3.4.81 \(\int \frac {x^3 \sqrt {1-c^2 x^2}}{(a+b \text {ArcSin}(c x))^2} \, dx\) [381]

Optimal. Leaf size=214 \[ -\frac {x^3 \left (1-c^2 x^2\right )}{b c (a+b \text {ArcSin}(c x))}+\frac {\cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{8 b^2 c^4}+\frac {3 \cos \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^4}-\frac {5 \cos \left (\frac {5 a}{b}\right ) \text {CosIntegral}\left (\frac {5 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^4}+\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{8 b^2 c^4}+\frac {3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^4}-\frac {5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^4} \]

[Out]

-x^3*(-c^2*x^2+1)/b/c/(a+b*arcsin(c*x))+1/8*Ci((a+b*arcsin(c*x))/b)*cos(a/b)/b^2/c^4+3/16*Ci(3*(a+b*arcsin(c*x
))/b)*cos(3*a/b)/b^2/c^4-5/16*Ci(5*(a+b*arcsin(c*x))/b)*cos(5*a/b)/b^2/c^4+1/8*Si((a+b*arcsin(c*x))/b)*sin(a/b
)/b^2/c^4+3/16*Si(3*(a+b*arcsin(c*x))/b)*sin(3*a/b)/b^2/c^4-5/16*Si(5*(a+b*arcsin(c*x))/b)*sin(5*a/b)/b^2/c^4

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Rubi [A]
time = 0.43, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4799, 4731, 4491, 3384, 3380, 3383} \begin {gather*} \frac {\cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{8 b^2 c^4}+\frac {3 \cos \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^4}-\frac {5 \cos \left (\frac {5 a}{b}\right ) \text {CosIntegral}\left (\frac {5 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^4}+\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{8 b^2 c^4}+\frac {3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^4}-\frac {5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \text {ArcSin}(c x))}{b}\right )}{16 b^2 c^4}-\frac {x^3 \left (1-c^2 x^2\right )}{b c (a+b \text {ArcSin}(c x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((x^3*(1 - c^2*x^2))/(b*c*(a + b*ArcSin[c*x]))) + (Cos[a/b]*CosIntegral[(a + b*ArcSin[c*x])/b])/(8*b^2*c^4) +
(3*Cos[(3*a)/b]*CosIntegral[(3*(a + b*ArcSin[c*x]))/b])/(16*b^2*c^4) - (5*Cos[(5*a)/b]*CosIntegral[(5*(a + b*
ArcSin[c*x]))/b])/(16*b^2*c^4) + (Sin[a/b]*SinIntegral[(a + b*ArcSin[c*x])/b])/(8*b^2*c^4) + (3*Sin[(3*a)/b]*S
inIntegral[(3*(a + b*ArcSin[c*x]))/b])/(16*b^2*c^4) - (5*Sin[(5*a)/b]*SinIntegral[(5*(a + b*ArcSin[c*x]))/b])/
(16*b^2*c^4)

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 4731

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/(b*c^(m + 1)), Subst[Int[x^n*Sin[-
a/b + x/b]^m*Cos[-a/b + x/b], x], x, a + b*ArcSin[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 4799

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[
(f*x)^m*Sqrt[1 - c^2*x^2]*(d + e*x^2)^p*((a + b*ArcSin[c*x])^(n + 1)/(b*c*(n + 1))), x] + (-Dist[f*(m/(b*c*(n
+ 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n +
1), x], x] + Dist[c*((m + 2*p + 1)/(b*f*(n + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 -
c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0]
&& LtQ[n, -1] && IGtQ[2*p, 0] && NeQ[m + 2*p + 1, 0] && IGtQ[m, -3]

Rubi steps

\begin {align*} \int \frac {x^3 \sqrt {1-c^2 x^2}}{\left (a+b \sin ^{-1}(c x)\right )^2} \, dx &=-\frac {x^3 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {3 \int \frac {x^2}{a+b \sin ^{-1}(c x)} \, dx}{b c}-\frac {(5 c) \int \frac {x^4}{a+b \sin ^{-1}(c x)} \, dx}{b}\\ &=-\frac {x^3 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {3 \text {Subst}\left (\int \frac {\cos (x) \sin ^2(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^4}-\frac {5 \text {Subst}\left (\int \frac {\cos (x) \sin ^4(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^4}\\ &=-\frac {x^3 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {3 \text {Subst}\left (\int \left (\frac {\cos (x)}{4 (a+b x)}-\frac {\cos (3 x)}{4 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c^4}-\frac {5 \text {Subst}\left (\int \left (\frac {\cos (x)}{8 (a+b x)}-\frac {3 \cos (3 x)}{16 (a+b x)}+\frac {\cos (5 x)}{16 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c^4}\\ &=-\frac {x^3 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {5 \text {Subst}\left (\int \frac {\cos (5 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^4}-\frac {5 \text {Subst}\left (\int \frac {\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^4}+\frac {3 \text {Subst}\left (\int \frac {\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}-\frac {3 \text {Subst}\left (\int \frac {\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}+\frac {15 \text {Subst}\left (\int \frac {\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^4}\\ &=-\frac {x^3 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {\left (5 \cos \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^4}+\frac {\left (3 \cos \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}-\frac {\left (3 \cos \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 )}{4 b c^4}+\frac {\left (15 \cos \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 b c^4}-\frac {\left (5 \cos \left (\frac {5 a}{b}\right )\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 b c^4}-\frac {\left (5 \sin \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^4}+\frac {\left (3 \sin \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}-\frac {\left (3 \sin \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 )}{4 b c^4}+\frac {\left (15 \sin \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 b c^4}-\frac {\left (5 \sin \left (\frac {5 a}{b}\right )\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 b c^4}\\ &=-\frac {x^3 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{8 b^2 c^4}+\frac {3 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b^2 c^4}-\frac {5 \cos \left (\frac {5 a}{b}\right ) \text {Ci}\left (\frac {5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b^2 c^4}+\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{8 b^2 c^4}+\frac {3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b^2 c^4}-\frac {5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b^2 c^4}\\ \end {align*}

