3.1.65 \(\int x (\pi +c^2 \pi x^2)^{3/2} (a+b \sinh ^{-1}(c x)) \, dx\) [65]

Optimal. Leaf size=77 \[ -\frac {b \pi ^{3/2} x}{5 c}-\frac {2}{15} b c \pi ^{3/2} x^3-\frac {1}{25} b c^3 \pi ^{3/2} x^5+\frac {\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 \pi } \]

[Out]

-1/5*b*Pi^(3/2)*x/c-2/15*b*c*Pi^(3/2)*x^3-1/25*b*c^3*Pi^(3/2)*x^5+1/5*(Pi*c^2*x^2+Pi)^(5/2)*(a+b*arcsinh(c*x))
/c^2/Pi

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Rubi [A]
time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {5798, 200} \begin {gather*} \frac {\left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 \pi c^2}-\frac {1}{25} \pi ^{3/2} b c^3 x^5-\frac {2}{15} \pi ^{3/2} b c x^3-\frac {\pi ^{3/2} b x}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(b*Pi^(3/2)*x)/c - (2*b*c*Pi^(3/2)*x^3)/15 - (b*c^3*Pi^(3/2)*x^5)/25 + ((Pi + c^2*Pi*x^2)^(5/2)*(a + b*Ar
cSinh[c*x]))/(5*c^2*Pi)

Rule 200

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

Rule 5798

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^
(p + 1)*((a + b*ArcSinh[c*x])^n/(2*e*(p + 1))), x] - Dist[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^p], Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e
, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 72, normalized size = 0.94 \begin {gather*} \frac {\pi ^{3/2} \left (15 a \left (1+c^2 x^2\right )^{5/2}-b c x \left (15+10 c^2 x^2+3 c^4 x^4\right )+15 b \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)\right )}{75 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Pi^(3/2)*(15*a*(1 + c^2*x^2)^(5/2) - b*c*x*(15 + 10*c^2*x^2 + 3*c^4*x^4) + 15*b*(1 + c^2*x^2)^(5/2)*ArcSinh[c
*x]))/(75*c^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 85, normalized size = 1.10 \begin {gather*} \frac {{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}} b \operatorname {arsinh}\left (c x\right )}{5 \, \pi c^{2}} + \frac {{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}} a}{5 \, \pi c^{2}} - \frac {{\left (3 \, \pi ^{\frac {5}{2}} c^{4} x^{5} + 10 \, \pi ^{\frac {5}{2}} c^{2} x^{3} + 15 \, \pi ^{\frac {5}{2}} x\right )} b}{75 \, \pi c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(pi + pi*c^2*x^2)^(5/2)*b*arcsinh(c*x)/(pi*c^2) + 1/5*(pi + pi*c^2*x^2)^(5/2)*a/(pi*c^2) - 1/75*(3*pi^(5/2
)*c^4*x^5 + 10*pi^(5/2)*c^2*x^3 + 15*pi^(5/2)*x)*b/(pi*c)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (61) = 122\).
time = 0.38, size = 167, normalized size = 2.17 \begin {gather*} \frac {15 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (\pi b c^{6} x^{6} + 3 \, \pi b c^{4} x^{4} + 3 \, \pi b c^{2} x^{2} + \pi b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + \sqrt {\pi + \pi c^{2} x^{2}} {\left (15 \, \pi a c^{6} x^{6} + 45 \, \pi a c^{4} x^{4} + 45 \, \pi a c^{2} x^{2} + 15 \, \pi a - {\left (3 \, \pi b c^{5} x^{5} + 10 \, \pi b c^{3} x^{3} + 15 \, \pi b c x\right )} \sqrt {c^{2} x^{2} + 1}\right )}}{75 \, {\left (c^{4} x^{2} + c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/75*(15*sqrt(pi + pi*c^2*x^2)*(pi*b*c^6*x^6 + 3*pi*b*c^4*x^4 + 3*pi*b*c^2*x^2 + pi*b)*log(c*x + sqrt(c^2*x^2
+ 1)) + sqrt(pi + pi*c^2*x^2)*(15*pi*a*c^6*x^6 + 45*pi*a*c^4*x^4 + 45*pi*a*c^2*x^2 + 15*pi*a - (3*pi*b*c^5*x^5
+ 10*pi*b*c^3*x^3 + 15*pi*b*c*x)*sqrt(c^2*x^2 + 1)))/(c^4*x^2 + c^2)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (70) = 140\).
time = 3.96, size = 221, normalized size = 2.87 \begin {gather*} \begin {cases} \frac {\pi ^{\frac {3}{2}} a c^{2} x^{4} \sqrt {c^{2} x^{2} + 1}}{5} + \frac {2 \pi ^{\frac {3}{2}} a x^{2} \sqrt {c^{2} x^{2} + 1}}{5} + \frac {\pi ^{\frac {3}{2}} a \sqrt {c^{2} x^{2} + 1}}{5 c^{2}} - \frac {\pi ^{\frac {3}{2}} b c^{3} x^{5}}{25} + \frac {\pi ^{\frac {3}{2}} b c^{2} x^{4} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{5} - \frac {2 \pi ^{\frac {3}{2}} b c x^{3}}{15} + \frac {2 \pi ^{\frac {3}{2}} b x^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{5} - \frac {\pi ^{\frac {3}{2}} b x}{5 c} + \frac {\pi ^{\frac {3}{2}} b \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{5 c^{2}} & \text {for}\: c \neq 0 \\\frac {\pi ^{\frac {3}{2}} a x^{2}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((pi**(3/2)*a*c**2*x**4*sqrt(c**2*x**2 + 1)/5 + 2*pi**(3/2)*a*x**2*sqrt(c**2*x**2 + 1)/5 + pi**(3/2)*
a*sqrt(c**2*x**2 + 1)/(5*c**2) - pi**(3/2)*b*c**3*x**5/25 + pi**(3/2)*b*c**2*x**4*sqrt(c**2*x**2 + 1)*asinh(c*
x)/5 - 2*pi**(3/2)*b*c*x**3/15 + 2*pi**(3/2)*b*x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/5 - pi**(3/2)*b*x/(5*c) + p
i**(3/2)*b*sqrt(c**2*x**2 + 1)*asinh(c*x)/(5*c**2), Ne(c, 0)), (pi**(3/2)*a*x**2/2, 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*(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\,\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*(a + b*asinh(c*x))*(Pi + Pi*c^2*x^2)^(3/2),x)

[Out]

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

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