3.1.92 \(\int \frac {x^3 (a+b \sinh ^{-1}(c x))}{(\pi +c^2 \pi x^2)^{3/2}} \, dx\) [92]

Optimal. Leaf size=86 \[ -\frac {b x}{c^3 \pi ^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{c^4 \pi \sqrt {\pi +c^2 \pi x^2}}+\frac {\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^4 \pi ^2}-\frac {b \text {ArcTan}(c x)}{c^4 \pi ^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {272, 45, 5804, 12, 396, 209} \begin {gather*} \frac {\sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^2 c^4}+\frac {a+b \sinh ^{-1}(c x)}{\pi c^4 \sqrt {\pi c^2 x^2+\pi }}-\frac {b \text {ArcTan}(c x)}{\pi ^{3/2} c^4}-\frac {b x}{\pi ^{3/2} c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*x)/(c^3*Pi^(3/2))) + (a + b*ArcSinh[c*x])/(c^4*Pi*Sqrt[Pi + c^2*Pi*x^2]) + (Sqrt[Pi + c^2*Pi*x^2]*(a + b*
ArcSinh[c*x]))/(c^4*Pi^2) - (b*ArcTan[c*x])/(c^4*Pi^(3/2))

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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 \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx &=\frac {a+b \sinh ^{-1}(c x)}{c^4 \pi ^{3/2} \sqrt {1+c^2 x^2}}+\frac {\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^4 \pi ^{3/2}}-\frac {(b c) \int \frac {2+c^2 x^2}{c^4+c^6 x^2} \, dx}{\pi ^{3/2}}\\ &=-\frac {b x}{c^3 \pi ^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{c^4 \pi ^{3/2} \sqrt {1+c^2 x^2}}+\frac {\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^4 \pi ^{3/2}}-\frac {(b c) \int \frac {1}{c^4+c^6 x^2} \, dx}{\pi ^{3/2}}\\ &=-\frac {b x}{c^3 \pi ^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{c^4 \pi ^{3/2} \sqrt {1+c^2 x^2}}+\frac {\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^4 \pi ^{3/2}}-\frac {b \tan ^{-1}(c x)}{c^4 \pi ^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 87, normalized size = 1.01 \begin {gather*} \frac {2 a+a c^2 x^2-b c x \sqrt {1+c^2 x^2}+b \left (2+c^2 x^2\right ) \sinh ^{-1}(c x)-b \sqrt {1+c^2 x^2} \text {ArcTan}(c x)}{c^4 \pi ^{3/2} \sqrt {1+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a + a*c^2*x^2 - b*c*x*Sqrt[1 + c^2*x^2] + b*(2 + c^2*x^2)*ArcSinh[c*x] - b*Sqrt[1 + c^2*x^2]*ArcTan[c*x])/(
c^4*Pi^(3/2)*Sqrt[1 + c^2*x^2])

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Maple [C] Result contains complex when optimal does not.
time = 5.15, size = 159, normalized size = 1.85

method	result	size
default	\(a \left (\frac {x^{2}}{\pi \,c^{2} \sqrt {\pi \,c^{2} x^{2}+\pi }}+\frac {2}{\pi \,c^{4} \sqrt {\pi \,c^{2} x^{2}+\pi }}\right )+\frac {b \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}}{\pi ^{\frac {3}{2}} c^{4}}-\frac {b x}{c^{3} \pi ^{\frac {3}{2}}}+\frac {b \arcsinh \left (c x \right )}{\pi ^{\frac {3}{2}} \sqrt {c^{2} x^{2}+1}\, c^{4}}+\frac {i b \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{c^{4} \pi ^{\frac {3}{2}}}-\frac {i b \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{c^{4} \pi ^{\frac {3}{2}}}\)	\(159\)

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)^(3/2),x,method=_RETURNVERBOSE)

[Out]

a*(x^2/Pi/c^2/(Pi*c^2*x^2+Pi)^(1/2)+2/Pi/c^4/(Pi*c^2*x^2+Pi)^(1/2))+b/Pi^(3/2)/c^4*arcsinh(c*x)*(c^2*x^2+1)^(1
/2)-b*x/c^3/Pi^(3/2)+b/Pi^(3/2)/(c^2*x^2+1)^(1/2)/c^4*arcsinh(c*x)+I*b/c^4/Pi^(3/2)*ln(c*x+(c^2*x^2+1)^(1/2)-I
)-I*b/c^4/Pi^(3/2)*ln(c*x+(c^2*x^2+1)^(1/2)+I)

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Maxima [A]
time = 0.53, size = 119, normalized size = 1.38 \begin {gather*} -b c {\left (\frac {x}{\pi ^{\frac {3}{2}} c^{4}} + \frac {\arctan \left (c x\right )}{\pi ^{\frac {3}{2}} c^{5}}\right )} + b {\left (\frac {x^{2}}{\pi \sqrt {\pi + \pi c^{2} x^{2}} c^{2}} + \frac {2}{\pi \sqrt {\pi + \pi c^{2} x^{2}} c^{4}}\right )} \operatorname {arsinh}\left (c x\right ) + a {\left (\frac {x^{2}}{\pi \sqrt {\pi + \pi c^{2} x^{2}} c^{2}} + \frac {2}{\pi \sqrt {\pi + \pi c^{2} 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)^(3/2),x, algorithm="maxima")

[Out]

-b*c*(x/(pi^(3/2)*c^4) + arctan(c*x)/(pi^(3/2)*c^5)) + b*(x^2/(pi*sqrt(pi + pi*c^2*x^2)*c^2) + 2/(pi*sqrt(pi +
pi*c^2*x^2)*c^4))*arcsinh(c*x) + a*(x^2/(pi*sqrt(pi + pi*c^2*x^2)*c^2) + 2/(pi*sqrt(pi + pi*c^2*x^2)*c^4))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (78) = 156\).
time = 0.42, size = 165, normalized size = 1.92 \begin {gather*} \frac {\sqrt {\pi } {\left (b c^{2} x^{2} + b\right )} \arctan \left (-\frac {2 \, \sqrt {\pi } \sqrt {\pi + \pi c^{2} x^{2}} \sqrt {c^{2} x^{2} + 1} c x}{\pi - \pi c^{4} x^{4}}\right ) + 2 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (b c^{2} x^{2} + 2 \, b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (a c^{2} x^{2} - \sqrt {c^{2} x^{2} + 1} b c x + 2 \, a\right )}}{2 \, {\left (\pi ^{2} c^{6} x^{2} + \pi ^{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)^(3/2),x, algorithm="fricas")

[Out]

1/2*(sqrt(pi)*(b*c^2*x^2 + b)*arctan(-2*sqrt(pi)*sqrt(pi + pi*c^2*x^2)*sqrt(c^2*x^2 + 1)*c*x/(pi - pi*c^4*x^4)
) + 2*sqrt(pi + pi*c^2*x^2)*(b*c^2*x^2 + 2*b)*log(c*x + sqrt(c^2*x^2 + 1)) + 2*sqrt(pi + pi*c^2*x^2)*(a*c^2*x^
2 - sqrt(c^2*x^2 + 1)*b*c*x + 2*a))/(pi^2*c^6*x^2 + pi^2*c^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a x^{3}}{c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b x^{3} \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {3}{2}}} \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)**(3/2),x)

[Out]

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

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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)^(3/2),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 \frac {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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