3.1144 \(\int \frac{(d+e x^2)^3 (a+b \tan ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=158 \[ -\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{b \left (c^2 d+e\right ) \left (c^4 d^2-10 c^2 d e+e^2\right ) \log \left (c^2 x^2+1\right )}{6 c^3}-\frac{1}{3} b c d^2 \log (x) \left (c^2 d-9 e\right )-\frac{b c d^3}{6 x^2}-\frac{b e^3 x^2}{6 c} \]

[Out]

-(b*c*d^3)/(6*x^2) - (b*e^3*x^2)/(6*c) - (d^3*(a + b*ArcTan[c*x]))/(3*x^3) - (3*d^2*e*(a + b*ArcTan[c*x]))/x +
3*d*e^2*x*(a + b*ArcTan[c*x]) + (e^3*x^3*(a + b*ArcTan[c*x]))/3 - (b*c*d^2*(c^2*d - 9*e)*Log[x])/3 + (b*(c^2*
d + e)*(c^4*d^2 - 10*c^2*d*e + e^2)*Log[1 + c^2*x^2])/(6*c^3)

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Rubi [A]  time = 0.264665, antiderivative size = 158, 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} \[ -\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{b \left (c^2 d+e\right ) \left (c^4 d^2-10 c^2 d e+e^2\right ) \log \left (c^2 x^2+1\right )}{6 c^3}-\frac{1}{3} b c d^2 \log (x) \left (c^2 d-9 e\right )-\frac{b c d^3}{6 x^2}-\frac{b e^3 x^2}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^4,x]

[Out]

-(b*c*d^3)/(6*x^2) - (b*e^3*x^2)/(6*c) - (d^3*(a + b*ArcTan[c*x]))/(3*x^3) - (3*d^2*e*(a + b*ArcTan[c*x]))/x +
3*d*e^2*x*(a + b*ArcTan[c*x]) + (e^3*x^3*(a + b*ArcTan[c*x]))/3 - (b*c*d^2*(c^2*d - 9*e)*Log[x])/3 + (b*(c^2*
d + e)*(c^4*d^2 - 10*c^2*d*e + e^2)*Log[1 + c^2*x^2])/(6*c^3)

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^4} \, dx &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac{-d^3-9 d^2 e x^2+9 d e^2 x^4+e^3 x^6}{3 x^3 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{3} (b c) \int \frac{-d^3-9 d^2 e x^2+9 d e^2 x^4+e^3 x^6}{x^3 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} (b c) \operatorname{Subst}\left (\int \frac{-d^3-9 d^2 e x+9 d e^2 x^2+e^3 x^3}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} (b c) \operatorname{Subst}\left (\int \left (\frac{e^3}{c^2}-\frac{d^3}{x^2}+\frac{d^2 \left (c^2 d-9 e\right )}{x}+\frac{\left (c^2 d+e\right ) \left (-c^4 d^2+10 c^2 d e-e^2\right )}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{b c d^3}{6 x^2}-\frac{b e^3 x^2}{6 c}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{3} b c d^2 \left (c^2 d-9 e\right ) \log (x)+\frac{b \left (c^2 d+e\right ) \left (c^4 d^2-10 c^2 d e+e^2\right ) \log \left (1+c^2 x^2\right )}{6 c^3}\\ \end{align*}

Mathematica [A]  time = 0.164613, size = 166, normalized size = 1.05 \[ \frac{1}{6} \left (-\frac{18 a d^2 e}{x}-\frac{2 a d^3}{x^3}+18 a d e^2 x+2 a e^3 x^3+\frac{b \left (-9 c^4 d^2 e+c^6 d^3-9 c^2 d e^2+e^3\right ) \log \left (c^2 x^2+1\right )}{c^3}-2 b c d^2 \log (x) \left (c^2 d-9 e\right )+\frac{2 b \tan ^{-1}(c x) \left (-9 d^2 e x^2-d^3+9 d e^2 x^4+e^3 x^6\right )}{x^3}-\frac{b c d^3}{x^2}-\frac{b e^3 x^2}{c}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^4,x]

[Out]

((-2*a*d^3)/x^3 - (b*c*d^3)/x^2 - (18*a*d^2*e)/x + 18*a*d*e^2*x - (b*e^3*x^2)/c + 2*a*e^3*x^3 + (2*b*(-d^3 - 9
*d^2*e*x^2 + 9*d*e^2*x^4 + e^3*x^6)*ArcTan[c*x])/x^3 - 2*b*c*d^2*(c^2*d - 9*e)*Log[x] + (b*(c^6*d^3 - 9*c^4*d^
2*e - 9*c^2*d*e^2 + e^3)*Log[1 + c^2*x^2])/c^3)/6

