3.1167 \(\int \frac{x (a+b \tan ^{-1}(c x))}{(d+e x^2)^3} \, dx\)

Optimal. Leaf size=131 \[ -\frac{a+b \tan ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}-\frac{b c \left (3 c^2 d-e\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 d^{3/2} \sqrt{e} \left (c^2 d-e\right )^2}-\frac{b c x}{8 d \left (c^2 d-e\right ) \left (d+e x^2\right )}+\frac{b c^4 \tan ^{-1}(c x)}{4 e \left (c^2 d-e\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.113401, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {4974, 414, 522, 203, 205} \[ -\frac{a+b \tan ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}-\frac{b c \left (3 c^2 d-e\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 d^{3/2} \sqrt{e} \left (c^2 d-e\right )^2}-\frac{b c x}{8 d \left (c^2 d-e\right ) \left (d+e x^2\right )}+\frac{b c^4 \tan ^{-1}(c x)}{4 e \left (c^2 d-e\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4974

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^(q +
1)*(a + b*ArcTan[c*x]))/(2*e*(q + 1)), x] - Dist[(b*c)/(2*e*(q + 1)), Int[(d + e*x^2)^(q + 1)/(1 + c^2*x^2), x
], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 414

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[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

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

Rule 203

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

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 \tan ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx &=-\frac{a+b \tan ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{(b c) \int \frac{1}{\left (1+c^2 x^2\right ) \left (d+e x^2\right )^2} \, dx}{4 e}\\ &=-\frac{b c x}{8 d \left (c^2 d-e\right ) \left (d+e x^2\right )}-\frac{a+b \tan ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{(b c) \int \frac{2 c^2 d-e-c^2 e x^2}{\left (1+c^2 x^2\right ) \left (d+e x^2\right )} \, dx}{8 d \left (c^2 d-e\right ) e}\\ &=-\frac{b c x}{8 d \left (c^2 d-e\right ) \left (d+e x^2\right )}-\frac{a+b \tan ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}-\frac{\left (b c \left (3 c^2 d-e\right )\right ) \int \frac{1}{d+e x^2} \, dx}{8 d \left (c^2 d-e\right )^2}+\frac{\left (b c^5\right ) \int \frac{1}{1+c^2 x^2} \, dx}{4 \left (c^2 d-e\right )^2 e}\\ &=-\frac{b c x}{8 d \left (c^2 d-e\right ) \left (d+e x^2\right )}+\frac{b c^4 \tan ^{-1}(c x)}{4 \left (c^2 d-e\right )^2 e}-\frac{a+b \tan ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}-\frac{b c \left (3 c^2 d-e\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 d^{3/2} \left (c^2 d-e\right )^2 \sqrt{e}}\\ \end{align*}

