3.122 \(\int \frac {x^2 (a+b \sin ^{-1}(c x))}{(d-c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=135 \[ \frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 c^3 d \sqrt {d-c^2 d x^2}} \]

[Out]

x*(a+b*arcsin(c*x))/c^2/d/(-c^2*d*x^2+d)^(1/2)-1/2*(a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/b/c^3/d/(-c^2*d*x^2+
d)^(1/2)+1/2*b*ln(-c^2*x^2+1)*(-c^2*x^2+1)^(1/2)/c^3/d/(-c^2*d*x^2+d)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {4703, 4643, 4641, 260} \[ -\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt {d-c^2 d x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 c^3 d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*ArcSin[c*x]))/(c^2*d*Sqrt[d - c^2*d*x^2]) - (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(2*b*c^3*d*Sqr
t[d - c^2*d*x^2]) + (b*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2])/(2*c^3*d*Sqrt[d - c^2*d*x^2])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 4641

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

Rule 4643

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 - c^2*x^2]/Sq
rt[d + e*x^2], Int[(a + b*ArcSin[c*x])^n/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*
d + e, 0] &&  !GtQ[d, 0]

Rule 4703

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

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{3/2}} \, dx &=\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {\int \frac {a+b \sin ^{-1}(c x)}{\sqrt {d-c^2 d x^2}} \, dx}{c^2 d}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int \frac {x}{1-c^2 x^2} \, dx}{c d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 c^3 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {1-c^2 x^2} \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{c^2 d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 c^3 d \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 160, normalized size = 1.19 \[ -\frac {a x \sqrt {-d \left (c^2 x^2-1\right )}}{c^2 d^2 \left (c^2 x^2-1\right )}+\frac {a \tan ^{-1}\left (\frac {c x \sqrt {-d \left (c^2 x^2-1\right )}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )}{c^3 d^{3/2}}+\frac {b \left (2 c x \sin ^{-1}(c x)-\sqrt {1-c^2 x^2} \left (\sin ^{-1}(c x)^2-2 \log \left (\sqrt {1-c^2 x^2}\right )\right )\right )}{2 c^3 d \sqrt {d \left (1-c^2 x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*x*Sqrt[-(d*(-1 + c^2*x^2))])/(c^2*d^2*(-1 + c^2*x^2))) + (a*ArcTan[(c*x*Sqrt[-(d*(-1 + c^2*x^2))])/(Sqrt[
d]*(-1 + c^2*x^2))])/(c^3*d^(3/2)) + (b*(2*c*x*ArcSin[c*x] - Sqrt[1 - c^2*x^2]*(ArcSin[c*x]^2 - 2*Log[Sqrt[1 -
c^2*x^2]])))/(2*c^3*d*Sqrt[d*(1 - c^2*x^2)])

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b x^{2} \arcsin \left (c x\right ) + a x^{2}\right )}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep^4-1)]index.cc index_m i_lex_is_greater Error: Bad Argument Value

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maple [C]  time = 0.28, size = 274, normalized size = 2.03 \[ \frac {a x}{c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}-\frac {a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{c^{2} d \sqrt {c^{2} d}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{2 d^{2} c^{3} \left (c^{2} x^{2}-1\right )}+\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )}{d^{2} c^{3} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{d^{2} c^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d^{2} c^{3} \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x/c^2/d/(-c^2*d*x^2+d)^(1/2)-a/c^2/d/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+1/2*b*(-d*(c
^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/c^3/(c^2*x^2-1)*arcsin(c*x)^2+I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^
(1/2)/d^2/c^3/(c^2*x^2-1)*arcsin(c*x)-b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)/d^2/c^2/(c^2*x^2-1)*x-b*(-d*(c^2*x^
2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/c^3/(c^2*x^2-1)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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