3.1148 \(\int \frac{(d+e x^2)^3 (a+b \tan ^{-1}(c x))}{x^8} \, dx\)

Optimal. Leaf size=224 \[ -\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac{d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{e^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{b c d \left (5 c^4 d^2-21 c^2 d e+35 e^2\right )}{70 x^2}+\frac{1}{70} b c \left (-21 c^4 d^2 e+5 c^6 d^3+35 c^2 d e^2-35 e^3\right ) \log \left (c^2 x^2+1\right )-\frac{1}{35} b c \log (x) \left (-21 c^4 d^2 e+5 c^6 d^3+35 c^2 d e^2-35 e^3\right )+\frac{b c d^2 \left (5 c^2 d-21 e\right )}{140 x^4}-\frac{b c d^3}{42 x^6} \]

[Out]

-(b*c*d^3)/(42*x^6) + (b*c*d^2*(5*c^2*d - 21*e))/(140*x^4) - (b*c*d*(5*c^4*d^2 - 21*c^2*d*e + 35*e^2))/(70*x^2
) - (d^3*(a + b*ArcTan[c*x]))/(7*x^7) - (3*d^2*e*(a + b*ArcTan[c*x]))/(5*x^5) - (d*e^2*(a + b*ArcTan[c*x]))/x^
3 - (e^3*(a + b*ArcTan[c*x]))/x - (b*c*(5*c^6*d^3 - 21*c^4*d^2*e + 35*c^2*d*e^2 - 35*e^3)*Log[x])/35 + (b*c*(5
*c^6*d^3 - 21*c^4*d^2*e + 35*c^2*d*e^2 - 35*e^3)*Log[1 + c^2*x^2])/70

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Rubi [A]  time = 0.327039, antiderivative size = 224, 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} \[ -\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac{d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{e^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{b c d \left (5 c^4 d^2-21 c^2 d e+35 e^2\right )}{70 x^2}+\frac{1}{70} b c \left (-21 c^4 d^2 e+5 c^6 d^3+35 c^2 d e^2-35 e^3\right ) \log \left (c^2 x^2+1\right )-\frac{1}{35} b c \log (x) \left (-21 c^4 d^2 e+5 c^6 d^3+35 c^2 d e^2-35 e^3\right )+\frac{b c d^2 \left (5 c^2 d-21 e\right )}{140 x^4}-\frac{b c d^3}{42 x^6} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^8,x]

[Out]

-(b*c*d^3)/(42*x^6) + (b*c*d^2*(5*c^2*d - 21*e))/(140*x^4) - (b*c*d*(5*c^4*d^2 - 21*c^2*d*e + 35*e^2))/(70*x^2
) - (d^3*(a + b*ArcTan[c*x]))/(7*x^7) - (3*d^2*e*(a + b*ArcTan[c*x]))/(5*x^5) - (d*e^2*(a + b*ArcTan[c*x]))/x^
3 - (e^3*(a + b*ArcTan[c*x]))/x - (b*c*(5*c^6*d^3 - 21*c^4*d^2*e + 35*c^2*d*e^2 - 35*e^3)*Log[x])/35 + (b*c*(5
*c^6*d^3 - 21*c^4*d^2*e + 35*c^2*d*e^2 - 35*e^3)*Log[1 + c^2*x^2])/70

