3.1.74 \(\int (\pi +c^2 \pi x^2)^{5/2} (a+b \sinh ^{-1}(c x)) \, dx\) [74]

Optimal. Leaf size=165 \[ -\frac {25}{96} b c \pi ^{5/2} x^2-\frac {5}{96} b c^3 \pi ^{5/2} x^4-\frac {b \pi ^{5/2} \left (1+c^2 x^2\right )^3}{36 c}+\frac {5}{16} \pi ^2 x \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{24} \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5 \pi ^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c} \]

[Out]

-25/96*b*c*Pi^(5/2)*x^2-5/96*b*c^3*Pi^(5/2)*x^4-1/36*b*Pi^(5/2)*(c^2*x^2+1)^3/c+5/24*Pi*x*(Pi*c^2*x^2+Pi)^(3/2
)*(a+b*arcsinh(c*x))+1/6*x*(Pi*c^2*x^2+Pi)^(5/2)*(a+b*arcsinh(c*x))+5/32*Pi^(5/2)*(a+b*arcsinh(c*x))^2/b/c+5/1
6*Pi^2*x*(a+b*arcsinh(c*x))*(Pi*c^2*x^2+Pi)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5786, 5785, 5783, 30, 14, 267} \begin {gather*} \frac {1}{6} x \left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{24} \pi x \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{16} \pi ^2 x \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac {5 \pi ^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c}-\frac {5}{96} \pi ^{5/2} b c^3 x^4-\frac {\pi ^{5/2} b \left (c^2 x^2+1\right )^3}{36 c}-\frac {25}{96} \pi ^{5/2} b c x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-25*b*c*Pi^(5/2)*x^2)/96 - (5*b*c^3*Pi^(5/2)*x^4)/96 - (b*Pi^(5/2)*(1 + c^2*x^2)^3)/(36*c) + (5*Pi^2*x*Sqrt[P
i + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/16 + (5*Pi*x*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/24 + (x*(Pi +
c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/6 + (5*Pi^(5/2)*(a + b*ArcSinh[c*x])^2)/(32*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 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -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*} \int \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac {1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} (5 \pi ) \int \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx-\frac {\left (b c \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \int x \left (1+c^2 x^2\right )^2 \, dx}{6 \sqrt {1+c^2 x^2}}\\ &=-\frac {b \pi ^2 \left (1+c^2 x^2\right )^{5/2} \sqrt {\pi +c^2 \pi x^2}}{36 c}+\frac {5}{24} \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} \left (5 \pi ^2\right ) \int \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx-\frac {\left (5 b c \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \int x \left (1+c^2 x^2\right ) \, dx}{24 \sqrt {1+c^2 x^2}}\\ &=-\frac {b \pi ^2 \left (1+c^2 x^2\right )^{5/2} \sqrt {\pi +c^2 \pi x^2}}{36 c}+\frac {5}{16} \pi ^2 x \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{24} \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {\left (5 \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{16 \sqrt {1+c^2 x^2}}-\frac {\left (5 b c \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \int \left (x+c^2 x^3\right ) \, dx}{24 \sqrt {1+c^2 x^2}}-\frac {\left (5 b c \pi ^2 \sqrt {\pi +c^2 \pi x^2}\right ) \int x \, dx}{16 \sqrt {1+c^2 x^2}}\\ &=-\frac {25 b c \pi ^2 x^2 \sqrt {\pi +c^2 \pi x^2}}{96 \sqrt {1+c^2 x^2}}-\frac {5 b c^3 \pi ^2 x^4 \sqrt {\pi +c^2 \pi x^2}}{96 \sqrt {1+c^2 x^2}}-\frac {b \pi ^2 \left (1+c^2 x^2\right )^{5/2} \sqrt {\pi +c^2 \pi x^2}}{36 c}+\frac {5}{16} \pi ^2 x \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5}{24} \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5 \pi ^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c \sqrt {1+c^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.26, size = 153, normalized size = 0.93 \begin {gather*} \frac {\pi ^{5/2} \left (1584 a c x \sqrt {1+c^2 x^2}+1248 a c^3 x^3 \sqrt {1+c^2 x^2}+384 a c^5 x^5 \sqrt {1+c^2 x^2}+360 b \sinh ^{-1}(c x)^2-270 b \cosh \left (2 \sinh ^{-1}(c x)\right )-27 b \cosh \left (4 \sinh ^{-1}(c x)\right )-2 b \cosh \left (6 \sinh ^{-1}(c x)\right )+12 \sinh ^{-1}(c x) \left (60 a+45 b \sinh \left (2 \sinh ^{-1}(c x)\right )+9 b \sinh \left (4 \sinh ^{-1}(c x)\right )+b \sinh \left (6 \sinh ^{-1}(c x)\right )\right )\right )}{2304 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Pi^(5/2)*(1584*a*c*x*Sqrt[1 + c^2*x^2] + 1248*a*c^3*x^3*Sqrt[1 + c^2*x^2] + 384*a*c^5*x^5*Sqrt[1 + c^2*x^2] +
360*b*ArcSinh[c*x]^2 - 270*b*Cosh[2*ArcSinh[c*x]] - 27*b*Cosh[4*ArcSinh[c*x]] - 2*b*Cosh[6*ArcSinh[c*x]] + 12
*ArcSinh[c*x]*(60*a + 45*b*Sinh[2*ArcSinh[c*x]] + 9*b*Sinh[4*ArcSinh[c*x]] + b*Sinh[6*ArcSinh[c*x]])))/(2304*c
)