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Maple [A]  time = 0.048, size = 213, normalized size = 1.4 \begin{align*}{\frac{a{e}^{3}{x}^{3}}{3}}+3\,ad{e}^{2}x-3\,{\frac{a{d}^{2}e}{x}}-{\frac{a{d}^{3}}{3\,{x}^{3}}}+{\frac{b\arctan \left ( cx \right ){e}^{3}{x}^{3}}{3}}+3\,b\arctan \left ( cx \right ) d{e}^{2}x-3\,{\frac{b{d}^{2}\arctan \left ( cx \right ) e}{x}}-{\frac{b{d}^{3}\arctan \left ( cx \right ) }{3\,{x}^{3}}}-{\frac{b{e}^{3}{x}^{2}}{6\,c}}+{\frac{b{c}^{3}{d}^{3}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{6}}-{\frac{3\,cb\ln \left ({c}^{2}{x}^{2}+1 \right ){d}^{2}e}{2}}-{\frac{3\,b\ln \left ({c}^{2}{x}^{2}+1 \right ) d{e}^{2}}{2\,c}}+{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ){e}^{3}}{6\,{c}^{3}}}-{\frac{{c}^{3}b{d}^{3}\ln \left ( cx \right ) }{3}}+3\,cb\ln \left ( cx \right ){d}^{2}e-{\frac{cb{d}^{3}}{6\,{x}^{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^4,x)

[Out]

1/3*a*e^3*x^3+3*a*d*e^2*x-3*a*d^2*e/x-1/3*a*d^3/x^3+1/3*b*arctan(c*x)*e^3*x^3+3*b*arctan(c*x)*d*e^2*x-3*b*arct
an(c*x)*d^2*e/x-1/3*b*arctan(c*x)*d^3/x^3-1/6*b*e^3*x^2/c+1/6*b*c^3*d^3*ln(c^2*x^2+1)-3/2*c*b*ln(c^2*x^2+1)*d^
2*e-3/2/c*b*ln(c^2*x^2+1)*d*e^2+1/6/c^3*b*ln(c^2*x^2+1)*e^3-1/3*c^3*b*d^3*ln(c*x)+3*c*b*ln(c*x)*d^2*e-1/6*b*c*
d^3/x^2

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Maxima [A]  time = 0.981411, size = 261, normalized size = 1.65 \begin{align*} \frac{1}{3} \, a e^{3} x^{3} + \frac{1}{6} \,{\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^{3} - \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^{2} e + \frac{1}{6} \,{\left (2 \, x^{3} \arctan \left (c x\right ) - c{\left (\frac{x^{2}}{c^{2}} - \frac{\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b e^{3} + 3 \, a d e^{2} x + \frac{3 \,{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d e^{2}}{2 \, c} - \frac{3 \, a d^{2} e}{x} - \frac{a d^{3}}{3 \, x^{3}} \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^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.72967, size = 433, normalized size = 2.74 \begin{align*} \frac{2 \, a c^{3} e^{3} x^{6} + 18 \, a c^{3} d e^{2} x^{4} - b c^{2} e^{3} x^{5} - b c^{4} d^{3} x - 18 \, a c^{3} d^{2} e x^{2} - 2 \, a c^{3} d^{3} +{\left (b c^{6} d^{3} - 9 \, b c^{4} d^{2} e - 9 \, b c^{2} d e^{2} + b e^{3}\right )} x^{3} \log \left (c^{2} x^{2} + 1\right ) - 2 \,{\left (b c^{6} d^{3} - 9 \, b c^{4} d^{2} e\right )} x^{3} \log \left (x\right ) + 2 \,{\left (b c^{3} e^{3} x^{6} + 9 \, b c^{3} d e^{2} x^{4} - 9 \, b c^{3} d^{2} e x^{2} - b c^{3} d^{3}\right )} \arctan \left (c x\right )}{6 \, c^{3} x^{3}} \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^4,x, algorithm="fricas")

[Out]

1/6*(2*a*c^3*e^3*x^6 + 18*a*c^3*d*e^2*x^4 - b*c^2*e^3*x^5 - b*c^4*d^3*x - 18*a*c^3*d^2*e*x^2 - 2*a*c^3*d^3 + (
b*c^6*d^3 - 9*b*c^4*d^2*e - 9*b*c^2*d*e^2 + b*e^3)*x^3*log(c^2*x^2 + 1) - 2*(b*c^6*d^3 - 9*b*c^4*d^2*e)*x^3*lo
g(x) + 2*(b*c^3*e^3*x^6 + 9*b*c^3*d*e^2*x^4 - 9*b*c^3*d^2*e*x^2 - b*c^3*d^3)*arctan(c*x))/(c^3*x^3)