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^8} \, dx &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{e^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-(b c) \int \frac{-5 d^3-21 d^2 e x^2-35 d e^2 x^4-35 e^3 x^6}{35 x^7 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{e^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{1}{35} (b c) \int \frac{-5 d^3-21 d^2 e x^2-35 d e^2 x^4-35 e^3 x^6}{x^7 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{e^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{1}{70} (b c) \operatorname{Subst}\left (\int \frac{-5 d^3-21 d^2 e x-35 d e^2 x^2-35 e^3 x^3}{x^4 \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{e^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{1}{70} (b c) \operatorname{Subst}\left (\int \left (-\frac{5 d^3}{x^4}+\frac{d^2 \left (5 c^2 d-21 e\right )}{x^3}-\frac{d \left (5 c^4 d^2-21 c^2 d e+35 e^2\right )}{x^2}+\frac{5 c^6 d^3-21 c^4 d^2 e+35 c^2 d e^2-35 e^3}{x}+\frac{-5 c^8 d^3+21 c^6 d^2 e-35 c^4 d e^2+35 c^2 e^3}{1+c^2 x}\right ) \, dx,x,x^2\right )\\ &=-\frac{b c d^3}{42 x^6}+\frac{b c d^2 \left (5 c^2 d-21 e\right )}{140 x^4}-\frac{b c d \left (5 c^4 d^2-21 c^2 d e+35 e^2\right )}{70 x^2}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{e^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{1}{35} b c \left (5 c^6 d^3-21 c^4 d^2 e+35 c^2 d e^2-35 e^3\right ) \log (x)+\frac{1}{70} b c \left (5 c^6 d^3-21 c^4 d^2 e+35 c^2 d e^2-35 e^3\right ) \log \left (1+c^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.171879, size = 230, normalized size = 1.03 \[ -\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac{d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{e^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{3}{20} b c d^2 e \left (-\frac{2 c^2}{x^2}+2 c^4 \log \left (c^2 x^2+1\right )-4 c^4 \log (x)+\frac{1}{x^4}\right )-\frac{1}{84} b c d^3 \left (\frac{6 c^4}{x^2}-\frac{3 c^2}{x^4}-6 c^6 \log \left (c^2 x^2+1\right )+12 c^6 \log (x)+\frac{2}{x^6}\right )-\frac{1}{2} b c d e^2 \left (-c^2 \log \left (c^2 x^2+1\right )+2 c^2 \log (x)+\frac{1}{x^2}\right )+\frac{1}{2} b c e^3 \left (2 \log (x)-\log \left (c^2 x^2+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^8,x]

[Out]

-(d^3*(a + b*ArcTan[c*x]))/(7*x^7) - (3*d^2*e*(a + b*ArcTan[c*x]))/(5*x^5) - (d*e^2*(a + b*ArcTan[c*x]))/x^3 -
(e^3*(a + b*ArcTan[c*x]))/x + (b*c*e^3*(2*Log[x] - Log[1 + c^2*x^2]))/2 - (b*c*d*e^2*(x^(-2) + 2*c^2*Log[x] -
c^2*Log[1 + c^2*x^2]))/2 - (3*b*c*d^2*e*(x^(-4) - (2*c^2)/x^2 - 4*c^4*Log[x] + 2*c^4*Log[1 + c^2*x^2]))/20 -
(b*c*d^3*(2/x^6 - (3*c^2)/x^4 + (6*c^4)/x^2 + 12*c^6*Log[x] - 6*c^6*Log[1 + c^2*x^2]))/84

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Maple [A]  time = 0.047, size = 290, normalized size = 1.3 \begin{align*} -{\frac{a{d}^{3}}{7\,{x}^{7}}}-{\frac{a{e}^{3}}{x}}-{\frac{3\,a{d}^{2}e}{5\,{x}^{5}}}-{\frac{ad{e}^{2}}{{x}^{3}}}-{\frac{b{d}^{3}\arctan \left ( cx \right ) }{7\,{x}^{7}}}-{\frac{b\arctan \left ( cx \right ){e}^{3}}{x}}-{\frac{3\,b{d}^{2}\arctan \left ( cx \right ) e}{5\,{x}^{5}}}-{\frac{\arctan \left ( cx \right ) bd{e}^{2}}{{x}^{3}}}+{\frac{{c}^{7}b\ln \left ({c}^{2}{x}^{2}+1 \right ){d}^{3}}{14}}-{\frac{3\,{c}^{5}b\ln \left ({c}^{2}{x}^{2}+1 \right ){d}^{2}e}{10}}+{\frac{{c}^{3}b\ln \left ({c}^{2}{x}^{2}+1 \right ) d{e}^{2}}{2}}-{\frac{cb\ln \left ({c}^{2}{x}^{2}+1 \right ){e}^{3}}{2}}-{\frac{{c}^{7}b{d}^{3}\ln \left ( cx \right ) }{7}}+{\frac{3\,{c}^{5}b\ln \left ( cx \right ){d}^{2}e}{5}}-{c}^{3}b\ln \left ( cx \right ) d{e}^{2}+cb\ln \left ( cx \right ){e}^{3}+{\frac{{c}^{3}b{d}^{3}}{28\,{x}^{4}}}-{\frac{3\,cb{d}^{2}e}{20\,{x}^{4}}}-{\frac{cb{d}^{3}}{42\,{x}^{6}}}-{\frac{{c}^{5}b{d}^{3}}{14\,{x}^{2}}}+{\frac{3\,{c}^{3}b{d}^{2}e}{10\,{x}^{2}}}-{\frac{bcd{e}^{2}}{2\,{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^8,x)

