3.1134 \(\int \frac{(d+e x^2)^2 (a+b \tan ^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=150 \[ -\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{2 d e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{1}{30} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log \left (c^2 x^2+1\right )+\frac{1}{15} b c \log (x) \left (3 c^4 d^2-10 c^2 d e+15 e^2\right )+\frac{b c d \left (3 c^2 d-10 e\right )}{30 x^2}-\frac{b c d^2}{20 x^4} \]

[Out]

-(b*c*d^2)/(20*x^4) + (b*c*d*(3*c^2*d - 10*e))/(30*x^2) - (d^2*(a + b*ArcTan[c*x]))/(5*x^5) - (2*d*e*(a + b*Ar
cTan[c*x]))/(3*x^3) - (e^2*(a + b*ArcTan[c*x]))/x + (b*c*(3*c^4*d^2 - 10*c^2*d*e + 15*e^2)*Log[x])/15 - (b*c*(
3*c^4*d^2 - 10*c^2*d*e + 15*e^2)*Log[1 + c^2*x^2])/30

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Rubi [A]  time = 0.185044, antiderivative size = 150, 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, 1251, 893} \[ -\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{2 d e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{1}{30} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log \left (c^2 x^2+1\right )+\frac{1}{15} b c \log (x) \left (3 c^4 d^2-10 c^2 d e+15 e^2\right )+\frac{b c d \left (3 c^2 d-10 e\right )}{30 x^2}-\frac{b c d^2}{20 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*d^2)/(20*x^4) + (b*c*d*(3*c^2*d - 10*e))/(30*x^2) - (d^2*(a + b*ArcTan[c*x]))/(5*x^5) - (2*d*e*(a + b*Ar
cTan[c*x]))/(3*x^3) - (e^2*(a + b*ArcTan[c*x]))/x + (b*c*(3*c^4*d^2 - 10*c^2*d*e + 15*e^2)*Log[x])/15 - (b*c*(
3*c^4*d^2 - 10*c^2*d*e + 15*e^2)*Log[1 + c^2*x^2])/30

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 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^2 \left (a+b \tan ^{-1}(c x)\right )}{x^6} \, dx &=-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{2 d e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-(b c) \int \frac{-3 d^2-10 d e x^2-15 e^2 x^4}{15 x^5 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{2 d e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{1}{15} (b c) \int \frac{-3 d^2-10 d e x^2-15 e^2 x^4}{x^5 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{2 d e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{1}{30} (b c) \operatorname{Subst}\left (\int \frac{-3 d^2-10 d e x-15 e^2 x^2}{x^3 \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{2 d e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{1}{30} (b c) \operatorname{Subst}\left (\int \left (-\frac{3 d^2}{x^3}+\frac{d \left (3 c^2 d-10 e\right )}{x^2}+\frac{-3 c^4 d^2+10 c^2 d e-15 e^2}{x}+\frac{3 c^6 d^2-10 c^4 d e+15 c^2 e^2}{1+c^2 x}\right ) \, dx,x,x^2\right )\\ &=-\frac{b c d^2}{20 x^4}+\frac{b c d \left (3 c^2 d-10 e\right )}{30 x^2}-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{2 d e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac{1}{15} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log (x)-\frac{1}{30} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log \left (1+c^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.128078, size = 149, normalized size = 0.99 \[ \frac{1}{60} \left (-\frac{12 d^2 \left (a+b \tan ^{-1}(c x)\right )}{x^5}-\frac{40 d e \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{60 e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-3 b c d^2 \left (-\frac{2 c^2}{x^2}+2 c^4 \log \left (c^2 x^2+1\right )-4 c^4 \log (x)+\frac{1}{x^4}\right )-20 b c d e \left (-c^2 \log \left (c^2 x^2+1\right )+2 c^2 \log (x)+\frac{1}{x^2}\right )+30 b c e^2 \left (2 \log (x)-\log \left (c^2 x^2+1\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-12*d^2*(a + b*ArcTan[c*x]))/x^5 - (40*d*e*(a + b*ArcTan[c*x]))/x^3 - (60*e^2*(a + b*ArcTan[c*x]))/x + 30*b*
c*e^2*(2*Log[x] - Log[1 + c^2*x^2]) - 20*b*c*d*e*(x^(-2) + 2*c^2*Log[x] - c^2*Log[1 + c^2*x^2]) - 3*b*c*d^2*(x
^(-4) - (2*c^2)/x^2 - 4*c^4*Log[x] + 2*c^4*Log[1 + c^2*x^2]))/60

