3.334 \(\int \frac {x^2 (1-c^2 x^2)^{5/2}}{a+b \sin ^{-1}(c x)} \, dx\)

Optimal. Leaf size=268 \[ \frac {\cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 b c^3}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Ci}\left (\frac {4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 b c^3}-\frac {\cos \left (\frac {6 a}{b}\right ) \text {Ci}\left (\frac {6 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 b c^3}-\frac {\cos \left (\frac {8 a}{b}\right ) \text {Ci}\left (\frac {8 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{128 b c^3}+\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 b c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 b c^3}-\frac {\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 b c^3}-\frac {\sin \left (\frac {8 a}{b}\right ) \text {Si}\left (\frac {8 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{128 b c^3}+\frac {5 \log \left (a+b \sin ^{-1}(c x)\right )}{128 b c^3} \]

[Out]

1/32*Ci(2*(a+b*arcsin(c*x))/b)*cos(2*a/b)/b/c^3-1/32*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^3-1/128*Ci(8*(a+b*arcsin(c*x))/b)*cos(8*a/b)/b/c^3+5/128*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^3-1/32*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^3-1/128*Si(8*(a+b*arcsin(c*x))/b)*sin(8*a/b)/b/c^3

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Rubi [A]  time = 0.53, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4723, 4406, 3303, 3299, 3302} \[ \frac {\cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{32 b c^3}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{32 b c^3}-\frac {\cos \left (\frac {6 a}{b}\right ) \text {CosIntegral}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{32 b c^3}-\frac {\cos \left (\frac {8 a}{b}\right ) \text {CosIntegral}\left (\frac {8 a}{b}+8 \sin ^{-1}(c x)\right )}{128 b c^3}+\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{32 b c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{32 b c^3}-\frac {\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{32 b c^3}-\frac {\sin \left (\frac {8 a}{b}\right ) \text {Si}\left (\frac {8 a}{b}+8 \sin ^{-1}(c x)\right )}{128 b c^3}+\frac {5 \log \left (a+b \sin ^{-1}(c x)\right )}{128 b c^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(1 - c^2*x^2)^(5/2))/(a + b*ArcSin[c*x]),x]

[Out]

(Cos[(2*a)/b]*CosIntegral[(2*a)/b + 2*ArcSin[c*x]])/(32*b*c^3) - (Cos[(4*a)/b]*CosIntegral[(4*a)/b + 4*ArcSin[
c*x]])/(32*b*c^3) - (Cos[(6*a)/b]*CosIntegral[(6*a)/b + 6*ArcSin[c*x]])/(32*b*c^3) - (Cos[(8*a)/b]*CosIntegral
[(8*a)/b + 8*ArcSin[c*x]])/(128*b*c^3) + (5*Log[a + b*ArcSin[c*x]])/(128*b*c^3) + (Sin[(2*a)/b]*SinIntegral[(2
*a)/b + 2*ArcSin[c*x]])/(32*b*c^3) - (Sin[(4*a)/b]*SinIntegral[(4*a)/b + 4*ArcSin[c*x]])/(32*b*c^3) - (Sin[(6*
a)/b]*SinIntegral[(6*a)/b + 6*ArcSin[c*x]])/(32*b*c^3) - (Sin[(8*a)/b]*SinIntegral[(8*a)/b + 8*ArcSin[c*x]])/(
128*b*c^3)

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 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^(
m + 1), Subst[Int[(a + b*x)^n*Sin[x]^m*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rubi steps