[Out]

-1/7*a*d^3/x^7-a*e^3/x-3/5*a*d^2*e/x^5-a*d*e^2/x^3-1/7*b*arctan(c*x)*d^3/x^7-b*arctan(c*x)*e^3/x-3/5*b*arctan(
c*x)*d^2*e/x^5-b*arctan(c*x)*d*e^2/x^3+1/14*c^7*b*ln(c^2*x^2+1)*d^3-3/10*c^5*b*ln(c^2*x^2+1)*d^2*e+1/2*c^3*b*l
n(c^2*x^2+1)*d*e^2-1/2*c*b*ln(c^2*x^2+1)*e^3-1/7*c^7*b*d^3*ln(c*x)+3/5*c^5*b*ln(c*x)*d^2*e-c^3*b*ln(c*x)*d*e^2
+c*b*ln(c*x)*e^3+1/28*c^3*b*d^3/x^4-3/20*c*b*d^2*e/x^4-1/42*b*c*d^3/x^6-1/14*c^5*b*d^3/x^2+3/10*c^3*b*d^2*e/x^
2-1/2*c*b*d*e^2/x^2

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Maxima [A]  time = 1.03388, size = 333, normalized size = 1.49 \begin{align*} \frac{1}{84} \,{\left ({\left (6 \, c^{6} \log \left (c^{2} x^{2} + 1\right ) - 6 \, c^{6} \log \left (x^{2}\right ) - \frac{6 \, c^{4} x^{4} - 3 \, c^{2} x^{2} + 2}{x^{6}}\right )} c - \frac{12 \, \arctan \left (c x\right )}{x^{7}}\right )} b d^{3} - \frac{3}{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} e + \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 e^{2} - \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^{3} - \frac{a e^{3}}{x} - \frac{a d e^{2}}{x^{3}} - \frac{3 \, a d^{2} e}{5 \, x^{5}} - \frac{a d^{3}}{7 \, x^{7}} \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^8,x, algorithm="maxima")

[Out]

1/84*((6*c^6*log(c^2*x^2 + 1) - 6*c^6*log(x^2) - (6*c^4*x^4 - 3*c^2*x^2 + 2)/x^6)*c - 12*arctan(c*x)/x^7)*b*d^
3 - 3/20*((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*e + 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*e^2 - 1/2*(c*(log(c^2*x^2 + 1) - lo
g(x^2)) + 2*arctan(c*x)/x)*b*e^3 - a*e^3/x - a*d*e^2/x^3 - 3/5*a*d^2*e/x^5 - 1/7*a*d^3/x^7

