3.1146 \(\int \frac{(d+e x^2)^3 (a+b \tan ^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=177 \[ -\frac{d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+e^3 x \left (a+b \tan ^{-1}(c x)\right )-\frac{b \left (-5 c^4 d^2 e+c^6 d^3+15 c^2 d e^2+5 e^3\right ) \log \left (c^2 x^2+1\right )}{10 c}+\frac{1}{5} b c d \log (x) \left (c^4 d^2-5 c^2 d e+15 e^2\right )+\frac{b c d^2 \left (c^2 d-5 e\right )}{10 x^2}-\frac{b c d^3}{20 x^4} \]

[Out]

-(b*c*d^3)/(20*x^4) + (b*c*d^2*(c^2*d - 5*e))/(10*x^2) - (d^3*(a + b*ArcTan[c*x]))/(5*x^5) - (d^2*e*(a + b*Arc
Tan[c*x]))/x^3 - (3*d*e^2*(a + b*ArcTan[c*x]))/x + e^3*x*(a + b*ArcTan[c*x]) + (b*c*d*(c^4*d^2 - 5*c^2*d*e + 1
5*e^2)*Log[x])/5 - (b*(c^6*d^3 - 5*c^4*d^2*e + 15*c^2*d*e^2 + 5*e^3)*Log[1 + c^2*x^2])/(10*c)

________________________________________________________________________________________

Rubi [A]  time = 0.285452, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {270, 4976, 12, 1799, 1620} \[ -\frac{d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+e^3 x \left (a+b \tan ^{-1}(c x)\right )-\frac{b \left (-5 c^4 d^2 e+c^6 d^3+15 c^2 d e^2+5 e^3\right ) \log \left (c^2 x^2+1\right )}{10 c}+\frac{1}{5} b c d \log (x) \left (c^4 d^2-5 c^2 d e+15 e^2\right )+\frac{b c d^2 \left (c^2 d-5 e\right )}{10 x^2}-\frac{b c d^3}{20 x^4} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

-(b*c*d^3)/(20*x^4) + (b*c*d^2*(c^2*d - 5*e))/(10*x^2) - (d^3*(a + b*ArcTan[c*x]))/(5*x^5) - (d^2*e*(a + b*Arc
Tan[c*x]))/x^3 - (3*d*e^2*(a + b*ArcTan[c*x]))/x + e^3*x*(a + b*ArcTan[c*x]) + (b*c*d*(c^4*d^2 - 5*c^2*d*e + 1
5*e^2)*Log[x])/5 - (b*(c^6*d^3 - 5*c^4*d^2*e + 15*c^2*d*e^2 + 5*e^3)*Log[1 + c^2*x^2])/(10*c)

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 4976

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 + c^
2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m +
2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] &&
!ILtQ[(m - 1)/2, 0]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1799

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^3 \left (a+b \tan ^{-1}(c x)\right )}{x^6} \, dx &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+e^3 x \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac{-d^3-5 d^2 e x^2-15 d e^2 x^4+5 e^3 x^6}{5 x^5 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+e^3 x \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{5} (b c) \int \frac{-d^3-5 d^2 e x^2-15 d e^2 x^4+5 e^3 x^6}{x^5 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+e^3 x \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{10} (b c) \operatorname{Subst}\left (\int \frac{-d^3-5 d^2 e x-15 d e^2 x^2+5 e^3 x^3}{x^3 \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+e^3 x \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{10} (b c) \operatorname{Subst}\left (\int \left (-\frac{d^3}{x^3}+\frac{d^2 \left (c^2 d-5 e\right )}{x^2}-\frac{d \left (c^4 d^2-5 c^2 d e+15 e^2\right )}{x}+\frac{c^6 d^3-5 c^4 d^2 e+15 c^2 d e^2+5 e^3}{1+c^2 x}\right ) \, dx,x,x^2\right )\\ &=-\frac{b c d^3}{20 x^4}+\frac{b c d^2 \left (c^2 d-5 e\right )}{10 x^2}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+e^3 x \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} b c d \left (c^4 d^2-5 c^2 d e+15 e^2\right ) \log (x)-\frac{b \left (c^6 d^3-5 c^4 d^2 e+15 c^2 d e^2+5 e^3\right ) \log \left (1+c^2 x^2\right )}{10 c}\\ \end{align*}

