3.4.27 \(\int \frac {x (1-c^2 x^2)^{3/2}}{a+b \text {ArcSin}(c x)} \, dx\) [327]

Optimal. Leaf size=183 \[ -\frac {\text {CosIntegral}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{8 b c^2}-\frac {3 \text {CosIntegral}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{16 b c^2}-\frac {\text {CosIntegral}\left (\frac {5 (a+b \text {ArcSin}(c x))}{b}\right ) \sin \left (\frac {5 a}{b}\right )}{16 b c^2}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{8 b c^2}+\frac {3 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b c^2}+\frac {\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b c^2} \]

[Out]

1/8*cos(a/b)*Si((a+b*arcsin(c*x))/b)/b/c^2+3/16*cos(3*a/b)*Si(3*(a+b*arcsin(c*x))/b)/b/c^2+1/16*cos(5*a/b)*Si(
5*(a+b*arcsin(c*x))/b)/b/c^2-1/8*Ci((a+b*arcsin(c*x))/b)*sin(a/b)/b/c^2-3/16*Ci(3*(a+b*arcsin(c*x))/b)*sin(3*a
/b)/b/c^2-1/16*Ci(5*(a+b*arcsin(c*x))/b)*sin(5*a/b)/b/c^2

________________________________________________________________________________________

Rubi [A]
time = 0.23, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {4809, 4491, 3384, 3380, 3383} \begin {gather*} -\frac {\sin \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{8 b c^2}-\frac {3 \sin \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b c^2}-\frac {\sin \left (\frac {5 a}{b}\right ) \text {CosIntegral}\left (\frac {5 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b c^2}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{8 b c^2}+\frac {3 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b c^2}+\frac {\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(CosIntegral[(a + b*ArcSin[c*x])/b]*Sin[a/b])/(b*c^2) - (3*CosIntegral[(3*(a + b*ArcSin[c*x]))/b]*Sin[(3*
a)/b])/(16*b*c^2) - (CosIntegral[(5*(a + b*ArcSin[c*x]))/b]*Sin[(5*a)/b])/(16*b*c^2) + (Cos[a/b]*SinIntegral[(
a + b*ArcSin[c*x])/b])/(8*b*c^2) + (3*Cos[(3*a)/b]*SinIntegral[(3*(a + b*ArcSin[c*x]))/b])/(16*b*c^2) + (Cos[(
5*a)/b]*SinIntegral[(5*(a + b*ArcSin[c*x]))/b])/(16*b*c^2)

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 \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 (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{c^2}\\ &=\frac {\text {Subst}\left (\int \left (\frac {\sin (x)}{8 (a+b x)}+\frac {3 \sin (3 x)}{16 (a+b x)}+\frac {\sin (5 x)}{16 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^2}\\ &=\frac {\text {Subst}\left (\int \frac {\sin (5 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 c^2}+\frac {\text {Subst}\left (\int \frac {\sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 c^2}+\frac {3 \text {Subst}\left (\int \frac {\sin (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 c^2}\\ &=\frac {\cos \left (\frac {a}{b}\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 c^2}+\frac {\left (3 \cos \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 c^2}+\frac {\cos \left (\frac {5 a}{b}\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 c^2}-\frac {\sin \left (\frac {a}{b}\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 c^2}-\frac {\left (3 \sin \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 c^2}-\frac {\sin \left (\frac {5 a}{b}\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 c^2}\\ &=-\frac {\text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c x)\right ) \sin \left (\frac {a}{b}\right )}{8 b c^2}-\frac {3 \text {Ci}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right ) \sin \left (\frac {3 a}{b}\right )}{16 b c^2}-\frac {\text {Ci}\left (\frac {5 a}{b}+5 \sin ^{-1}(c x)\right ) \sin \left (\frac {5 a}{b}\right )}{16 b c^2}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{8 b c^2}+\frac {3 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b c^2}+\frac {\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b c^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.34, size = 136, normalized size = 0.74 \begin {gather*} \frac {-2 \text {CosIntegral}\left (\frac {a}{b}+\text {ArcSin}(c x)\right ) \sin \left (\frac {a}{b}\right )-3 \text {CosIntegral}\left (3 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right ) \sin \left (\frac {3 a}{b}\right )-\text {CosIntegral}\left (5 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right ) \sin \left (\frac {5 a}{b}\right )+2 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\text {ArcSin}(c x)\right )+3 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )}{16 b c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*CosIntegral[a/b + ArcSin[c*x]]*Sin[a/b] - 3*CosIntegral[3*(a/b + ArcSin[c*x])]*Sin[(3*a)/b] - CosIntegral[
5*(a/b + ArcSin[c*x])]*Sin[(5*a)/b] + 2*Cos[a/b]*SinIntegral[a/b + ArcSin[c*x]] + 3*Cos[(3*a)/b]*SinIntegral[3
*(a/b + ArcSin[c*x])] + Cos[(5*a)/b]*SinIntegral[5*(a/b + ArcSin[c*x])])/(16*b*c^2)

