3.4.50 \(\int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))} \, dx\) [350]

Optimal. Leaf size=144 \[ -\frac {\cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b c^5}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{8 b c^5}+\frac {3 \log (a+b \text {ArcSin}(c x))}{8 b c^5}-\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b c^5}+\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{8 b c^5} \]

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4809, 3393, 3384, 3380, 3383} \begin {gather*} -\frac {\cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b c^5}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{8 b c^5}-\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b c^5}+\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{8 b c^5}+\frac {3 \log (a+b \text {ArcSin}(c x))}{8 b c^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/(Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])),x]

[Out]

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

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 3393

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 4809

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

Rubi steps

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

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Mathematica [A]
time = 0.16, size = 108, normalized size = 0.75 \begin {gather*} \frac {-4 \cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+\cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+3 \log (a+b \text {ArcSin}(c x))-4 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )}{8 b c^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/(Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])),x]

[Out]

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

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Maple [A]
time = 0.09, size = 111, normalized size = 0.77

method	result	size
default	\(\frac {\sinIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )+\cosineIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )-4 \sinIntegral \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )-4 \cosineIntegral \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )+3 \ln \left (a +b \arcsin \left (c x \right )\right )}{8 c^{5} b}\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a+b*arcsin(c*x))/(-c^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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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^4/(a+b*arcsin(c*x))/(-c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

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

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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^4/(a+b*arcsin(c*x))/(-c^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\sqrt {- \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(a+b*asin(c*x))/(-c**2*x**2+1)**(1/2),x)

[Out]

Integral(x**4/(sqrt(-(c*x - 1)*(c*x + 1))*(a + b*asin(c*x))), x)

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Giac [A]
time = 0.46, size = 254, normalized size = 1.76 \begin {gather*} \frac {\cos \left (\frac {a}{b}\right )^{4} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c^{5}} + \frac {\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 )}{b c^{5}} - \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c^{5}} - \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b c^{5}} - \frac {\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 )}{2 \, b c^{5}} - \frac {\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 )}{b c^{5}} + \frac {\operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{8 \, b c^{5}} + \frac {\operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{2 \, b c^{5}} + \frac {3 \, \log \left (b \arcsin \left (c x\right ) + a\right )}{8 \, b c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b*arcsin(c*x))/(-c^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

cos(a/b)^4*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^5) + cos(a/b)^3*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*x
))/(b*c^5) - cos(a/b)^2*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^5) - cos(a/b)^2*cos_integral(2*a/b + 2*arcsin
(c*x))/(b*c^5) - 1/2*cos(a/b)*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*x))/(b*c^5) - cos(a/b)*sin(a/b)*sin_int
egral(2*a/b + 2*arcsin(c*x))/(b*c^5) + 1/8*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^5) + 1/2*cos_integral(2*a/
b + 2*arcsin(c*x))/(b*c^5) + 3/8*log(b*arcsin(c*x) + a)/(b*c^5)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4}{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {1-c^2\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((a + b*asin(c*x))*(1 - c^2*x^2)^(1/2)),x)

[Out]

int(x^4/((a + b*asin(c*x))*(1 - c^2*x^2)^(1/2)), x)

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