\(\int \frac {x^3 (a+b \text {arcsinh}(c x))}{(\pi +c^2 \pi x^2)^{5/2}} \, dx\) [103]

Optimal result

Integrand size = 26, antiderivative size = 105 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=-\frac {b x}{6 c^3 \pi ^{5/2} \left (1+c^2 x^2\right )}+\frac {a+b \text {arcsinh}(c x)}{3 c^4 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{c^4 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {5 b \arctan (c x)}{6 c^4 \pi ^{5/2}} \]

[Out]

-1/6*b*x/c^3/Pi^(5/2)/(c^2*x^2+1)+1/3*(a+b*arcsinh(c*x))/c^4/Pi/(Pi*c^2*x^2+Pi)^(3/2)+5/6*b*arctan(c*x)/c^4/Pi
^(5/2)+(-a-b*arcsinh(c*x))/c^4/Pi^2/(Pi*c^2*x^2+Pi)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {272, 45, 5804, 12, 393, 209} \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=-\frac {a+b \text {arcsinh}(c x)}{\pi ^2 c^4 \sqrt {\pi c^2 x^2+\pi }}+\frac {a+b \text {arcsinh}(c x)}{3 \pi c^4 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {5 b \arctan (c x)}{6 \pi ^{5/2} c^4}-\frac {b x}{6 \pi ^{5/2} c^3 \left (c^2 x^2+1\right )} \]

[In]

Int[(x^3*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(5/2),x]

[Out]

-1/6*(b*x)/(c^3*Pi^(5/2)*(1 + c^2*x^2)) + (a + b*ArcSinh[c*x])/(3*c^4*Pi*(Pi + c^2*Pi*x^2)^(3/2)) - (a + b*Arc
Sinh[c*x])/(c^4*Pi^2*Sqrt[Pi + c^2*Pi*x^2]) + (5*b*ArcTan[c*x])/(6*c^4*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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 209

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

Rule 272

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

Rule 393

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

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 c^4 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{c^4 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\left (b c \sqrt {\pi }\right ) \int \frac {-2-3 c^2 x^2}{3 c^4 \pi ^3 \left (1+c^2 x^2\right )^2} \, dx \\ & = \frac {a+b \text {arcsinh}(c x)}{3 c^4 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{c^4 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {b \int \frac {-2-3 c^2 x^2}{\left (1+c^2 x^2\right )^2} \, dx}{3 c^3 \pi ^{5/2}} \\ & = -\frac {b x}{6 c^3 \pi ^{5/2} \left (1+c^2 x^2\right )}+\frac {a+b \text {arcsinh}(c x)}{3 c^4 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{c^4 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {(5 b) \int \frac {1}{1+c^2 x^2} \, dx}{6 c^3 \pi ^{5/2}} \\ & = -\frac {b x}{6 c^3 \pi ^{5/2} \left (1+c^2 x^2\right )}+\frac {a+b \text {arcsinh}(c x)}{3 c^4 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{c^4 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {5 b \arctan (c x)}{6 c^4 \pi ^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\frac {-4 a-6 a c^2 x^2-b c x \sqrt {1+c^2 x^2}-2 b \left (2+3 c^2 x^2\right ) \text {arcsinh}(c x)+5 b \left (1+c^2 x^2\right )^{3/2} \arctan (c x)}{6 c^4 \pi ^{5/2} \left (1+c^2 x^2\right )^{3/2}} \]

[In]

Integrate[(x^3*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(5/2),x]

[Out]

(-4*a - 6*a*c^2*x^2 - b*c*x*Sqrt[1 + c^2*x^2] - 2*b*(2 + 3*c^2*x^2)*ArcSinh[c*x] + 5*b*(1 + c^2*x^2)^(3/2)*Arc
Tan[c*x])/(6*c^4*Pi^(5/2)*(1 + c^2*x^2)^(3/2))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.50

method	result	size
default	\(a \left (-\frac {x^{2}}{\pi \,c^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}-\frac {2}{3 \pi \,c^{4} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}\right )+b \left (-\frac {6 \,\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}+c x \sqrt {c^{2} x^{2}+1}+4 \,\operatorname {arcsinh}\left (c x \right )}{6 \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} c^{4}}+\frac {5 i \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{6 c^{4} \pi ^{\frac {5}{2}}}-\frac {5 i \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{6 c^{4} \pi ^{\frac {5}{2}}}\right )\)	\(157\)
parts	\(a \left (-\frac {x^{2}}{\pi \,c^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}-\frac {2}{3 \pi \,c^{4} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}\right )+b \left (-\frac {6 \,\operatorname {arcsinh}\left (c x \right ) c^{2} x^{2}+c x \sqrt {c^{2} x^{2}+1}+4 \,\operatorname {arcsinh}\left (c x \right )}{6 \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} c^{4}}+\frac {5 i \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{6 c^{4} \pi ^{\frac {5}{2}}}-\frac {5 i \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{6 c^{4} \pi ^{\frac {5}{2}}}\right )\)	\(157\)

