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

Optimal result

Integrand size = 23, antiderivative size = 111 \[ \int \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=-\frac {5}{16} b c \pi ^{3/2} x^2-\frac {1}{16} b c^3 \pi ^{3/2} x^4+\frac {3}{8} \pi x \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))+\frac {1}{4} x \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x))+\frac {3 \pi ^{3/2} (a+b \text {arcsinh}(c x))^2}{16 b c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {5786, 5785, 5783, 30, 14} \[ \int \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {1}{4} x \left (\pi c^2 x^2+\pi \right )^{3/2} (a+b \text {arcsinh}(c x))+\frac {3}{8} \pi x \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))+\frac {3 \pi ^{3/2} (a+b \text {arcsinh}(c x))^2}{16 b c}-\frac {1}{16} \pi ^{3/2} b c^3 x^4-\frac {5}{16} \pi ^{3/2} b c x^2 \]

[In]

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

[Out]

(-5*b*c*Pi^(3/2)*x^2)/16 - (b*c^3*Pi^(3/2)*x^4)/16 + (3*Pi*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/8 + (
x*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/4 + (3*Pi^(3/2)*(a + b*ArcSinh[c*x])^2)/(16*b*c)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

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

Rule 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[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 5785

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] + (Dist[(1/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]], Int[(a + b*ArcSinh[c*x])^
n/Sqrt[1 + c^2*x^2], x], x] - Dist[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 5786

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] + (Dist[2*d*(p/(2*p + 1)), Int[(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^
n, x], x] - Dist[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*A
rcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0]

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00 \[ \int \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {\pi ^{3/2} \left (80 a c x \sqrt {1+c^2 x^2}+32 a c^3 x^3 \sqrt {1+c^2 x^2}+24 b \text {arcsinh}(c x)^2-16 b \cosh (2 \text {arcsinh}(c x))-b \cosh (4 \text {arcsinh}(c x))+4 \text {arcsinh}(c x) (12 a+8 b \sinh (2 \text {arcsinh}(c x))+b \sinh (4 \text {arcsinh}(c x)))\right )}{128 c} \]

[In]

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

[Out]

(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] + 24*b*ArcSinh[c*x]^2 - 16*b*Cosh[2*Arc
Sinh[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)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.37

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

[In]

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

[Out]

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(c*x)*(c^2*x^2+1)^(1/2)*x^3*c^3-c^4*x^4+10*arcsinh(c*x)
*c*x*(c^2*x^2+1)^(1/2)-5*c^2*x^2+3*arcsinh(c*x)^2-4)/c

Fricas [F]

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

[In]

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

[Out]

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 = 1.58 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.67 \[ \int \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\begin {cases} \frac {\pi ^{\frac {3}{2}} a c^{2} x^{3} \sqrt {c^{2} x^{2} + 1}}{4} + \frac {5 \pi ^{\frac {3}{2}} a x \sqrt {c^{2} x^{2} + 1}}{8} + \frac {3 \pi ^{\frac {3}{2}} a \operatorname {asinh}{\left (c x \right )}}{8 c} - \frac {\pi ^{\frac {3}{2}} b c^{3} x^{4}}{16} + \frac {\pi ^{\frac {3}{2}} b c^{2} x^{3} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{4} - \frac {5 \pi ^{\frac {3}{2}} b c x^{2}}{16} + \frac {5 \pi ^{\frac {3}{2}} b x \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{8} + \frac {3 \pi ^{\frac {3}{2}} b \operatorname {asinh}^{2}{\left (c x \right )}}{16 c} & \text {for}\: c \neq 0 \\\pi ^{\frac {3}{2}} a x & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((pi**(3/2)*a*c**2*x**3*sqrt(c**2*x**2 + 1)/4 + 5*pi**(3/2)*a*x*sqrt(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. \[ \int \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

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

[In]

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

[Out]

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