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

Optimal. Leaf size=86 \[ \frac {-a-b \sin ^{-1}(c x)}{2 e \left (d+e x^2\right )}+\frac {b c \tan ^{-1}\left (\frac {x \sqrt {c^2 d+e}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{2 \sqrt {d} e \sqrt {c^2 d+e}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 83, normalized size of antiderivative = 0.97, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.15, size = 87, normalized size = 1.01 \[ -\frac {\frac {a}{d+e x^2}-\frac {b c \tan ^{-1}\left (\frac {x \sqrt {c^2 d+e}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{\sqrt {d} \sqrt {c^2 d+e}}+\frac {b \sin ^{-1}(c x)}{d+e x^2}}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(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]))/e

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fricas [B]  time = 0.65, size = 395, normalized size = 4.59 \[ \left [-\frac {4 \, a c^{2} d^{2} + 4 \, a d e + {\left (b c e x^{2} + b c d\right )} \sqrt {-c^{2} d^{2} - d e} \log \left (\frac {{\left (8 \, c^{4} d^{2} + 8 \, c^{2} d e + e^{2}\right )} x^{4} - 2 \, {\left (4 \, c^{2} d^{2} + 3 \, d e\right )} x^{2} - 4 \, \sqrt {-c^{2} d^{2} - d e} \sqrt {-c^{2} x^{2} + 1} {\left ({\left (2 \, c^{2} d + e\right )} x^{3} - d x\right )} + d^{2}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}\right ) + 4 \, {\left (b c^{2} d^{2} + b d e\right )} \arcsin \left (c x\right )}{8 \, {\left (c^{2} d^{3} e + d^{2} e^{2} + {\left (c^{2} d^{2} e^{2} + d e^{3}\right )} x^{2}\right )}}, -\frac {2 \, a c^{2} d^{2} + 2 \, a d e + {\left (b c e x^{2} + b c d\right )} \sqrt {c^{2} d^{2} + d e} \arctan \left (\frac {\sqrt {c^{2} d^{2} + d e} \sqrt {-c^{2} x^{2} + 1} {\left ({\left (2 \, c^{2} d + e\right )} x^{2} - d\right )}}{2 \, {\left ({\left (c^{4} d^{2} + c^{2} d e\right )} x^{3} - {\left (c^{2} d^{2} + d e\right )} x\right )}}\right ) + 2 \, {\left (b c^{2} d^{2} + b d e\right )} \arcsin \left (c x\right )}{4 \, {\left (c^{2} d^{3} e + d^{2} e^{2} + {\left (c^{2} d^{2} e^{2} + d e^{3}\right )} x^{2}\right )}}\right ] \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x}{{\left (e x^{2} + d\right )}^{2}}\,{d x} \]

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)

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maple [B]  time = 0.04, size = 414, normalized size = 4.81 \[ -\frac {c^{2} a}{2 e \left (c^{2} e \,x^{2}+c^{2} d \right )}-\frac {c^{2} b \arcsin \left (c x \right )}{2 e \left (c^{2} e \,x^{2}+c^{2} d \right )}+\frac {c^{2} b \ln \left (\frac {\frac {2 c^{2} d +2 e}{e}+\frac {2 \sqrt {-c^{2} e d}\, \left (c x +\frac {\sqrt {-c^{2} e d}}{e}\right )}{e}+2 \sqrt {\frac {c^{2} d +e}{e}}\, \sqrt {-\left (c x +\frac {\sqrt {-c^{2} e d}}{e}\right )^{2}+\frac {2 \sqrt {-c^{2} e d}\, \left (c x +\frac {\sqrt {-c^{2} e d}}{e}\right )}{e}+\frac {c^{2} d +e}{e}}}{c x +\frac {\sqrt {-c^{2} e d}}{e}}\right )}{4 e \sqrt {-c^{2} e d}\, \sqrt {\frac {c^{2} d +e}{e}}}-\frac {c^{2} b \ln \left (\frac {\frac {2 c^{2} d +2 e}{e}-\frac {2 \sqrt {-c^{2} e d}\, \left (c x -\frac {\sqrt {-c^{2} e d}}{e}\right )}{e}+2 \sqrt {\frac {c^{2} d +e}{e}}\, \sqrt {-\left (c x -\frac {\sqrt {-c^{2} e d}}{e}\right )^{2}-\frac {2 \sqrt {-c^{2} e d}\, \left (c x -\frac {\sqrt {-c^{2} e d}}{e}\right )}{e}+\frac {c^{2} d +e}{e}}}{c x -\frac {\sqrt {-c^{2} e d}}{e}}\right )}{4 e \sqrt {-c^{2} e d}\, \sqrt {\frac {c^{2} d +e}{e}}} \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left ({\left (c e^{2} x^{2} + c d e\right )} \int \frac {e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )}}{c^{4} e^{2} x^{6} - c^{2} d e x^{2} + {\left (c^{4} d e - c^{2} e^{2}\right )} x^{4} - {\left (c^{2} e^{2} x^{4} + {\left (c^{2} d e - e^{2}\right )} x^{2} - d e\right )} {\left (c x + 1\right )} {\left (c x - 1\right )}}\,{d x} + \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\right )} b}{2 \, {\left (e^{2} x^{2} + d e\right )}} - \frac {a}{2 \, {\left (e^{2} x^{2} + d e\right )}} \]

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]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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