3.643 \(\int \frac{x (a+b \sin ^{-1}(c x))}{(d+e x^2)^3} \, dx\)
Optimal. Leaf size=133 \[ -\frac{a+b \sin ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{b c \left (2 c^2 d+e\right ) \tan ^{-1}\left (\frac{x \sqrt{c^2 d+e}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{8 d^{3/2} e \left (c^2 d+e\right )^{3/2}}+\frac{b c x \sqrt{1-c^2 x^2}}{8 d \left (c^2 d+e\right ) \left (d+e x^2\right )} \]
[Out]
(b*c*x*Sqrt[1 - c^2*x^2])/(8*d*(c^2*d + e)*(d + e*x^2)) - (a + b*ArcSin[c*x])/(4*e*(d + e*x^2)^2) + (b*c*(2*c^
2*d + e)*ArcTan[(Sqrt[c^2*d + e]*x)/(Sqrt[d]*Sqrt[1 - c^2*x^2])])/(8*d^(3/2)*e*(c^2*d + e)^(3/2))
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Rubi [A] time = 0.0953057, antiderivative size = 133, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used =
{4729, 382, 377, 205} \[ -\frac{a+b \sin ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{b c \left (2 c^2 d+e\right ) \tan ^{-1}\left (\frac{x \sqrt{c^2 d+e}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{8 d^{3/2} e \left (c^2 d+e\right )^{3/2}}+\frac{b c x \sqrt{1-c^2 x^2}}{8 d \left (c^2 d+e\right ) \left (d+e x^2\right )} \]
Antiderivative was successfully verified.
[In]
Int[(x*(a + b*ArcSin[c*x]))/(d + e*x^2)^3,x]
[Out]
(b*c*x*Sqrt[1 - c^2*x^2])/(8*d*(c^2*d + e)*(d + e*x^2)) - (a + b*ArcSin[c*x])/(4*e*(d + e*x^2)^2) + (b*c*(2*c^
2*d + e)*ArcTan[(Sqrt[c^2*d + e]*x)/(Sqrt[d]*Sqrt[1 - c^2*x^2])])/(8*d^(3/2)*e*(c^2*d + e)^(3/2))
Rule 4729
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^(p + 1
)*(a + b*ArcSin[c*x]))/(2*e*(p + 1)), x] - Dist[(b*c)/(2*e*(p + 1)), Int[(d + e*x^2)^(p + 1)/Sqrt[1 - c^2*x^2]
, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[c^2*d + e, 0] && NeQ[p, -1]
Rule 382
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] || !LtQ[q, -1]) && NeQ[p, -1]
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 \left (a+b \sin ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx &=-\frac{a+b \sin ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{(b c) \int \frac{1}{\sqrt{1-c^2 x^2} \left (d+e x^2\right )^2} \, dx}{4 e}\\ &=\frac{b c x \sqrt{1-c^2 x^2}}{8 d \left (c^2 d+e\right ) \left (d+e x^2\right )}-\frac{a+b \sin ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{\left (b c \left (2 c^2 d+e\right )\right ) \int \frac{1}{\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 d \left (c^2 d+e\right ) \left (d+e x^2\right )}-\frac{a+b \sin ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{\left (b c \left (2 c^2 d+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 d e \left (c^2 d+e\right )}\\ &=\frac{b c x \sqrt{1-c^2 x^2}}{8 d \left (c^2 d+e\right ) \left (d+e x^2\right )}-\frac{a+b \sin ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{b c \left (2 c^2 d+e\right ) \tan ^{-1}\left (\frac{\sqrt{c^2 d+e} x}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{8 d^{3/2} e \left (c^2 d+e\right )^{3/2}}\\ \end{align*}
Mathematica [A] time = 0.57626, size = 141, normalized size = 1.06 \[ \frac{1}{8} \left (\frac{\frac{b c x \sqrt{1-c^2 x^2} \left (d+e x^2\right )}{d \left (c^2 d+e\right )}-\frac{2 a}{e}}{\left (d+e x^2\right )^2}+\frac{b c \left (2 c^2 d+e\right ) \tan ^{-1}\left (\frac{x \sqrt{c^2 d+e}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{d^{3/2} e \left (c^2 d+e\right )^{3/2}}-\frac{2 b \sin ^{-1}(c x)}{e \left (d+e x^2\right )^2}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(x*(a + b*ArcSin[c*x]))/(d + e*x^2)^3,x]
[Out]
