3.374 \(\int \frac {(c-a^2 c x^2)^3}{\sin ^{-1}(a x)^2} \, dx\)

Optimal. Leaf size=95 \[ -\frac {c^3 \left (1-a^2 x^2\right )^{7/2}}{a \sin ^{-1}(a x)}-\frac {35 c^3 \text {Si}\left (\sin ^{-1}(a x)\right )}{64 a}-\frac {63 c^3 \text {Si}\left (3 \sin ^{-1}(a x)\right )}{64 a}-\frac {35 c^3 \text {Si}\left (5 \sin ^{-1}(a x)\right )}{64 a}-\frac {7 c^3 \text {Si}\left (7 \sin ^{-1}(a x)\right )}{64 a} \]

[Out]

-c^3*(-a^2*x^2+1)^(7/2)/a/arcsin(a*x)-35/64*c^3*Si(arcsin(a*x))/a-63/64*c^3*Si(3*arcsin(a*x))/a-35/64*c^3*Si(5
*arcsin(a*x))/a-7/64*c^3*Si(7*arcsin(a*x))/a

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Rubi [A]  time = 0.17, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4659, 4723, 4406, 3299} \[ -\frac {c^3 \left (1-a^2 x^2\right )^{7/2}}{a \sin ^{-1}(a x)}-\frac {35 c^3 \text {Si}\left (\sin ^{-1}(a x)\right )}{64 a}-\frac {63 c^3 \text {Si}\left (3 \sin ^{-1}(a x)\right )}{64 a}-\frac {35 c^3 \text {Si}\left (5 \sin ^{-1}(a x)\right )}{64 a}-\frac {7 c^3 \text {Si}\left (7 \sin ^{-1}(a x)\right )}{64 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c^3*(1 - a^2*x^2)^(7/2))/(a*ArcSin[a*x])) - (35*c^3*SinIntegral[ArcSin[a*x]])/(64*a) - (63*c^3*SinIntegral[
3*ArcSin[a*x]])/(64*a) - (35*c^3*SinIntegral[5*ArcSin[a*x]])/(64*a) - (7*c^3*SinIntegral[7*ArcSin[a*x]])/(64*a
)

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 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 4659

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(Sqrt[1 - c^2*x^2]*
(d + e*x^2)^p*(a + b*ArcSin[c*x])^(n + 1))/(b*c*(n + 1)), x] + Dist[(c*(2*p + 1)*d^IntPart[p]*(d + e*x^2)^Frac
Part[p])/(b*(n + 1)*(1 - c^2*x^2)^FracPart[p]), Int[x*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x],
x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1]

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 {\left (c-a^2 c x^2\right )^3}{\sin ^{-1}(a x)^2} \, dx &=-\frac {c^3 \left (1-a^2 x^2\right )^{7/2}}{a \sin ^{-1}(a x)}-\left (7 a c^3\right ) \int \frac {x \left (1-a^2 x^2\right )^{5/2}}{\sin ^{-1}(a x)} \, dx\\ &=-\frac {c^3 \left (1-a^2 x^2\right )^{7/2}}{a \sin ^{-1}(a x)}-\frac {\left (7 c^3\right ) \operatorname {Subst}\left (\int \frac {\cos ^6(x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=-\frac {c^3 \left (1-a^2 x^2\right )^{7/2}}{a \sin ^{-1}(a x)}-\frac {\left (7 c^3\right ) \operatorname {Subst}\left (\int \left (\frac {5 \sin (x)}{64 x}+\frac {9 \sin (3 x)}{64 x}+\frac {5 \sin (5 x)}{64 x}+\frac {\sin (7 x)}{64 x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=-\frac {c^3 \left (1-a^2 x^2\right )^{7/2}}{a \sin ^{-1}(a x)}-\frac {\left (7 c^3\right ) \operatorname {Subst}\left (\int \frac {\sin (7 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{64 a}-\frac {\left (35 c^3\right ) \operatorname {Subst}\left (\int \frac {\sin (x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{64 a}-\frac {\left (35 c^3\right ) \operatorname {Subst}\left (\int \frac {\sin (5 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{64 a}-\frac {\left (63 c^3\right ) \operatorname {Subst}\left (\int \frac {\sin (3 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{64 a}\\ &=-\frac {c^3 \left (1-a^2 x^2\right )^{7/2}}{a \sin ^{-1}(a x)}-\frac {35 c^3 \text {Si}\left (\sin ^{-1}(a x)\right )}{64 a}-\frac {63 c^3 \text {Si}\left (3 \sin ^{-1}(a x)\right )}{64 a}-\frac {35 c^3 \text {Si}\left (5 \sin ^{-1}(a x)\right )}{64 a}-\frac {7 c^3 \text {Si}\left (7 \sin ^{-1}(a x)\right )}{64 a}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 83, normalized size = 0.87 \[ -\frac {c^3 \left (64 \left (1-a^2 x^2\right )^{7/2}+35 \sin ^{-1}(a x) \text {Si}\left (\sin ^{-1}(a x)\right )+63 \sin ^{-1}(a x) \text {Si}\left (3 \sin ^{-1}(a x)\right )+35 \sin ^{-1}(a x) \text {Si}\left (5 \sin ^{-1}(a x)\right )+7 \sin ^{-1}(a x) \text {Si}\left (7 \sin ^{-1}(a x)\right )\right )}{64 a \sin ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/64*(c^3*(64*(1 - a^2*x^2)^(7/2) + 35*ArcSin[a*x]*SinIntegral[ArcSin[a*x]] + 63*ArcSin[a*x]*SinIntegral[3*Ar
cSin[a*x]] + 35*ArcSin[a*x]*SinIntegral[5*ArcSin[a*x]] + 7*ArcSin[a*x]*SinIntegral[7*ArcSin[a*x]]))/(a*ArcSin[
a*x])

