\(\int \frac {a+b \text {arcsinh}(c x)}{x^4 (\pi +c^2 \pi x^2)^{5/2}} \, dx\) [120]

Optimal result

Integrand size = 26, antiderivative size = 206 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=-\frac {b c}{6 \pi ^{5/2} x^2}+\frac {b c^3}{6 \pi ^{5/2} \left (1+c^2 x^2\right )}+\frac {a+b \text {arcsinh}(c x)}{3 \pi x^3 \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {2 (a+b \text {arcsinh}(c x))}{\pi ^2 x^3 \sqrt {\pi +c^2 \pi x^2}}-\frac {8 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{3 \pi ^3 x^3}+\frac {16 c^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{3 \pi ^3 x}-\frac {8 b c^3 \log (x)}{3 \pi ^{5/2}}-\frac {4 b c^3 \log \left (1+c^2 x^2\right )}{3 \pi ^{5/2}} \] Output:

-1/6*b*c/Pi^(5/2)/x^2+1/6*b*c^3/Pi^(5/2)/(c^2*x^2+1)+1/3*(a+b*arcsinh(c*x) 
)/Pi/x^3/(Pi*c^2*x^2+Pi)^(3/2)+2*(a+b*arcsinh(c*x))/Pi^2/x^3/(Pi*c^2*x^2+P 
i)^(1/2)-8/3*(Pi*c^2*x^2+Pi)^(1/2)*(a+b*arcsinh(c*x))/Pi^3/x^3+16/3*c^2*(P 
i*c^2*x^2+Pi)^(1/2)*(a+b*arcsinh(c*x))/Pi^3/x-8/3*b*c^3*ln(x)/Pi^(5/2)-4/3 
*b*c^3*ln(c^2*x^2+1)/Pi^(5/2)
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.16 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\frac {-2 a+12 a c^2 x^2+48 a c^4 x^4+32 a c^6 x^6-b c x \sqrt {1+c^2 x^2}-32 b c^3 x^3 \sqrt {1+c^2 x^2}-32 b c^5 x^5 \sqrt {1+c^2 x^2}+2 b \left (-1+6 c^2 x^2+24 c^4 x^4+16 c^6 x^6\right ) \text {arcsinh}(c x)-16 b c^3 x^3 \left (1+c^2 x^2\right )^{3/2} \log (x)-8 b c^3 x^3 \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )-8 b c^5 x^5 \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{6 \pi ^{5/2} x^3 \left (1+c^2 x^2\right )^{3/2}} \] Input:

Integrate[(a + b*ArcSinh[c*x])/(x^4*(Pi + c^2*Pi*x^2)^(5/2)),x]
 

Output:

(-2*a + 12*a*c^2*x^2 + 48*a*c^4*x^4 + 32*a*c^6*x^6 - b*c*x*Sqrt[1 + c^2*x^ 
2] - 32*b*c^3*x^3*Sqrt[1 + c^2*x^2] - 32*b*c^5*x^5*Sqrt[1 + c^2*x^2] + 2*b 
*(-1 + 6*c^2*x^2 + 24*c^4*x^4 + 16*c^6*x^6)*ArcSinh[c*x] - 16*b*c^3*x^3*(1 
 + c^2*x^2)^(3/2)*Log[x] - 8*b*c^3*x^3*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2] 
- 8*b*c^5*x^5*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2])/(6*Pi^(5/2)*x^3*(1 + c^2 
*x^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {6219, 27, 2331, 2123, 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.

