3.4.75 \(\int \frac {(c-a^2 c x^2)^2}{\text {ArcSin}(a x)^2} \, dx\) [375]

Optimal. Leaf size=78 \[ -\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \text {ArcSin}(a x)}-\frac {5 c^2 \text {Si}(\text {ArcSin}(a x))}{8 a}-\frac {15 c^2 \text {Si}(3 \text {ArcSin}(a x))}{16 a}-\frac {5 c^2 \text {Si}(5 \text {ArcSin}(a x))}{16 a} \]

[Out]

-c^2*(-a^2*x^2+1)^(5/2)/a/arcsin(a*x)-5/8*c^2*Si(arcsin(a*x))/a-15/16*c^2*Si(3*arcsin(a*x))/a-5/16*c^2*Si(5*ar
csin(a*x))/a

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Rubi [A]
time = 0.11, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4751, 4809, 4491, 3380} \begin {gather*} -\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \text {ArcSin}(a x)}-\frac {5 c^2 \text {Si}(\text {ArcSin}(a x))}{8 a}-\frac {15 c^2 \text {Si}(3 \text {ArcSin}(a x))}{16 a}-\frac {5 c^2 \text {Si}(5 \text {ArcSin}(a x))}{16 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - a^2*c*x^2)^2/ArcSin[a*x]^2,x]

[Out]

-((c^2*(1 - a^2*x^2)^(5/2))/(a*ArcSin[a*x])) - (5*c^2*SinIntegral[ArcSin[a*x]])/(8*a) - (15*c^2*SinIntegral[3*
ArcSin[a*x]])/(16*a) - (5*c^2*SinIntegral[5*ArcSin[a*x]])/(16*a)

Rule 3380

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

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 4751

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[Sqrt[1 - c^2*x^2]*(
d + e*x^2)^p*((a + b*ArcSin[c*x])^(n + 1)/(b*c*(n + 1))), x] + Dist[c*((2*p + 1)/(b*(n + 1)))*Simp[(d + e*x^2)
^p/(1 - c^2*x^2)^p], Int[x*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e
, p}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -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*} \int \frac {\left (c-a^2 c x^2\right )^2}{\sin ^{-1}(a x)^2} \, dx &=-\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \sin ^{-1}(a x)}-\left (5 a c^2\right ) \int \frac {x \left (1-a^2 x^2\right )^{3/2}}{\sin ^{-1}(a x)} \, dx\\ &=-\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \sin ^{-1}(a x)}-\frac {\left (5 c^2\right ) \text {Subst}\left (\int \frac {\cos ^4(x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=-\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \sin ^{-1}(a x)}-\frac {\left (5 c^2\right ) \text {Subst}\left (\int \left (\frac {\sin (x)}{8 x}+\frac {3 \sin (3 x)}{16 x}+\frac {\sin (5 x)}{16 x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=-\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \sin ^{-1}(a x)}-\frac {\left (5 c^2\right ) \text {Subst}\left (\int \frac {\sin (5 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{16 a}-\frac {\left (5 c^2\right ) \text {Subst}\left (\int \frac {\sin (x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{8 a}-\frac {\left (15 c^2\right ) \text {Subst}\left (\int \frac {\sin (3 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{16 a}\\ &=-\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \sin ^{-1}(a x)}-\frac {5 c^2 \text {Si}\left (\sin ^{-1}(a x)\right )}{8 a}-\frac {15 c^2 \text {Si}\left (3 \sin ^{-1}(a x)\right )}{16 a}-\frac {5 c^2 \text {Si}\left (5 \sin ^{-1}(a x)\right )}{16 a}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 70, normalized size = 0.90 \begin {gather*} -\frac {c^2 \left (16 \left (1-a^2 x^2\right )^{5/2}+10 \text {ArcSin}(a x) \text {Si}(\text {ArcSin}(a x))+15 \text {ArcSin}(a x) \text {Si}(3 \text {ArcSin}(a x))+5 \text {ArcSin}(a x) \text {Si}(5 \text {ArcSin}(a x))\right )}{16 a \text {ArcSin}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c - a^2*c*x^2)^2/ArcSin[a*x]^2,x]

