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

Optimal. Leaf size=153 \[ \frac{x^4 \left (a+b \sin ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}+\frac{b c \left (2 c^2 d+3 e\right ) \tan ^{-1}\left (\frac{x \sqrt{c^2 d+e}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{8 \sqrt{d} e^2 \left (c^2 d+e\right )^{3/2}}-\frac{b c x \sqrt{1-c^2 x^2}}{8 e \left (c^2 d+e\right ) \left (d+e x^2\right )}-\frac{b \sin ^{-1}(c x)}{4 d e^2} \]

[Out]

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

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Rubi [A]  time = 0.193865, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {264, 4731, 12, 470, 523, 216, 377, 205} \[ \frac{x^4 \left (a+b \sin ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}+\frac{b c \left (2 c^2 d+3 e\right ) \tan ^{-1}\left (\frac{x \sqrt{c^2 d+e}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{8 \sqrt{d} e^2 \left (c^2 d+e\right )^{3/2}}-\frac{b c x \sqrt{1-c^2 x^2}}{8 e \left (c^2 d+e\right ) \left (d+e x^2\right )}-\frac{b \sin ^{-1}(c x)}{4 d e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 4731

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 -
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (GtQ[p, 0] ||
(IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 470

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

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.537112, size = 152, normalized size = 0.99 \[ \frac{-\frac{2 a \left (d+2 e x^2\right )+\frac{b c e x \sqrt{1-c^2 x^2} \left (d+e x^2\right )}{c^2 d+e}}{\left (d+e x^2\right )^2}+\frac{b c \left (2 c^2 d+3 e\right ) \tan ^{-1}\left (\frac{x \sqrt{c^2 d+e}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{\sqrt{d} \left (c^2 d+e\right )^{3/2}}-\frac{2 b \sin ^{-1}(c x) \left (d+2 e x^2\right )}{\left (d+e x^2\right )^2}}{8 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(((b*c*e*x*Sqrt[1 - c^2*x^2]*(d + e*x^2))/(c^2*d + e) + 2*a*(d + 2*e*x^2))/(d + e*x^2)^2) - (2*b*(d + 2*e*x^
2)*ArcSin[c*x])/(d + e*x^2)^2 + (b*c*(2*c^2*d + 3*e)*ArcTan[(Sqrt[c^2*d + e]*x)/(Sqrt[d]*Sqrt[1 - c^2*x^2])])/
(Sqrt[d]*(c^2*d + e)^(3/2)))/(8*e^2)

