3.1142 \(\int \frac{(d+e x^2)^3 (a+b \tan ^{-1}(c x))}{x^2} \, dx\)

Optimal. Leaf size=160 \[ 3 d^2 e x \left (a+b \tan ^{-1}(c x)\right )-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+d e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e^3 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{b \left (15 c^4 d^2 e+5 c^6 d^3-5 c^2 d e^2+e^3\right ) \log \left (c^2 x^2+1\right )}{10 c^5}-\frac{b e^2 x^2 \left (5 c^2 d-e\right )}{10 c^3}+b c d^3 \log (x)-\frac{b e^3 x^4}{20 c} \]

[Out]

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

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Rubi [A]  time = 0.258139, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {270, 4976, 1799, 1620} \[ 3 d^2 e x \left (a+b \tan ^{-1}(c x)\right )-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+d e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{5} e^3 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{b \left (15 c^4 d^2 e+5 c^6 d^3-5 c^2 d e^2+e^3\right ) \log \left (c^2 x^2+1\right )}{10 c^5}-\frac{b e^2 x^2 \left (5 c^2 d-e\right )}{10 c^3}+b c d^3 \log (x)-\frac{b e^3 x^4}{20 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.149035, size = 169, normalized size = 1.06 \[ \frac{1}{20} \left (60 a d^2 e x-\frac{20 a d^3}{x}+20 a d e^2 x^3+4 a e^3 x^5-\frac{2 b \left (15 c^4 d^2 e+5 c^6 d^3-5 c^2 d e^2+e^3\right ) \log \left (c^2 x^2+1\right )}{c^5}+\frac{2 b e^2 x^2 \left (e-5 c^2 d\right )}{c^3}+\frac{4 b \tan ^{-1}(c x) \left (15 d^2 e x^2-5 d^3+5 d e^2 x^4+e^3 x^6\right )}{x}+20 b c d^3 \log (x)-\frac{b e^3 x^4}{c}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.044, size = 211, normalized size = 1.3 \begin{align*}{\frac{a{x}^{5}{e}^{3}}{5}}+a{x}^{3}d{e}^{2}+3\,a{d}^{2}ex-{\frac{a{d}^{3}}{x}}+{\frac{b\arctan \left ( cx \right ){x}^{5}{e}^{3}}{5}}+b\arctan \left ( cx \right ){x}^{3}d{e}^{2}+3\,b\arctan \left ( cx \right ){d}^{2}ex-{\frac{b{d}^{3}\arctan \left ( cx \right ) }{x}}-{\frac{b{e}^{3}{x}^{4}}{20\,c}}-{\frac{b{x}^{2}d{e}^{2}}{2\,c}}+{\frac{b{e}^{3}{x}^{2}}{10\,{c}^{3}}}-{\frac{bc{d}^{3}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2}}-{\frac{3\,b\ln \left ({c}^{2}{x}^{2}+1 \right ){d}^{2}e}{2\,c}}+{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ) d{e}^{2}}{2\,{c}^{3}}}-{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ){e}^{3}}{10\,{c}^{5}}}+cb{d}^{3}\ln \left ( cx \right ) \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^2,x)

[Out]

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

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Maxima [A]  time = 0.965819, size = 266, normalized size = 1.66 \begin{align*} \frac{1}{5} \, a e^{3} x^{5} + a d e^{2} x^{3} - \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 d^{3} + \frac{1}{2} \,{\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 d e^{2} + \frac{1}{20} \,{\left (4 \, x^{5} \arctan \left (c x\right ) - c{\left (\frac{c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac{2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b e^{3} + 3 \, a d^{2} e x + \frac{3 \,{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{2} e}{2 \, c} - \frac{a d^{3}}{x} \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^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.95102, size = 446, normalized size = 2.79 \begin{align*} \frac{4 \, a c^{5} e^{3} x^{6} + 20 \, a c^{5} d e^{2} x^{4} - b c^{4} e^{3} x^{5} + 20 \, b c^{6} d^{3} x \log \left (x\right ) + 60 \, a c^{5} d^{2} e x^{2} - 20 \, a c^{5} d^{3} - 2 \,{\left (5 \, b c^{4} d e^{2} - b c^{2} e^{3}\right )} x^{3} - 2 \,{\left (5 \, b c^{6} d^{3} + 15 \, b c^{4} d^{2} e - 5 \, b c^{2} d e^{2} + b e^{3}\right )} x \log \left (c^{2} x^{2} + 1\right ) + 4 \,{\left (b c^{5} e^{3} x^{6} + 5 \, b c^{5} d e^{2} x^{4} + 15 \, b c^{5} d^{2} e x^{2} - 5 \, b c^{5} d^{3}\right )} \arctan \left (c x\right )}{20 \, c^{5} x} \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^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.54261, size = 258, normalized size = 1.61 \begin{align*} \begin{cases} - \frac{a d^{3}}{x} + 3 a d^{2} e x + a d e^{2} x^{3} + \frac{a e^{3} x^{5}}{5} + b c d^{3} \log{\left (x \right )} - \frac{b c d^{3} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2} - \frac{b d^{3} \operatorname{atan}{\left (c x \right )}}{x} + 3 b d^{2} e x \operatorname{atan}{\left (c x \right )} + b d e^{2} x^{3} \operatorname{atan}{\left (c x \right )} + \frac{b e^{3} x^{5} \operatorname{atan}{\left (c x \right )}}{5} - \frac{3 b d^{2} e \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2 c} - \frac{b d e^{2} x^{2}}{2 c} - \frac{b e^{3} x^{4}}{20 c} + \frac{b d e^{2} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2 c^{3}} + \frac{b e^{3} x^{2}}{10 c^{3}} - \frac{b e^{3} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{10 c^{5}} & \text{for}\: c \neq 0 \\a \left (- \frac{d^{3}}{x} + 3 d^{2} e x + d e^{2} x^{3} + \frac{e^{3} x^{5}}{5}\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**2,x)

[Out]

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

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Giac [A]  time = 1.11454, size = 325, normalized size = 2.03 \begin{align*} \frac{4 \, b c^{5} x^{6} \arctan \left (c x\right ) e^{3} + 4 \, a c^{5} x^{6} e^{3} + 20 \, b c^{5} d x^{4} \arctan \left (c x\right ) e^{2} + 20 \, a c^{5} d x^{4} e^{2} + 60 \, b c^{5} d^{2} x^{2} \arctan \left (c x\right ) e - 10 \, b c^{6} d^{3} x \log \left (c^{2} x^{2} + 1\right ) + 20 \, b c^{6} d^{3} x \log \left (x\right ) - b c^{4} x^{5} e^{3} + 60 \, a c^{5} d^{2} x^{2} e - 20 \, b c^{5} d^{3} \arctan \left (c x\right ) - 10 \, b c^{4} d x^{3} e^{2} - 30 \, b c^{4} d^{2} x e \log \left (c^{2} x^{2} + 1\right ) - 20 \, a c^{5} d^{3} + 2 \, b c^{2} x^{3} e^{3} + 10 \, b c^{2} d x e^{2} \log \left (c^{2} x^{2} + 1\right ) - 2 \, b x e^{3} \log \left (c^{2} x^{2} + 1\right )}{20 \, c^{5} x} \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^2,x, algorithm="giac")

[Out]

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