3.1.55 \(\int x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \sinh ^{-1}(c x)) \, dx\) [55]

Optimal. Leaf size=109 \[ \frac {2 b \sqrt {\pi } x}{15 c^3}-\frac {b \sqrt {\pi } x^3}{45 c}-\frac {1}{25} b c \sqrt {\pi } x^5-\frac {\left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 \pi }+\frac {\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4 \pi ^2} \]

[Out]

-1/3*(Pi*c^2*x^2+Pi)^(3/2)*(a+b*arcsinh(c*x))/c^4/Pi+1/5*(Pi*c^2*x^2+Pi)^(5/2)*(a+b*arcsinh(c*x))/c^4/Pi^2+2/1
5*b*x*Pi^(1/2)/c^3-1/45*b*x^3*Pi^(1/2)/c-1/25*b*c*x^5*Pi^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45, 5804, 12} \begin {gather*} \frac {\left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 \pi ^2 c^4}-\frac {\left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^4}+\frac {2 \sqrt {\pi } b x}{15 c^3}-\frac {1}{25} \sqrt {\pi } b c x^5-\frac {\sqrt {\pi } b x^3}{45 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*Sqrt[Pi]*x)/(15*c^3) - (b*Sqrt[Pi]*x^3)/(45*c) - (b*c*Sqrt[Pi]*x^5)/25 - ((Pi + c^2*Pi*x^2)^(3/2)*(a + b*
ArcSinh[c*x]))/(3*c^4*Pi) + ((Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/(5*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 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 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=-\frac {\sqrt {\pi } \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4}+\frac {\sqrt {\pi } \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4}-\left (b c \sqrt {\pi }\right ) \int \frac {-2+c^2 x^2+3 c^4 x^4}{15 c^4} \, dx\\ &=-\frac {\sqrt {\pi } \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4}+\frac {\sqrt {\pi } \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4}-\frac {\left (b \sqrt {\pi }\right ) \int \left (-2+c^2 x^2+3 c^4 x^4\right ) \, dx}{15 c^3}\\ &=\frac {2 b \sqrt {\pi } x}{15 c^3}-\frac {b \sqrt {\pi } x^3}{45 c}-\frac {1}{25} b c \sqrt {\pi } x^5-\frac {\sqrt {\pi } \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4}+\frac {\sqrt {\pi } \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 106, normalized size = 0.97 \begin {gather*} \frac {\sqrt {\pi } \left (15 a \sqrt {1+c^2 x^2} \left (-2+c^2 x^2+3 c^4 x^4\right )+b \left (30 c x-5 c^3 x^3-9 c^5 x^5\right )+15 b \sqrt {1+c^2 x^2} \left (-2+c^2 x^2+3 c^4 x^4\right ) \sinh ^{-1}(c x)\right )}{225 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Pi]*(15*a*Sqrt[1 + c^2*x^2]*(-2 + c^2*x^2 + 3*c^4*x^4) + b*(30*c*x - 5*c^3*x^3 - 9*c^5*x^5) + 15*b*Sqrt[
1 + c^2*x^2]*(-2 + c^2*x^2 + 3*c^4*x^4)*ArcSinh[c*x]))/(225*c^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arcsinh(c*x))*(Pi*c^2*x^2+Pi)^(1/2),x)

[Out]

int(x^3*(a+b*arcsinh(c*x))*(Pi*c^2*x^2+Pi)^(1/2),x)

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Maxima [A]
time = 0.27, size = 134, normalized size = 1.23 \begin {gather*} \frac {1}{15} \, b {\left (\frac {3 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x^{2}}{\pi c^{2}} - \frac {2 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}}{\pi c^{4}}\right )} \operatorname {arsinh}\left (c x\right ) + \frac {1}{15} \, a {\left (\frac {3 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x^{2}}{\pi c^{2}} - \frac {2 \, {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}}{\pi c^{4}}\right )} - \frac {{\left (9 \, \sqrt {\pi } c^{4} x^{5} + 5 \, \sqrt {\pi } c^{2} x^{3} - 30 \, \sqrt {\pi } x\right )} b}{225 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsinh(c*x))*(pi*c^2*x^2+pi)^(1/2),x, algorithm="maxima")

[Out]