Mathematica [A]  time = 0.172457, size = 184, normalized size = 1.04 \[ \frac{1}{20} \left (-\frac{20 a d^2 e}{x^3}-\frac{4 a d^3}{x^5}-\frac{60 a d e^2}{x}+20 a e^3 x-\frac{2 b \left (-5 c^4 d^2 e+c^6 d^3+15 c^2 d e^2+5 e^3\right ) \log \left (c^2 x^2+1\right )}{c}+4 b c d \log (x) \left (c^4 d^2-5 c^2 d e+15 e^2\right )+\frac{2 b c d^2 \left (c^2 d-5 e\right )}{x^2}-\frac{4 b \tan ^{-1}(c x) \left (5 d^2 e x^2+d^3+15 d e^2 x^4-5 e^3 x^6\right )}{x^5}-\frac{b c d^3}{x^4}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

((-4*a*d^3)/x^5 - (b*c*d^3)/x^4 - (20*a*d^2*e)/x^3 + (2*b*c*d^2*(c^2*d - 5*e))/x^2 - (60*a*d*e^2)/x + 20*a*e^3
*x - (4*b*(d^3 + 5*d^2*e*x^2 + 15*d*e^2*x^4 - 5*e^3*x^6)*ArcTan[c*x])/x^5 + 4*b*c*d*(c^4*d^2 - 5*c^2*d*e + 15*
e^2)*Log[x] - (2*b*(c^6*d^3 - 5*c^4*d^2*e + 15*c^2*d*e^2 + 5*e^3)*Log[1 + c^2*x^2])/c)/20

________________________________________________________________________________________

Maple [A]  time = 0.047, size = 236, normalized size = 1.3 \begin{align*} a{e}^{3}x-3\,{\frac{ad{e}^{2}}{x}}-{\frac{a{d}^{3}}{5\,{x}^{5}}}-{\frac{a{d}^{2}e}{{x}^{3}}}+b\arctan \left ( cx \right ){e}^{3}x-3\,{\frac{\arctan \left ( cx \right ) bd{e}^{2}}{x}}-{\frac{b{d}^{3}\arctan \left ( cx \right ) }{5\,{x}^{5}}}-{\frac{b{d}^{2}\arctan \left ( cx \right ) e}{{x}^{3}}}-{\frac{{c}^{5}b\ln \left ({c}^{2}{x}^{2}+1 \right ){d}^{3}}{10}}+{\frac{{c}^{3}b\ln \left ({c}^{2}{x}^{2}+1 \right ){d}^{2}e}{2}}-{\frac{3\,cb\ln \left ({c}^{2}{x}^{2}+1 \right ) d{e}^{2}}{2}}-{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ){e}^{3}}{2\,c}}+{\frac{{c}^{3}b{d}^{3}}{10\,{x}^{2}}}-{\frac{cb{d}^{2}e}{2\,{x}^{2}}}-{\frac{cb{d}^{3}}{20\,{x}^{4}}}+{\frac{{c}^{5}b{d}^{3}\ln \left ( cx \right ) }{5}}-{c}^{3}b\ln \left ( cx \right ){d}^{2}e+3\,cb\ln \left ( cx \right ) d{e}^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^3*(a+b*arctan(c*x))/x^6,x)

[Out]

a*e^3*x-3*a*d*e^2/x-1/5*a*d^3/x^5-a*d^2*e/x^3+b*arctan(c*x)*e^3*x-3*b*arctan(c*x)*d*e^2/x-1/5*b*arctan(c*x)*d^
3/x^5-b*arctan(c*x)*d^2*e/x^3-1/10*c^5*b*ln(c^2*x^2+1)*d^3+1/2*c^3*b*ln(c^2*x^2+1)*d^2*e-3/2*c*b*ln(c^2*x^2+1)
*d*e^2-1/2/c*b*ln(c^2*x^2+1)*e^3+1/10*b*c^3*d^3/x^2-1/2*c*b*d^2*e/x^2-1/20*b*c*d^3/x^4+1/5*c^5*b*d^3*ln(c*x)-c
^3*b*ln(c*x)*d^2*e+3*c*b*ln(c*x)*d*e^2

