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

Optimal. Leaf size=206 \[ \frac {15 \cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 b c}+\frac {3 \cos \left (\frac {4 a}{b}\right ) \text {Ci}\left (\frac {4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b c}+\frac {\cos \left (\frac {6 a}{b}\right ) \text {Ci}\left (\frac {6 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 b c}+\frac {15 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 b c}+\frac {3 \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b c}+\frac {\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 b c}+\frac {5 \log \left (a+b \sin ^{-1}(c x)\right )}{16 b c} \]

[Out]

15/32*Ci(2*(a+b*arcsin(c*x))/b)*cos(2*a/b)/b/c+3/16*Ci(4*(a+b*arcsin(c*x))/b)*cos(4*a/b)/b/c+1/32*Ci(6*(a+b*ar
csin(c*x))/b)*cos(6*a/b)/b/c+5/16*ln(a+b*arcsin(c*x))/b/c+15/32*Si(2*(a+b*arcsin(c*x))/b)*sin(2*a/b)/b/c+3/16*
Si(4*(a+b*arcsin(c*x))/b)*sin(4*a/b)/b/c+1/32*Si(6*(a+b*arcsin(c*x))/b)*sin(6*a/b)/b/c

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Rubi [A]  time = 0.32, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4661, 3312, 3303, 3299, 3302} \[ \frac {15 \cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{32 b c}+\frac {3 \cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{16 b c}+\frac {\cos \left (\frac {6 a}{b}\right ) \text {CosIntegral}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{32 b c}+\frac {15 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{32 b c}+\frac {3 \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{16 b c}+\frac {\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{32 b c}+\frac {5 \log \left (a+b \sin ^{-1}(c x)\right )}{16 b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*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*ArcS
in[c*x]])/(16*b*c) + (Cos[(6*a)/b]*CosIntegral[(6*a)/b + 6*ArcSin[c*x]])/(32*b*c) + (5*Log[a + b*ArcSin[c*x]])
/(16*b*c) + (15*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]])/(16*b*c) + (Sin[(6*a)/b]*SinIntegral[(6*a)/b + 6*ArcSin[c*x]])/(32*b*c)

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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 4661

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c, Subst[Int[(
a + b*x)^n*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] && I
GtQ[2*p, 0] && (IntegerQ[p] || GtQ[d, 0])

Rubi steps

\begin {align*} \int \frac {\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)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{c}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {5}{16 (a+b x)}+\frac {15 \cos (2 x)}{32 (a+b x)}+\frac {3 \cos (4 x)}{16 (a+b x)}+\frac {\cos (6 x)}{32 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}\\ &=\frac {5 \log \left (a+b \sin ^{-1}(c x)\right )}{16 b c}+\frac {\operatorname {Subst}\left (\int \frac {\cos (6 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c}+\frac {3 \operatorname {Subst}\left (\int \frac {\cos (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 c}+\frac {15 \operatorname {Subst}\left (\int \frac {\cos (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c}\\ &=\frac {5 \log \left (a+b \sin ^{-1}(c x)\right )}{16 b c}+\frac {\left (15 \cos \left (\frac {2 a}{b}\right )\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}+\frac {\left (3 \cos \left (\frac {4 a}{b}\right )\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 )}{16 c}+\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}+\frac {\left (15 \sin \left (\frac {2 a}{b}\right )\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}+\frac {\left (3 \sin \left (\frac {4 a}{b}\right )\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 )}{16 c}+\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}\\ &=\frac {15 \cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{32 b c}+\frac {3 \cos \left (\frac {4 a}{b}\right ) \text {Ci}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{16 b c}+\frac {\cos \left (\frac {6 a}{b}\right ) \text {Ci}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{32 b c}+\frac {5 \log \left (a+b \sin ^{-1}(c x)\right )}{16 b c}+\frac {15 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{32 b c}+\frac {3 \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{16 b c}+\frac {\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 a}{b}+6 \sin ^{-1}(c x)\right )}{32 b c}\\ \end {align*}

