\(\int \frac {x^3 (a+b \text {arcsinh}(c x))}{(\pi +c^2 \pi x^2)^{5/2}} \, dx\) [113]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\frac {-4 a-6 a c^2 x^2-b c x \sqrt {1+c^2 x^2}-2 b \left (2+3 c^2 x^2\right ) \text {arcsinh}(c x)+5 b \left (1+c^2 x^2\right )^{3/2} \arctan (c x)}{6 c^4 \pi ^{5/2} \left (1+c^2 x^2\right )^{3/2}} \] Input:

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

Output:

(-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)*ArcSin 
h[c*x] + 5*b*(1 + c^2*x^2)^(3/2)*ArcTan[c*x])/(6*c^4*Pi^(5/2)*(1 + c^2*x^2 
)^(3/2))
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6219, 27, 298, 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.

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

\(\Big \downarrow \) 6219

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 298

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

(a + b*ArcSinh[c*x])/(3*c^4*Pi*(Pi + c^2*Pi*x^2)^(3/2)) - (a + b*ArcSinh[c 
*x])/(c^4*Pi^2*Sqrt[Pi + c^2*Pi*x^2]) + (b*(-1/2*x/(1 + c^2*x^2) + (5*ArcT 
an[c*x])/(2*c)))/(3*c^3*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 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 298

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

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 = 1.10 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.50

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

Input:

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

Output:

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*arcsinh(x*c)*c^2*x^2+(c^2*x^2+1)^(1/2)* 
x*c+4*arcsinh(x*c))/c^4+5/6*I/Pi^(5/2)/c^4*ln(x*c+(c^2*x^2+1)^(1/2)+I)-5/6 
*I/Pi^(5/2)/c^4*ln(x*c+(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 = 0.12 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.78 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=-\frac {5 \, \sqrt {\pi } {\left (b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b\right )} \arctan \left (-\frac {2 \, \sqrt {\pi } \sqrt {\pi + \pi c^{2} x^{2}} \sqrt {c^{2} x^{2} + 1} c x}{\pi - \pi c^{4} x^{4}}\right ) + 4 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (3 \, b c^{2} x^{2} + 2 \, b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (6 \, a c^{2} x^{2} + \sqrt {c^{2} x^{2} + 1} b c x + 4 \, a\right )}}{12 \, {\left (\pi ^{3} c^{8} x^{4} + 2 \, \pi ^{3} c^{6} x^{2} + \pi ^{3} c^{4}\right )}} \] Input:

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

Output:

-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 + pi* 
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]

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

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

Output:

(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)

Time = 0.12 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.31 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=-\frac {1}{6} \, b c {\left (\frac {x}{\pi ^{\frac {5}{2}} c^{6} x^{2} + \pi ^{\frac {5}{2}} c^{4}} - \frac {5 \, \arctan \left (c x\right )}{\pi ^{\frac {5}{2}} c^{5}}\right )} - \frac {1}{3} \, b {\left (\frac {3 \, x^{2}}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{2}} + \frac {2}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{4}}\right )} \operatorname {arsinh}\left (c x\right ) - \frac {1}{3} \, a {\left (\frac {3 \, x^{2}}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{2}} + \frac {2}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{4}}\right )} \] Input:

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

Output:

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

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 3*sqrt(c**2*x**2 + 1)*a*c**2*x**2 - 2*sqrt(c**2*x**2 + 1)*a + 3*int((a 
sinh(c*x)*x**3)/(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**4 + 6*int((asinh(c*x)*x**3)/(sq 
rt(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*x**2 + 3*int((asinh(c*x)*x**3)/(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** 
4)/(3*sqrt(pi)*c**4*pi**2*(c**4*x**4 + 2*c**2*x**2 + 1))