^6 \sqrt {\pi c^2 x^2+\pi }}-\frac {a+b \text {arcsinh}(c x)}{3 \pi c^6 \left (\pi c^2 x^2+\pi \right )^{3/2}}\)

\(\Big \downarrow \) 299

\(\displaystyle -\frac {b \left (\frac {1}{2} \left (11 \int \frac {1}{c^2 x^2+1}dx+6 x\right )-\frac {x}{2 \left (c^2 x^2+1\right )}\right )}{3 \pi ^{5/2} c^5}+\frac {\sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{\pi ^3 c^6}+\frac {2 (a+b \text {arcsinh}(c x))}{\pi ^2 c^6 \sqrt {\pi c^2 x^2+\pi }}-\frac {a+b \text {arcsinh}(c x)}{3 \pi c^6 \left (\pi c^2 x^2+\pi \right )^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{\pi ^3 c^6}+\frac {2 (a+b \text {arcsinh}(c x))}{\pi ^2 c^6 \sqrt {\pi c^2 x^2+\pi }}-\frac {a+b \text {arcsinh}(c x)}{3 \pi c^6 \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {b \left (\frac {1}{2} \left (\frac {11 \arctan (c x)}{c}+6 x\right )-\frac {x}{2 \left (c^2 x^2+1\right )}\right )}{3 \pi ^{5/2} c^5}\)

Input:

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

Output:

-1/3*(a + b*ArcSinh[c*x])/(c^6*Pi*(Pi + c^2*Pi*x^2)^(3/2)) + (2*(a + b*Arc 
Sinh[c*x]))/(c^6*Pi^2*Sqrt[Pi + c^2*Pi*x^2]) + (Sqrt[Pi + c^2*Pi*x^2]*(a + 
 b*ArcSinh[c*x]))/(c^6*Pi^3) - (b*(-1/2*x/(1 + c^2*x^2) + (6*x + (11*ArcTa 
n[c*x])/c)/2))/(3*c^5*Pi^(5/2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 
, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x 
, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q 
 + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), 
x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 
2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
 

rule 6219

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])
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.09 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.62

method	result	size
default	\(a \left (\frac {x^{4}}{\pi \,c^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}-\frac {4 \left (-\frac {x^{2}}{\pi \,c^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}-\frac {2}{3 \pi \,c^{4} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}\right )}{c^{2}}\right )+\frac {b \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right )}{\pi ^{\frac {5}{2}} c^{6}}-\frac {b x}{c^{5} \pi ^{\frac {5}{2}}}+\frac {2 b \,\operatorname {arcsinh}\left (x c \right ) x^{2}}{\pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} c^{4}}+\frac {b x}{6 c^{5} \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )}+\frac {5 b \,\operatorname {arcsinh}\left (x c \right )}{3 \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} c^{6}}+\frac {11 i b \ln \left (x c +\sqrt {c^{2} x^{2}+1}-i\right )}{6 \pi ^{\frac {5}{2}} c^{6}}-\frac {11 i b \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right )}{6 \pi ^{\frac {5}{2}} c^{6}}\)	\(237\)
parts	\(a \left (\frac {x^{4}}{\pi \,c^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}-\frac {4 \left (-\frac {x^{2}}{\pi \,c^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}-\frac {2}{3 \pi \,c^{4} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}\right )}{c^{2}}\right )+\frac {b \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right )}{\pi ^{\frac {5}{2}} c^{6}}-\frac {b x}{c^{5} \pi ^{\frac {5}{2}}}+\frac {2 b \,\operatorname {arcsinh}\left (x c \right ) x^{2}}{\pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} c^{4}}+\frac {b x}{6 c^{5} \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )}+\frac {5 b \,\operatorname {arcsinh}\left (x c \right )}{3 \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} c^{6}}+\frac {11 i b \ln \left (x c +\sqrt {c^{2} x^{2}+1}-i\right )}{6 \pi ^{\frac {5}{2}} c^{6}}-\frac {11 i b \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right )}{6 \pi ^{\frac {5}{2}} c^{6}}\)	\(237\)

Input:

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

Output:

a*(x^4/Pi/c^2/(Pi*c^2*x^2+Pi)^(3/2)-4/c^2*(-x^2/Pi/c^2/(Pi*c^2*x^2+Pi)^(3/ 
2)-2/3/Pi/c^4/(Pi*c^2*x^2+Pi)^(3/2)))+b/Pi^(5/2)/c^6*(c^2*x^2+1)^(1/2)*arc 
sinh(x*c)-b*x/c^5/Pi^(5/2)+2*b/Pi^(5/2)/(c^2*x^2+1)^(3/2)/c^4*arcsinh(x*c) 
*x^2+1/6*b*x/c^5/Pi^(5/2)/(c^2*x^2+1)+5/3*b/Pi^(5/2)/(c^2*x^2+1)^(3/2)/c^6 
*arcsinh(x*c)+11/6*I*b/Pi^(5/2)/c^6*ln(x*c+(c^2*x^2+1)^(1/2)-I)-11/6*I*b/P 
i^(5/2)/c^6*ln(x*c+(c^2*x^2+1)^(1/2)+I)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.49 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\frac {11 \, \sqrt {\pi } {\left (b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b\right )} \arctan \left (-\frac {2 \, \sqrt {\pi } \sqrt {\pi + \pi c^{2} x^{2}} \sqrt {c^{2} x^{2} + 1} c x}{\pi - \pi c^{4} x^{4}}\right ) + 4 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (3 \, b c^{4} x^{4} + 12 \, b c^{2} x^{2} + 8 \, b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (6 \, a c^{4} x^{4} + 24 \, a c^{2} x^{2} - {\left (6 \, b c^{3} x^{3} + 5 \, b c x\right )} \sqrt {c^{2} x^{2} + 1} + 16 \, a\right )}}{12 \, {\left (\pi ^{3} c^{10} x^{4} + 2 \, \pi ^{3} c^{8} x^{2} + \pi ^{3} c^{6}\right )}} \] Input:

