3.1.71 \(\int x^3 (\pi +c^2 \pi x^2)^{5/2} (a+b \sinh ^{-1}(c x)) \, dx\) [71]

Optimal. Leaf size=141 \[ \frac {2 b \pi ^{5/2} x}{63 c^3}-\frac {b \pi ^{5/2} x^3}{189 c}-\frac {1}{21} b c \pi ^{5/2} x^5-\frac {19}{441} b c^3 \pi ^{5/2} x^7-\frac {1}{81} b c^5 \pi ^{5/2} x^9-\frac {\left (\pi +c^2 \pi x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^4 \pi }+\frac {\left (\pi +c^2 \pi x^2\right )^{9/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^4 \pi ^2} \]

[Out]

2/63*b*Pi^(5/2)*x/c^3-1/189*b*Pi^(5/2)*x^3/c-1/21*b*c*Pi^(5/2)*x^5-19/441*b*c^3*Pi^(5/2)*x^7-1/81*b*c^5*Pi^(5/
2)*x^9-1/7*(Pi*c^2*x^2+Pi)^(7/2)*(a+b*arcsinh(c*x))/c^4/Pi+1/9*(Pi*c^2*x^2+Pi)^(9/2)*(a+b*arcsinh(c*x))/c^4/Pi
^2

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Rubi [A]
time = 0.11, antiderivative size = 141, 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 )^{9/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 \pi ^2 c^4}-\frac {\left (\pi c^2 x^2+\pi \right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 \pi c^4}-\frac {1}{81} \pi ^{5/2} b c^5 x^9-\frac {19}{441} \pi ^{5/2} b c^3 x^7+\frac {2 \pi ^{5/2} b x}{63 c^3}-\frac {1}{21} \pi ^{5/2} b c x^5-\frac {\pi ^{5/2} b x^3}{189 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x]),x]

[Out]

(2*b*Pi^(5/2)*x)/(63*c^3) - (b*Pi^(5/2)*x^3)/(189*c) - (b*c*Pi^(5/2)*x^5)/21 - (19*b*c^3*Pi^(5/2)*x^7)/441 - (
b*c^5*Pi^(5/2)*x^9)/81 - ((Pi + c^2*Pi*x^2)^(7/2)*(a + b*ArcSinh[c*x]))/(7*c^4*Pi) + ((Pi + c^2*Pi*x^2)^(9/2)*
(a + b*ArcSinh[c*x]))/(9*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 )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=-\frac {\pi ^{5/2} \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^4}+\frac {\pi ^{5/2} \left (1+c^2 x^2\right )^{9/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^4}-\left (b c \pi ^{5/2}\right ) \int \frac {\left (1+c^2 x^2\right )^3 \left (-2+7 c^2 x^2\right )}{63 c^4} \, dx\\ &=-\frac {\pi ^{5/2} \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^4}+\frac {\pi ^{5/2} \left (1+c^2 x^2\right )^{9/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^4}-\frac {\left (b \pi ^{5/2}\right ) \int \left (1+c^2 x^2\right )^3 \left (-2+7 c^2 x^2\right ) \, dx}{63 c^3}\\ &=-\frac {\pi ^{5/2} \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^4}+\frac {\pi ^{5/2} \left (1+c^2 x^2\right )^{9/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^4}-\frac {\left (b \pi ^{5/2}\right ) \int \left (-2+c^2 x^2+15 c^4 x^4+19 c^6 x^6+7 c^8 x^8\right ) \, dx}{63 c^3}\\ &=\frac {2 b \pi ^{5/2} x}{63 c^3}-\frac {b \pi ^{5/2} x^3}{189 c}-\frac {1}{21} b c \pi ^{5/2} x^5-\frac {19}{441} b c^3 \pi ^{5/2} x^7-\frac {1}{81} b c^5 \pi ^{5/2} x^9-\frac {\pi ^{5/2} \left (1+c^2 x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^4}+\frac {\pi ^{5/2} \left (1+c^2 x^2\right )^{9/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^4}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 108, normalized size = 0.77 \begin {gather*} \frac {\pi ^{5/2} \left (63 a \left (1+c^2 x^2\right )^{7/2} \left (-2+7 c^2 x^2\right )-b c x \left (-126+21 c^2 x^2+189 c^4 x^4+171 c^6 x^6+49 c^8 x^8\right )+63 b \left (1+c^2 x^2\right )^{7/2} \left (-2+7 c^2 x^2\right ) \sinh ^{-1}(c x)\right )}{3969 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x]),x]

[Out]

(Pi^(5/2)*(63*a*(1 + c^2*x^2)^(7/2)*(-2 + 7*c^2*x^2) - b*c*x*(-126 + 21*c^2*x^2 + 189*c^4*x^4 + 171*c^6*x^6 +
49*c^8*x^8) + 63*b*(1 + c^2*x^2)^(7/2)*(-2 + 7*c^2*x^2)*ArcSinh[c*x]))/(3969*c^4)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{3} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}} \left (a +b \arcsinh \left (c x \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(Pi*c^2*x^2+Pi)^(5/2)*(a+b*arcsinh(c*x)),x)