[Out]

-1/16*(c^2*(16*(1 - a^2*x^2)^(5/2) + 10*ArcSin[a*x]*SinIntegral[ArcSin[a*x]] + 15*ArcSin[a*x]*SinIntegral[3*Ar
cSin[a*x]] + 5*ArcSin[a*x]*SinIntegral[5*ArcSin[a*x]]))/(a*ArcSin[a*x])

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Maple [A]
time = 0.10, size = 83, normalized size = 1.06

method	result	size
derivativedivides	\(-\frac {c^{2} \left (10 \sinIntegral \left (\arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+15 \sinIntegral \left (3 \arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+5 \sinIntegral \left (5 \arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+10 \sqrt {-a^{2} x^{2}+1}+5 \cos \left (3 \arcsin \left (a x \right )\right )+\cos \left (5 \arcsin \left (a x \right )\right )\right )}{16 a \arcsin \left (a x \right )}\)	\(83\)
default	\(-\frac {c^{2} \left (10 \sinIntegral \left (\arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+15 \sinIntegral \left (3 \arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+5 \sinIntegral \left (5 \arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+10 \sqrt {-a^{2} x^{2}+1}+5 \cos \left (3 \arcsin \left (a x \right )\right )+\cos \left (5 \arcsin \left (a x \right )\right )\right )}{16 a \arcsin \left (a x \right )}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/a*c^2*(10*Si(arcsin(a*x))*arcsin(a*x)+15*Si(3*arcsin(a*x))*arcsin(a*x)+5*Si(5*arcsin(a*x))*arcsin(a*x)+1
0*(-a^2*x^2+1)^(1/2)+5*cos(3*arcsin(a*x))+cos(5*arcsin(a*x)))/arcsin(a*x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))*integrate(5*(a^3*c^2*x^3 - a*c^2*x)*sqrt(a*x + 1)*sqrt(-a*x + 1)
/arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1)), x) - (a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2)*sqrt(a*x + 1)*sqrt(-a*x
+ 1))/(a*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1)))

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} c^{2} \left (\int \left (- \frac {2 a^{2} x^{2}}{\operatorname {asin}^{2}{\left (a x \right )}}\right )\, dx + \int \frac {a^{4} x^{4}}{\operatorname {asin}^{2}{\left (a x \right )}}\, dx + \int \frac {1}{\operatorname {asin}^{2}{\left (a x \right )}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**2*(Integral(-2*a**2*x**2/asin(a*x)**2, x) + Integral(a**4*x**4/asin(a*x)**2, x) + Integral(asin(a*x)**(-2),
x))

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Giac [A]
time = 0.48, size = 81, normalized size = 1.04 \begin {gather*} -\frac {{\left (a^{2} x^{2} - 1\right )}^{2} \sqrt {-a^{2} x^{2} + 1} c^{2}}{a \arcsin \left (a x\right )} - \frac {5 \, c^{2} \operatorname {Si}\left (5 \, \arcsin \left (a x\right )\right )}{16 \, a} - \frac {15 \, c^{2} \operatorname {Si}\left (3 \, \arcsin \left (a x\right )\right )}{16 \, a} - \frac {5 \, c^{2} \operatorname {Si}\left (\arcsin \left (a x\right )\right )}{8 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*x^2 - 1)^2*sqrt(-a^2*x^2 + 1)*c^2/(a*arcsin(a*x)) - 5/16*c^2*sin_integral(5*arcsin(a*x))/a - 15/16*c^2*s
in_integral(3*arcsin(a*x))/a - 5/8*c^2*sin_integral(arcsin(a*x))/a

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c-a^2\,c\,x^2\right )}^2}{{\mathrm {asin}\left (a\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - a^2*c*x^2)^2/asin(a*x)^2,x)

[Out]

int((c - a^2*c*x^2)^2/asin(a*x)^2, x)

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