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

Output:

1/12*(11*sqrt(pi)*(b*c^4*x^4 + 2*b*c^2*x^2 + b)*arctan(-2*sqrt(pi)*sqrt(pi 
 + pi*c^2*x^2)*sqrt(c^2*x^2 + 1)*c*x/(pi - pi*c^4*x^4)) + 4*sqrt(pi + pi*c 
^2*x^2)*(3*b*c^4*x^4 + 12*b*c^2*x^2 + 8*b)*log(c*x + sqrt(c^2*x^2 + 1)) + 
2*sqrt(pi + pi*c^2*x^2)*(6*a*c^4*x^4 + 24*a*c^2*x^2 - (6*b*c^3*x^3 + 5*b*c 
*x)*sqrt(c^2*x^2 + 1) + 16*a))/(pi^3*c^10*x^4 + 2*pi^3*c^8*x^2 + pi^3*c^6)
 

Sympy [F]

\[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\frac {\int \frac {a x^{5}}{c^{4} x^{4} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b x^{5} \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {5}{2}}} \] Input:

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

Output:

(Integral(a*x**5/(c**4*x**4*sqrt(c**2*x**2 + 1) + 2*c**2*x**2*sqrt(c**2*x* 
*2 + 1) + sqrt(c**2*x**2 + 1)), x) + Integral(b*x**5*asinh(c*x)/(c**4*x**4 
*sqrt(c**2*x**2 + 1) + 2*c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2 + 
1)), x))/pi**(5/2)
 

Maxima [F]

\[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{5}}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

1/3*b*((3*sqrt(pi)*c^4*x^4 + 12*sqrt(pi)*c^2*x^2 + 8*sqrt(pi))*log(c*x + s 
qrt(c^2*x^2 + 1))/((pi^3*c^8*x^2 + pi^3*c^6)*sqrt(c^2*x^2 + 1)) + 3*integr 
ate(1/3*(3*sqrt(pi)*c^4*x^4 + 12*sqrt(pi)*c^2*x^2 + 8*sqrt(pi))/(pi^3*c^11 
*x^6 + 2*pi^3*c^9*x^4 + pi^3*c^7*x^2 + (pi^3*c^10*x^5 + 2*pi^3*c^8*x^3 + p 
i^3*c^6*x)*sqrt(c^2*x^2 + 1)), x) - 3*integrate(1/3*(3*sqrt(pi)*c^4*x^4 + 
12*sqrt(pi)*c^2*x^2 + 8*sqrt(pi))/((pi^3*c^8*x^3 + pi^3*c^6*x)*sqrt(c^2*x^ 
2 + 1)), x)) + 1/3*a*(3*x^4/(pi*(pi + pi*c^2*x^2)^(3/2)*c^2) + 12*x^2/(pi* 
(pi + pi*c^2*x^2)^(3/2)*c^4) + 8/(pi*(pi + pi*c^2*x^2)^(3/2)*c^6))
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\frac {3 \sqrt {c^{2} x^{2}+1}\, a \,c^{4} x^{4}+12 \sqrt {c^{2} x^{2}+1}\, a \,c^{2} x^{2}+8 \sqrt {c^{2} x^{2}+1}\, a +3 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{5}}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{4}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}}d x \right ) b \,c^{10} x^{4}+6 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{5}}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{4}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}}d x \right ) b \,c^{8} x^{2}+3 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{5}}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{4}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}}d x \right ) b \,c^{6}}{3 \sqrt {\pi }\, c^{6} \pi ^{2} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )} \] Input:

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

Output:

(3*sqrt(c**2*x**2 + 1)*a*c**4*x**4 + 12*sqrt(c**2*x**2 + 1)*a*c**2*x**2 + 
8*sqrt(c**2*x**2 + 1)*a + 3*int((asinh(c*x)*x**5)/(sqrt(c**2*x**2 + 1)*c** 
4*x**4 + 2*sqrt(c**2*x**2 + 1)*c**2*x**2 + sqrt(c**2*x**2 + 1)),x)*b*c**10 
*x**4 + 6*int((asinh(c*x)*x**5)/(sqrt(c**2*x**2 + 1)*c**4*x**4 + 2*sqrt(c* 
*2*x**2 + 1)*c**2*x**2 + sqrt(c**2*x**2 + 1)),x)*b*c**8*x**2 + 3*int((asin 
h(c*x)*x**5)/(sqrt(c**2*x**2 + 1)*c**4*x**4 + 2*sqrt(c**2*x**2 + 1)*c**2*x 
**2 + sqrt(c**2*x**2 + 1)),x)*b*c**6)/(3*sqrt(pi)*c**6*pi**2*(c**4*x**4 + 
2*c**2*x**2 + 1))