\(\displaystyle \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (\pi c^2 x^2+\pi \right )^{5/2}} \, dx\)

\(\Big \downarrow \) 6219

\(\displaystyle -\sqrt {\pi } b c \int -\frac {-16 c^6 x^6-24 c^4 x^4-6 c^2 x^2+1}{3 \pi ^3 x^3 \left (c^2 x^2+1\right )^2}dx+\frac {2 c^2 (a+b \text {arcsinh}(c x))}{\pi x \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{3 \pi x^3 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi ^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b c \int \frac {-16 c^6 x^6-24 c^4 x^4-6 c^2 x^2+1}{x^3 \left (c^2 x^2+1\right )^2}dx}{3 \pi ^{5/2}}+\frac {2 c^2 (a+b \text {arcsinh}(c x))}{\pi x \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{3 \pi x^3 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi ^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}\)

\(\Big \downarrow \) 2331

\(\displaystyle \frac {b c \int \frac {-16 c^6 x^6-24 c^4 x^4-6 c^2 x^2+1}{x^4 \left (c^2 x^2+1\right )^2}dx^2}{6 \pi ^{5/2}}+\frac {2 c^2 (a+b \text {arcsinh}(c x))}{\pi x \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{3 \pi x^3 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi ^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}\)

\(\Big \downarrow \) 2123

\(\displaystyle \frac {b c \int \left (-\frac {8 c^4}{c^2 x^2+1}-\frac {c^4}{\left (c^2 x^2+1\right )^2}-\frac {8 c^2}{x^2}+\frac {1}{x^4}\right )dx^2}{6 \pi ^{5/2}}+\frac {2 c^2 (a+b \text {arcsinh}(c x))}{\pi x \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{3 \pi x^3 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi ^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 c^2 (a+b \text {arcsinh}(c x))}{\pi x \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{3 \pi x^3 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi ^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {b c \left (\frac {c^2}{c^2 x^2+1}-8 c^2 \log \left (x^2\right )-8 c^2 \log \left (c^2 x^2+1\right )-\frac {1}{x^2}\right )}{6 \pi ^{5/2}}\)

Input:

Int[(a + b*ArcSinh[c*x])/(x^4*(Pi + c^2*Pi*x^2)^(5/2)),x]
 

Output:

-1/3*(a + b*ArcSinh[c*x])/(Pi*x^3*(Pi + c^2*Pi*x^2)^(3/2)) + (2*c^2*(a + b 
*ArcSinh[c*x]))/(Pi*x*(Pi + c^2*Pi*x^2)^(3/2)) + (8*c^4*x*(a + b*ArcSinh[c 
*x]))/(3*Pi*(Pi + c^2*Pi*x^2)^(3/2)) + (16*c^4*x*(a + b*ArcSinh[c*x]))/(3* 
Pi^2*Sqrt[Pi + c^2*Pi*x^2]) + (b*c*(-x^(-2) + c^2/(1 + c^2*x^2) - 8*c^2*Lo 
g[x^2] - 8*c^2*Log[1 + c^2*x^2]))/(6*Pi^(5/2))
 

Defintions of rubi rules used

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2123

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] 
:> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c 
, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
 

rule 2331

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/2   S 
ubst[Int[x^((m - 1)/2)*SubstFor[x^2, Pq, x]*(a + b*x)^p, x], x, x^2], x] /; 
 FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1154\) vs. \(2(176)=352\).

Time = 1.19 (sec) , antiderivative size = 1155, normalized size of antiderivative = 5.61

method	result	size
default	\(\text {Expression too large to display}\)	\(1155\)
parts	\(\text {Expression too large to display}\)	\(1155\)

Input:

int((a+b*arcsinh(x*c))/x^4/(Pi*c^2*x^2+Pi)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

