3.4.12 \(\int \frac {(c-a^2 c x^2)^2}{\text {ArcSin}(a x)} \, dx\) [312]

Optimal. Leaf size=50 \[ \frac {5 c^2 \text {CosIntegral}(\text {ArcSin}(a x))}{8 a}+\frac {5 c^2 \text {CosIntegral}(3 \text {ArcSin}(a x))}{16 a}+\frac {c^2 \text {CosIntegral}(5 \text {ArcSin}(a x))}{16 a} \]

[Out]

5/8*c^2*Ci(arcsin(a*x))/a+5/16*c^2*Ci(3*arcsin(a*x))/a+1/16*c^2*Ci(5*arcsin(a*x))/a

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Rubi [A]
time = 0.07, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4753, 3393, 3383} \begin {gather*} \frac {5 c^2 \text {CosIntegral}(\text {ArcSin}(a x))}{8 a}+\frac {5 c^2 \text {CosIntegral}(3 \text {ArcSin}(a x))}{16 a}+\frac {c^2 \text {CosIntegral}(5 \text {ArcSin}(a x))}{16 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - a^2*c*x^2)^2/ArcSin[a*x],x]

[Out]

(5*c^2*CosIntegral[ArcSin[a*x]])/(8*a) + (5*c^2*CosIntegral[3*ArcSin[a*x]])/(16*a) + (c^2*CosIntegral[5*ArcSin
[a*x]])/(16*a)

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 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 4753

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(1/(b*c))*Simp[(d
+ e*x^2)^p/(1 - c^2*x^2)^p], Subst[Int[x^n*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, 0]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 34, normalized size = 0.68 \begin {gather*} \frac {c^2 (10 \text {CosIntegral}(\text {ArcSin}(a x))+5 \text {CosIntegral}(3 \text {ArcSin}(a x))+\text {CosIntegral}(5 \text {ArcSin}(a x)))}{16 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(10*CosIntegral[ArcSin[a*x]] + 5*CosIntegral[3*ArcSin[a*x]] + CosIntegral[5*ArcSin[a*x]]))/(16*a)

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Maple [A]
time = 0.10, size = 33, normalized size = 0.66

method	result	size
derivativedivides	\(\frac {c^{2} \left (10 \cosineIntegral \left (\arcsin \left (a x \right )\right )+5 \cosineIntegral \left (3 \arcsin \left (a x \right )\right )+\cosineIntegral \left (5 \arcsin \left (a x \right )\right )\right )}{16 a}\)	\(33\)
default	\(\frac {c^{2} \left (10 \cosineIntegral \left (\arcsin \left (a x \right )\right )+5 \cosineIntegral \left (3 \arcsin \left (a x \right )\right )+\cosineIntegral \left (5 \arcsin \left (a x \right )\right )\right )}{16 a}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/a*c^2*(10*Ci(arcsin(a*x))+5*Ci(3*arcsin(a*x))+Ci(5*arcsin(a*x)))

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

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} c^{2} \left (\int \left (- \frac {2 a^{2} x^{2}}{\operatorname {asin}{\left (a x \right )}}\right )\, dx + \int \frac {a^{4} x^{4}}{\operatorname {asin}{\left (a x \right )}}\, dx + \int \frac {1}{\operatorname {asin}{\left (a x \right )}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.46, size = 44, normalized size = 0.88 \begin {gather*} \frac {c^{2} \operatorname {Ci}\left (5 \, \arcsin \left (a x\right )\right )}{16 \, a} + \frac {5 \, c^{2} \operatorname {Ci}\left (3 \, \arcsin \left (a x\right )\right )}{16 \, a} + \frac {5 \, c^{2} \operatorname {Ci}\left (\arcsin \left (a x\right )\right )}{8 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*c^2*cos_integral(5*arcsin(a*x))/a + 5/16*c^2*cos_integral(3*arcsin(a*x))/a + 5/8*c^2*cos_integral(arcsin(
a*x))/a

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (c-a^2\,c\,x^2\right )}^2}{\mathrm {asin}\left (a\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - a^2*c*x^2)^2/asin(a*x),x)

[Out]

int((c - a^2*c*x^2)^2/asin(a*x), x)

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