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Fricas [A]  time = 2.00347, size = 567, normalized size = 2.53 \begin{align*} -\frac{420 \, a e^{3} x^{6} - 6 \,{\left (5 \, b c^{7} d^{3} - 21 \, b c^{5} d^{2} e + 35 \, b c^{3} d e^{2} - 35 \, b c e^{3}\right )} x^{7} \log \left (c^{2} x^{2} + 1\right ) + 12 \,{\left (5 \, b c^{7} d^{3} - 21 \, b c^{5} d^{2} e + 35 \, b c^{3} d e^{2} - 35 \, b c e^{3}\right )} x^{7} \log \left (x\right ) + 420 \, a d e^{2} x^{4} + 10 \, b c d^{3} x + 252 \, a d^{2} e x^{2} + 6 \,{\left (5 \, b c^{5} d^{3} - 21 \, b c^{3} d^{2} e + 35 \, b c d e^{2}\right )} x^{5} + 60 \, a d^{3} - 3 \,{\left (5 \, b c^{3} d^{3} - 21 \, b c d^{2} e\right )} x^{3} + 12 \,{\left (35 \, b e^{3} x^{6} + 35 \, b d e^{2} x^{4} + 21 \, b d^{2} e x^{2} + 5 \, b d^{3}\right )} \arctan \left (c x\right )}{420 \, x^{7}} \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^8,x, algorithm="fricas")

[Out]

-1/420*(420*a*e^3*x^6 - 6*(5*b*c^7*d^3 - 21*b*c^5*d^2*e + 35*b*c^3*d*e^2 - 35*b*c*e^3)*x^7*log(c^2*x^2 + 1) +
12*(5*b*c^7*d^3 - 21*b*c^5*d^2*e + 35*b*c^3*d*e^2 - 35*b*c*e^3)*x^7*log(x) + 420*a*d*e^2*x^4 + 10*b*c*d^3*x +
252*a*d^2*e*x^2 + 6*(5*b*c^5*d^3 - 21*b*c^3*d^2*e + 35*b*c*d*e^2)*x^5 + 60*a*d^3 - 3*(5*b*c^3*d^3 - 21*b*c*d^2
*e)*x^3 + 12*(35*b*e^3*x^6 + 35*b*d*e^2*x^4 + 21*b*d^2*e*x^2 + 5*b*d^3)*arctan(c*x))/x^7

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Sympy [A]  time = 6.39111, size = 362, normalized size = 1.62 \begin{align*} \begin{cases} - \frac{a d^{3}}{7 x^{7}} - \frac{3 a d^{2} e}{5 x^{5}} - \frac{a d e^{2}}{x^{3}} - \frac{a e^{3}}{x} - \frac{b c^{7} d^{3} \log{\left (x \right )}}{7} + \frac{b c^{7} d^{3} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{14} - \frac{b c^{5} d^{3}}{14 x^{2}} + \frac{3 b c^{5} d^{2} e \log{\left (x \right )}}{5} - \frac{3 b c^{5} d^{2} e \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{10} + \frac{b c^{3} d^{3}}{28 x^{4}} + \frac{3 b c^{3} d^{2} e}{10 x^{2}} - b c^{3} d e^{2} \log{\left (x \right )} + \frac{b c^{3} d e^{2} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2} - \frac{b c d^{3}}{42 x^{6}} - \frac{3 b c d^{2} e}{20 x^{4}} - \frac{b c d e^{2}}{2 x^{2}} + b c e^{3} \log{\left (x \right )} - \frac{b c e^{3} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{2} - \frac{b d^{3} \operatorname{atan}{\left (c x \right )}}{7 x^{7}} - \frac{3 b d^{2} e \operatorname{atan}{\left (c x \right )}}{5 x^{5}} - \frac{b d e^{2} \operatorname{atan}{\left (c x \right )}}{x^{3}} - \frac{b e^{3} \operatorname{atan}{\left (c x \right )}}{x} & \text{for}\: c \neq 0 \\a \left (- \frac{d^{3}}{7 x^{7}} - \frac{3 d^{2} e}{5 x^{5}} - \frac{d e^{2}}{x^{3}} - \frac{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**8,x)

[Out]

