3.1.63 \(\int x^3 (\pi +c^2 \pi x^2)^{3/2} (a+b \sinh ^{-1}(c x)) \, dx\) [63]

Optimal. Leaf size=125 \[ \frac {2 b \pi ^{3/2} x}{35 c^3}-\frac {b \pi ^{3/2} x^3}{105 c}-\frac {8}{175} b c \pi ^{3/2} x^5-\frac {1}{49} b c^3 \pi ^{3/2} x^7-\frac {\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4 \pi }+\frac {\left (\pi +c^2 \pi x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^4 \pi ^2} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {272, 45, 5804, 12, 380} \begin {gather*} \frac {\left (\pi c^2 x^2+\pi \right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 \pi ^2 c^4}-\frac {\left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 \pi c^4}-\frac {1}{49} \pi ^{3/2} b c^3 x^7+\frac {2 \pi ^{3/2} b x}{35 c^3}-\frac {8}{175} \pi ^{3/2} b c x^5-\frac {\pi ^{3/2} b x^3}{105 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 380

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rule 5804

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]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]], Int[
SimplifyIntegrand[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])

Rubi steps

\begin {align*} \int x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=-\frac {\pi ^{3/2} \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4}+\frac {\pi ^{3/2} \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^4}-\left (b c \pi ^{3/2}\right ) \int \frac {\left (1+c^2 x^2\right )^2 \left (-2+5 c^2 x^2\right )}{35 c^4} \, dx\\ &=-\frac {\pi ^{3/2} \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4}+\frac {\pi ^{3/2} \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^4}-\frac {\left (b \pi ^{3/2}\right ) \int \left (1+c^2 x^2\right )^2 \left (-2+5 c^2 x^2\right ) \, dx}{35 c^3}\\ &=-\frac {\pi ^{3/2} \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4}+\frac {\pi ^{3/2} \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^4}-\frac {\left (b \pi ^{3/2}\right ) \int \left (-2+c^2 x^2+8 c^4 x^4+5 c^6 x^6\right ) \, dx}{35 c^3}\\ &=\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 {\pi ^{3/2} \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4}+\frac {\pi ^{3/2} \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^4}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 100, normalized size = 0.80 \begin {gather*} \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 ) \sinh ^{-1}(c x)\right )}{3675 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{3} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}} \left (a +b \arcsinh \left (c x \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 145, normalized size = 1.16 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*(p
i + 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
*pi^(3/2)*c^4*x^5 + 35*pi^(3/2)*c^2*x^3 - 210*pi^(3/2)*x)*b/c^3

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Fricas [A]
time = 0.39, size = 199, normalized size = 1.59 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (117) = 234\).
time = 11.90, size = 301, normalized size = 2.41 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

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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(x^3*(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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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