\(\int \frac {a+b \text {arcsinh}(c x)}{x^4 (\pi +c^2 \pi x^2)^{5/2}} \, dx\) [110]

Optimal result

Integrand size = 26, antiderivative size = 208 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=-\frac {b c}{6 \pi ^{5/2} x^2}+\frac {b c^3}{6 \pi ^{5/2} \left (1+c^2 x^2\right )}-\frac {a+b \text {arcsinh}(c x)}{3 \pi x^3 \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {2 c^2 (a+b \text {arcsinh}(c x))}{\pi x \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {8 b c^3 \log (x)}{3 \pi ^{5/2}}-\frac {4 b c^3 \log \left (1+c^2 x^2\right )}{3 \pi ^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {277, 198, 197, 5804, 12, 1813, 1634} \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\frac {2 c^2 (a+b \text {arcsinh}(c x))}{\pi x \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{3 \pi x^3 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi ^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {8 b c^3 \log (x)}{3 \pi ^{5/2}}+\frac {b c^3}{6 \pi ^{5/2} \left (c^2 x^2+1\right )}-\frac {4 b c^3 \log \left (c^2 x^2+1\right )}{3 \pi ^{5/2}}-\frac {b c}{6 \pi ^{5/2} x^2} \]

[In]

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

[Out]

-1/6*(b*c)/(Pi^(5/2)*x^2) + (b*c^3)/(6*Pi^(5/2)*(1 + c^2*x^2)) - (a + b*ArcSinh[c*x])/(3*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*P
i*(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]) - (8*b*c^3*Log[x])
/(3*Pi^(5/2)) - (4*b*c^3*Log[1 + c^2*x^2])/(3*Pi^(5/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 1634

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]) &&
GtQ[Expon[Px, x], 2]

Rule 1813

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[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 5804

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]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]], Int[
SimplifyIntegrand[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])

Rubi steps \begin{align*} \text {integral}& = -\frac {a+b \text {arcsinh}(c x)}{3 \pi x^3 \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {2 c^2 (a+b \text {arcsinh}(c x))}{\pi x \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\left (b c \sqrt {\pi }\right ) \int \frac {-1+6 c^2 x^2+24 c^4 x^4+16 c^6 x^6}{3 \pi ^3 x^3 \left (1+c^2 x^2\right )^2} \, dx \\ & = -\frac {a+b \text {arcsinh}(c x)}{3 \pi x^3 \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {2 c^2 (a+b \text {arcsinh}(c x))}{\pi x \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {(b c) \int \frac {-1+6 c^2 x^2+24 c^4 x^4+16 c^6 x^6}{x^3 \left (1+c^2 x^2\right )^2} \, dx}{3 \pi ^{5/2}} \\ & = -\frac {a+b \text {arcsinh}(c x)}{3 \pi x^3 \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {2 c^2 (a+b \text {arcsinh}(c x))}{\pi x \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {(b c) \text {Subst}\left (\int \frac {-1+6 c^2 x+24 c^4 x^2+16 c^6 x^3}{x^2 \left (1+c^2 x\right )^2} \, dx,x,x^2\right )}{6 \pi ^{5/2}} \\ & = -\frac {a+b \text {arcsinh}(c x)}{3 \pi x^3 \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {2 c^2 (a+b \text {arcsinh}(c x))}{\pi x \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {(b c) \text {Subst}\left (\int \left (-\frac {1}{x^2}+\frac {8 c^2}{x}+\frac {c^4}{\left (1+c^2 x\right )^2}+\frac {8 c^4}{1+c^2 x}\right ) \, dx,x,x^2\right )}{6 \pi ^{5/2}} \\ & = -\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 c^2 (a+b \text {arcsinh}(c x))}{\pi x \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {8 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {16 c^4 x (a+b \text {arcsinh}(c x))}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\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}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.15 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\frac {-2 a+12 a c^2 x^2+48 a c^4 x^4+32 a c^6 x^6-b c x \sqrt {1+c^2 x^2}-32 b c^3 x^3 \sqrt {1+c^2 x^2}-32 b c^5 x^5 \sqrt {1+c^2 x^2}+2 b \left (-1+6 c^2 x^2+24 c^4 x^4+16 c^6 x^6\right ) \text {arcsinh}(c x)-16 b c^3 x^3 \left (1+c^2 x^2\right )^{3/2} \log (x)-8 b c^3 x^3 \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )-8 b c^5 x^5 \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{6 \pi ^{5/2} x^3 \left (1+c^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

(-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]*Lo
g[1 + c^2*x^2])/(6*Pi^(5/2)*x^3*(1 + c^2*x^2)^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1154\) vs. \(2(181)=362\).

Time = 0.17 (sec) , antiderivative size = 1155, normalized size of antiderivative = 5.55

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

[In]

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

[Out]

-2*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)*c^3+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^7-2*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)*x^2*c^5+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(c*x)+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+
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+16/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^
2+1)^2*arcsinh(c*x)*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^4*c^7-128/3*b/Pi^(5/2)/(12*
c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^12*c^15-512/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^10*c^
13-256*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^8*c^11-512/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/
(c^2*x^2+1)^2*x^6*c^9+a*(-1/3/Pi/x^3/(Pi*c^2*x^2+Pi)^(3/2)-2*c^2*(-1/Pi/x/(Pi*c^2*x^2+Pi)^(3/2)-4*c^2*(1/3/Pi*
x/(Pi*c^2*x^2+Pi)^(3/2)+2/3/Pi^2*x/(Pi*c^2*x^2+Pi)^(1/2))))+32/3*b*c^3/Pi^(5/2)*arcsinh(c*x)-8/3*b*c^3/Pi^(5/2
)*ln((c*x+(c^2*x^2+1)^(1/2))^4-1)-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(c*x)*c^
7-160/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^2*arcsinh(c*x)*c^5-64*b/Pi^(5/2)/(12*c^4*x^4+12*c
^2*x^2-1)/(c^2*x^2+1)^2*x^8*arcsinh(c*x)*c^11-192*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^6*arcsi
nh(c*x)*c^9+64*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^(3/2)*x^7*arcsinh(c*x)*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(c*x)*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(c*x)*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(c*x)*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(c*x)*c^2

Fricas [F]

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

[In]

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

[Out]

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]

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

[In]

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

[Out]

(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)

none

Time = 0.22 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.13 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=-\frac {1}{6} \, b c {\left (\frac {8 \, c^{2} \log \left (c^{2} x^{2} + 1\right )}{\pi ^{\frac {5}{2}}} + \frac {16 \, c^{2} \log \left (x\right )}{\pi ^{\frac {5}{2}}} + \frac {1}{\pi ^{\frac {5}{2}} c^{2} x^{4} + \pi ^{\frac {5}{2}} x^{2}}\right )} + \frac {1}{3} \, {\left (\frac {8 \, c^{4} x}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}} + \frac {16 \, c^{4} x}{\pi ^{2} \sqrt {\pi + \pi c^{2} x^{2}}} + \frac {6 \, c^{2}}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x} - \frac {1}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x^{3}}\right )} b \operatorname {arsinh}\left (c x\right ) + \frac {1}{3} \, {\left (\frac {8 \, c^{4} x}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}} + \frac {16 \, c^{4} x}{\pi ^{2} \sqrt {\pi + \pi c^{2} x^{2}}} + \frac {6 \, c^{2}}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x} - \frac {1}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x^{3}}\right )} a \]

[In]

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

[Out]

-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]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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