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Mathematica [A]
time = 0.37, size = 175, normalized size = 0.82 \begin {gather*} \frac {-\frac {16 b c^3 x^3}{a+b \text {ArcSin}(c x)}+\frac {16 b c^5 x^5}{a+b \text {ArcSin}(c x)}+2 \cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\text {ArcSin}(c x)\right )+3 \cos \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (3 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )-5 \cos \left (\frac {5 a}{b}\right ) \text {CosIntegral}\left (5 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\text {ArcSin}(c x)\right )+3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )-5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )}{16 b^2 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-16*b*c^3*x^3)/(a + b*ArcSin[c*x]) + (16*b*c^5*x^5)/(a + b*ArcSin[c*x]) + 2*Cos[a/b]*CosIntegral[a/b + ArcSi
n[c*x]] + 3*Cos[(3*a)/b]*CosIntegral[3*(a/b + ArcSin[c*x])] - 5*Cos[(5*a)/b]*CosIntegral[5*(a/b + ArcSin[c*x])
] + 2*Sin[a/b]*SinIntegral[a/b + ArcSin[c*x]] + 3*Sin[(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c*x])] - 5*Sin[(5*a
)/b]*SinIntegral[5*(a/b + ArcSin[c*x])])/(16*b^2*c^4)

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Maple [A]
time = 0.11, size = 340, normalized size = 1.59

method	result	size
default	\(-\frac {5 \arcsin \left (c x \right ) \sinIntegral \left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) b +5 \arcsin \left (c x \right ) \cosineIntegral \left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) b -2 \arcsin \left (c x \right ) \sinIntegral \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b -2 \arcsin \left (c x \right ) \cosineIntegral \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b -3 \arcsin \left (c x \right ) \sinIntegral \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) b -3 \arcsin \left (c x \right ) \cosineIntegral \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) b +5 \sinIntegral \left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) a +5 \cosineIntegral \left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) a -2 \sinIntegral \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a -2 \cosineIntegral \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a -3 \sinIntegral \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) a -3 \cosineIntegral \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) a +2 x b c -\sin \left (5 \arcsin \left (c x \right )\right ) b +\sin \left (3 \arcsin \left (c x \right )\right ) b}{16 c^{4} \left (a +b \arcsin \left (c x \right )\right ) b^{2}}\)	\(340\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/c^4*(5*arcsin(c*x)*Si(5*arcsin(c*x)+5*a/b)*sin(5*a/b)*b+5*arcsin(c*x)*Ci(5*arcsin(c*x)+5*a/b)*cos(5*a/b)
*b-2*arcsin(c*x)*Si(arcsin(c*x)+a/b)*sin(a/b)*b-2*arcsin(c*x)*Ci(arcsin(c*x)+a/b)*cos(a/b)*b-3*arcsin(c*x)*Si(
3*arcsin(c*x)+3*a/b)*sin(3*a/b)*b-3*arcsin(c*x)*Ci(3*arcsin(c*x)+3*a/b)*cos(3*a/b)*b+5*Si(5*arcsin(c*x)+5*a/b)
*sin(5*a/b)*a+5*Ci(5*arcsin(c*x)+5*a/b)*cos(5*a/b)*a-2*Si(arcsin(c*x)+a/b)*sin(a/b)*a-2*Ci(arcsin(c*x)+a/b)*co
s(a/b)*a-3*Si(3*arcsin(c*x)+3*a/b)*sin(3*a/b)*a-3*Ci(3*arcsin(c*x)+3*a/b)*cos(3*a/b)*a+2*x*b*c-sin(5*arcsin(c*
x))*b+sin(3*arcsin(c*x))*b)/(a+b*arcsin(c*x))/b^2

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

[Out]

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

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \sqrt {- \left (c x - 1\right ) \left (c x + 1\right )}}{\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3\,\sqrt {1-c^2\,x^2}}{{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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