32/3*b/Pi^(5/2)*c^3*arcsinh(x*c)-8/3*b/Pi^(5/2)*c^3*ln((x*c+(c^2*x^2+1)^(1 
/2))^4-1)+16/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*arcsinh( 
x*c)*c^3-128/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^12*c^1 
5+1/6*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)/x^2*c+1/3*b/Pi^(5/2 
)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^(3/2)/x^3*arcsinh(x*c)-2*b/Pi^(5/2 
)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)*x^2*c^5+128/3*b/Pi^(5/2)/(12*c^4*x 
^4+12*c^2*x^2-1)/(c^2*x^2+1)*x^4*c^7+128*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2 
-1)/(c^2*x^2+1)*x^6*c^9+128/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^ 
2+1)*x^10*c^13+128*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)*x^8*c^ 
11-2*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)*c^3+64*b/Pi^(5/2)/(1 
2*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^(3/2)*x^7*arcsinh(x*c)*c^10+160*b/Pi^( 
5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^(3/2)*x^5*arcsinh(x*c)*c^8+344/ 
3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^(3/2)*x^3*arcsinh(x*c)* 
c^6+12*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^(3/2)*x*arcsinh(x* 
c)*c^4-6*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^(3/2)/x*arcsinh( 
x*c)*c^2-160/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^2*arcs 
inh(x*c)*c^5-560/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^4* 
arcsinh(x*c)*c^7-192*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^ 
6*arcsinh(x*c)*c^9-64*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x 
^8*arcsinh(x*c)*c^11-256*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+...
 

Fricas [F]

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

integrate((a+b*arcsinh(c*x))/x^4/(pi*c^2*x^2+pi)^(5/2),x, algorithm="frica 
s")
 

Output:

integral(sqrt(pi + pi*c^2*x^2)*(b*arcsinh(c*x) + a)/(pi^3*c^6*x^10 + 3*pi^ 
3*c^4*x^8 + 3*pi^3*c^2*x^6 + pi^3*x^4), x)
                                                                                    
                                                                                    
 

Sympy [F]

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

integrate((a+b*asinh(c*x))/x**4/(pi*c**2*x**2+pi)**(5/2),x)
 

Output:

(Integral(a/(c**4*x**8*sqrt(c**2*x**2 + 1) + 2*c**2*x**6*sqrt(c**2*x**2 + 
1) + x**4*sqrt(c**2*x**2 + 1)), x) + Integral(b*asinh(c*x)/(c**4*x**8*sqrt 
(c**2*x**2 + 1) + 2*c**2*x**6*sqrt(c**2*x**2 + 1) + x**4*sqrt(c**2*x**2 + 
1)), x))/pi**(5/2)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.15 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=-\frac {1}{6} \, b c {\left (\frac {8 \, c^{2} \log \left (c^{2} x^{2} + 1\right )}{\pi ^{\frac {5}{2}}} + \frac {16 \, c^{2} \log \left (x\right )}{\pi ^{\frac {5}{2}}} + \frac {1}{\pi ^{\frac {5}{2}} c^{2} x^{4} + \pi ^{\frac {5}{2}} x^{2}}\right )} + \frac {1}{3} \, {\left (\frac {8 \, c^{4} x}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}} + \frac {16 \, c^{4} x}{\pi ^{2} \sqrt {\pi + \pi c^{2} x^{2}}} + \frac {6 \, c^{2}}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x} - \frac {1}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x^{3}}\right )} b \operatorname {arsinh}\left (c x\right ) + \frac {1}{3} \, {\left (\frac {8 \, c^{4} x}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}} + \frac {16 \, c^{4} x}{\pi ^{2} \sqrt {\pi + \pi c^{2} x^{2}}} + \frac {6 \, c^{2}}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x} - \frac {1}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x^{3}}\right )} a \] Input:

integrate((a+b*arcsinh(c*x))/x^4/(pi*c^2*x^2+pi)^(5/2),x, algorithm="maxim 
a")
 

Output:

-1/6*b*c*(8*c^2*log(c^2*x^2 + 1)/pi^(5/2) + 16*c^2*log(x)/pi^(5/2) + 1/(pi 
^(5/2)*c^2*x^4 + pi^(5/2)*x^2)) + 1/3*(8*c^4*x/(pi*(pi + pi*c^2*x^2)^(3/2) 
) + 16*c^4*x/(pi^2*sqrt(pi + pi*c^2*x^2)) + 6*c^2/(pi*(pi + pi*c^2*x^2)^(3 
/2)*x) - 1/(pi*(pi + pi*c^2*x^2)^(3/2)*x^3))*b*arcsinh(c*x) + 1/3*(8*c^4*x 
/(pi*(pi + pi*c^2*x^2)^(3/2)) + 16*c^4*x/(pi^2*sqrt(pi + pi*c^2*x^2)) + 6* 
c^2/(pi*(pi + pi*c^2*x^2)^(3/2)*x) - 1/(pi*(pi + pi*c^2*x^2)^(3/2)*x^3))*a
 

Giac [F]

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

integrate((a+b*arcsinh(c*x))/x^4/(pi*c^2*x^2+pi)^(5/2),x, algorithm="giac" 
)
 

Output:

integrate((b*arcsinh(c*x) + a)/((pi + pi*c^2*x^2)^(5/2)*x^4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{x^4\,{\left (\Pi \,c^2\,x^2+\Pi \right )}^{5/2}} \,d x \] Input:

int((a + b*asinh(c*x))/(x^4*(Pi + Pi*c^2*x^2)^(5/2)),x)
 

Output:

int((a + b*asinh(c*x))/(x^4*(Pi + Pi*c^2*x^2)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\frac {16 \sqrt {c^{2} x^{2}+1}\, a \,c^{6} x^{6}+24 \sqrt {c^{2} x^{2}+1}\, a \,c^{4} x^{4}+6 \sqrt {c^{2} x^{2}+1}\, a \,c^{2} x^{2}-\sqrt {c^{2} x^{2}+1}\, a +3 \left (\int \frac {\mathit {asinh} \left (c x \right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{8}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{6}+\sqrt {c^{2} x^{2}+1}\, x^{4}}d x \right ) b \,c^{4} x^{7}+6 \left (\int \frac {\mathit {asinh} \left (c x \right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{8}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{6}+\sqrt {c^{2} x^{2}+1}\, x^{4}}d x \right ) b \,c^{2} x^{5}+3 \left (\int \frac {\mathit {asinh} \left (c x \right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{8}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{6}+\sqrt {c^{2} x^{2}+1}\, x^{4}}d x \right ) b \,x^{3}-16 a \,c^{7} x^{7}-32 a \,c^{5} x^{5}-16 a \,c^{3} x^{3}}{3 \sqrt {\pi }\, \pi ^{2} x^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )} \] Input:

int((a+b*asinh(c*x))/x^4/(Pi*c^2*x^2+Pi)^(5/2),x)
 

Output:

(16*sqrt(c**2*x**2 + 1)*a*c**6*x**6 + 24*sqrt(c**2*x**2 + 1)*a*c**4*x**4 + 
 6*sqrt(c**2*x**2 + 1)*a*c**2*x**2 - sqrt(c**2*x**2 + 1)*a + 3*int(asinh(c 
*x)/(sqrt(c**2*x**2 + 1)*c**4*x**8 + 2*sqrt(c**2*x**2 + 1)*c**2*x**6 + sqr 
t(c**2*x**2 + 1)*x**4),x)*b*c**4*x**7 + 6*int(asinh(c*x)/(sqrt(c**2*x**2 + 
 1)*c**4*x**8 + 2*sqrt(c**2*x**2 + 1)*c**2*x**6 + sqrt(c**2*x**2 + 1)*x**4 
),x)*b*c**2*x**5 + 3*int(asinh(c*x)/(sqrt(c**2*x**2 + 1)*c**4*x**8 + 2*sqr 
t(c**2*x**2 + 1)*c**2*x**6 + sqrt(c**2*x**2 + 1)*x**4),x)*b*x**3 - 16*a*c* 
*7*x**7 - 32*a*c**5*x**5 - 16*a*c**3*x**3)/(3*sqrt(pi)*pi**2*x**3*(c**4*x* 
*4 + 2*c**2*x**2 + 1))