\(\int (\pi +c^2 \pi x^2)^{3/2} (a+b \text {arcsinh}(c x)) \, dx\) [71]

Optimal result

Integrand size = 23, antiderivative size = 119 \[ \int \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=-\frac {3}{16} b c \pi ^{3/2} x^2-\frac {b \pi ^{3/2} \left (1+c^2 x^2\right )^2}{16 c}+\frac {3}{8} \pi x \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))+\frac {1}{4} x \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x))+\frac {3 \pi ^{3/2} (a+b \text {arcsinh}(c x))^2}{16 b c} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.93 \[ \int \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {\pi ^{3/2} \left (80 a c x \sqrt {1+c^2 x^2}+32 a c^3 x^3 \sqrt {1+c^2 x^2}+24 b \text {arcsinh}(c x)^2-16 b \cosh (2 \text {arcsinh}(c x))-b \cosh (4 \text {arcsinh}(c x))+4 \text {arcsinh}(c x) (12 a+8 b \sinh (2 \text {arcsinh}(c x))+b \sinh (4 \text {arcsinh}(c x)))\right )}{128 c} \] Input:

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

Output:

(Pi^(3/2)*(80*a*c*x*Sqrt[1 + c^2*x^2] + 32*a*c^3*x^3*Sqrt[1 + c^2*x^2] + 2 
4*b*ArcSinh[c*x]^2 - 16*b*Cosh[2*ArcSinh[c*x]] - b*Cosh[4*ArcSinh[c*x]] + 
4*ArcSinh[c*x]*(12*a + 8*b*Sinh[2*ArcSinh[c*x]] + b*Sinh[4*ArcSinh[c*x]])) 
)/(128*c)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6201, 244, 2009, 6200, 15, 6198}

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.

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

\(\Big \downarrow \) 6201

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

\(\Big \downarrow \) 244

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 6200

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 6198

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009

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

rule 6198

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_ 
Symbol] :> Simp[(1/(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*( 
a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c 
^2*d] && NeQ[n, -1]
 

rule 6200

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_ 
Symbol] :> Simp[x*Sqrt[d + e*x^2]*((a + b*ArcSinh[c*x])^n/2), x] + (Simp[(1 
/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]]   Int[(a + b*ArcSinh[c*x])^n/Sq 
rt[1 + c^2*x^2], x], x] - Simp[b*c*(n/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2* 
x^2]]   Int[x*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e 
}, x] && EqQ[e, c^2*d] && GtQ[n, 0]
 

rule 6201

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), 
x_Symbol] :> Simp[x*(d + e*x^2)^p*((a + b*ArcSinh[c*x])^n/(2*p + 1)), x] + 
(Simp[2*d*(p/(2*p + 1))   Int[(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^n, x 
], x] - Simp[b*c*(n/(2*p + 1))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p]   Int[x* 
(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, 
b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0]
 

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.28

method	result	size
default	\(\frac {x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}} a}{4}+\frac {3 a \pi x \sqrt {\pi \,c^{2} x^{2}+\pi }}{8}+\frac {3 a \,\pi ^{2} \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{8 \sqrt {\pi \,c^{2}}}+\frac {b \,\pi ^{\frac {3}{2}} \left (4 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}-c^{4} x^{4}+10 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x c -5 c^{2} x^{2}+3 \operatorname {arcsinh}\left (x c \right )^{2}-4\right )}{16 c}\)	\(152\)
parts	\(\frac {x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}} a}{4}+\frac {3 a \pi x \sqrt {\pi \,c^{2} x^{2}+\pi }}{8}+\frac {3 a \,\pi ^{2} \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{8 \sqrt {\pi \,c^{2}}}+\frac {b \,\pi ^{\frac {3}{2}} \left (4 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}-c^{4} x^{4}+10 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x c -5 c^{2} x^{2}+3 \operatorname {arcsinh}\left (x c \right )^{2}-4\right )}{16 c}\)	\(152\)

Input:

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

Output:

1/4*x*(Pi*c^2*x^2+Pi)^(3/2)*a+3/8*a*Pi*x*(Pi*c^2*x^2+Pi)^(1/2)+3/8*a*Pi^2* 
ln(Pi*c^2*x/(Pi*c^2)^(1/2)+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)+1/16*b*Pi 
^(3/2)*(4*arcsinh(x*c)*(c^2*x^2+1)^(1/2)*x^3*c^3-c^4*x^4+10*arcsinh(x*c)*( 
c^2*x^2+1)^(1/2)*x*c-5*c^2*x^2+3*arcsinh(x*c)^2-4)/c
 

Fricas [F]

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

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

Output:

integral(sqrt(pi + pi*c^2*x^2)*(pi*a*c^2*x^2 + pi*a + (pi*b*c^2*x^2 + pi*b 
)*arcsinh(c*x)), x)
 

Sympy [A] (verification not implemented)

Time = 1.66 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.55 \[ \int \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\begin {cases} \frac {\pi ^{\frac {3}{2}} a c^{2} x^{3} \sqrt {c^{2} x^{2} + 1}}{4} + \frac {5 \pi ^{\frac {3}{2}} a x \sqrt {c^{2} x^{2} + 1}}{8} + \frac {3 \pi ^{\frac {3}{2}} a \operatorname {asinh}{\left (c x \right )}}{8 c} - \frac {\pi ^{\frac {3}{2}} b c^{3} x^{4}}{16} + \frac {\pi ^{\frac {3}{2}} b c^{2} x^{3} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{4} - \frac {5 \pi ^{\frac {3}{2}} b c x^{2}}{16} + \frac {5 \pi ^{\frac {3}{2}} b x \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{8} + \frac {3 \pi ^{\frac {3}{2}} b \operatorname {asinh}^{2}{\left (c x \right )}}{16 c} & \text {for}\: c \neq 0 \\\pi ^{\frac {3}{2}} a x & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((pi**(3/2)*a*c**2*x**3*sqrt(c**2*x**2 + 1)/4 + 5*pi**(3/2)*a*x*s 
qrt(c**2*x**2 + 1)/8 + 3*pi**(3/2)*a*asinh(c*x)/(8*c) - pi**(3/2)*b*c**3*x 
**4/16 + pi**(3/2)*b*c**2*x**3*sqrt(c**2*x**2 + 1)*asinh(c*x)/4 - 5*pi**(3 
/2)*b*c*x**2/16 + 5*pi**(3/2)*b*x*sqrt(c**2*x**2 + 1)*asinh(c*x)/8 + 3*pi* 
*(3/2)*b*asinh(c*x)**2/(16*c), Ne(c, 0)), (pi**(3/2)*a*x, True))
 

Maxima [F(-2)]

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

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

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(pi)*pi*(2*sqrt(c**2*x**2 + 1)*a*c**3*x**3 + 5*sqrt(c**2*x**2 + 1)*a* 
c*x + 8*int(sqrt(c**2*x**2 + 1)*asinh(c*x)*x**2,x)*b*c**3 + 8*int(sqrt(c** 
2*x**2 + 1)*asinh(c*x),x)*b*c + 3*log(sqrt(c**2*x**2 + 1) + c*x)*a))/(8*c)