3.131 \(\int \frac{x^4 (a+b \sin ^{-1}(c x))}{(d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=212 \[ -\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^5 d^2 \sqrt{d-c^2 d x^2}}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{b}{6 c^5 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}-\frac{2 b \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c^5 d^2 \sqrt{d-c^2 d x^2}} \]

[Out]

-b/(6*c^5*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) + (x^3*(a + b*ArcSin[c*x]))/(3*c^2*d*(d - c^2*d*x^2)^(3/2
)) - (x*(a + b*ArcSin[c*x]))/(c^4*d^2*Sqrt[d - c^2*d*x^2]) + (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(2*b*c^
5*d^2*Sqrt[d - c^2*d*x^2]) - (2*b*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2])/(3*c^5*d^2*Sqrt[d - c^2*d*x^2])

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Rubi [A]  time = 0.30235, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {4703, 4643, 4641, 260, 266, 43} \[ -\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2 \sqrt{d-c^2 d x^2}}+\frac{\sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^5 d^2 \sqrt{d-c^2 d x^2}}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{b}{6 c^5 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}-\frac{2 b \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c^5 d^2 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

-b/(6*c^5*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) + (x^3*(a + b*ArcSin[c*x]))/(3*c^2*d*(d - c^2*d*x^2)^(3/2
)) - (x*(a + b*ArcSin[c*x]))/(c^4*d^2*Sqrt[d - c^2*d*x^2]) + (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(2*b*c^
5*d^2*Sqrt[d - c^2*d*x^2]) - (2*b*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2])/(3*c^5*d^2*Sqrt[d - c^2*d*x^2])

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]

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 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 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x^4 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{5/2}} \, dx &=\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{\int \frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{3/2}} \, dx}{c^2 d}-\frac{\left (b \sqrt{1-c^2 x^2}\right ) \int \frac{x^3}{\left (1-c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2 \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^4 d^2}+\frac{\left (b \sqrt{1-c^2 x^2}\right ) \int \frac{x}{1-c^2 x^2} \, dx}{c^3 d^2 \sqrt{d-c^2 d x^2}}-\frac{\left (b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{x}{\left (1-c^2 x\right )^2} \, dx,x,x^2\right )}{6 c d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2 \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^5 d^2 \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^4 d^2 \sqrt{d-c^2 d x^2}}-\frac{\left (b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{c^2 \left (-1+c^2 x\right )^2}+\frac{1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{6 c d^2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b}{6 c^5 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2 \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^5 d^2 \sqrt{d-c^2 d x^2}}-\frac{2 b \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c^5 d^2 \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.423924, size = 213, normalized size = 1. \[ \frac{\sqrt{d} \left (-8 a c^3 x^3+6 a c x+b \sqrt{1-c^2 x^2}+4 b \left (1-c^2 x^2\right )^{3/2} \log \left (1-c^2 x^2\right )\right )-6 a \left (c^2 x^2-1\right ) \sqrt{d-c^2 d x^2} \tan ^{-1}\left (\frac{c x \sqrt{d-c^2 d x^2}}{\sqrt{d} \left (c^2 x^2-1\right )}\right )-3 b \sqrt{d} \left (1-c^2 x^2\right )^{3/2} \sin ^{-1}(c x)^2+2 b \sqrt{d} \sin \left (3 \sin ^{-1}(c x)\right ) \sin ^{-1}(c x)}{6 c^5 d^{5/2} \left (c^2 x^2-1\right ) \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

(-3*b*Sqrt[d]*(1 - c^2*x^2)^(3/2)*ArcSin[c*x]^2 - 6*a*(-1 + c^2*x^2)*Sqrt[d - c^2*d*x^2]*ArcTan[(c*x*Sqrt[d -
c^2*d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))] + Sqrt[d]*(6*a*c*x - 8*a*c^3*x^3 + b*Sqrt[1 - c^2*x^2] + 4*b*(1 - c^2*x^
2)^(3/2)*Log[1 - c^2*x^2]) + 2*b*Sqrt[d]*ArcSin[c*x]*Sin[3*ArcSin[c*x]])/(6*c^5*d^(5/2)*(-1 + c^2*x^2)*Sqrt[d
- c^2*d*x^2])

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Maple [C]  time = 0.277, size = 1510, normalized size = 7.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x)

[Out]

1/3*a*x^3/c^2/d/(-c^2*d*x^2+d)^(3/2)-a/c^4/d^2*x/(-c^2*d*x^2+d)^(1/2)+a/c^4/d^2/(c^2*d)^(1/2)*arctan((c^2*d)^(
1/2)*x/(-c^2*d*x^2+d)^(1/2))-8/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+
16)*(-c^2*x^2+1)*x^5-64/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^5
*arcsin(c*x)*(-c^2*x^2+1)^(1/2)+22/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*
x^2+16)*x^5+32*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)*c^2*arcsin(c*x)*
x^7-4*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c*(-c^2*x^2+1)^(1/2)*x^4+
181/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^2*arcsin(c*x)*x^3+13/2*
b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^3*x^2*(-c^2*x^2+1)^(1/2)-16*b
*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^4*arcsin(c*x)*x+4/3*b*(-d*(c^2
*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^5/d^3/(c^2*x^2-1)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)-8/3*I*b*(-d*(c^2*x^2-
1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)*c^2*x^7+32*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*
c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)*c*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^6-8/3*I*b*(-d*(c^2*x^2-1))^(1
/2)*(-c^2*x^2+1)^(1/2)/c^5/d^3/(c^2*x^2-1)*arcsin(c*x)-20/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*
x^6+118*c^4*x^4-71*c^2*x^2+16)/c^2*x^3-1/2*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^5/d^3/(c^2*x^2-1)*arc
sin(c*x)^2+220/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^3*arcsin(c
*x)*(-c^2*x^2+1)^(1/2)*x^2-2*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/
c^4*(-c^2*x^2+1)*x-84*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c*arcsi
n(c*x)*(-c^2*x^2+1)^(1/2)*x^4+2*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+1
6)/c^4*x-8/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^5*(-c^2*x^2+1)^(
1/2)-76*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)*arcsin(c*x)*x^5+14/3*I*
b*(-d*(c^2*x^2-1))^(1/2)/d^3/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^2*(-c^2*x^2+1)*x^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (b x^{4} \arcsin \left (c x\right ) + a x^{4}\right )} \sqrt{-c^{2} d x^{2} + d}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(-(b*x^4*arcsin(c*x) + a*x^4)*sqrt(-c^2*d*x^2 + d)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3)
, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \left (a + b \operatorname{asin}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(5/2),x)

[Out]

Integral(x**4*(a + b*asin(c*x))/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arcsin \left (c x\right ) + a\right )} x^{4}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)*x^4/(-c^2*d*x^2 + d)^(5/2), x)