3.4.25 \(\int \frac {(1-a^2-2 a b x-b^2 x^2)^{3/2}}{\text {ArcSin}(a+b x)^3} \, dx\) [325]

Optimal. Leaf size=90 \[ -\frac {\left (1-(a+b x)^2\right )^2}{2 b \text {ArcSin}(a+b x)^2}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{b \text {ArcSin}(a+b x)}-\frac {\text {CosIntegral}(2 \text {ArcSin}(a+b x))}{b}-\frac {\text {CosIntegral}(4 \text {ArcSin}(a+b x))}{b} \]

[Out]

-1/2*(1-(b*x+a)^2)^2/b/arcsin(b*x+a)^2+2*(b*x+a)*(1-(b*x+a)^2)^(3/2)/b/arcsin(b*x+a)-Ci(2*arcsin(b*x+a))/b-Ci(
4*arcsin(b*x+a))/b

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Rubi [A]
time = 0.20, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4891, 4751, 4799, 4753, 3393, 3383, 4809, 4491} \begin {gather*} -\frac {\text {CosIntegral}(2 \text {ArcSin}(a+b x))}{b}-\frac {\text {CosIntegral}(4 \text {ArcSin}(a+b x))}{b}-\frac {\left (1-(a+b x)^2\right )^2}{2 b \text {ArcSin}(a+b x)^2}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{b \text {ArcSin}(a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(1 - (a + b*x)^2)^2/(b*ArcSin[a + b*x]^2) + (2*(a + b*x)*(1 - (a + b*x)^2)^(3/2))/(b*ArcSin[a + b*x]) - C
osIntegral[2*ArcSin[a + b*x]]/b - CosIntegral[4*ArcSin[a + b*x]]/b

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 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 4751

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)/(b*(n + 1)))*Simp[(d + e*x^2)
^p/(1 - c^2*x^2)^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 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]

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]

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]

Rule 4891

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> Di
st[1/d, Subst[Int[(-C/d^2 + (C/d^2)*x^2)^p*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A, B
, C, n, p}, x] && EqQ[B*(1 - c^2) + 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rubi steps

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

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Mathematica [A]
time = 0.27, size = 110, normalized size = 1.22 \begin {gather*} -\frac {\frac {\left (-1+a^2+2 a b x+b^2 x^2\right ) \left (-1+a^2+2 a b x+b^2 x^2+4 (a+b x) \sqrt {1-a^2-2 a b x-b^2 x^2} \text {ArcSin}(a+b x)\right )}{\text {ArcSin}(a+b x)^2}+2 \text {CosIntegral}(2 \text {ArcSin}(a+b x))+2 \text {CosIntegral}(4 \text {ArcSin}(a+b x))}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.42, size = 108, normalized size = 1.20

method	result	size
default	\(-\frac {16 \cosineIntegral \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}+16 \cosineIntegral \left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}-4 \sin \left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )-8 \sin \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+4 \cos \left (2 \arcsin \left (b x +a \right )\right )+\cos \left (4 \arcsin \left (b x +a \right )\right )+3}{16 b \arcsin \left (b x +a \right )^{2}}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/b*(16*Ci(2*arcsin(b*x+a))*arcsin(b*x+a)^2+16*Ci(4*arcsin(b*x+a))*arcsin(b*x+a)^2-4*sin(4*arcsin(b*x+a))*
arcsin(b*x+a)-8*sin(2*arcsin(b*x+a))*arcsin(b*x+a)+4*cos(2*arcsin(b*x+a))+cos(4*arcsin(b*x+a))+3)/arcsin(b*x+a
)^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((-b^2*x^2-2*a*b*x-a^2+1)^(3/2)/arcsin(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/2*(b^4*x^4 + 4*a*b^3*x^3 + 2*(3*a^2 - 1)*b^2*x^2 + a^4 - 2*b*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x -
a + 1))^2*integrate(2*(4*b^2*x^2 + 8*a*b*x + 4*a^2 - 1)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)/arctan2(b*x + a,
sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)), x) + 4*(a^3 - a)*b*x + 4*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2 - 1)*b*
x - a)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)) - 2*a^2 + 1
)/(b*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^2)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.49, size = 101, normalized size = 1.12 \begin {gather*} \frac {2 \, {\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} {\left (b x + a\right )}}{b \arcsin \left (b x + a\right )} - \frac {\operatorname {Ci}\left (4 \, \arcsin \left (b x + a\right )\right )}{b} - \frac {\operatorname {Ci}\left (2 \, \arcsin \left (b x + a\right )\right )}{b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2}}{2 \, b \arcsin \left (b x + a\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)*(b*x + a)/(b*arcsin(b*x + a)) - cos_integral(4*arcsin(b*x + a))/b - cos
_integral(2*arcsin(b*x + a))/b - 1/2*(b^2*x^2 + 2*a*b*x + a^2 - 1)^2/(b*arcsin(b*x + a)^2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (-a^2-2\,a\,b\,x-b^2\,x^2+1\right )}^{3/2}}{{\mathrm {asin}\left (a+b\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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