\begin {align*} \int \frac {x^2 \left (1-c^2 x^2\right )^{5/2}}{a+b \sin ^{-1}(c x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\cos ^6(x) \sin ^2(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{c^3}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {5}{128 (a+b x)}+\frac {\cos (2 x)}{32 (a+b x)}-\frac {\cos (4 x)}{32 (a+b x)}-\frac {\cos (6 x)}{32 (a+b x)}-\frac {\cos (8 x)}{128 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^3}\\ &=\frac {5 \log \left (a+b \sin ^{-1}(c x)\right )}{128 b c^3}-\frac {\operatorname {Subst}\left (\int \frac {\cos (8 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{128 c^3}+\frac {\operatorname {Subst}\left (\int \frac {\cos (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c^3}-\frac {\operatorname {Subst}\left (\int \frac {\cos (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c^3}-\frac {\operatorname {Subst}\left (\int \frac {\cos (6 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c^3}\\ &=\frac {5 \log \left (a+b \sin ^{-1}(c x)\right )}{128 b c^3}+\frac {\cos \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 )}{32 c^3}-\frac {\cos \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 )}{32 c^3}-\frac {\cos \left (\frac {6 a}{b}\right ) \operatorname {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 {\cos \left (\frac {8 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {8 a}{b}+8 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{128 c^3}+\frac {\sin \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 )}{32 c^3}-\frac {\sin \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 )}{32 c^3}-\frac {\sin \left (\frac {6 a}{b}\right ) \operatorname {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 {\sin \left (\frac {8 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {8 a}{b}+8 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{128 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 )}{32 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 {\cos \left (\frac {8 a}{b}\right ) \text {Ci}\left (\frac {8 a}{b}+8 \sin ^{-1}(c x)\right )}{128 b c^3}+\frac {5 \log \left (a+b \sin ^{-1}(c x)\right )}{128 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 )}{32 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}-\frac {\sin \left (\frac {8 a}{b}\right ) \text {Si}\left (\frac {8 a}{b}+8 \sin ^{-1}(c x)\right )}{128 b c^3}\\ \end {align*}

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Mathematica [A]  time = 1.14, size = 209, normalized size = 0.78 \[ -\frac {-4 \cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+4 \cos \left (\frac {4 a}{b}\right ) \text {Ci}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+4 \cos \left (\frac {6 a}{b}\right ) \text {Ci}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+\cos \left (\frac {8 a}{b}\right ) \text {Ci}\left (8 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-4 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+4 \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+4 \sin \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+\sin \left (\frac {8 a}{b}\right ) \text {Si}\left (8 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+11 \log \left (a+b \sin ^{-1}(c x)\right )-16 \log \left (8 \left (a+b \sin ^{-1}(c x)\right )\right )}{128 b c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(1 - c^2*x^2)^(5/2))/(a + b*ArcSin[c*x]),x]

[Out]

-1/128*(-4*Cos[(2*a)/b]*CosIntegral[2*(a/b + ArcSin[c*x])] + 4*Cos[(4*a)/b]*CosIntegral[4*(a/b + ArcSin[c*x])]
+ 4*Cos[(6*a)/b]*CosIntegral[6*(a/b + ArcSin[c*x])] + Cos[(8*a)/b]*CosIntegral[8*(a/b + ArcSin[c*x])] + 11*Lo
g[a + b*ArcSin[c*x]] - 16*Log[8*(a + b*ArcSin[c*x])] - 4*Sin[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] + 4*S
in[(4*a)/b]*SinIntegral[4*(a/b + ArcSin[c*x])] + 4*Sin[(6*a)/b]*SinIntegral[6*(a/b + ArcSin[c*x])] + Sin[(8*a)
/b]*SinIntegral[8*(a/b + ArcSin[c*x])])/(b*c^3)

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{4} x^{6} - 2 \, c^{2} x^{4} + x^{2}\right )} \sqrt {-c^{2} x^{2} + 1}}{b \arcsin \left (c x\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

integral((c^4*x^6 - 2*c^2*x^4 + x^2)*sqrt(-c^2*x^2 + 1)/(b*arcsin(c*x) + a), x)

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giac [B]  time = 0.59, size = 757, normalized size = 2.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