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Maple [B]  time = 0.018, size = 1055, normalized size = 6.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^2*a/e^2/(c^2*e*x^2+c^2*d)+1/4*c^4*a/e^2*d/(c^2*e*x^2+c^2*d)^2-1/2*c^2*b*arcsin(c*x)/e^2/(c^2*e*x^2+c^2*
d)+1/4*c^4*b*arcsin(c*x)/e^2*d/(c^2*e*x^2+c^2*d)^2+3/16*c^2*b/e^2/(-c^2*e*d)^(1/2)/((c^2*d+e)/e)^(1/2)*ln((2*(
c^2*d+e)/e+2*(-c^2*e*d)^(1/2)/e*(c*x+(-c^2*e*d)^(1/2)/e)+2*((c^2*d+e)/e)^(1/2)*(-(c*x+(-c^2*e*d)^(1/2)/e)^2+2*
(-c^2*e*d)^(1/2)/e*(c*x+(-c^2*e*d)^(1/2)/e)+(c^2*d+e)/e)^(1/2))/(c*x+(-c^2*e*d)^(1/2)/e))-1/16*c^2*b/e^2/(c^2*
d+e)/(c*x+(-c^2*e*d)^(1/2)/e)*(-(c*x+(-c^2*e*d)^(1/2)/e)^2+2*(-c^2*e*d)^(1/2)/e*(c*x+(-c^2*e*d)^(1/2)/e)+(c^2*
d+e)/e)^(1/2)+1/16*c^2*b/e^3*(-c^2*e*d)^(1/2)/(c^2*d+e)/((c^2*d+e)/e)^(1/2)*ln((2*(c^2*d+e)/e+2*(-c^2*e*d)^(1/
2)/e*(c*x+(-c^2*e*d)^(1/2)/e)+2*((c^2*d+e)/e)^(1/2)*(-(c*x+(-c^2*e*d)^(1/2)/e)^2+2*(-c^2*e*d)^(1/2)/e*(c*x+(-c
^2*e*d)^(1/2)/e)+(c^2*d+e)/e)^(1/2))/(c*x+(-c^2*e*d)^(1/2)/e))-3/16*c^2*b/e^2/(-c^2*e*d)^(1/2)/((c^2*d+e)/e)^(
1/2)*ln((2*(c^2*d+e)/e-2*(-c^2*e*d)^(1/2)/e*(c*x-(-c^2*e*d)^(1/2)/e)+2*((c^2*d+e)/e)^(1/2)*(-(c*x-(-c^2*e*d)^(
1/2)/e)^2-2*(-c^2*e*d)^(1/2)/e*(c*x-(-c^2*e*d)^(1/2)/e)+(c^2*d+e)/e)^(1/2))/(c*x-(-c^2*e*d)^(1/2)/e))-1/16*c^2
*b/e^2/(c^2*d+e)/(c*x-(-c^2*e*d)^(1/2)/e)*(-(c*x-(-c^2*e*d)^(1/2)/e)^2-2*(-c^2*e*d)^(1/2)/e*(c*x-(-c^2*e*d)^(1
/2)/e)+(c^2*d+e)/e)^(1/2)-1/16*c^2*b/e^3*(-c^2*e*d)^(1/2)/(c^2*d+e)/((c^2*d+e)/e)^(1/2)*ln((2*(c^2*d+e)/e-2*(-
c^2*e*d)^(1/2)/e*(c*x-(-c^2*e*d)^(1/2)/e)+2*((c^2*d+e)/e)^(1/2)*(-(c*x-(-c^2*e*d)^(1/2)/e)^2-2*(-c^2*e*d)^(1/2
)/e*(c*x-(-c^2*e*d)^(1/2)/e)+(c^2*d+e)/e)^(1/2))/(c*x-(-c^2*e*d)^(1/2)/e))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (2 \, e x^{2} + d\right )} a}{4 \,{\left (e^{4} x^{4} + 2 \, d e^{3} x^{2} + d^{2} e^{2}\right )}} - \frac{{\left ({\left (2 \, e x^{2} + d\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) +{\left (e^{4} x^{4} + 2 \, d e^{3} x^{2} + d^{2} e^{2}\right )} \int \frac{{\left (2 \, c e x^{2} + c d\right )} e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (-c x + 1\right )\right )}}{c^{4} e^{4} x^{8} - c^{2} d^{2} e^{2} x^{2} +{\left (2 \, c^{4} d e^{3} - c^{2} e^{4}\right )} x^{6} +{\left (c^{4} d^{2} e^{2} - 2 \, c^{2} d e^{3}\right )} x^{4} -{\left (c^{2} e^{4} x^{6} +{\left (2 \, c^{2} d e^{3} - e^{4}\right )} x^{4} - d^{2} e^{2} +{\left (c^{2} d^{2} e^{2} - 2 \, d e^{3}\right )} x^{2}\right )}{\left (c x + 1\right )}{\left (c x - 1\right )}}\,{d x}\right )} b}{4 \,{\left (e^{4} x^{4} + 2 \, d e^{3} x^{2} + d^{2} e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*e*x^2 + d)*a/(e^4*x^4 + 2*d*e^3*x^2 + d^2*e^2) - 1/4*((2*e*x^2 + d)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c
*x + 1)) + 4*(e^4*x^4 + 2*d*e^3*x^2 + d^2*e^2)*integrate(1/4*(2*c*e*x^2 + c*d)*e^(1/2*log(c*x + 1) + 1/2*log(-
c*x + 1))/(c^4*e^4*x^8 - c^2*d^2*e^2*x^2 + (2*c^4*d*e^3 - c^2*e^4)*x^6 + (c^4*d^2*e^2 - 2*c^2*d*e^3)*x^4 + (c^
2*e^4*x^6 + (2*c^2*d*e^3 - e^4)*x^4 - d^2*e^2 + (c^2*d^2*e^2 - 2*d*e^3)*x^2)*e^(log(c*x + 1) + log(-c*x + 1)))
, x))*b/(e^4*x^4 + 2*d*e^3*x^2 + d^2*e^2)

