3.409 \(\int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \sin ^{-1}(c x))^2} \, dx\)

Optimal. Leaf size=141 \[ -\frac {\sin \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{b^2 c^5}+\frac {\sin \left (\frac {4 a}{b}\right ) \text {Ci}\left (\frac {4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{2 b^2 c^5}+\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{b^2 c^5}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{2 b^2 c^5}-\frac {x^4}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

[Out]

-x^4/b/c/(a+b*arcsin(c*x))+cos(2*a/b)*Si(2*(a+b*arcsin(c*x))/b)/b^2/c^5-1/2*cos(4*a/b)*Si(4*(a+b*arcsin(c*x))/
b)/b^2/c^5-Ci(2*(a+b*arcsin(c*x))/b)*sin(2*a/b)/b^2/c^5+1/2*Ci(4*(a+b*arcsin(c*x))/b)*sin(4*a/b)/b^2/c^5

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Rubi [A]  time = 0.36, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4719, 4635, 4406, 3303, 3299, 3302} \[ -\frac {\sin \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{b^2 c^5}+\frac {\sin \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{2 b^2 c^5}+\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{b^2 c^5}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{2 b^2 c^5}-\frac {x^4}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^4/(b*c*(a + b*ArcSin[c*x]))) - (CosIntegral[(2*a)/b + 2*ArcSin[c*x]]*Sin[(2*a)/b])/(b^2*c^5) + (CosIntegra
l[(4*a)/b + 4*ArcSin[c*x]]*Sin[(4*a)/b])/(2*b^2*c^5) + (Cos[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSin[c*x]])/(b^
2*c^5) - (Cos[(4*a)/b]*SinIntegral[(4*a)/b + 4*ArcSin[c*x]])/(2*b^2*c^5)

Rule 3299

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 3302

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 3303

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 4406

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 4635

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

Rule 4719

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

Rubi steps

\begin {align*} \int \frac {x^4}{\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2} \, dx &=-\frac {x^4}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {4 \int \frac {x^3}{a+b \sin ^{-1}(c x)} \, dx}{b c}\\ &=-\frac {x^4}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {4 \operatorname {Subst}\left (\int \frac {\cos (x) \sin ^3(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^5}\\ &=-\frac {x^4}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {4 \operatorname {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^5}\\ &=-\frac {x^4}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {\sin (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{2 b c^5}+\frac {\operatorname {Subst}\left (\int \frac {\sin (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^5}\\ &=-\frac {x^4}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {\cos \left (\frac {2 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^5}-\frac {\cos \left (\frac {4 a}{b}\right ) \operatorname {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^5}-\frac {\sin \left (\frac {2 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^5}+\frac {\sin \left (\frac {4 a}{b}\right ) \operatorname {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^5}\\ &=-\frac {x^4}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {\text {Ci}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right ) \sin \left (\frac {2 a}{b}\right )}{b^2 c^5}+\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^5}+\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{b^2 c^5}-\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^5}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 117, normalized size = 0.83 \[ \frac {-\frac {2 b c^4 x^4}{a+b \sin ^{-1}(c x)}-2 \sin \left (\frac {2 a}{b}\right ) \text {Ci}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+\sin \left (\frac {4 a}{b}\right ) \text {Ci}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+2 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )}{2 b^2 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*b*c^4*x^4)/(a + b*ArcSin[c*x]) - 2*CosIntegral[2*(a/b + ArcSin[c*x])]*Sin[(2*a)/b] + CosIntegral[4*(a/b +
ArcSin[c*x])]*Sin[(4*a)/b] + 2*Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] - Cos[(4*a)/b]*SinIntegral[4*(
a/b + ArcSin[c*x])])/(2*b^2*c^5)

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fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} x^{2} + 1} x^{4}}{a^{2} c^{2} x^{2} + {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arcsin \left (c x\right )^{2} - a^{2} + 2 \, {\left (a b c^{2} x^{2} - a b\right )} \arcsin \left (c x\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.90, size = 876, normalized size = 6.21 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b*arcsin(c*x))^2/(-c^2*x^2+1)^(1/2),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^5*arcsin(c*x) + a*b^2*c^5) - 4*
b*arcsin(c*x)*cos(a/b)^4*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) + 4*a*cos(a/b)^
3*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - 4*a*cos(a/b)^4*sin_integral
(4*a/b + 4*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - 2*b*arcsin(c*x)*cos(a/b)*cos_integral(4*a/b + 4*ar
csin(c*x))*sin(a/b)/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - 2*b*arcsin(c*x)*cos(a/b)*cos_integral(2*a/b + 2*arcsin
(c*x))*sin(a/b)/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) + 4*b*arcsin(c*x)*cos(a/b)^2*sin_integral(4*a/b + 4*arcsin(c
*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) + 2*b*arcsin(c*x)*cos(a/b)^2*sin_integral(2*a/b + 2*arcsin(c*x))/(b^3*c
^5*arcsin(c*x) + a*b^2*c^5) - 2*a*cos(a/b)*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c^5*arcsin(c*x) +
a*b^2*c^5) - 2*a*cos(a/b)*cos_integral(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) + 4*
a*cos(a/b)^2*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) + 2*a*cos(a/b)^2*sin_integr
al(2*a/b + 2*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - (c^2*x^2 - 1)^2*b/(b^3*c^5*arcsin(c*x) + a*b^2*c
^5) - 1/2*b*arcsin(c*x)*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - b*arcsin(c*x)*
sin_integral(2*a/b + 2*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - 2*(c^2*x^2 - 1)*b/(b^3*c^5*arcsin(c*x)
+ a*b^2*c^5) - 1/2*a*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - a*sin_integral(2
*a/b + 2*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - b/(b^3*c^5*arcsin(c*x) + a*b^2*c^5)

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maple [A]  time = 0.11, size = 250, normalized size = 1.77 \[ -\frac {4 \arcsin \left (c x \right ) \Si \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b -4 \arcsin \left (c x \right ) \Ci \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) b -8 \Si \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) \arcsin \left (c x \right ) b +8 \Ci \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) \arcsin \left (c x \right ) b +4 \Si \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -4 \Ci \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) a -8 \Si \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a +8 \Ci \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a +\cos \left (4 \arcsin \left (c x \right )\right ) b -4 \cos \left (2 \arcsin \left (c x \right )\right ) b +3 b}{8 c^{5} b^{2} \left (a +b \arcsin \left (c x \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a+b*arcsin(c*x))^2/(-c^2*x^2+1)^(1/2),x)

[Out]

-1/8/c^5*(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-8*Si(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*arcsin(c*x)*b+8*Ci(2*arcsin(c*x)+2*a/b)*sin(2*a/b)*arcsin(c*x)*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-8*Si(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*a
+8*Ci(2*arcsin(c*x)+2*a/b)*sin(2*a/b)*a+cos(4*arcsin(c*x))*b-4*cos(2*arcsin(c*x))*b+3*b)/b^2/(a+b*arcsin(c*x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {x^{4} - \frac {4 \, {\left (b^{2} c \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a b c\right )} \int \frac {x^{3}}{b \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a}\,{d x}}{b c}}{b^{2} c \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + a b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x^4 - 4*(b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c)*integrate(x^3/(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {1-c^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {- \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4/(sqrt(-(c*x - 1)*(c*x + 1))*(a + b*asin(c*x))**2), x)

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