-cos(a/b)^8*cos_integral(8*a/b + 8*arcsin(c*x))/(b*c^3) - cos(a/b)^7*sin(a/b)*sin_integral(8*a/b + 8*arcsin(c*
x))/(b*c^3) + 2*cos(a/b)^6*cos_integral(8*a/b + 8*arcsin(c*x))/(b*c^3) - cos(a/b)^6*cos_integral(6*a/b + 6*arc
sin(c*x))/(b*c^3) + 3/2*cos(a/b)^5*sin(a/b)*sin_integral(8*a/b + 8*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) - 5/4*cos(a/b)^4*cos_integral(8*a/b + 8*arcsin(c*x))/(b*c^3) + 3/2
*cos(a/b)^4*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^3) - 1/4*cos(a/b)^4*cos_integral(4*a/b + 4*arcsin(c*x))/(
b*c^3) - 5/8*cos(a/b)^3*sin(a/b)*sin_integral(8*a/b + 8*arcsin(c*x))/(b*c^3) + cos(a/b)^3*sin(a/b)*sin_integra
l(6*a/b + 6*arcsin(c*x))/(b*c^3) - 1/4*cos(a/b)^3*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*x))/(b*c^3) + 1/4*c
os(a/b)^2*cos_integral(8*a/b + 8*arcsin(c*x))/(b*c^3) - 9/16*cos(a/b)^2*cos_integral(6*a/b + 6*arcsin(c*x))/(b
*c^3) + 1/4*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*ar
csin(c*x))/(b*c^3) + 1/16*cos(a/b)*sin(a/b)*sin_integral(8*a/b + 8*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) + 1/8*cos(a/b)*sin(a/b)*sin_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) - 1/128*cos_integral(8*a/b + 8*arcsi
n(c*x))/(b*c^3) + 1/32*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^3) - 1/32*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) + 5/128*log(b*arcsin(c*x) + a)/(b*c^3)

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maple [A]  time = 0.08, size = 251, normalized size = 0.94 \[ -\frac {\Ci \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )}{32 c^{3} b}+\frac {\Si \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{32 c^{3} b}-\frac {\Si \left (8 \arcsin \left (c x \right )+\frac {8 a}{b}\right ) \sin \left (\frac {8 a}{b}\right )}{128 c^{3} b}-\frac {\Ci \left (8 \arcsin \left (c x \right )+\frac {8 a}{b}\right ) \cos \left (\frac {8 a}{b}\right )}{128 c^{3} b}+\frac {\Ci \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )}{32 c^{3} b}-\frac {\Si \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right )}{32 c^{3} b}-\frac {\Ci \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right )}{32 c^{3} b}-\frac {\Si \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{32 c^{3} b}+\frac {5 \ln \left (a +b \arcsin \left (c x \right )\right )}{128 b \,c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x)),x)

[Out]

-1/32/c^3/b*Ci(4*arcsin(c*x)+4*a/b)*cos(4*a/b)+1/32/c^3/b*Si(2*arcsin(c*x)+2*a/b)*sin(2*a/b)-1/128/c^3/b*Si(8*
arcsin(c*x)+8*a/b)*sin(8*a/b)-1/128/c^3/b*Ci(8*arcsin(c*x)+8*a/b)*cos(8*a/b)+1/32/c^3/b*Ci(2*arcsin(c*x)+2*a/b
)*cos(2*a/b)-1/32/c^3/b*Si(6*arcsin(c*x)+6*a/b)*sin(6*a/b)-1/32/c^3/b*Ci(6*arcsin(c*x)+6*a/b)*cos(6*a/b)-1/32/
c^3/b*Si(4*arcsin(c*x)+4*a/b)*sin(4*a/b)+5/128*ln(a+b*arcsin(c*x))/b/c^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}} x^{2}}{b \arcsin \left (c x\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

integrate((-c^2*x^2 + 1)^(5/2)*x^2/(b*arcsin(c*x) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2\,{\left (1-c^2\,x^2\right )}^{5/2}}{a+b\,\mathrm {asin}\left (c\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(1 - c^2*x^2)^(5/2))/(a + b*asin(c*x)),x)

[Out]

int((x^2*(1 - c^2*x^2)^(5/2))/(a + b*asin(c*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}{a + b \operatorname {asin}{\left (c x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(-c**2*x**2+1)**(5/2)/(a+b*asin(c*x)),x)

[Out]

Integral(x**2*(-(c*x - 1)*(c*x + 1))**(5/2)/(a + b*asin(c*x)), x)

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