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Fricas [B]  time = 3.99899, size = 1891, normalized size = 12.36 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/32*(8*a*c^4*d^4 + 16*a*c^2*d^3*e + 8*a*d^2*e^2 + 16*(a*c^4*d^3*e + 2*a*c^2*d^2*e^2 + a*d*e^3)*x^2 + (2*b*c
^3*d^3 + 3*b*c*d^2*e + (2*b*c^3*d*e^2 + 3*b*c*e^3)*x^4 + 2*(2*b*c^3*d^2*e + 3*b*c*d*e^2)*x^2)*sqrt(-c^2*d^2 -
d*e)*log(((8*c^4*d^2 + 8*c^2*d*e + e^2)*x^4 - 2*(4*c^2*d^2 + 3*d*e)*x^2 - 4*sqrt(-c^2*d^2 - d*e)*sqrt(-c^2*x^2
+ 1)*((2*c^2*d + e)*x^3 - d*x) + d^2)/(e^2*x^4 + 2*d*e*x^2 + d^2)) + 8*(b*c^4*d^4 + 2*b*c^2*d^3*e + b*d^2*e^2
+ 2*(b*c^4*d^3*e + 2*b*c^2*d^2*e^2 + b*d*e^3)*x^2)*arcsin(c*x) + 4*sqrt(-c^2*x^2 + 1)*((b*c^3*d^2*e^2 + b*c*d
*e^3)*x^3 + (b*c^3*d^3*e + b*c*d^2*e^2)*x))/(c^4*d^5*e^2 + 2*c^2*d^4*e^3 + d^3*e^4 + (c^4*d^3*e^4 + 2*c^2*d^2*
e^5 + d*e^6)*x^4 + 2*(c^4*d^4*e^3 + 2*c^2*d^3*e^4 + d^2*e^5)*x^2), -1/16*(4*a*c^4*d^4 + 8*a*c^2*d^3*e + 4*a*d^
2*e^2 + 8*(a*c^4*d^3*e + 2*a*c^2*d^2*e^2 + a*d*e^3)*x^2 + (2*b*c^3*d^3 + 3*b*c*d^2*e + (2*b*c^3*d*e^2 + 3*b*c*
e^3)*x^4 + 2*(2*b*c^3*d^2*e + 3*b*c*d*e^2)*x^2)*sqrt(c^2*d^2 + d*e)*arctan(1/2*sqrt(c^2*d^2 + d*e)*sqrt(-c^2*x
^2 + 1)*((2*c^2*d + e)*x^2 - d)/((c^4*d^2 + c^2*d*e)*x^3 - (c^2*d^2 + d*e)*x)) + 4*(b*c^4*d^4 + 2*b*c^2*d^3*e
+ b*d^2*e^2 + 2*(b*c^4*d^3*e + 2*b*c^2*d^2*e^2 + b*d*e^3)*x^2)*arcsin(c*x) + 2*sqrt(-c^2*x^2 + 1)*((b*c^3*d^2*
e^2 + b*c*d*e^3)*x^3 + (b*c^3*d^3*e + b*c*d^2*e^2)*x))/(c^4*d^5*e^2 + 2*c^2*d^4*e^3 + d^3*e^4 + (c^4*d^3*e^4 +
2*c^2*d^2*e^5 + d*e^6)*x^4 + 2*(c^4*d^4*e^3 + 2*c^2*d^3*e^4 + d^2*e^5)*x^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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^{3}}{{\left (e x^{2} + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsin(c*x) + a)*x^3/(e*x^2 + d)^3, x)