(((-2*a)/e + (b*c*x*Sqrt[1 - c^2*x^2]*(d + e*x^2))/(d*(c^2*d + e)))/(d + e*x^2)^2 - (2*b*ArcSin[c*x])/(e*(d +
e*x^2)^2) + (b*c*(2*c^2*d + e)*ArcTan[(Sqrt[c^2*d + e]*x)/(Sqrt[d]*Sqrt[1 - c^2*x^2])])/(d^(3/2)*e*(c^2*d + e)
^(3/2)))/8
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Maple [B] time = 0.012, size = 1017, normalized size = 7.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*(a+b*arcsin(c*x))/(e*x^2+d)^3,x)
[Out]
-1/4*c^4*a/e/(c^2*e*x^2+c^2*d)^2-1/4*c^4*b/e/(c^2*e*x^2+c^2*d)^2*arcsin(c*x)+1/16*c^2*b/e/d/(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^2/d*(-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))+1/16*c^2*b/e/d/(-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/d/(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^2/d*(-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))-1/16*c^2*b/e/d/(-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))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left ({\left (c e^{3} x^{4} + 2 \, c d e^{2} x^{2} + c d^{2} e\right )} \int \frac{e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (-c x + 1\right )\right )}}{c^{4} e^{3} x^{8} - c^{2} d^{2} e x^{2} +{\left (2 \, c^{4} d e^{2} - c^{2} e^{3}\right )} x^{6} +{\left (c^{4} d^{2} e - 2 \, c^{2} d e^{2}\right )} x^{4} -{\left (c^{2} e^{3} x^{6} +{\left (2 \, c^{2} d e^{2} - e^{3}\right )} x^{4} - d^{2} e +{\left (c^{2} d^{2} e - 2 \, d e^{2}\right )} x^{2}\right )}{\left (c x + 1\right )}{\left (c x - 1\right )}}\,{d x} + \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )\right )} b}{4 \,{\left (e^{3} x^{4} + 2 \, d e^{2} x^{2} + d^{2} e\right )}} - \frac{a}{4 \,{\left (e^{3} x^{4} + 2 \, d e^{2} x^{2} + d^{2} e\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(a+b*arcsin(c*x))/(e*x^2+d)^3,x, algorithm="maxima")
[Out]
-1/4*(4*(c*e^3*x^4 + 2*c*d*e^2*x^2 + c*d^2*e)*integrate(1/4*e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1))/(c^4*e^3*
x^8 - c^2*d^2*e*x^2 + (2*c^4*d*e^2 - c^2*e^3)*x^6 + (c^4*d^2*e - 2*c^2*d*e^2)*x^4 + (c^2*e^3*x^6 + (2*c^2*d*e^
2 - e^3)*x^4 - d^2*e + (c^2*d^2*e - 2*d*e^2)*x^2)*e^(log(c*x + 1) + log(-c*x + 1))), x) + arctan2(c*x, sqrt(c*
x + 1)*sqrt(-c*x + 1)))*b/(e^3*x^4 + 2*d*e^2*x^2 + d^2*e) - 1/4*a/(e^3*x^4 + 2*d*e^2*x^2 + d^2*e)
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Fricas [B] time = 3.90244, size = 1604, normalized size = 12.06 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(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 + (2*b*c^3*d^3 + b*c*d^2*e + (2*b*c^3*d*e^2 + b*c*e^3)*x^4
+ 2*(2*b*c^3*d^2*e + 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)*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^6*e + 2*c^2*d^5*e^2 + d^4*e^3 + (c^4*d^4*e^3 + 2*c^2*d^3
*e^4 + d^2*e^5)*x^4 + 2*(c^4*d^5*e^2 + 2*c^2*d^4*e^3 + d^3*e^4)*x^2), -1/16*(4*a*c^4*d^4 + 8*a*c^2*d^3*e + 4*a
*d^2*e^2 + (2*b*c^3*d^3 + b*c*d^2*e + (2*b*c^3*d*e^2 + b*c*e^3)*x^4 + 2*(2*b*c^3*d^2*e + 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)*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^6*e + 2*c^2*d^5*e^2 + d^4*e^3 + (c^4*d^4*
e^3 + 2*c^2*d^3*e^4 + d^2*e^5)*x^4 + 2*(c^4*d^5*e^2 + 2*c^2*d^4*e^3 + d^3*e^4)*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*(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}{{\left (e x^{2} + d\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(a+b*arcsin(c*x))/(e*x^2+d)^3,x, algorithm="giac")
[Out]
integrate((b*arcsin(c*x) + a)*x/(e*x^2 + d)^3, x)