Piecewise((-a*d**3/(7*x**7) - 3*a*d**2*e/(5*x**5) - a*d*e**2/x**3 - a*e**3/x - b*c**7*d**3*log(x)/7 + b*c**7*d
**3*log(x**2 + c**(-2))/14 - b*c**5*d**3/(14*x**2) + 3*b*c**5*d**2*e*log(x)/5 - 3*b*c**5*d**2*e*log(x**2 + c**
(-2))/10 + b*c**3*d**3/(28*x**4) + 3*b*c**3*d**2*e/(10*x**2) - b*c**3*d*e**2*log(x) + b*c**3*d*e**2*log(x**2 +
c**(-2))/2 - b*c*d**3/(42*x**6) - 3*b*c*d**2*e/(20*x**4) - b*c*d*e**2/(2*x**2) + b*c*e**3*log(x) - b*c*e**3*l
og(x**2 + c**(-2))/2 - b*d**3*atan(c*x)/(7*x**7) - 3*b*d**2*e*atan(c*x)/(5*x**5) - b*d*e**2*atan(c*x)/x**3 - b
*e**3*atan(c*x)/x, Ne(c, 0)), (a*(-d**3/(7*x**7) - 3*d**2*e/(5*x**5) - d*e**2/x**3 - e**3/x), True))

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Giac [A]  time = 1.10426, size = 405, normalized size = 1.81 \begin{align*} \frac{30 \, b c^{7} d^{3} x^{7} \log \left (c^{2} x^{2} + 1\right ) - 60 \, b c^{7} d^{3} x^{7} \log \left (x\right ) - 126 \, b c^{5} d^{2} x^{7} e \log \left (c^{2} x^{2} + 1\right ) + 252 \, b c^{5} d^{2} x^{7} e \log \left (x\right ) - 30 \, b c^{5} d^{3} x^{5} + 210 \, b c^{3} d x^{7} e^{2} \log \left (c^{2} x^{2} + 1\right ) - 420 \, b c^{3} d x^{7} e^{2} \log \left (x\right ) + 126 \, b c^{3} d^{2} x^{5} e - 210 \, b c x^{7} e^{3} \log \left (c^{2} x^{2} + 1\right ) + 420 \, b c x^{7} e^{3} \log \left (x\right ) + 15 \, b c^{3} d^{3} x^{3} - 420 \, b x^{6} \arctan \left (c x\right ) e^{3} - 210 \, b c d x^{5} e^{2} - 420 \, a x^{6} e^{3} - 420 \, b d x^{4} \arctan \left (c x\right ) e^{2} - 63 \, b c d^{2} x^{3} e - 420 \, a d x^{4} e^{2} - 252 \, b d^{2} x^{2} \arctan \left (c x\right ) e - 10 \, b c d^{3} x - 252 \, a d^{2} x^{2} e - 60 \, b d^{3} \arctan \left (c x\right ) - 60 \, a d^{3}}{420 \, x^{7}} \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^8,x, algorithm="giac")

[Out]

1/420*(30*b*c^7*d^3*x^7*log(c^2*x^2 + 1) - 60*b*c^7*d^3*x^7*log(x) - 126*b*c^5*d^2*x^7*e*log(c^2*x^2 + 1) + 25
2*b*c^5*d^2*x^7*e*log(x) - 30*b*c^5*d^3*x^5 + 210*b*c^3*d*x^7*e^2*log(c^2*x^2 + 1) - 420*b*c^3*d*x^7*e^2*log(x
) + 126*b*c^3*d^2*x^5*e - 210*b*c*x^7*e^3*log(c^2*x^2 + 1) + 420*b*c*x^7*e^3*log(x) + 15*b*c^3*d^3*x^3 - 420*b
*x^6*arctan(c*x)*e^3 - 210*b*c*d*x^5*e^2 - 420*a*x^6*e^3 - 420*b*d*x^4*arctan(c*x)*e^2 - 63*b*c*d^2*x^3*e - 42
0*a*d*x^4*e^2 - 252*b*d^2*x^2*arctan(c*x)*e - 10*b*c*d^3*x - 252*a*d^2*x^2*e - 60*b*d^3*arctan(c*x) - 60*a*d^3
)/x^7