________________________________________________________________________________________

Maple [A]
time = 0.09, size = 139, normalized size = 0.76

method	result	size
default	\(\frac {\sinIntegral \left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right )-\cosineIntegral \left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right )+3 \sinIntegral \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )-3 \cosineIntegral \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )+2 \sinIntegral \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-2 \cosineIntegral \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )}{16 c^{2} b}\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/c^2*(Si(5*arcsin(c*x)+5*a/b)*cos(5*a/b)-Ci(5*arcsin(c*x)+5*a/b)*sin(5*a/b)+3*Si(3*arcsin(c*x)+3*a/b)*cos(
3*a/b)-3*Ci(3*arcsin(c*x)+3*a/b)*sin(3*a/b)+2*Si(arcsin(c*x)+a/b)*cos(a/b)-2*Ci(arcsin(c*x)+a/b)*sin(a/b))/b

________________________________________________________________________________________

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*(-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/(b*arcsin(c*x) + a), x)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (171) = 342\).
time = 0.46, size = 360, normalized size = 1.97 \begin {gather*} -\frac {\cos \left (\frac {a}{b}\right )^{4} \operatorname {Ci}\left (\frac {5 \, a}{b} + 5 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b c^{2}} + \frac {\cos \left (\frac {a}{b}\right )^{5} \operatorname {Si}\left (\frac {5 \, a}{b} + 5 \, \arcsin \left (c x\right )\right )}{b c^{2}} + \frac {3 \, \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {5 \, a}{b} + 5 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{4 \, b c^{2}} - \frac {3 \, \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{4 \, b c^{2}} - \frac {5 \, \cos \left (\frac {a}{b}\right )^{3} \operatorname {Si}\left (\frac {5 \, a}{b} + 5 \, \arcsin \left (c x\right )\right )}{4 \, b c^{2}} + \frac {3 \, \cos \left (\frac {a}{b}\right )^{3} \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right )}{4 \, b c^{2}} - \frac {\operatorname {Ci}\left (\frac {5 \, a}{b} + 5 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{16 \, b c^{2}} + \frac {3 \, \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{16 \, b c^{2}} - \frac {\operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{8 \, b c^{2}} + \frac {5 \, \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {5 \, a}{b} + 5 \, \arcsin \left (c x\right )\right )}{16 \, b c^{2}} - \frac {9 \, \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right )}{16 \, b c^{2}} + \frac {\cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{8 \, b c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(a/b)^4*cos_integral(5*a/b + 5*arcsin(c*x))*sin(a/b)/(b*c^2) + cos(a/b)^5*sin_integral(5*a/b + 5*arcsin(c*
x))/(b*c^2) + 3/4*cos(a/b)^2*cos_integral(5*a/b + 5*arcsin(c*x))*sin(a/b)/(b*c^2) - 3/4*cos(a/b)^2*cos_integra
l(3*a/b + 3*arcsin(c*x))*sin(a/b)/(b*c^2) - 5/4*cos(a/b)^3*sin_integral(5*a/b + 5*arcsin(c*x))/(b*c^2) + 3/4*c
os(a/b)^3*sin_integral(3*a/b + 3*arcsin(c*x))/(b*c^2) - 1/16*cos_integral(5*a/b + 5*arcsin(c*x))*sin(a/b)/(b*c
^2) + 3/16*cos_integral(3*a/b + 3*arcsin(c*x))*sin(a/b)/(b*c^2) - 1/8*cos_integral(a/b + arcsin(c*x))*sin(a/b)
/(b*c^2) + 5/16*cos(a/b)*sin_integral(5*a/b + 5*arcsin(c*x))/(b*c^2) - 9/16*cos(a/b)*sin_integral(3*a/b + 3*ar
csin(c*x))/(b*c^2) + 1/8*cos(a/b)*sin_integral(a/b + arcsin(c*x))/(b*c^2)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,{\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*(1 - c^2*x^2)^(3/2))/(a + b*asin(c*x)),x)

[Out]

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

________________________________________________________________________________________