1/15*b*(3*(pi + pi*c^2*x^2)^(3/2)*x^2/(pi*c^2) - 2*(pi + pi*c^2*x^2)^(3/2)/(pi*c^4))*arcsinh(c*x) + 1/15*a*(3*
(pi + pi*c^2*x^2)^(3/2)*x^2/(pi*c^2) - 2*(pi + pi*c^2*x^2)^(3/2)/(pi*c^4)) - 1/225*(9*sqrt(pi)*c^4*x^5 + 5*sqr
t(pi)*c^2*x^3 - 30*sqrt(pi)*x)*b/c^3

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Fricas [A]
time = 0.47, size = 158, normalized size = 1.45 \begin {gather*} \frac {15 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (3 \, b c^{6} x^{6} + 4 \, b c^{4} x^{4} - b c^{2} x^{2} - 2 \, b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + \sqrt {\pi + \pi c^{2} x^{2}} {\left (45 \, a c^{6} x^{6} + 60 \, a c^{4} x^{4} - 15 \, a c^{2} x^{2} - {\left (9 \, b c^{5} x^{5} + 5 \, b c^{3} x^{3} - 30 \, b c x\right )} \sqrt {c^{2} x^{2} + 1} - 30 \, a\right )}}{225 \, {\left (c^{6} x^{2} + c^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsinh(c*x))*(pi*c^2*x^2+pi)^(1/2),x, algorithm="fricas")

[Out]

1/225*(15*sqrt(pi + pi*c^2*x^2)*(3*b*c^6*x^6 + 4*b*c^4*x^4 - b*c^2*x^2 - 2*b)*log(c*x + sqrt(c^2*x^2 + 1)) + s
qrt(pi + pi*c^2*x^2)*(45*a*c^6*x^6 + 60*a*c^4*x^4 - 15*a*c^2*x^2 - (9*b*c^5*x^5 + 5*b*c^3*x^3 - 30*b*c*x)*sqrt
(c^2*x^2 + 1) - 30*a))/(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. 221 vs. \(2 (100) = 200\).
time = 1.06, size = 221, normalized size = 2.03 \begin {gather*} \begin {cases} \frac {\sqrt {\pi } a x^{4} \sqrt {c^{2} x^{2} + 1}}{5} + \frac {\sqrt {\pi } a x^{2} \sqrt {c^{2} x^{2} + 1}}{15 c^{2}} - \frac {2 \sqrt {\pi } a \sqrt {c^{2} x^{2} + 1}}{15 c^{4}} - \frac {\sqrt {\pi } b c x^{5}}{25} + \frac {\sqrt {\pi } b x^{4} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{5} - \frac {\sqrt {\pi } b x^{3}}{45 c} + \frac {\sqrt {\pi } b x^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{15 c^{2}} + \frac {2 \sqrt {\pi } b x}{15 c^{3}} - \frac {2 \sqrt {\pi } b \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{15 c^{4}} & \text {for}\: c \neq 0 \\\frac {\sqrt {\pi } a x^{4}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*asinh(c*x))*(pi*c**2*x**2+pi)**(1/2),x)

[Out]

Piecewise((sqrt(pi)*a*x**4*sqrt(c**2*x**2 + 1)/5 + sqrt(pi)*a*x**2*sqrt(c**2*x**2 + 1)/(15*c**2) - 2*sqrt(pi)*
a*sqrt(c**2*x**2 + 1)/(15*c**4) - sqrt(pi)*b*c*x**5/25 + sqrt(pi)*b*x**4*sqrt(c**2*x**2 + 1)*asinh(c*x)/5 - sq
rt(pi)*b*x**3/(45*c) + sqrt(pi)*b*x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/(15*c**2) + 2*sqrt(pi)*b*x/(15*c**3) - 2
*sqrt(pi)*b*sqrt(c**2*x**2 + 1)*asinh(c*x)/(15*c**4), Ne(c, 0)), (sqrt(pi)*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*(a+b*arcsinh(c*x))*(pi*c^2*x^2+pi)^(1/2),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 )\,\sqrt {\Pi \,c^2\,x^2+\Pi } \,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)^(1/2),x)

[Out]

int(x^3*(a + b*asinh(c*x))*(Pi + Pi*c^2*x^2)^(1/2), x)

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