\(\int x^3 (\pi +c^2 \pi x^2)^{3/2} (a+b \text {arcsinh}(c x)) \, dx\) [66]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [B] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 26, antiderivative size = 125 \[ \int x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {2 b \pi ^{3/2} x}{35 c^3}-\frac {b \pi ^{3/2} x^3}{105 c}-\frac {8}{175} b c \pi ^{3/2} x^5-\frac {1}{49} b c^3 \pi ^{3/2} x^7-\frac {\left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x))}{5 c^4 \pi }+\frac {\left (\pi +c^2 \pi x^2\right )^{7/2} (a+b \text {arcsinh}(c x))}{7 c^4 \pi ^2} \] Output:

2/35*b*Pi^(3/2)*x/c^3-1/105*b*Pi^(3/2)*x^3/c-8/175*b*c*Pi^(3/2)*x^5-1/49*b 
*c^3*Pi^(3/2)*x^7-1/5*(Pi*c^2*x^2+Pi)^(5/2)*(a+b*arcsinh(c*x))/c^4/Pi+1/7* 
(Pi*c^2*x^2+Pi)^(7/2)*(a+b*arcsinh(c*x))/c^4/Pi^2
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.80 \[ \int x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {\pi ^{3/2} \left (105 a \left (1+c^2 x^2\right )^{5/2} \left (-2+5 c^2 x^2\right )-b c x \left (-210+35 c^2 x^2+168 c^4 x^4+75 c^6 x^6\right )+105 b \left (1+c^2 x^2\right )^{5/2} \left (-2+5 c^2 x^2\right ) \text {arcsinh}(c x)\right )}{3675 c^4} \] Input:

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

Output:

(Pi^(3/2)*(105*a*(1 + c^2*x^2)^(5/2)*(-2 + 5*c^2*x^2) - b*c*x*(-210 + 35*c 
^2*x^2 + 168*c^4*x^4 + 75*c^6*x^6) + 105*b*(1 + c^2*x^2)^(5/2)*(-2 + 5*c^2 
*x^2)*ArcSinh[c*x]))/(3675*c^4)
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6219, 27, 290, 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.

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

\(\Big \downarrow \) 6219

\(\displaystyle -\sqrt {\pi } b c \int -\frac {\pi \left (2-5 c^2 x^2\right ) \left (c^2 x^2+1\right )^2}{35 c^4}dx+\frac {\left (\pi c^2 x^2+\pi \right )^{7/2} (a+b \text {arcsinh}(c x))}{7 \pi ^2 c^4}-\frac {\left (\pi c^2 x^2+\pi \right )^{5/2} (a+b \text {arcsinh}(c x))}{5 \pi c^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\pi ^{3/2} b \int \left (2-5 c^2 x^2\right ) \left (c^2 x^2+1\right )^2dx}{35 c^3}+\frac {\left (\pi c^2 x^2+\pi \right )^{7/2} (a+b \text {arcsinh}(c x))}{7 \pi ^2 c^4}-\frac {\left (\pi c^2 x^2+\pi \right )^{5/2} (a+b \text {arcsinh}(c x))}{5 \pi c^4}\)

\(\Big \downarrow \) 290

\(\displaystyle \frac {\pi ^{3/2} b \int \left (-5 c^6 x^6-8 c^4 x^4-c^2 x^2+2\right )dx}{35 c^3}+\frac {\left (\pi c^2 x^2+\pi \right )^{7/2} (a+b \text {arcsinh}(c x))}{7 \pi ^2 c^4}-\frac {\left (\pi c^2 x^2+\pi \right )^{5/2} (a+b \text {arcsinh}(c x))}{5 \pi c^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (\pi c^2 x^2+\pi \right )^{7/2} (a+b \text {arcsinh}(c x))}{7 \pi ^2 c^4}-\frac {\left (\pi c^2 x^2+\pi \right )^{5/2} (a+b \text {arcsinh}(c x))}{5 \pi c^4}+\frac {\pi ^{3/2} b \left (-\frac {5}{7} c^6 x^7-\frac {8 c^4 x^5}{5}-\frac {c^2 x^3}{3}+2 x\right )}{35 c^3}\)

Input:

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

Output:

(b*Pi^(3/2)*(2*x - (c^2*x^3)/3 - (8*c^4*x^5)/5 - (5*c^6*x^7)/7))/(35*c^3) 
- ((Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/(5*c^4*Pi) + ((Pi + c^2*P 
i*x^2)^(7/2)*(a + b*ArcSinh[c*x]))/(7*c^4*Pi^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 290
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I 
nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d 
}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 [A] (verified)

