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

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

[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)

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Rubi [A]  time = 0.143125, antiderivative size = 127, normalized size of antiderivative = 1.02, 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 = {266, 43, 5732, 12, 373} \[ \frac{\pi ^{3/2} \left (c^2 x^2+1\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^4}-\frac{\pi ^{3/2} \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^4}-\frac{1}{49} \pi ^{3/2} b c^3 x^7+\frac{2 \pi ^{3/2} b x}{35 c^3}-\frac{8}{175} \pi ^{3/2} b c x^5-\frac{\pi ^{3/2} b x^3}{105 c} \]

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 - (P
i^(3/2)*(1 + c^2*x^2)^(5/2)*(a + b*ArcSinh[c*x]))/(5*c^4) + (Pi^(3/2)*(1 + c^2*x^2)^(7/2)*(a + b*ArcSinh[c*x])
)/(7*c^4)

Rule 266

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 43

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 5732

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x
^m*(1 + c^2*x^2)^p, x]}, Dist[d^p*(a + b*ArcSinh[c*x]), u, x] - Dist[b*c*d^p, Int[SimplifyIntegrand[u/Sqrt[1 +
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegerQ[p - 1/2] && (IGtQ[(m + 1)/2,
0] || ILtQ[(m + 2*p + 3)/2, 0]) && NeQ[p, -2^(-1)] && GtQ[d, 0]

Rule 12

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

Rule 373

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]

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*}

Mathematica [A]  time = 0.1717, size = 100, normalized size = 0.8 \[ \frac{\pi ^{3/2} \left (105 a \left (5 c^2 x^2-2\right ) \left (c^2 x^2+1\right )^{5/2}-b c x \left (75 c^6 x^6+168 c^4 x^4+35 c^2 x^2-210\right )+105 b \left (5 c^2 x^2-2\right ) \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)\right )}{3675 c^4} \]

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 [A]  time = 0.092, size = 195, normalized size = 1.6 \begin{align*} a \left ({\frac{{x}^{2}}{7\,\pi \,{c}^{2}} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{5}{2}}}}-{\frac{2}{35\,\pi \,{c}^{4}} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{5}{2}}}} \right ) +{\frac{b{\pi }^{{\frac{3}{2}}}}{3675\,{c}^{4}} \left ( 525\,{\it Arcsinh} \left ( cx \right ){c}^{8}{x}^{8}+1365\,{\it Arcsinh} \left ( cx \right ){c}^{6}{x}^{6}-75\,{c}^{7}{x}^{7}\sqrt{{c}^{2}{x}^{2}+1}+945\,{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{4}-168\,{c}^{5}{x}^{5}\sqrt{{c}^{2}{x}^{2}+1}-105\,{\it Arcsinh} \left ( cx \right ){c}^{2}{x}^{2}-35\,{c}^{3}{x}^{3}\sqrt{{c}^{2}{x}^{2}+1}-210\,{\it Arcsinh} \left ( cx \right ) +210\,cx\sqrt{{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \end{align*}

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]

a*(1/7*x^2*(Pi*c^2*x^2+Pi)^(5/2)/Pi/c^2-2/35/Pi/c^4*(Pi*c^2*x^2+Pi)^(5/2))+1/3675*b/c^4*Pi^(3/2)/(c^2*x^2+1)^(
1/2)*(525*arcsinh(c*x)*c^8*x^8+1365*arcsinh(c*x)*c^6*x^6-75*c^7*x^7*(c^2*x^2+1)^(1/2)+945*arcsinh(c*x)*c^4*x^4
-168*c^5*x^5*(c^2*x^2+1)^(1/2)-105*arcsinh(c*x)*c^2*x^2-35*c^3*x^3*(c^2*x^2+1)^(1/2)-210*arcsinh(c*x)+210*c*x*
(c^2*x^2+1)^(1/2))

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Maxima [A]  time = 1.20039, size = 196, normalized size = 1.57 \begin{align*} \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{align*}

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 = 2.39489, size = 485, normalized size = 3.88 \begin{align*} \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{align*}

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 [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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

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: NotImplementedError