Mathematica [A]  time = 1.02572, size = 131, normalized size = 1. \[ \frac{1}{8} \left (-\frac{\frac{2 a}{e}+\frac{b c x \left (d+e x^2\right )}{d \left (c^2 d-e\right )}}{\left (d+e x^2\right )^2}-\frac{b c \left (3 c^2 d-e\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{d^{3/2} \sqrt{e} \left (e-c^2 d\right )^2}+\frac{2 b \tan ^{-1}(c x) \left (\frac{c^4}{\left (e-c^2 d\right )^2}-\frac{1}{\left (d+e x^2\right )^2}\right )}{e}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 216, normalized size = 1.7 \begin{align*} -{\frac{a{c}^{4}}{4\,e \left ({c}^{2}e{x}^{2}+{c}^{2}d \right ) ^{2}}}-{\frac{b{c}^{4}\arctan \left ( cx \right ) }{4\,e \left ({c}^{2}e{x}^{2}+{c}^{2}d \right ) ^{2}}}-{\frac{{c}^{5}bx}{8\, \left ({c}^{2}d-e \right ) ^{2} \left ({c}^{2}e{x}^{2}+{c}^{2}d \right ) }}+{\frac{{c}^{3}bex}{8\, \left ({c}^{2}d-e \right ) ^{2}d \left ({c}^{2}e{x}^{2}+{c}^{2}d \right ) }}-{\frac{3\,{c}^{3}b}{8\, \left ({c}^{2}d-e \right ) ^{2}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{bce}{8\, \left ({c}^{2}d-e \right ) ^{2}d}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{b{c}^{4}\arctan \left ( cx \right ) }{4\, \left ({c}^{2}d-e \right ) ^{2}e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arctan(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*arctan(c*x)-1/8*c^5*b/(c^2*d-e)^2*x/(c^2*e*x^
2+c^2*d)+1/8*c^3*b*e/(c^2*d-e)^2*x/d/(c^2*e*x^2+c^2*d)-3/8*c^3*b/(c^2*d-e)^2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2
))+1/8*c*b*e/(c^2*d-e)^2/d/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))+1/4*b*c^4*arctan(c*x)/(c^2*d-e)^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*arctan(c*x))/(e*x^2+d)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.52233, size = 1288, normalized size = 9.83 \begin{align*} \left [-\frac{4 \, a c^{4} d^{4} - 8 \, a c^{2} d^{3} e + 4 \, a d^{2} e^{2} + 2 \,{\left (b c^{3} d^{2} e^{2} - b c d e^{3}\right )} x^{3} -{\left (3 \, b c^{3} d^{3} - b c d^{2} e +{\left (3 \, b c^{3} d e^{2} - b c e^{3}\right )} x^{4} + 2 \,{\left (3 \, b c^{3} d^{2} e - b c d e^{2}\right )} x^{2}\right )} \sqrt{-d e} \log \left (\frac{e x^{2} - 2 \, \sqrt{-d e} x - d}{e x^{2} + d}\right ) + 2 \,{\left (b c^{3} d^{3} e - b c d^{2} e^{2}\right )} x - 4 \,{\left (b c^{4} d^{2} e^{2} x^{4} + 2 \, b c^{4} d^{3} e x^{2} + 2 \, b c^{2} d^{3} e - b d^{2} e^{2}\right )} \arctan \left (c x\right )}{16 \,{\left (c^{4} d^{6} e - 2 \, c^{2} d^{5} e^{2} + d^{4} e^{3} +{\left (c^{4} d^{4} e^{3} - 2 \, c^{2} d^{3} e^{4} + d^{2} e^{5}\right )} x^{4} + 2 \,{\left (c^{4} d^{5} e^{2} - 2 \, c^{2} d^{4} e^{3} + d^{3} e^{4}\right )} x^{2}\right )}}, -\frac{2 \, a c^{4} d^{4} - 4 \, a c^{2} d^{3} e + 2 \, a d^{2} e^{2} +{\left (b c^{3} d^{2} e^{2} - b c d e^{3}\right )} x^{3} +{\left (3 \, b c^{3} d^{3} - b c d^{2} e +{\left (3 \, b c^{3} d e^{2} - b c e^{3}\right )} x^{4} + 2 \,{\left (3 \, b c^{3} d^{2} e - b c d e^{2}\right )} x^{2}\right )} \sqrt{d e} \arctan \left (\frac{\sqrt{d e} x}{d}\right ) +{\left (b c^{3} d^{3} e - b c d^{2} e^{2}\right )} x - 2 \,{\left (b c^{4} d^{2} e^{2} x^{4} + 2 \, b c^{4} d^{3} e x^{2} + 2 \, b c^{2} d^{3} e - b d^{2} e^{2}\right )} \arctan \left (c x\right )}{8 \,{\left (c^{4} d^{6} e - 2 \, c^{2} d^{5} e^{2} + d^{4} e^{3} +{\left (c^{4} d^{4} e^{3} - 2 \, c^{2} d^{3} e^{4} + d^{2} e^{5}\right )} x^{4} + 2 \,{\left (c^{4} d^{5} e^{2} - 2 \, c^{2} d^{4} e^{3} + d^{3} e^{4}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-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^2*e^2 - b*c*d*e^3)*x^3 - (3*b*c^3*d^3 - b*c*d^2
*e + (3*b*c^3*d*e^2 - b*c*e^3)*x^4 + 2*(3*b*c^3*d^2*e - b*c*d*e^2)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x
- d)/(e*x^2 + d)) + 2*(b*c^3*d^3*e - b*c*d^2*e^2)*x - 4*(b*c^4*d^2*e^2*x^4 + 2*b*c^4*d^3*e*x^2 + 2*b*c^2*d^3*
e - b*d^2*e^2)*arctan(c*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), -1/8*(2*a*c^4*d^4 - 4*a*c^2*d^3*e + 2*a*d^2*e^2 + (b*c^3*d^
2*e^2 - b*c*d*e^3)*x^3 + (3*b*c^3*d^3 - b*c*d^2*e + (3*b*c^3*d*e^2 - b*c*e^3)*x^4 + 2*(3*b*c^3*d^2*e - b*c*d*e
^2)*x^2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) + (b*c^3*d^3*e - b*c*d^2*e^2)*x - 2*(b*c^4*d^2*e^2*x^4 + 2*b*c^4*d^3*
e*x^2 + 2*b*c^2*d^3*e - b*d^2*e^2)*arctan(c*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*atan(c*x))/(e*x**2+d)**3,x)