[In]

int(x^3*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(5/2),x,method=_RETURNVERBOSE)

[Out]

a*(-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*(-1/6/Pi^(5/2)/(c^2*x^2+1)^(3/2)*(6*a
rcsinh(c*x)*c^2*x^2+c*x*(c^2*x^2+1)^(1/2)+4*arcsinh(c*x))/c^4+5/6*I/c^4/Pi^(5/2)*ln(c*x+(c^2*x^2+1)^(1/2)+I)-5
/6*I/c^4/Pi^(5/2)*ln(c*x+(c^2*x^2+1)^(1/2)-I))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (91) = 182\).

Time = 0.30 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.78 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=-\frac {5 \, \sqrt {\pi } {\left (b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b\right )} \arctan \left (-\frac {2 \, \sqrt {\pi } \sqrt {\pi + \pi c^{2} x^{2}} \sqrt {c^{2} x^{2} + 1} c x}{\pi - \pi c^{4} x^{4}}\right ) + 4 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (3 \, b c^{2} x^{2} + 2 \, b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (6 \, a c^{2} x^{2} + \sqrt {c^{2} x^{2} + 1} b c x + 4 \, a\right )}}{12 \, {\left (\pi ^{3} c^{8} x^{4} + 2 \, \pi ^{3} c^{6} x^{2} + \pi ^{3} c^{4}\right )}} \]

[In]

integrate(x^3*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(5*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^2*x^2 + 2*b)*log(c*x + sqrt(c^2*x^2 + 1)) + 2*sqrt(pi + p
i*c^2*x^2)*(6*a*c^2*x^2 + sqrt(c^2*x^2 + 1)*b*c*x + 4*a))/(pi^3*c^8*x^4 + 2*pi^3*c^6*x^2 + pi^3*c^4)

Sympy [F]

\[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\frac {\int \frac {a x^{3}}{c^{4} x^{4} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b x^{3} \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {5}{2}}} \]

[In]

integrate(x**3*(a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(5/2),x)

[Out]

(Integral(a*x**3/(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**3*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 [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.31 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=-\frac {1}{6} \, b c {\left (\frac {x}{\pi ^{\frac {5}{2}} c^{6} x^{2} + \pi ^{\frac {5}{2}} c^{4}} - \frac {5 \, \arctan \left (c x\right )}{\pi ^{\frac {5}{2}} c^{5}}\right )} - \frac {1}{3} \, b {\left (\frac {3 \, x^{2}}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{2}} + \frac {2}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{4}}\right )} \operatorname {arsinh}\left (c x\right ) - \frac {1}{3} \, a {\left (\frac {3 \, x^{2}}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{2}} + \frac {2}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{4}}\right )} \]

[In]

integrate(x^3*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(5/2),x, algorithm="maxima")

[Out]

-1/6*b*c*(x/(pi^(5/2)*c^6*x^2 + pi^(5/2)*c^4) - 5*arctan(c*x)/(pi^(5/2)*c^5)) - 1/3*b*(3*x^2/(pi*(pi + pi*c^2*
x^2)^(3/2)*c^2) + 2/(pi*(pi + pi*c^2*x^2)^(3/2)*c^4))*arcsinh(c*x) - 1/3*a*(3*x^2/(pi*(pi + pi*c^2*x^2)^(3/2)*
c^2) + 2/(pi*(pi + pi*c^2*x^2)^(3/2)*c^4))

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^3*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (\Pi \,c^2\,x^2+\Pi \right )}^{5/2}} \,d x \]

[In]

int((x^3*(a + b*asinh(c*x)))/(Pi + Pi*c^2*x^2)^(5/2),x)

[Out]

int((x^3*(a + b*asinh(c*x)))/(Pi + Pi*c^2*x^2)^(5/2), x)