3.4.26 \(\int \frac {x^2 (1-c^2 x^2)^{3/2}}{a+b \text {ArcSin}(c x)} \, dx\) [326]

Optimal. Leaf size=206 \[ \frac {\cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{32 b c^3}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b c^3}-\frac {\cos \left (\frac {6 a}{b}\right ) \text {CosIntegral}\left (\frac {6 (a+b \text {ArcSin}(c x))}{b}\right )}{32 b c^3}+\frac {\log (a+b \text {ArcSin}(c x))}{16 b c^3}+\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{32 b c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b c^3}-\frac {\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \text {ArcSin}(c x))}{b}\right )}{32 b c^3} \]

[Out]

1/32*Ci(2*(a+b*arcsin(c*x))/b)*cos(2*a/b)/b/c^3-1/16*Ci(4*(a+b*arcsin(c*x))/b)*cos(4*a/b)/b/c^3-1/32*Ci(6*(a+b
*arcsin(c*x))/b)*cos(6*a/b)/b/c^3+1/16*ln(a+b*arcsin(c*x))/b/c^3+1/32*Si(2*(a+b*arcsin(c*x))/b)*sin(2*a/b)/b/c
^3-1/16*Si(4*(a+b*arcsin(c*x))/b)*sin(4*a/b)/b/c^3-1/32*Si(6*(a+b*arcsin(c*x))/b)*sin(6*a/b)/b/c^3

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Rubi [A]
time = 0.26, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4809, 4491, 3384, 3380, 3383} \begin {gather*} \frac {\cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{32 b c^3}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b c^3}-\frac {\cos \left (\frac {6 a}{b}\right ) \text {CosIntegral}\left (\frac {6 (a+b \text {ArcSin}(c x))}{b}\right )}{32 b c^3}+\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{32 b c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b c^3}-\frac {\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \text {ArcSin}(c x))}{b}\right )}{32 b c^3}+\frac {\log (a+b \text {ArcSin}(c x))}{16 b c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(1 - c^2*x^2)^(3/2))/(a + b*ArcSin[c*x]),x]

[Out]

(Cos[(2*a)/b]*CosIntegral[(2*(a + b*ArcSin[c*x]))/b])/(32*b*c^3) - (Cos[(4*a)/b]*CosIntegral[(4*(a + b*ArcSin[
c*x]))/b])/(16*b*c^3) - (Cos[(6*a)/b]*CosIntegral[(6*(a + b*ArcSin[c*x]))/b])/(32*b*c^3) + Log[a + b*ArcSin[c*
x]]/(16*b*c^3) + (Sin[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c*x]))/b])/(32*b*c^3) - (Sin[(4*a)/b]*SinIntegral[
(4*(a + b*ArcSin[c*x]))/b])/(16*b*c^3) - (Sin[(6*a)/b]*SinIntegral[(6*(a + b*ArcSin[c*x]))/b])/(32*b*c^3)

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 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 {x^2 \left (1-c^2 x^2\right )^{3/2}}{a+b \sin ^{-1}(c x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\cos ^4(x) \sin ^2(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{c^3}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{16 (a+b x)}+\frac {\cos (2 x)}{32 (a+b x)}-\frac {\cos (4 x)}{16 (a+b x)}-\frac {\cos (6 x)}{32 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^3}\\ &=\frac {\log \left (a+b \sin ^{-1}(c x)\right )}{16 b c^3}+\frac {\text {Subst}\left (\int \frac {\cos (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c^3}-\frac {\text {Subst}\left (\int \frac {\cos (6 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c^3}-\frac {\text {Subst}\left (\int \frac {\cos (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 c^3}\\ &=\frac {\log \left (a+b \sin ^{-1}(c x)\right )}{16 b c^3}+\frac {\cos \left (\frac {2 a}{b}\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c^3}-\frac {\cos \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 )}{16 c^3}-\frac {\cos \left (\frac {6 a}{b}\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c^3}+\frac {\sin \left (\frac {2 a}{b}\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c^3}-\frac {\sin \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 )}{16 c^3}-\frac {\sin \left (\frac {6 a}{b}\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c^3}\\ &=\frac {\cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{32 b c^3}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Ci}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{16 b c^3}-\frac {\cos \left (\frac {6 a}{b}\right ) \text {Ci}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{32 b c^3}+\frac {\log \left (a+b \sin ^{-1}(c x)\right )}{16 b c^3}+\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{32 b c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{16 b c^3}-\frac {\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{32 b c^3}\\ \end {align*}

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Mathematica [A]
time = 0.42, size = 165, normalized size = 0.80 \begin {gather*} -\frac {-\cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+2 \cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+\cos \left (\frac {6 a}{b}\right ) \text {CosIntegral}\left (6 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+2 \log (a+b \text {ArcSin}(c x))-4 \log (8 (a+b \text {ArcSin}(c x)))-\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+2 \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )}{32 b c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(1 - c^2*x^2)^(3/2))/(a + b*ArcSin[c*x]),x]

[Out]

-1/32*(-(Cos[(2*a)/b]*CosIntegral[2*(a/b + ArcSin[c*x])]) + 2*Cos[(4*a)/b]*CosIntegral[4*(a/b + ArcSin[c*x])]
+ Cos[(6*a)/b]*CosIntegral[6*(a/b + ArcSin[c*x])] + 2*Log[a + b*ArcSin[c*x]] - 4*Log[8*(a + b*ArcSin[c*x])] -
Sin[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] + 2*Sin[(4*a)/b]*SinIntegral[4*(a/b + ArcSin[c*x])] + Sin[(6*a
)/b]*SinIntegral[6*(a/b + ArcSin[c*x])])/(b*c^3)