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {a^{6} c^{3} x^{6} - 3 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} - c^{3}}{\arcsin \left (a x\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a^6*c^3*x^6 - 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 - c^3)/arcsin(a*x)^2, x)

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giac [A]  time = 0.47, size = 95, normalized size = 1.00 \[ \frac {{\left (a^{2} x^{2} - 1\right )}^{3} \sqrt {-a^{2} x^{2} + 1} c^{3}}{a \arcsin \left (a x\right )} - \frac {7 \, c^{3} \operatorname {Si}\left (7 \, \arcsin \left (a x\right )\right )}{64 \, a} - \frac {35 \, c^{3} \operatorname {Si}\left (5 \, \arcsin \left (a x\right )\right )}{64 \, a} - \frac {63 \, c^{3} \operatorname {Si}\left (3 \, \arcsin \left (a x\right )\right )}{64 \, a} - \frac {35 \, c^{3} \operatorname {Si}\left (\arcsin \left (a x\right )\right )}{64 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*x^2 - 1)^3*sqrt(-a^2*x^2 + 1)*c^3/(a*arcsin(a*x)) - 7/64*c^3*sin_integral(7*arcsin(a*x))/a - 35/64*c^3*si
n_integral(5*arcsin(a*x))/a - 63/64*c^3*sin_integral(3*arcsin(a*x))/a - 35/64*c^3*sin_integral(arcsin(a*x))/a

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maple [A]  time = 0.17, size = 105, normalized size = 1.11 \[ -\frac {c^{3} \left (35 \Si \left (\arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+63 \Si \left (3 \arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+35 \Si \left (5 \arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+7 \Si \left (7 \arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+35 \sqrt {-a^{2} x^{2}+1}+\cos \left (7 \arcsin \left (a x \right )\right )+21 \cos \left (3 \arcsin \left (a x \right )\right )+7 \cos \left (5 \arcsin \left (a x \right )\right )\right )}{64 a \arcsin \left (a x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*c*x^2+c)^3/arcsin(a*x)^2,x)

[Out]

-1/64/a*c^3*(35*Si(arcsin(a*x))*arcsin(a*x)+63*Si(3*arcsin(a*x))*arcsin(a*x)+35*Si(5*arcsin(a*x))*arcsin(a*x)+
7*Si(7*arcsin(a*x))*arcsin(a*x)+35*(-a^2*x^2+1)^(1/2)+cos(7*arcsin(a*x))+21*cos(3*arcsin(a*x))+7*cos(5*arcsin(
a*x)))/arcsin(a*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {7 \, a \arctan \left (a x, \sqrt {a x + 1} \sqrt {-a x + 1}\right ) \int \frac {{\left (a^{5} c^{3} x^{5} - 2 \, a^{3} c^{3} x^{3} + a c^{3} x\right )} \sqrt {a x + 1} \sqrt {-a x + 1}}{\arctan \left (a x, \sqrt {a x + 1} \sqrt {-a x + 1}\right )}\,{d x} - {\left (a^{6} c^{3} x^{6} - 3 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} - c^{3}\right )} \sqrt {a x + 1} \sqrt {-a x + 1}}{a \arctan \left (a x, \sqrt {a x + 1} \sqrt {-a x + 1}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))*integrate(7*(a^5*c^3*x^5 - 2*a^3*c^3*x^3 + a*c^3*x)*sqrt(a*x +
1)*sqrt(-a*x + 1)/arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1)), x) - (a^6*c^3*x^6 - 3*a^4*c^3*x^4 + 3*a^2*c^3*x^
2 - c^3)*sqrt(a*x + 1)*sqrt(-a*x + 1))/(a*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c-a^2\,c\,x^2\right )}^3}{{\mathrm {asin}\left (a\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - a^2*c*x^2)^3/asin(a*x)^2,x)

[Out]

int((c - a^2*c*x^2)^3/asin(a*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - c^{3} \left (\int \frac {3 a^{2} x^{2}}{\operatorname {asin}^{2}{\left (a x \right )}}\, dx + \int \left (- \frac {3 a^{4} x^{4}}{\operatorname {asin}^{2}{\left (a x \right )}}\right )\, dx + \int \frac {a^{6} x^{6}}{\operatorname {asin}^{2}{\left (a x \right )}}\, dx + \int \left (- \frac {1}{\operatorname {asin}^{2}{\left (a x \right )}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c**3*(Integral(3*a**2*x**2/asin(a*x)**2, x) + Integral(-3*a**4*x**4/asin(a*x)**2, x) + Integral(a**6*x**6/asi
n(a*x)**2, x) + Integral(-1/asin(a*x)**2, x))

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