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Sympy [A]  time = 4.60556, size = 272, normalized size = 1.72 \begin{align*} \begin{cases} - \frac{a d^{3}}{3 x^{3}} - \frac{3 a d^{2} e}{x} + 3 a d e^{2} x + \frac{a e^{3} x^{3}}{3} - \frac{b c^{3} d^{3} \log{\left (x \right )}}{3} + \frac{b c^{3} d^{3} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{6} - \frac{b c d^{3}}{6 x^{2}} + 3 b c d^{2} e \log{\left (x \right )} - \frac{3 b c d^{2} e \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2} - \frac{b d^{3} \operatorname{atan}{\left (c x \right )}}{3 x^{3}} - \frac{3 b d^{2} e \operatorname{atan}{\left (c x \right )}}{x} + 3 b d e^{2} x \operatorname{atan}{\left (c x \right )} + \frac{b e^{3} x^{3} \operatorname{atan}{\left (c x \right )}}{3} - \frac{3 b d e^{2} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2 c} - \frac{b e^{3} x^{2}}{6 c} + \frac{b e^{3} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{6 c^{3}} & \text{for}\: c \neq 0 \\a \left (- \frac{d^{3}}{3 x^{3}} - \frac{3 d^{2} e}{x} + 3 d e^{2} x + \frac{e^{3} x^{3}}{3}\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**4,x)

[Out]

Piecewise((-a*d**3/(3*x**3) - 3*a*d**2*e/x + 3*a*d*e**2*x + a*e**3*x**3/3 - b*c**3*d**3*log(x)/3 + b*c**3*d**3
*log(x**2 + c**(-2))/6 - b*c*d**3/(6*x**2) + 3*b*c*d**2*e*log(x) - 3*b*c*d**2*e*log(x**2 + c**(-2))/2 - b*d**3
*atan(c*x)/(3*x**3) - 3*b*d**2*e*atan(c*x)/x + 3*b*d*e**2*x*atan(c*x) + b*e**3*x**3*atan(c*x)/3 - 3*b*d*e**2*l
og(x**2 + c**(-2))/(2*c) - b*e**3*x**2/(6*c) + b*e**3*log(x**2 + c**(-2))/(6*c**3), Ne(c, 0)), (a*(-d**3/(3*x*
*3) - 3*d**2*e/x + 3*d*e**2*x + e**3*x**3/3), True))

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Giac [A]  time = 1.10084, size = 340, normalized size = 2.15 \begin{align*} \frac{b c^{6} d^{3} x^{3} \log \left (c^{2} x^{2} + 1\right ) - 2 \, b c^{6} d^{3} x^{3} \log \left (x\right ) + 2 \, b c^{3} x^{6} \arctan \left (c x\right ) e^{3} - 9 \, b c^{4} d^{2} x^{3} e \log \left (c^{2} x^{2} + 1\right ) + 18 \, b c^{4} d^{2} x^{3} e \log \left (x\right ) + 2 \, a c^{3} x^{6} e^{3} + 18 \, b c^{3} d x^{4} \arctan \left (c x\right ) e^{2} + 18 \, a c^{3} d x^{4} e^{2} - 18 \, b c^{3} d^{2} x^{2} \arctan \left (c x\right ) e - b c^{4} d^{3} x - b c^{2} x^{5} e^{3} - 18 \, a c^{3} d^{2} x^{2} e - 9 \, b c^{2} d x^{3} e^{2} \log \left (c^{2} x^{2} + 1\right ) - 2 \, b c^{3} d^{3} \arctan \left (c x\right ) - 2 \, a c^{3} d^{3} + b x^{3} e^{3} \log \left (c^{2} x^{2} + 1\right )}{6 \, c^{3} x^{3}} \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^4,x, algorithm="giac")

[Out]

1/6*(b*c^6*d^3*x^3*log(c^2*x^2 + 1) - 2*b*c^6*d^3*x^3*log(x) + 2*b*c^3*x^6*arctan(c*x)*e^3 - 9*b*c^4*d^2*x^3*e
*log(c^2*x^2 + 1) + 18*b*c^4*d^2*x^3*e*log(x) + 2*a*c^3*x^6*e^3 + 18*b*c^3*d*x^4*arctan(c*x)*e^2 + 18*a*c^3*d*
x^4*e^2 - 18*b*c^3*d^2*x^2*arctan(c*x)*e - b*c^4*d^3*x - b*c^2*x^5*e^3 - 18*a*c^3*d^2*x^2*e - 9*b*c^2*d*x^3*e^
2*log(c^2*x^2 + 1) - 2*b*c^3*d^3*arctan(c*x) - 2*a*c^3*d^3 + b*x^3*e^3*log(c^2*x^2 + 1))/(c^3*x^3)