Time = 1.17 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.56

method result size
default \(a \left (\frac {x^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}{7 \pi \,c^{2}}-\frac {2 \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}{35 \pi \,c^{4}}\right )+\frac {b \,\pi ^{\frac {3}{2}} \left (525 \,\operatorname {arcsinh}\left (x c \right ) x^{8} c^{8}+1365 \,\operatorname {arcsinh}\left (x c \right ) x^{6} c^{6}-75 x^{7} c^{7} \sqrt {c^{2} x^{2}+1}+945 \,\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}-168 \sqrt {c^{2} x^{2}+1}\, x^{5} c^{5}-105 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}-35 \sqrt {c^{2} x^{2}+1}\, c^{3} x^{3}-210 \,\operatorname {arcsinh}\left (x c \right )+210 \sqrt {c^{2} x^{2}+1}\, x c \right )}{3675 c^{4} \sqrt {c^{2} x^{2}+1}}\) \(195\)
parts \(a \left (\frac {x^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}{7 \pi \,c^{2}}-\frac {2 \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}{35 \pi \,c^{4}}\right )+\frac {b \,\pi ^{\frac {3}{2}} \left (525 \,\operatorname {arcsinh}\left (x c \right ) x^{8} c^{8}+1365 \,\operatorname {arcsinh}\left (x c \right ) x^{6} c^{6}-75 x^{7} c^{7} \sqrt {c^{2} x^{2}+1}+945 \,\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}-168 \sqrt {c^{2} x^{2}+1}\, x^{5} c^{5}-105 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}-35 \sqrt {c^{2} x^{2}+1}\, c^{3} x^{3}-210 \,\operatorname {arcsinh}\left (x c \right )+210 \sqrt {c^{2} x^{2}+1}\, x c \right )}{3675 c^{4} \sqrt {c^{2} x^{2}+1}}\) \(195\)
orering \(\frac {\left (325 c^{8} x^{8}+866 c^{6} x^{6}+553 c^{4} x^{4}-420 c^{2} x^{2}-280\right ) \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}{1225 c^{4} \left (c^{2} x^{2}+1\right )^{2}}-\frac {\left (75 c^{6} x^{6}+168 c^{4} x^{4}+35 c^{2} x^{2}-210\right ) \left (3 x^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )+3 x^{4} \sqrt {\pi \,c^{2} x^{2}+\pi }\, \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right ) \pi \,c^{2}+\frac {x^{3} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}} b c}{\sqrt {c^{2} x^{2}+1}}\right )}{3675 x^{2} c^{4} \left (c^{2} x^{2}+1\right )}\) \(201\)

Input:

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

Output:

a*(1/7*x^2*(Pi*c^2*x^2+Pi)^(5/2)/Pi/c^2-2/35/Pi/c^4*(Pi*c^2*x^2+Pi)^(5/2)) 
+1/3675*b*Pi^(3/2)/c^4/(c^2*x^2+1)^(1/2)*(525*arcsinh(x*c)*x^8*c^8+1365*ar 
csinh(x*c)*x^6*c^6-75*x^7*c^7*(c^2*x^2+1)^(1/2)+945*arcsinh(x*c)*c^4*x^4-1 
68*(c^2*x^2+1)^(1/2)*x^5*c^5-105*arcsinh(x*c)*c^2*x^2-35*(c^2*x^2+1)^(1/2) 
*c^3*x^3-210*arcsinh(x*c)+210*(c^2*x^2+1)^(1/2)*x*c)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.59 \[ \int x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {105 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (5 \, \pi b c^{8} x^{8} + 13 \, \pi b c^{6} x^{6} + 9 \, \pi b c^{4} x^{4} - \pi b c^{2} x^{2} - 2 \, \pi b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + \sqrt {\pi + \pi c^{2} x^{2}} {\left (525 \, \pi a c^{8} x^{8} + 1365 \, \pi a c^{6} x^{6} + 945 \, \pi a c^{4} x^{4} - 105 \, \pi a c^{2} x^{2} - 210 \, \pi a - {\left (75 \, \pi b c^{7} x^{7} + 168 \, \pi b c^{5} x^{5} + 35 \, \pi b c^{3} x^{3} - 210 \, \pi b c x\right )} \sqrt {c^{2} x^{2} + 1}\right )}}{3675 \, {\left (c^{6} x^{2} + c^{4}\right )}} \] Input:

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

Output:

1/3675*(105*sqrt(pi + pi*c^2*x^2)*(5*pi*b*c^8*x^8 + 13*pi*b*c^6*x^6 + 9*pi 
*b*c^4*x^4 - pi*b*c^2*x^2 - 2*pi*b)*log(c*x + sqrt(c^2*x^2 + 1)) + sqrt(pi 
 + pi*c^2*x^2)*(525*pi*a*c^8*x^8 + 1365*pi*a*c^6*x^6 + 945*pi*a*c^4*x^4 - 
105*pi*a*c^2*x^2 - 210*pi*a - (75*pi*b*c^7*x^7 + 168*pi*b*c^5*x^5 + 35*pi* 
b*c^3*x^3 - 210*pi*b*c*x)*sqrt(c^2*x^2 + 1)))/(c^6*x^2 + c^4)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (117) = 234\).

