3.4.82 \(\int \frac {x^2 \sqrt {1-c^2 x^2}}{(a+b \text {ArcSin}(c x))^2} \, dx\) [382]

Optimal. Leaf size=94 \[ -\frac {x^2 \left (1-c^2 x^2\right )}{b c (a+b \text {ArcSin}(c x))}-\frac {\text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{2 b^2 c^3}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b^2 c^3} \]

[Out]

-x^2*(-c^2*x^2+1)/b/c/(a+b*arcsin(c*x))+1/2*cos(4*a/b)*Si(4*(a+b*arcsin(c*x))/b)/b^2/c^3-1/2*Ci(4*(a+b*arcsin(
c*x))/b)*sin(4*a/b)/b^2/c^3

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Rubi [A]
time = 0.30, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4799, 4731, 4491, 12, 3384, 3380, 3383} \begin {gather*} -\frac {\sin \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b^2 c^3}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b^2 c^3}-\frac {x^2 \left (1-c^2 x^2\right )}{b c (a+b \text {ArcSin}(c x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((x^2*(1 - c^2*x^2))/(b*c*(a + b*ArcSin[c*x]))) - (CosIntegral[(4*(a + b*ArcSin[c*x]))/b]*Sin[(4*a)/b])/(2*b^
2*c^3) + (Cos[(4*a)/b]*SinIntegral[(4*(a + b*ArcSin[c*x]))/b])/(2*b^2*c^3)

Rule 12

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

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 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 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[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 4731

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

Rule 4799

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[
(f*x)^m*Sqrt[1 - c^2*x^2]*(d + e*x^2)^p*((a + b*ArcSin[c*x])^(n + 1)/(b*c*(n + 1))), x] + (-Dist[f*(m/(b*c*(n
+ 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n +
1), x], x] + Dist[c*((m + 2*p + 1)/(b*f*(n + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 -
c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0]
&& LtQ[n, -1] && IGtQ[2*p, 0] && NeQ[m + 2*p + 1, 0] && IGtQ[m, -3]

Rubi steps

\begin {align*} \int \frac {x^2 \sqrt {1-c^2 x^2}}{\left (a+b \sin ^{-1}(c x)\right )^2} \, dx &=-\frac {x^2 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {2 \int \frac {x}{a+b \sin ^{-1}(c x)} \, dx}{b c}-\frac {(4 c) \int \frac {x^3}{a+b \sin ^{-1}(c x)} \, dx}{b}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {2 \text {Subst}\left (\int \frac {\cos (x) \sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^3}-\frac {4 \text {Subst}\left (\int \frac {\cos (x) \sin ^3(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^3}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {2 \text {Subst}\left (\int \frac {\sin (2 x)}{2 (a+b x)} \, dx,x,\sin ^{-1}(c x)\right )}{b c^3}-\frac {4 \text {Subst}\left (\int \left (\frac {\sin (2 x)}{4 (a+b x)}-\frac {\sin (4 x)}{8 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c^3}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {\text {Subst}\left (\int \frac {\sin (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{2 b c^3}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{2 b c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{2 b c^3}\\ &=-\frac {x^2 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {\text {Ci}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right ) \sin \left (\frac {4 a}{b}\right )}{2 b^2 c^3}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{2 b^2 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 82, normalized size = 0.87 \begin {gather*} \frac {\frac {2 b c^2 x^2 \left (-1+c^2 x^2\right )}{a+b \text {ArcSin}(c x)}-\text {CosIntegral}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right ) \sin \left (\frac {4 a}{b}\right )+\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )}{2 b^2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b*c^2*x^2*(-1 + c^2*x^2))/(a + b*ArcSin[c*x]) - CosIntegral[4*(a/b + ArcSin[c*x])]*Sin[(4*a)/b] + Cos[(4*a
)/b]*SinIntegral[4*(a/b + ArcSin[c*x])])/(2*b^2*c^3)