[Out]

Timed out

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Giac [B]  time = 2.04461, size = 518, normalized size = 3.95 \begin{align*} -\frac{{\left (3 \, b c^{3} d - b c e\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{1}{2}\right )}}{8 \,{\left (c^{4} d^{3} - 2 \, c^{2} d^{2} e + d e^{2}\right )} \sqrt{d}} - \frac{\pi b c^{4} d x^{4} e^{2} \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (x\right ) + 2 \, \pi b c^{4} d^{2} x^{2} e \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (x\right ) - b c^{4} d x^{4} \arctan \left (c x\right ) e^{2} - 2 \, b c^{4} d^{2} x^{2} \arctan \left (c x\right ) e + \pi b c^{4} d^{3} \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (x\right ) + b c^{3} d x^{3} e^{2} + 2 \, a c^{4} d^{3} + b c^{3} d^{2} x e - 2 \, b c^{2} d^{2} \arctan \left (c x\right ) e - b c x^{3} e^{3} - 4 \, a c^{2} d^{2} e - b c d x e^{2} + b d \arctan \left (c x\right ) e^{2} + 2 \, a d e^{2}}{4 \,{\left (c^{4} d^{3} x^{4} e^{3} + 2 \, c^{4} d^{4} x^{2} e^{2} + c^{4} d^{5} e - 2 \, c^{2} d^{2} x^{4} e^{4} - 4 \, c^{2} d^{3} x^{2} e^{3} - 2 \, c^{2} d^{4} e^{2} + d x^{4} e^{5} + 2 \, d^{2} x^{2} e^{4} + d^{3} e^{3}\right )}} - \frac{b c x^{3} e^{2} + 2 \, a c^{2} d^{2} + b c d x e - 2 \, a d e}{8 \,{\left (c^{2} d^{2} e - d e^{2}\right )}{\left (x^{2} e + d\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*b*c^3*d - b*c*e)*arctan(x*e^(1/2)/sqrt(d))*e^(-1/2)/((c^4*d^3 - 2*c^2*d^2*e + d*e^2)*sqrt(d)) - 1/4*(p
i*b*c^4*d*x^4*e^2*sgn(c)*sgn(x) + 2*pi*b*c^4*d^2*x^2*e*sgn(c)*sgn(x) - b*c^4*d*x^4*arctan(c*x)*e^2 - 2*b*c^4*d
^2*x^2*arctan(c*x)*e + pi*b*c^4*d^3*sgn(c)*sgn(x) + b*c^3*d*x^3*e^2 + 2*a*c^4*d^3 + b*c^3*d^2*x*e - 2*b*c^2*d^
2*arctan(c*x)*e - b*c*x^3*e^3 - 4*a*c^2*d^2*e - b*c*d*x*e^2 + b*d*arctan(c*x)*e^2 + 2*a*d*e^2)/(c^4*d^3*x^4*e^
3 + 2*c^4*d^4*x^2*e^2 + c^4*d^5*e - 2*c^2*d^2*x^4*e^4 - 4*c^2*d^3*x^2*e^3 - 2*c^2*d^4*e^2 + d*x^4*e^5 + 2*d^2*
x^2*e^4 + d^3*e^3) - 1/8*(b*c*x^3*e^2 + 2*a*c^2*d^2 + b*c*d*x*e - 2*a*d*e)/((c^2*d^2*e - d*e^2)*(x^2*e + d)^2)