Time = 9.32 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.41 \[ \int x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\begin {cases} \frac {\pi ^{\frac {3}{2}} a c^{2} x^{6} \sqrt {c^{2} x^{2} + 1}}{7} + \frac {8 \pi ^{\frac {3}{2}} a x^{4} \sqrt {c^{2} x^{2} + 1}}{35} + \frac {\pi ^{\frac {3}{2}} a x^{2} \sqrt {c^{2} x^{2} + 1}}{35 c^{2}} - \frac {2 \pi ^{\frac {3}{2}} a \sqrt {c^{2} x^{2} + 1}}{35 c^{4}} - \frac {\pi ^{\frac {3}{2}} b c^{3} x^{7}}{49} + \frac {\pi ^{\frac {3}{2}} b c^{2} x^{6} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{7} - \frac {8 \pi ^{\frac {3}{2}} b c x^{5}}{175} + \frac {8 \pi ^{\frac {3}{2}} b x^{4} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{35} - \frac {\pi ^{\frac {3}{2}} b x^{3}}{105 c} + \frac {\pi ^{\frac {3}{2}} b x^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{35 c^{2}} + \frac {2 \pi ^{\frac {3}{2}} b x}{35 c^{3}} - \frac {2 \pi ^{\frac {3}{2}} b \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{35 c^{4}} & \text {for}\: c \neq 0 \\\frac {\pi ^{\frac {3}{2}} a x^{4}}{4} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((pi**(3/2)*a*c**2*x**6*sqrt(c**2*x**2 + 1)/7 + 8*pi**(3/2)*a*x** 
4*sqrt(c**2*x**2 + 1)/35 + pi**(3/2)*a*x**2*sqrt(c**2*x**2 + 1)/(35*c**2) 
- 2*pi**(3/2)*a*sqrt(c**2*x**2 + 1)/(35*c**4) - pi**(3/2)*b*c**3*x**7/49 + 
 pi**(3/2)*b*c**2*x**6*sqrt(c**2*x**2 + 1)*asinh(c*x)/7 - 8*pi**(3/2)*b*c* 
x**5/175 + 8*pi**(3/2)*b*x**4*sqrt(c**2*x**2 + 1)*asinh(c*x)/35 - pi**(3/2 
)*b*x**3/(105*c) + pi**(3/2)*b*x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/(35*c** 
2) + 2*pi**(3/2)*b*x/(35*c**3) - 2*pi**(3/2)*b*sqrt(c**2*x**2 + 1)*asinh(c 
*x)/(35*c**4), Ne(c, 0)), (pi**(3/2)*a*x**4/4, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.16 \[ \int x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {1}{35} \, {\left (\frac {5 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}} x^{2}}{\pi c^{2}} - \frac {2 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}}}{\pi c^{4}}\right )} b \operatorname {arsinh}\left (c x\right ) + \frac {1}{35} \, {\left (\frac {5 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}} x^{2}}{\pi c^{2}} - \frac {2 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}}}{\pi c^{4}}\right )} a - \frac {{\left (75 \, \pi ^{\frac {3}{2}} c^{6} x^{7} + 168 \, \pi ^{\frac {3}{2}} c^{4} x^{5} + 35 \, \pi ^{\frac {3}{2}} c^{2} x^{3} - 210 \, \pi ^{\frac {3}{2}} x\right )} b}{3675 \, c^{3}} \] Input:

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

Output:

1/35*(5*(pi + pi*c^2*x^2)^(5/2)*x^2/(pi*c^2) - 2*(pi + pi*c^2*x^2)^(5/2)/( 
pi*c^4))*b*arcsinh(c*x) + 1/35*(5*(pi + pi*c^2*x^2)^(5/2)*x^2/(pi*c^2) - 2 
*(pi + pi*c^2*x^2)^(5/2)/(pi*c^4))*a - 1/3675*(75*pi^(3/2)*c^6*x^7 + 168*p 
i^(3/2)*c^4*x^5 + 35*pi^(3/2)*c^2*x^3 - 210*pi^(3/2)*x)*b/c^3
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

\[ \int x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {\sqrt {\pi }\, \pi \left (5 \sqrt {c^{2} x^{2}+1}\, a \,c^{6} x^{6}+8 \sqrt {c^{2} x^{2}+1}\, a \,c^{4} x^{4}+\sqrt {c^{2} x^{2}+1}\, a \,c^{2} x^{2}-2 \sqrt {c^{2} x^{2}+1}\, a +35 \left (\int \sqrt {c^{2} x^{2}+1}\, \mathit {asinh} \left (c x \right ) x^{5}d x \right ) b \,c^{6}+35 \left (\int \sqrt {c^{2} x^{2}+1}\, \mathit {asinh} \left (c x \right ) x^{3}d x \right ) b \,c^{4}\right )}{35 c^{4}} \] Input:

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

Output:

(sqrt(pi)*pi*(5*sqrt(c**2*x**2 + 1)*a*c**6*x**6 + 8*sqrt(c**2*x**2 + 1)*a* 
c**4*x**4 + sqrt(c**2*x**2 + 1)*a*c**2*x**2 - 2*sqrt(c**2*x**2 + 1)*a + 35 
*int(sqrt(c**2*x**2 + 1)*asinh(c*x)*x**5,x)*b*c**6 + 35*int(sqrt(c**2*x**2 
 + 1)*asinh(c*x)*x**3,x)*b*c**4))/(35*c**4)