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

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

[Out]

(15*Cos[(2*a)/b]*CosIntegral[(2*(a + b*ArcSin[c*x]))/b])/(32*b*c) + (3*Cos[(4*a)/b]*CosIntegral[(4*(a + b*ArcS
in[c*x]))/b])/(16*b*c) + (Cos[(6*a)/b]*CosIntegral[(6*(a + b*ArcSin[c*x]))/b])/(32*b*c) + (5*Log[a + b*ArcSin[
c*x]])/(16*b*c) + (15*Sin[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c*x]))/b])/(32*b*c) + (3*Sin[(4*a)/b]*SinInteg
ral[(4*(a + b*ArcSin[c*x]))/b])/(16*b*c) + (Sin[(6*a)/b]*SinIntegral[(6*(a + b*ArcSin[c*x]))/b])/(32*b*c)

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Rubi [A]  time = 0.320551, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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])

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 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 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]

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*}

Mathematica [A]  time = 0.682498, size = 165, normalized size = 0.8 \[ \frac{15 \cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+6 \cos \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+\cos \left (\frac{6 a}{b}\right ) \text{CosIntegral}\left (6 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+15 \sin \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+6 \sin \left (\frac{4 a}{b}\right ) \text{Si}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+\sin \left (\frac{6 a}{b}\right ) \text{Si}\left (6 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+18 \log \left (a+b \sin ^{-1}(c x)\right )-8 \log \left (8 \left (a+b \sin ^{-1}(c x)\right )\right )}{32 b c} \]

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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Maple [A]  time = 0.044, size = 193, normalized size = 0.9 \begin{align*}{\frac{15}{32\,cb}{\it Si} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) }+{\frac{15}{32\,cb}{\it Ci} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) }+{\frac{1}{32\,cb}{\it Si} \left ( 6\,\arcsin \left ( cx \right ) +6\,{\frac{a}{b}} \right ) \sin \left ( 6\,{\frac{a}{b}} \right ) }+{\frac{1}{32\,cb}{\it Ci} \left ( 6\,\arcsin \left ( cx \right ) +6\,{\frac{a}{b}} \right ) \cos \left ( 6\,{\frac{a}{b}} \right ) }+{\frac{3}{16\,cb}{\it Si} \left ( 4\,\arcsin \left ( cx \right ) +4\,{\frac{a}{b}} \right ) \sin \left ( 4\,{\frac{a}{b}} \right ) }+{\frac{3}{16\,cb}{\it Ci} \left ( 4\,\arcsin \left ( cx \right ) +4\,{\frac{a}{b}} \right ) \cos \left ( 4\,{\frac{a}{b}} \right ) }+{\frac{5\,\ln \left ( a+b\arcsin \left ( cx \right ) \right ) }{16\,cb}} \end{align*}

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]

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)+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)+5/16*ln(a+b*arcsin(c*x))/b/c

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{5}{2}}}{b \arcsin \left (c x\right ) + a}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [B]  time = 1.41033, size = 637, normalized size = 3.09 \begin{align*} \text{result too large to display} \end{align*}

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)