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Maple [A]  time = 0.049, size = 186, normalized size = 1.2 \begin{align*} -{\frac{a{e}^{2}}{x}}-{\frac{a{d}^{2}}{5\,{x}^{5}}}-{\frac{2\,aed}{3\,{x}^{3}}}-{\frac{b\arctan \left ( cx \right ){e}^{2}}{x}}-{\frac{b{d}^{2}\arctan \left ( cx \right ) }{5\,{x}^{5}}}-{\frac{2\,b\arctan \left ( cx \right ) ed}{3\,{x}^{3}}}-{\frac{{c}^{5}b\ln \left ({c}^{2}{x}^{2}+1 \right ){d}^{2}}{10}}+{\frac{{c}^{3}b\ln \left ({c}^{2}{x}^{2}+1 \right ) ed}{3}}-{\frac{cb\ln \left ({c}^{2}{x}^{2}+1 \right ){e}^{2}}{2}}+{\frac{{c}^{5}b{d}^{2}\ln \left ( cx \right ) }{5}}-{\frac{2\,{c}^{3}b\ln \left ( cx \right ) de}{3}}+cb\ln \left ( cx \right ){e}^{2}+{\frac{{c}^{3}b{d}^{2}}{10\,{x}^{2}}}-{\frac{bced}{3\,{x}^{2}}}-{\frac{cb{d}^{2}}{20\,{x}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.982012, size = 224, normalized size = 1.49 \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^{2} + \frac{1}{3} \,{\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 e - \frac{1}{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 e^{2} - \frac{a e^{2}}{x} - \frac{2 \, a d e}{3 \, x^{3}} - \frac{a d^{2}}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 1.49382, size = 379, normalized size = 2.53 \begin{align*} -\frac{60 \, a e^{2} x^{4} + 2 \,{\left (3 \, b c^{5} d^{2} - 10 \, b c^{3} d e + 15 \, b c e^{2}\right )} x^{5} \log \left (c^{2} x^{2} + 1\right ) - 4 \,{\left (3 \, b c^{5} d^{2} - 10 \, b c^{3} d e + 15 \, b c e^{2}\right )} x^{5} \log \left (x\right ) + 3 \, b c d^{2} x + 40 \, a d e x^{2} - 2 \,{\left (3 \, b c^{3} d^{2} - 10 \, b c d e\right )} x^{3} + 12 \, a d^{2} + 4 \,{\left (15 \, b e^{2} x^{4} + 10 \, b d e x^{2} + 3 \, b d^{2}\right )} \arctan \left (c x\right )}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*a*e^2*x^4 + 2*(3*b*c^5*d^2 - 10*b*c^3*d*e + 15*b*c*e^2)*x^5*log(c^2*x^2 + 1) - 4*(3*b*c^5*d^2 - 10*b
*c^3*d*e + 15*b*c*e^2)*x^5*log(x) + 3*b*c*d^2*x + 40*a*d*e*x^2 - 2*(3*b*c^3*d^2 - 10*b*c*d*e)*x^3 + 12*a*d^2 +
4*(15*b*e^2*x^4 + 10*b*d*e*x^2 + 3*b*d^2)*arctan(c*x))/x^5

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Sympy [A]  time = 3.43594, size = 235, normalized size = 1.57 \begin{align*} \begin{cases} - \frac{a d^{2}}{5 x^{5}} - \frac{2 a d e}{3 x^{3}} - \frac{a e^{2}}{x} + \frac{b c^{5} d^{2} \log{\left (x \right )}}{5} - \frac{b c^{5} d^{2} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{10} + \frac{b c^{3} d^{2}}{10 x^{2}} - \frac{2 b c^{3} d e \log{\left (x \right )}}{3} + \frac{b c^{3} d e \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{3} - \frac{b c d^{2}}{20 x^{4}} - \frac{b c d e}{3 x^{2}} + b c e^{2} \log{\left (x \right )} - \frac{b c e^{2} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2} - \frac{b d^{2} \operatorname{atan}{\left (c x \right )}}{5 x^{5}} - \frac{2 b d e \operatorname{atan}{\left (c x \right )}}{3 x^{3}} - \frac{b e^{2} \operatorname{atan}{\left (c x \right )}}{x} & \text{for}\: c \neq 0 \\a \left (- \frac{d^{2}}{5 x^{5}} - \frac{2 d e}{3 x^{3}} - \frac{e^{2}}{x}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.09799, size = 265, normalized size = 1.77 \begin{align*} -\frac{6 \, b c^{5} d^{2} x^{5} \log \left (c^{2} x^{2} + 1\right ) - 12 \, b c^{5} d^{2} x^{5} \log \left (x\right ) - 20 \, b c^{3} d x^{5} e \log \left (c^{2} x^{2} + 1\right ) + 40 \, b c^{3} d x^{5} e \log \left (x\right ) - 6 \, b c^{3} d^{2} x^{3} + 30 \, b c x^{5} e^{2} \log \left (c^{2} x^{2} + 1\right ) - 60 \, b c x^{5} e^{2} \log \left (x\right ) + 60 \, b x^{4} \arctan \left (c x\right ) e^{2} + 20 \, b c d x^{3} e + 60 \, a x^{4} e^{2} + 40 \, b d x^{2} \arctan \left (c x\right ) e + 3 \, b c d^{2} x + 40 \, a d x^{2} e + 12 \, b d^{2} \arctan \left (c x\right ) + 12 \, a d^{2}}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(6*b*c^5*d^2*x^5*log(c^2*x^2 + 1) - 12*b*c^5*d^2*x^5*log(x) - 20*b*c^3*d*x^5*e*log(c^2*x^2 + 1) + 40*b*c
^3*d*x^5*e*log(x) - 6*b*c^3*d^2*x^3 + 30*b*c*x^5*e^2*log(c^2*x^2 + 1) - 60*b*c*x^5*e^2*log(x) + 60*b*x^4*arcta
n(c*x)*e^2 + 20*b*c*d*x^3*e + 60*a*x^4*e^2 + 40*b*d*x^2*arctan(c*x)*e + 3*b*c*d^2*x + 40*a*d*x^2*e + 12*b*d^2*
arctan(c*x) + 12*a*d^2)/x^5