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

Optimal. Leaf size=83 \[ \frac{b c \tan ^{-1}\left (\frac{x \sqrt{c^2 d+e}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{2 \sqrt{d} e \sqrt{c^2 d+e}}-\frac{a+b \sin ^{-1}(c x)}{2 e \left (d+e x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0594234, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4729, 377, 205} \[ \frac{b c \tan ^{-1}\left (\frac{x \sqrt{c^2 d+e}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{2 \sqrt{d} e \sqrt{c^2 d+e}}-\frac{a+b \sin ^{-1}(c x)}{2 e \left (d+e x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.146604, size = 87, normalized size = 1.05 \[ -\frac{\frac{a}{d+e x^2}-\frac{b c \tan ^{-1}\left (\frac{x \sqrt{c^2 d+e}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{\sqrt{d} \sqrt{c^2 d+e}}+\frac{b \sin ^{-1}(c x)}{d+e x^2}}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.032, size = 414, normalized size = 5. \begin{align*} -{\frac{{c}^{2}a}{2\,e \left ({c}^{2}e{x}^{2}+{c}^{2}d \right ) }}-{\frac{{c}^{2}b\arcsin \left ( cx \right ) }{2\,e \left ({c}^{2}e{x}^{2}+{c}^{2}d \right ) }}+{\frac{{c}^{2}b}{4\,e}\ln \left ({ \left ( 2\,{\frac{{c}^{2}d+e}{e}}+2\,{\frac{\sqrt{-{c}^{2}ed}}{e} \left ( cx+{\frac{\sqrt{-{c}^{2}ed}}{e}} \right ) }+2\,\sqrt{{\frac{{c}^{2}d+e}{e}}}\sqrt{- \left ( cx+{\frac{\sqrt{-{c}^{2}ed}}{e}} \right ) ^{2}+2\,{\frac{\sqrt{-{c}^{2}ed}}{e} \left ( cx+{\frac{\sqrt{-{c}^{2}ed}}{e}} \right ) }+{\frac{{c}^{2}d+e}{e}}} \right ) \left ( cx+{\frac{1}{e}\sqrt{-{c}^{2}ed}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{c}^{2}ed}}}{\frac{1}{\sqrt{{\frac{{c}^{2}d+e}{e}}}}}}-{\frac{{c}^{2}b}{4\,e}\ln \left ({ \left ( 2\,{\frac{{c}^{2}d+e}{e}}-2\,{\frac{\sqrt{-{c}^{2}ed}}{e} \left ( cx-{\frac{\sqrt{-{c}^{2}ed}}{e}} \right ) }+2\,\sqrt{{\frac{{c}^{2}d+e}{e}}}\sqrt{- \left ( cx-{\frac{\sqrt{-{c}^{2}ed}}{e}} \right ) ^{2}-2\,{\frac{\sqrt{-{c}^{2}ed}}{e} \left ( cx-{\frac{\sqrt{-{c}^{2}ed}}{e}} \right ) }+{\frac{{c}^{2}d+e}{e}}} \right ) \left ( cx-{\frac{1}{e}\sqrt{-{c}^{2}ed}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{c}^{2}ed}}}{\frac{1}{\sqrt{{\frac{{c}^{2}d+e}{e}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^2*a/e/(c^2*e*x^2+c^2*d)-1/2*c^2*b/e/(c^2*e*x^2+c^2*d)*arcsin(c*x)+1/4*c^2*b/e/(-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
/4*c^2*b/e/(-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(-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*(a+b*arcsin(c*x))/(e*x^2+d)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.71617, size = 830, normalized size = 10. \begin{align*} \left [-\frac{4 \, a c^{2} d^{2} + 4 \, a d e +{\left (b c e x^{2} + b c d\right )} \sqrt{-c^{2} d^{2} - d e} \log \left (\frac{{\left (8 \, c^{4} d^{2} + 8 \, c^{2} d e + e^{2}\right )} x^{4} - 2 \,{\left (4 \, c^{2} d^{2} + 3 \, d e\right )} x^{2} - 4 \, \sqrt{-c^{2} d^{2} - d e} \sqrt{-c^{2} x^{2} + 1}{\left ({\left (2 \, c^{2} d + e\right )} x^{3} - d x\right )} + d^{2}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}\right ) + 4 \,{\left (b c^{2} d^{2} + b d e\right )} \arcsin \left (c x\right )}{8 \,{\left (c^{2} d^{3} e + d^{2} e^{2} +{\left (c^{2} d^{2} e^{2} + d e^{3}\right )} x^{2}\right )}}, -\frac{2 \, a c^{2} d^{2} + 2 \, a d e +{\left (b c e x^{2} + b c d\right )} \sqrt{c^{2} d^{2} + d e} \arctan \left (\frac{\sqrt{c^{2} d^{2} + d e} \sqrt{-c^{2} x^{2} + 1}{\left ({\left (2 \, c^{2} d + e\right )} x^{2} - d\right )}}{2 \,{\left ({\left (c^{4} d^{2} + c^{2} d e\right )} x^{3} -{\left (c^{2} d^{2} + d e\right )} x\right )}}\right ) + 2 \,{\left (b c^{2} d^{2} + b d e\right )} \arcsin \left (c x\right )}{4 \,{\left (c^{2} d^{3} e + d^{2} e^{2} +{\left (c^{2} d^{2} e^{2} + d e^{3}\right )} x^{2}\right )}}\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)^2,x, algorithm="fricas")

[Out]

[-1/8*(4*a*c^2*d^2 + 4*a*d*e + (b*c*e*x^2 + b*c*d)*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)) + 4*(b*c^2*d^2 + b*d*e)*arcsin(c*x))/(c^2*d^3*e + d^2*e^2 + (c^2*d^2*e^2 + d*e^3)*x^2
), -1/4*(2*a*c^2*d^2 + 2*a*d*e + (b*c*e*x^2 + b*c*d)*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)) + 2*(b*c^2*d^2 + b*d*e)*ar
csin(c*x))/(c^2*d^3*e + d^2*e^2 + (c^2*d^2*e^2 + d*e^3)*x^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(a + b*asin(c*x))/(d + e*x**2)**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}{{\left (e x^{2} + d\right )}^{2}}\,{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)^2,x, algorithm="giac")

[Out]

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