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Mathematica [A]  time = 0.90, size = 165, normalized size = 0.80 \[ \frac {15 \cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+6 \cos \left (\frac {4 a}{b}\right ) \text {Ci}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+\cos \left (\frac {6 a}{b}\right ) \text {Ci}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+15 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+6 \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+18 \log \left (a+b \sin ^{-1}(c x)\right )-8 \log \left (8 \left (a+b \sin ^{-1}(c x)\right )\right )}{32 b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*Cos[(2*a)/b]*CosIntegral[2*(a/b + ArcSin[c*x])] + 6*Cos[(4*a)/b]*CosIntegral[4*(a/b + ArcSin[c*x])] + Cos[
(6*a)/b]*CosIntegral[6*(a/b + ArcSin[c*x])] + 18*Log[a + b*ArcSin[c*x]] - 8*Log[8*(a + b*ArcSin[c*x])] + 15*Si
n[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] + 6*Sin[(4*a)/b]*SinIntegral[4*(a/b + ArcSin[c*x])] + Sin[(6*a)/
b]*SinIntegral[6*(a/b + ArcSin[c*x])])/(32*b*c)

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fricas [F]  time = 2.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \sqrt {-c^{2} x^{2} + 1}}{b \arcsin \left (c x\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.69, size = 472, normalized size = 2.29 \[ \frac {\cos \left (\frac {a}{b}\right )^{6} \operatorname {Ci}\left (\frac {6 \, a}{b} + 6 \, \arcsin \left (c x\right )\right )}{b c} + \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} - \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} + \frac {3 \, \cos \left (\frac {a}{b}\right )^{4} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{2 \, b c} - \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} + \frac {3 \, \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} + \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} - \frac {3 \, \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{2 \, b c} + \frac {15 \, \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{16 \, b c} + \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} - \frac {3 \, \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} + \frac {15 \, \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} - \frac {\operatorname {Ci}\left (\frac {6 \, a}{b} + 6 \, \arcsin \left (c x\right )\right )}{32 \, b c} + \frac {3 \, \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{16 \, b c} - \frac {15 \, \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{32 \, b c} + \frac {5 \, \log \left (b \arcsin \left (c x\right ) + a\right )}{16 \, b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos(a/b)^6*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c) + cos(a/b)^5*sin(a/b)*sin_integral(6*a/b + 6*arcsin(c*x))
/(b*c) - 3/2*cos(a/b)^4*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c) + 3/2*cos(a/b)^4*cos_integral(4*a/b + 4*arcs
in(c*x))/(b*c) - cos(a/b)^3*sin(a/b)*sin_integral(6*a/b + 6*arcsin(c*x))/(b*c) + 3/2*cos(a/b)^3*sin(a/b)*sin_i
ntegral(4*a/b + 4*arcsin(c*x))/(b*c) + 9/16*cos(a/b)^2*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c) - 3/2*cos(a/b
)^2*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c) + 15/16*cos(a/b)^2*cos_integral(2*a/b + 2*arcsin(c*x))/(b*c) + 3
/16*cos(a/b)*sin(a/b)*sin_integral(6*a/b + 6*arcsin(c*x))/(b*c) - 3/4*cos(a/b)*sin(a/b)*sin_integral(4*a/b + 4
*arcsin(c*x))/(b*c) + 15/16*cos(a/b)*sin(a/b)*sin_integral(2*a/b + 2*arcsin(c*x))/(b*c) - 1/32*cos_integral(6*
a/b + 6*arcsin(c*x))/(b*c) + 3/16*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c) - 15/32*cos_integral(2*a/b + 2*arc
sin(c*x))/(b*c) + 5/16*log(b*arcsin(c*x) + a)/(b*c)

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maple [A]  time = 0.08, size = 193, normalized size = 0.94 \[ \frac {\Si \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right )}{32 c b}+\frac {\Ci \left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right )}{32 c b}+\frac {3 \Si \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{16 c b}+\frac {3 \Ci \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )}{16 c b}+\frac {15 \Si \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{32 c b}+\frac {15 \Ci \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )}{32 c b}+\frac {5 \ln \left (a +b \arcsin \left (c x \right )\right )}{16 b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/c/b*Si(6*arcsin(c*x)+6*a/b)*sin(6*a/b)+1/32/c/b*Ci(6*arcsin(c*x)+6*a/b)*cos(6*a/b)+3/16/c/b*Si(4*arcsin(c
*x)+4*a/b)*sin(4*a/b)+3/16/c/b*Ci(4*arcsin(c*x)+4*a/b)*cos(4*a/b)+15/32/c/b*Si(2*arcsin(c*x)+2*a/b)*sin(2*a/b)
+15/32/c/b*Ci(2*arcsin(c*x)+2*a/b)*cos(2*a/b)+5/16*ln(a+b*arcsin(c*x))/b/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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 \[ \int \frac {\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}{a + b \operatorname {asin}{\left (c x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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