________________________________________________________________________________________

Maxima [A]  time = 0.998314, size = 281, normalized size = 1.59 \begin{align*} -\frac{1}{20} \,{\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac{2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac{4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d^{3} + \frac{1}{2} \,{\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac{1}{x^{2}}\right )} c - \frac{2 \, \arctan \left (c x\right )}{x^{3}}\right )} b d^{2} e - \frac{3}{2} \,{\left (c{\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac{2 \, \arctan \left (c x\right )}{x}\right )} b d e^{2} + a e^{3} x + \frac{{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b e^{3}}{2 \, c} - \frac{3 \, a d e^{2}}{x} - \frac{a d^{2} e}{x^{3}} - \frac{a d^{3}}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arctan(c*x))/x^6,x, algorithm="maxima")

[Out]

-1/20*((2*c^4*log(c^2*x^2 + 1) - 2*c^4*log(x^2) - (2*c^2*x^2 - 1)/x^4)*c + 4*arctan(c*x)/x^5)*b*d^3 + 1/2*((c^
2*log(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c - 2*arctan(c*x)/x^3)*b*d^2*e - 3/2*(c*(log(c^2*x^2 + 1) - log(x^2
)) + 2*arctan(c*x)/x)*b*d*e^2 + a*e^3*x + 1/2*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*e^3/c - 3*a*d*e^2/x - a
*d^2*e/x^3 - 1/5*a*d^3/x^5

________________________________________________________________________________________

Fricas [A]  time = 1.78263, size = 473, normalized size = 2.67 \begin{align*} \frac{20 \, a c e^{3} x^{6} - 60 \, a c d e^{2} x^{4} - b c^{2} d^{3} x - 20 \, a c d^{2} e x^{2} - 2 \,{\left (b c^{6} d^{3} - 5 \, b c^{4} d^{2} e + 15 \, b c^{2} d e^{2} + 5 \, b e^{3}\right )} x^{5} \log \left (c^{2} x^{2} + 1\right ) + 4 \,{\left (b c^{6} d^{3} - 5 \, b c^{4} d^{2} e + 15 \, b c^{2} d e^{2}\right )} x^{5} \log \left (x\right ) - 4 \, a c d^{3} + 2 \,{\left (b c^{4} d^{3} - 5 \, b c^{2} d^{2} e\right )} x^{3} + 4 \,{\left (5 \, b c e^{3} x^{6} - 15 \, b c d e^{2} x^{4} - 5 \, b c d^{2} e x^{2} - b c d^{3}\right )} \arctan \left (c x\right )}{20 \, c x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arctan(c*x))/x^6,x, algorithm="fricas")

[Out]

1/20*(20*a*c*e^3*x^6 - 60*a*c*d*e^2*x^4 - b*c^2*d^3*x - 20*a*c*d^2*e*x^2 - 2*(b*c^6*d^3 - 5*b*c^4*d^2*e + 15*b
*c^2*d*e^2 + 5*b*e^3)*x^5*log(c^2*x^2 + 1) + 4*(b*c^6*d^3 - 5*b*c^4*d^2*e + 15*b*c^2*d*e^2)*x^5*log(x) - 4*a*c
*d^3 + 2*(b*c^4*d^3 - 5*b*c^2*d^2*e)*x^3 + 4*(5*b*c*e^3*x^6 - 15*b*c*d*e^2*x^4 - 5*b*c*d^2*e*x^2 - b*c*d^3)*ar
ctan(c*x))/(c*x^5)

________________________________________________________________________________________

Sympy [A]  time = 4.91528, size = 289, normalized size = 1.63 \begin{align*} \begin{cases} - \frac{a d^{3}}{5 x^{5}} - \frac{a d^{2} e}{x^{3}} - \frac{3 a d e^{2}}{x} + a e^{3} x + \frac{b c^{5} d^{3} \log{\left (x \right )}}{5} - \frac{b c^{5} d^{3} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{10} + \frac{b c^{3} d^{3}}{10 x^{2}} - b c^{3} d^{2} e \log{\left (x \right )} + \frac{b c^{3} d^{2} e \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2} - \frac{b c d^{3}}{20 x^{4}} - \frac{b c d^{2} e}{2 x^{2}} + 3 b c d e^{2} \log{\left (x \right )} - \frac{3 b c d e^{2} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2} - \frac{b d^{3} \operatorname{atan}{\left (c x \right )}}{5 x^{5}} - \frac{b d^{2} e \operatorname{atan}{\left (c x \right )}}{x^{3}} - \frac{3 b d e^{2} \operatorname{atan}{\left (c x \right )}}{x} + b e^{3} x \operatorname{atan}{\left (c x \right )} - \frac{b e^{3} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2 c} & \text{for}\: c \neq 0 \\a \left (- \frac{d^{3}}{5 x^{5}} - \frac{d^{2} e}{x^{3}} - \frac{3 d e^{2}}{x} + e^{3} x\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**3*(a+b*atan(c*x))/x**6,x)

