\(\int \frac {x^4}{\sqrt {1-a^2 x^2} \arcsin (a x)} \, dx\) [341]

Optimal result

Integrand size = 24, antiderivative size = 41 \[ \int \frac {x^4}{\sqrt {1-a^2 x^2} \arcsin (a x)} \, dx=-\frac {\operatorname {CosIntegral}(2 \arcsin (a x))}{2 a^5}+\frac {\operatorname {CosIntegral}(4 \arcsin (a x))}{8 a^5}+\frac {3 \log (\arcsin (a x))}{8 a^5} \]

[Out]

-1/2*Ci(2*arcsin(a*x))/a^5+1/8*Ci(4*arcsin(a*x))/a^5+3/8*ln(arcsin(a*x))/a^5

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4809, 3393, 3383} \[ \int \frac {x^4}{\sqrt {1-a^2 x^2} \arcsin (a x)} \, dx=-\frac {\operatorname {CosIntegral}(2 \arcsin (a x))}{2 a^5}+\frac {\operatorname {CosIntegral}(4 \arcsin (a x))}{8 a^5}+\frac {3 \log (\arcsin (a x))}{8 a^5} \]

[In]

Int[x^4/(Sqrt[1 - a^2*x^2]*ArcSin[a*x]),x]

[Out]

-1/2*CosIntegral[2*ArcSin[a*x]]/a^5 + CosIntegral[4*ArcSin[a*x]]/(8*a^5) + (3*Log[ArcSin[a*x]])/(8*a^5)

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 4809

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\sin ^4(x)}{x} \, dx,x,\arcsin (a x)\right )}{a^5} \\ & = \frac {\text {Subst}\left (\int \left (\frac {3}{8 x}-\frac {\cos (2 x)}{2 x}+\frac {\cos (4 x)}{8 x}\right ) \, dx,x,\arcsin (a x)\right )}{a^5} \\ & = \frac {3 \log (\arcsin (a x))}{8 a^5}+\frac {\text {Subst}\left (\int \frac {\cos (4 x)}{x} \, dx,x,\arcsin (a x)\right )}{8 a^5}-\frac {\text {Subst}\left (\int \frac {\cos (2 x)}{x} \, dx,x,\arcsin (a x)\right )}{2 a^5} \\ & = -\frac {\operatorname {CosIntegral}(2 \arcsin (a x))}{2 a^5}+\frac {\operatorname {CosIntegral}(4 \arcsin (a x))}{8 a^5}+\frac {3 \log (\arcsin (a x))}{8 a^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80 \[ \int \frac {x^4}{\sqrt {1-a^2 x^2} \arcsin (a x)} \, dx=-\frac {4 \operatorname {CosIntegral}(2 \arcsin (a x))-\operatorname {CosIntegral}(4 \arcsin (a x))-3 \log (\arcsin (a x))}{8 a^5} \]

[In]

Integrate[x^4/(Sqrt[1 - a^2*x^2]*ArcSin[a*x]),x]

[Out]

-1/8*(4*CosIntegral[2*ArcSin[a*x]] - CosIntegral[4*ArcSin[a*x]] - 3*Log[ArcSin[a*x]])/a^5

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.73

method	result	size
default	\(\frac {3 \ln \left (\arcsin \left (a x \right )\right )-4 \,\operatorname {Ci}\left (2 \arcsin \left (a x \right )\right )+\operatorname {Ci}\left (4 \arcsin \left (a x \right )\right )}{8 a^{5}}\)	\(30\)

[In]

int(x^4/arcsin(a*x)/(-a^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/8*(3*ln(arcsin(a*x))-4*Ci(2*arcsin(a*x))+Ci(4*arcsin(a*x)))/a^5

Fricas [F]

\[ \int \frac {x^4}{\sqrt {1-a^2 x^2} \arcsin (a x)} \, dx=\int { \frac {x^{4}}{\sqrt {-a^{2} x^{2} + 1} \arcsin \left (a x\right )} \,d x } \]

[In]

integrate(x^4/arcsin(a*x)/(-a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)*x^4/((a^2*x^2 - 1)*arcsin(a*x)), x)

Sympy [F]

\[ \int \frac {x^4}{\sqrt {1-a^2 x^2} \arcsin (a x)} \, dx=\int \frac {x^{4}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname {asin}{\left (a x \right )}}\, dx \]

[In]

integrate(x**4/asin(a*x)/(-a**2*x**2+1)**(1/2),x)

[Out]

Integral(x**4/(sqrt(-(a*x - 1)*(a*x + 1))*asin(a*x)), x)

Maxima [F]

\[ \int \frac {x^4}{\sqrt {1-a^2 x^2} \arcsin (a x)} \, dx=\int { \frac {x^{4}}{\sqrt {-a^{2} x^{2} + 1} \arcsin \left (a x\right )} \,d x } \]

[In]

integrate(x^4/arcsin(a*x)/(-a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4/(sqrt(-a^2*x^2 + 1)*arcsin(a*x)), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \frac {x^4}{\sqrt {1-a^2 x^2} \arcsin (a x)} \, dx=\frac {\operatorname {Ci}\left (4 \, \arcsin \left (a x\right )\right )}{8 \, a^{5}} - \frac {\operatorname {Ci}\left (2 \, \arcsin \left (a x\right )\right )}{2 \, a^{5}} + \frac {3 \, \log \left (\arcsin \left (a x\right )\right )}{8 \, a^{5}} \]

[In]

integrate(x^4/arcsin(a*x)/(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

1/8*cos_integral(4*arcsin(a*x))/a^5 - 1/2*cos_integral(2*arcsin(a*x))/a^5 + 3/8*log(arcsin(a*x))/a^5

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\sqrt {1-a^2 x^2} \arcsin (a x)} \, dx=\int \frac {x^4}{\mathrm {asin}\left (a\,x\right )\,\sqrt {1-a^2\,x^2}} \,d x \]

[In]

int(x^4/(asin(a*x)*(1 - a^2*x^2)^(1/2)),x)

[Out]

int(x^4/(asin(a*x)*(1 - a^2*x^2)^(1/2)), x)