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Maple [A]
time = 0.09, size = 157, normalized size = 0.76

method	result	size
default	\(\frac {\sinIntegral \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )+\cosineIntegral \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )-\sinIntegral \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right )-\cosineIntegral \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right )-2 \sinIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )-2 \cosineIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )+2 \ln \left (a +b \arcsin \left (c x \right )\right )}{32 c^{3} b}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(-c^2*x^2+1)^(3/2)/(a+b*arcsin(c*x)),x,method=_RETURNVERBOSE)

[Out]

1/32/c^3*(Si(2*arcsin(c*x)+2*a/b)*sin(2*a/b)+Ci(2*arcsin(c*x)+2*a/b)*cos(2*a/b)-Si(6*arcsin(c*x)+6*a/b)*sin(6*
a/b)-Ci(6*arcsin(c*x)+6*a/b)*cos(6*a/b)-2*Si(4*arcsin(c*x)+4*a/b)*sin(4*a/b)-2*Ci(4*arcsin(c*x)+4*a/b)*cos(4*a
/b)+2*ln(a+b*arcsin(c*x)))/b

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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)^(3/2)/(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

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

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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)^(3/2)/(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(-c**2*x**2+1)**(3/2)/(a+b*asin(c*x)),x)

[Out]

Integral(x**2*(-(c*x - 1)*(c*x + 1))**(3/2)/(a + b*asin(c*x)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 473 vs. \(2 (192) = 384\).
time = 0.46, size = 473, normalized size = 2.30 \begin {gather*} -\frac {\cos \left (\frac {a}{b}\right )^{6} \operatorname {Ci}\left (\frac {6 \, a}{b} + 6 \, \arcsin \left (c x\right )\right )}{b c^{3}} - \frac {\cos \left (\frac {a}{b}\right )^{5} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {6 \, a}{b} + 6 \, \arcsin \left (c x\right )\right )}{b c^{3}} + \frac {3 \, \cos \left (\frac {a}{b}\right )^{4} \operatorname {Ci}\left (\frac {6 \, a}{b} + 6 \, \arcsin \left (c x\right )\right )}{2 \, b c^{3}} - \frac {\cos \left (\frac {a}{b}\right )^{4} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{2 \, b c^{3}} + \frac {\cos \left (\frac {a}{b}\right )^{3} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {6 \, a}{b} + 6 \, \arcsin \left (c x\right )\right )}{b c^{3}} - \frac {\cos \left (\frac {a}{b}\right )^{3} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{2 \, b c^{3}} - \frac {9 \, \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {6 \, a}{b} + 6 \, \arcsin \left (c x\right )\right )}{16 \, b c^{3}} + \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{2 \, b c^{3}} + \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{16 \, b c^{3}} - \frac {3 \, \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {6 \, a}{b} + 6 \, \arcsin \left (c x\right )\right )}{16 \, b c^{3}} + \frac {\cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{4 \, b c^{3}} + \frac {\cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{16 \, b c^{3}} + \frac {\operatorname {Ci}\left (\frac {6 \, a}{b} + 6 \, \arcsin \left (c x\right )\right )}{32 \, b c^{3}} - \frac {\operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{16 \, b c^{3}} - \frac {\operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{32 \, b c^{3}} + \frac {\log \left (b \arcsin \left (c x\right ) + a\right )}{16 \, b c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-c^2*x^2+1)^(3/2)/(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

-cos(a/b)^6*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^3) - cos(a/b)^5*sin(a/b)*sin_integral(6*a/b + 6*arcsin(c*
x))/(b*c^3) + 3/2*cos(a/b)^4*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^3) - 1/2*cos(a/b)^4*cos_integral(4*a/b +
4*arcsin(c*x))/(b*c^3) + cos(a/b)^3*sin(a/b)*sin_integral(6*a/b + 6*arcsin(c*x))/(b*c^3) - 1/2*cos(a/b)^3*sin
(a/b)*sin_integral(4*a/b + 4*arcsin(c*x))/(b*c^3) - 9/16*cos(a/b)^2*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^3
) + 1/2*cos(a/b)^2*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^3) + 1/16*cos(a/b)^2*cos_integral(2*a/b + 2*arcsin
(c*x))/(b*c^3) - 3/16*cos(a/b)*sin(a/b)*sin_integral(6*a/b + 6*arcsin(c*x))/(b*c^3) + 1/4*cos(a/b)*sin(a/b)*si
n_integral(4*a/b + 4*arcsin(c*x))/(b*c^3) + 1/16*cos(a/b)*sin(a/b)*sin_integral(2*a/b + 2*arcsin(c*x))/(b*c^3)
+ 1/32*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^3) - 1/16*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^3) - 1/32*
cos_integral(2*a/b + 2*arcsin(c*x))/(b*c^3) + 1/16*log(b*arcsin(c*x) + a)/(b*c^3)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2\,{\left (1-c^2\,x^2\right )}^{3/2}}{a+b\,\mathrm {asin}\left (c\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(1 - c^2*x^2)^(3/2))/(a + b*asin(c*x)),x)

[Out]

int((x^2*(1 - c^2*x^2)^(3/2))/(a + b*asin(c*x)), x)

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