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

Optimal. Leaf size=146 \[ -\frac{a+b \tan ^{-1}(c x)}{2 e (d+e x)^2}-\frac{b c^3 d \log \left (c^2 x^2+1\right )}{2 \left (c^2 d^2+e^2\right )^2}-\frac{b c}{2 \left (c^2 d^2+e^2\right ) (d+e x)}+\frac{b c^3 d \log (d+e x)}{\left (c^2 d^2+e^2\right )^2}+\frac{b c^2 (c d-e) (c d+e) \tan ^{-1}(c x)}{2 e \left (c^2 d^2+e^2\right )^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.123111, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {4862, 710, 801, 635, 203, 260} \[ -\frac{a+b \tan ^{-1}(c x)}{2 e (d+e x)^2}-\frac{b c^3 d \log \left (c^2 x^2+1\right )}{2 \left (c^2 d^2+e^2\right )^2}-\frac{b c}{2 \left (c^2 d^2+e^2\right ) (d+e x)}+\frac{b c^3 d \log (d+e x)}{\left (c^2 d^2+e^2\right )^2}+\frac{b c^2 (c d-e) (c d+e) \tan ^{-1}(c x)}{2 e \left (c^2 d^2+e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4862

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

Rule 710

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 +
a*e^2)), x] + Dist[c/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*(d - e*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d,
e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

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 260

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

Rubi steps

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

Mathematica [A]  time = 0.327461, size = 192, normalized size = 1.32 \[ -\frac{2 \left (a+b \tan ^{-1}(c x)\right )+\frac{b c (d+e x) \left (2 e \left (c^2 d^2+e^2\right )-\left (c^2 d \left (\sqrt{-c^2} d-2 e\right )-\sqrt{-c^2} e^2\right ) \log \left (1-\sqrt{-c^2} x\right ) (d+e x)-\left (\sqrt{-c^2} e^2-c^2 d \left (\sqrt{-c^2} d+2 e\right )\right ) \log \left (\sqrt{-c^2} x+1\right ) (d+e x)-4 c^2 d e (d+e x) \log (d+e x)\right )}{\left (c^2 d^2+e^2\right )^2}}{4 e (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.034, size = 184, normalized size = 1.3 \begin{align*} -{\frac{{c}^{2}a}{2\, \left ( ecx+dc \right ) ^{2}e}}-{\frac{{c}^{2}b\arctan \left ( cx \right ) }{2\, \left ( ecx+dc \right ) ^{2}e}}-{\frac{{c}^{2}b}{ \left ( 2\,{c}^{2}{d}^{2}+2\,{e}^{2} \right ) \left ( ecx+dc \right ) }}+{\frac{b{c}^{3}d\ln \left ( ecx+dc \right ) }{ \left ({c}^{2}{d}^{2}+{e}^{2} \right ) ^{2}}}+{\frac{b{c}^{4}\arctan \left ( cx \right ){d}^{2}}{2\,e \left ({c}^{2}{d}^{2}+{e}^{2} \right ) ^{2}}}-{\frac{b{c}^{3}d\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\, \left ({c}^{2}{d}^{2}+{e}^{2} \right ) ^{2}}}-{\frac{{c}^{2}be\arctan \left ( cx \right ) }{2\, \left ({c}^{2}{d}^{2}+{e}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.46999, size = 289, normalized size = 1.98 \begin{align*} -\frac{1}{2} \,{\left ({\left (\frac{c^{2} d \log \left (c^{2} x^{2} + 1\right )}{c^{4} d^{4} + 2 \, c^{2} d^{2} e^{2} + e^{4}} - \frac{2 \, c^{2} d \log \left (e x + d\right )}{c^{4} d^{4} + 2 \, c^{2} d^{2} e^{2} + e^{4}} - \frac{{\left (c^{4} d^{2} - c^{2} e^{2}\right )} \arctan \left (c x\right )}{{\left (c^{4} d^{4} e + 2 \, c^{2} d^{2} e^{3} + e^{5}\right )} c} + \frac{1}{c^{2} d^{3} + d e^{2} +{\left (c^{2} d^{2} e + e^{3}\right )} x}\right )} c + \frac{\arctan \left (c x\right )}{e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e}\right )} b - \frac{a}{2 \,{\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((c^2*d*log(c^2*x^2 + 1)/(c^4*d^4 + 2*c^2*d^2*e^2 + e^4) - 2*c^2*d*log(e*x + d)/(c^4*d^4 + 2*c^2*d^2*e^2
+ e^4) - (c^4*d^2 - c^2*e^2)*arctan(c*x)/((c^4*d^4*e + 2*c^2*d^2*e^3 + e^5)*c) + 1/(c^2*d^3 + d*e^2 + (c^2*d^2
*e + e^3)*x))*c + arctan(c*x)/(e^3*x^2 + 2*d*e^2*x + d^2*e))*b - 1/2*a/(e^3*x^2 + 2*d*e^2*x + d^2*e)