[Out]

Piecewise((-a*d**3/(5*x**5) - a*d**2*e/x**3 - 3*a*d*e**2/x + a*e**3*x + b*c**5*d**3*log(x)/5 - b*c**5*d**3*log
(x**2 + c**(-2))/10 + b*c**3*d**3/(10*x**2) - b*c**3*d**2*e*log(x) + b*c**3*d**2*e*log(x**2 + c**(-2))/2 - b*c
*d**3/(20*x**4) - b*c*d**2*e/(2*x**2) + 3*b*c*d*e**2*log(x) - 3*b*c*d*e**2*log(x**2 + c**(-2))/2 - b*d**3*atan
(c*x)/(5*x**5) - b*d**2*e*atan(c*x)/x**3 - 3*b*d*e**2*atan(c*x)/x + b*e**3*x*atan(c*x) - b*e**3*log(x**2 + c**
(-2))/(2*c), Ne(c, 0)), (a*(-d**3/(5*x**5) - d**2*e/x**3 - 3*d*e**2/x + e**3*x), True))

________________________________________________________________________________________

Giac [A]  time = 1.10351, size = 359, normalized size = 2.03 \begin{align*} -\frac{2 \, b c^{6} d^{3} x^{5} \log \left (c^{2} x^{2} + 1\right ) - 4 \, b c^{6} d^{3} x^{5} \log \left (x\right ) - 10 \, b c^{4} d^{2} x^{5} e \log \left (c^{2} x^{2} + 1\right ) + 20 \, b c^{4} d^{2} x^{5} e \log \left (x\right ) - 2 \, b c^{4} d^{3} x^{3} + 30 \, b c^{2} d x^{5} e^{2} \log \left (c^{2} x^{2} + 1\right ) - 60 \, b c^{2} d x^{5} e^{2} \log \left (x\right ) - 20 \, b c x^{6} \arctan \left (c x\right ) e^{3} - 20 \, a c x^{6} e^{3} + 60 \, b c d x^{4} \arctan \left (c x\right ) e^{2} + 10 \, b c^{2} d^{2} x^{3} e + 60 \, a c d x^{4} e^{2} + 20 \, b c d^{2} x^{2} \arctan \left (c x\right ) e + 10 \, b x^{5} e^{3} \log \left (c^{2} x^{2} + 1\right ) + b c^{2} d^{3} x + 20 \, a c d^{2} x^{2} e + 4 \, b c d^{3} \arctan \left (c x\right ) + 4 \, a c d^{3}}{20 \, c x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arctan(c*x))/x^6,x, algorithm="giac")

[Out]

-1/20*(2*b*c^6*d^3*x^5*log(c^2*x^2 + 1) - 4*b*c^6*d^3*x^5*log(x) - 10*b*c^4*d^2*x^5*e*log(c^2*x^2 + 1) + 20*b*
c^4*d^2*x^5*e*log(x) - 2*b*c^4*d^3*x^3 + 30*b*c^2*d*x^5*e^2*log(c^2*x^2 + 1) - 60*b*c^2*d*x^5*e^2*log(x) - 20*
b*c*x^6*arctan(c*x)*e^3 - 20*a*c*x^6*e^3 + 60*b*c*d*x^4*arctan(c*x)*e^2 + 10*b*c^2*d^2*x^3*e + 60*a*c*d*x^4*e^
2 + 20*b*c*d^2*x^2*arctan(c*x)*e + 10*b*x^5*e^3*log(c^2*x^2 + 1) + b*c^2*d^3*x + 20*a*c*d^2*x^2*e + 4*b*c*d^3*
arctan(c*x) + 4*a*c*d^3)/(c*x^5)