[Out]

int(x^3*(Pi*c^2*x^2+Pi)^(5/2)*(a+b*arcsinh(c*x)),x)

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Maxima [A]
time = 0.29, size = 156, normalized size = 1.11 \begin {gather*} \frac {1}{63} \, {\left (\frac {7 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {7}{2}} x^{2}}{\pi c^{2}} - \frac {2 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {7}{2}}}{\pi c^{4}}\right )} b \operatorname {arsinh}\left (c x\right ) + \frac {1}{63} \, {\left (\frac {7 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {7}{2}} x^{2}}{\pi c^{2}} - \frac {2 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {7}{2}}}{\pi c^{4}}\right )} a - \frac {{\left (49 \, \pi ^{\frac {5}{2}} c^{8} x^{9} + 171 \, \pi ^{\frac {5}{2}} c^{6} x^{7} + 189 \, \pi ^{\frac {5}{2}} c^{4} x^{5} + 21 \, \pi ^{\frac {5}{2}} c^{2} x^{3} - 126 \, \pi ^{\frac {5}{2}} x\right )} b}{3969 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(pi*c^2*x^2+pi)^(5/2)*(a+b*arcsinh(c*x)),x, algorithm="maxima")

[Out]

1/63*(7*(pi + pi*c^2*x^2)^(7/2)*x^2/(pi*c^2) - 2*(pi + pi*c^2*x^2)^(7/2)/(pi*c^4))*b*arcsinh(c*x) + 1/63*(7*(p
i + pi*c^2*x^2)^(7/2)*x^2/(pi*c^2) - 2*(pi + pi*c^2*x^2)^(7/2)/(pi*c^4))*a - 1/3969*(49*pi^(5/2)*c^8*x^9 + 171
*pi^(5/2)*c^6*x^7 + 189*pi^(5/2)*c^4*x^5 + 21*pi^(5/2)*c^2*x^3 - 126*pi^(5/2)*x)*b/c^3

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (113) = 226\).
time = 0.38, size = 263, normalized size = 1.87 \begin {gather*} \frac {63 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (7 \, \pi ^{2} b c^{10} x^{10} + 26 \, \pi ^{2} b c^{8} x^{8} + 34 \, \pi ^{2} b c^{6} x^{6} + 16 \, \pi ^{2} b c^{4} x^{4} - \pi ^{2} b c^{2} x^{2} - 2 \, \pi ^{2} b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + \sqrt {\pi + \pi c^{2} x^{2}} {\left (441 \, \pi ^{2} a c^{10} x^{10} + 1638 \, \pi ^{2} a c^{8} x^{8} + 2142 \, \pi ^{2} a c^{6} x^{6} + 1008 \, \pi ^{2} a c^{4} x^{4} - 63 \, \pi ^{2} a c^{2} x^{2} - 126 \, \pi ^{2} a - {\left (49 \, \pi ^{2} b c^{9} x^{9} + 171 \, \pi ^{2} b c^{7} x^{7} + 189 \, \pi ^{2} b c^{5} x^{5} + 21 \, \pi ^{2} b c^{3} x^{3} - 126 \, \pi ^{2} b c x\right )} \sqrt {c^{2} x^{2} + 1}\right )}}{3969 \, {\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)^(5/2)*(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

1/3969*(63*sqrt(pi + pi*c^2*x^2)*(7*pi^2*b*c^10*x^10 + 26*pi^2*b*c^8*x^8 + 34*pi^2*b*c^6*x^6 + 16*pi^2*b*c^4*x
^4 - pi^2*b*c^2*x^2 - 2*pi^2*b)*log(c*x + sqrt(c^2*x^2 + 1)) + sqrt(pi + pi*c^2*x^2)*(441*pi^2*a*c^10*x^10 + 1
638*pi^2*a*c^8*x^8 + 2142*pi^2*a*c^6*x^6 + 1008*pi^2*a*c^4*x^4 - 63*pi^2*a*c^2*x^2 - 126*pi^2*a - (49*pi^2*b*c
^9*x^9 + 171*pi^2*b*c^7*x^7 + 189*pi^2*b*c^5*x^5 + 21*pi^2*b*c^3*x^3 - 126*pi^2*b*c*x)*sqrt(c^2*x^2 + 1)))/(c^
6*x^2 + c^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(pi*c**2*x**2+pi)**(5/2)*(a+b*asinh(c*x)),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3060 deep

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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)^(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

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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 )}^{5/2} \,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)^(5/2),x)

[Out]

int(x^3*(a + b*asinh(c*x))*(Pi + Pi*c^2*x^2)^(5/2), x)

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