________________________________________________________________________________________

Fricas [B]  time = 3.64017, size = 637, normalized size = 4.36 \begin{align*} -\frac{a c^{4} d^{4} + b c^{3} d^{3} e + 2 \, a c^{2} d^{2} e^{2} + b c d e^{3} + a e^{4} +{\left (b c^{3} d^{2} e^{2} + b c e^{4}\right )} x +{\left (3 \, b c^{2} d^{2} e^{2} + b e^{4} -{\left (b c^{4} d^{2} e^{2} - b c^{2} e^{4}\right )} x^{2} - 2 \,{\left (b c^{4} d^{3} e - b c^{2} d e^{3}\right )} x\right )} \arctan \left (c x\right ) +{\left (b c^{3} d e^{3} x^{2} + 2 \, b c^{3} d^{2} e^{2} x + b c^{3} d^{3} e\right )} \log \left (c^{2} x^{2} + 1\right ) - 2 \,{\left (b c^{3} d e^{3} x^{2} + 2 \, b c^{3} d^{2} e^{2} x + b c^{3} d^{3} e\right )} \log \left (e x + d\right )}{2 \,{\left (c^{4} d^{6} e + 2 \, c^{2} d^{4} e^{3} + d^{2} e^{5} +{\left (c^{4} d^{4} e^{3} + 2 \, c^{2} d^{2} e^{5} + e^{7}\right )} x^{2} + 2 \,{\left (c^{4} d^{5} e^{2} + 2 \, c^{2} d^{3} e^{4} + d e^{6}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*c^4*d^4 + b*c^3*d^3*e + 2*a*c^2*d^2*e^2 + b*c*d*e^3 + a*e^4 + (b*c^3*d^2*e^2 + b*c*e^4)*x + (3*b*c^2*d
^2*e^2 + b*e^4 - (b*c^4*d^2*e^2 - b*c^2*e^4)*x^2 - 2*(b*c^4*d^3*e - b*c^2*d*e^3)*x)*arctan(c*x) + (b*c^3*d*e^3
*x^2 + 2*b*c^3*d^2*e^2*x + b*c^3*d^3*e)*log(c^2*x^2 + 1) - 2*(b*c^3*d*e^3*x^2 + 2*b*c^3*d^2*e^2*x + b*c^3*d^3*
e)*log(e*x + d))/(c^4*d^6*e + 2*c^2*d^4*e^3 + d^2*e^5 + (c^4*d^4*e^3 + 2*c^2*d^2*e^5 + e^7)*x^2 + 2*(c^4*d^5*e
^2 + 2*c^2*d^3*e^4 + d*e^6)*x)

________________________________________________________________________________________

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((a+b*atan(c*x))/(e*x+d)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.63408, size = 489, normalized size = 3.35 \begin{align*} \frac{b c^{4} d^{2} x^{2} \arctan \left (c x\right ) e^{2} + 2 \, b c^{4} d^{3} x \arctan \left (c x\right ) e - a c^{4} d^{4} - b c^{3} d x^{2} e^{3} \log \left (c^{2} x^{2} + 1\right ) - 2 \, b c^{3} d^{2} x e^{2} \log \left (c^{2} x^{2} + 1\right ) - b c^{3} d^{3} e \log \left (c^{2} x^{2} + 1\right ) + 2 \, b c^{3} d x^{2} e^{3} \log \left ({\left | x e + d \right |}\right ) + 4 \, b c^{3} d^{2} x e^{2} \log \left ({\left | x e + d \right |}\right ) + 2 \, b c^{3} d^{3} e \log \left ({\left | x e + d \right |}\right ) - b c^{3} d^{2} x e^{2} - b c^{3} d^{3} e - b c^{2} x^{2} \arctan \left (c x\right ) e^{4} - 2 \, b c^{2} d x \arctan \left (c x\right ) e^{3} - 3 \, b c^{2} d^{2} \arctan \left (c x\right ) e^{2} - 2 \, a c^{2} d^{2} e^{2} - b c x e^{4} - b c d e^{3} - b \arctan \left (c x\right ) e^{4} - a e^{4}}{2 \,{\left (c^{4} d^{4} x^{2} e^{3} + 2 \, c^{4} d^{5} x e^{2} + c^{4} d^{6} e + 2 \, c^{2} d^{2} x^{2} e^{5} + 4 \, c^{2} d^{3} x e^{4} + 2 \, c^{2} d^{4} e^{3} + x^{2} e^{7} + 2 \, d x e^{6} + d^{2} e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*c^4*d^2*x^2*arctan(c*x)*e^2 + 2*b*c^4*d^3*x*arctan(c*x)*e - a*c^4*d^4 - b*c^3*d*x^2*e^3*log(c^2*x^2 + 1
) - 2*b*c^3*d^2*x*e^2*log(c^2*x^2 + 1) - b*c^3*d^3*e*log(c^2*x^2 + 1) + 2*b*c^3*d*x^2*e^3*log(abs(x*e + d)) +
4*b*c^3*d^2*x*e^2*log(abs(x*e + d)) + 2*b*c^3*d^3*e*log(abs(x*e + d)) - b*c^3*d^2*x*e^2 - b*c^3*d^3*e - b*c^2*
x^2*arctan(c*x)*e^4 - 2*b*c^2*d*x*arctan(c*x)*e^3 - 3*b*c^2*d^2*arctan(c*x)*e^2 - 2*a*c^2*d^2*e^2 - b*c*x*e^4
- b*c*d*e^3 - b*arctan(c*x)*e^4 - a*e^4)/(c^4*d^4*x^2*e^3 + 2*c^4*d^5*x*e^2 + c^4*d^6*e + 2*c^2*d^2*x^2*e^5 +
4*c^2*d^3*x*e^4 + 2*c^2*d^4*e^3 + x^2*e^7 + 2*d*x*e^6 + d^2*e^5)