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Maple [A]
time = 0.10, size = 136, normalized size = 1.45

method	result	size
default	\(\frac {4 \arcsin \left (c x \right ) \sinIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b -4 \arcsin \left (c x \right ) \cosineIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) b +4 \sinIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -4 \cosineIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) a +\cos \left (4 \arcsin \left (c x \right )\right ) b -b}{8 c^{3} \left (a +b \arcsin \left (c x \right )\right ) b^{2}}\)	\(136\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/c^3*(4*arcsin(c*x)*Si(4*arcsin(c*x)+4*a/b)*cos(4*a/b)*b-4*arcsin(c*x)*Ci(4*arcsin(c*x)+4*a/b)*sin(4*a/b)*b
+4*Si(4*arcsin(c*x)+4*a/b)*cos(4*a/b)*a-4*Ci(4*arcsin(c*x)+4*a/b)*sin(4*a/b)*a+cos(4*arcsin(c*x))*b-b)/(a+b*ar
csin(c*x))/b^2

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sqrt(-(c*x - 1)*(c*x + 1))/(a + b*asin(c*x))**2, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 563 vs. \(2 (88) = 176\).
time = 0.51, size = 563, normalized size = 5.99 \begin {gather*} -\frac {4 \, b \arcsin \left (c x\right ) \cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{3} c^{3} \arcsin \left (c x\right ) + a b^{2} c^{3}} + \frac {4 \, b \arcsin \left (c x\right ) \cos \left (\frac {a}{b}\right )^{4} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b^{3} c^{3} \arcsin \left (c x\right ) + a b^{2} c^{3}} - \frac {4 \, a \cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{3} c^{3} \arcsin \left (c x\right ) + a b^{2} c^{3}} + \frac {4 \, a \cos \left (\frac {a}{b}\right )^{4} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b^{3} c^{3} \arcsin \left (c x\right ) + a b^{2} c^{3}} + \frac {2 \, b \arcsin \left (c x\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{3} c^{3} \arcsin \left (c x\right ) + a b^{2} c^{3}} - \frac {4 \, b \arcsin \left (c x\right ) \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b^{3} c^{3} \arcsin \left (c x\right ) + a b^{2} c^{3}} + \frac {2 \, a \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{3} c^{3} \arcsin \left (c x\right ) + a b^{2} c^{3}} - \frac {4 \, a \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b^{3} c^{3} \arcsin \left (c x\right ) + a b^{2} c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b}{b^{3} c^{3} \arcsin \left (c x\right ) + a b^{2} c^{3}} + \frac {b \arcsin \left (c x\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{2 \, {\left (b^{3} c^{3} \arcsin \left (c x\right ) + a b^{2} c^{3}\right )}} + \frac {{\left (c^{2} x^{2} - 1\right )} b}{b^{3} c^{3} \arcsin \left (c x\right ) + a b^{2} c^{3}} + \frac {a \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{2 \, {\left (b^{3} c^{3} \arcsin \left (c x\right ) + a b^{2} c^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*b*arcsin(c*x)*cos(a/b)^3*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 4
*b*arcsin(c*x)*cos(a/b)^4*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 4*a*cos(a/b)
^3*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 4*a*cos(a/b)^4*sin_integra
l(4*a/b + 4*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 2*b*arcsin(c*x)*cos(a/b)*cos_integral(4*a/b + 4*a
rcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 4*b*arcsin(c*x)*cos(a/b)^2*sin_integral(4*a/b + 4*arc
sin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 2*a*cos(a/b)*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c
^3*arcsin(c*x) + a*b^2*c^3) - 4*a*cos(a/b)^2*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*
c^3) + (c^2*x^2 - 1)^2*b/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 1/2*b*arcsin(c*x)*sin_integral(4*a/b + 4*arcsin(c
*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + (c^2*x^2 - 1)*b/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 1/2*a*sin_integra
l(4*a/b + 4*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(1 - c^2*x^2)^(1/2))/(a + b*asin(c*x))^2,x)

[Out]

int((x^2*(1 - c^2*x^2)^(1/2))/(a + b*asin(c*x))^2, x)

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