________________________________________________________________________________________

Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}} \left (a +b \arcsinh \left (c x \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(pi^2*a*c^4*x^4 + 2*pi^2*a*c^2*x^2 + pi^2*a + (pi^2*b*c^4*x^4 + 2*pi^2*b*c^2*x^
2 + pi^2*b)*arcsinh(c*x)), x)

________________________________________________________________________________________

Sympy [A]
time = 21.67, size = 265, normalized size = 1.61 \begin {gather*} \begin {cases} \frac {\pi ^{\frac {5}{2}} a c^{4} x^{5} \sqrt {c^{2} x^{2} + 1}}{6} + \frac {13 \pi ^{\frac {5}{2}} a c^{2} x^{3} \sqrt {c^{2} x^{2} + 1}}{24} + \frac {11 \pi ^{\frac {5}{2}} a x \sqrt {c^{2} x^{2} + 1}}{16} + \frac {5 \pi ^{\frac {5}{2}} a \operatorname {asinh}{\left (c x \right )}}{16 c} - \frac {\pi ^{\frac {5}{2}} b c^{5} x^{6}}{36} + \frac {\pi ^{\frac {5}{2}} b c^{4} x^{5} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{6} - \frac {13 \pi ^{\frac {5}{2}} b c^{3} x^{4}}{96} + \frac {13 \pi ^{\frac {5}{2}} b c^{2} x^{3} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{24} - \frac {11 \pi ^{\frac {5}{2}} b c x^{2}}{32} + \frac {11 \pi ^{\frac {5}{2}} b x \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{16} + \frac {5 \pi ^{\frac {5}{2}} b \operatorname {asinh}^{2}{\left (c x \right )}}{32 c} & \text {for}\: c \neq 0 \\\pi ^{\frac {5}{2}} a x & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((pi**(5/2)*a*c**4*x**5*sqrt(c**2*x**2 + 1)/6 + 13*pi**(5/2)*a*c**2*x**3*sqrt(c**2*x**2 + 1)/24 + 11*
pi**(5/2)*a*x*sqrt(c**2*x**2 + 1)/16 + 5*pi**(5/2)*a*asinh(c*x)/(16*c) - pi**(5/2)*b*c**5*x**6/36 + pi**(5/2)*
b*c**4*x**5*sqrt(c**2*x**2 + 1)*asinh(c*x)/6 - 13*pi**(5/2)*b*c**3*x**4/96 + 13*pi**(5/2)*b*c**2*x**3*sqrt(c**
2*x**2 + 1)*asinh(c*x)/24 - 11*pi**(5/2)*b*c*x**2/32 + 11*pi**(5/2)*b*x*sqrt(c**2*x**2 + 1)*asinh(c*x)/16 + 5*
pi**(5/2)*b*asinh(c*x)**2/(32*c), Ne(c, 0)), (pi**(5/2)*a*x, True))

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((pi*c^2*x^2+pi)^(5/2)*(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]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (\Pi \,c^2\,x^2+\Pi \right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________