3.1136 \(\int \frac {(d+e x^2)^2 (a+b \tan ^{-1}(c x))}{x^8} \, dx\)

Optimal. Leaf size=186 \[ -\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+\frac {b c d \left (5 c^2 d-14 e\right )}{140 x^4}-\frac {b c \left (15 c^4 d^2-42 c^2 d e+35 e^2\right )}{210 x^2}+\frac {1}{210} b c^3 \left (15 c^4 d^2-42 c^2 d e+35 e^2\right ) \log \left (c^2 x^2+1\right )-\frac {1}{105} b c^3 \log (x) \left (15 c^4 d^2-42 c^2 d e+35 e^2\right )-\frac {b c d^2}{42 x^6} \]

[Out]

-1/42*b*c*d^2/x^6+1/140*b*c*d*(5*c^2*d-14*e)/x^4-1/210*b*c*(15*c^4*d^2-42*c^2*d*e+35*e^2)/x^2-1/7*d^2*(a+b*arc
tan(c*x))/x^7-2/5*d*e*(a+b*arctan(c*x))/x^5-1/3*e^2*(a+b*arctan(c*x))/x^3-1/105*b*c^3*(15*c^4*d^2-42*c^2*d*e+3
5*e^2)*ln(x)+1/210*b*c^3*(15*c^4*d^2-42*c^2*d*e+35*e^2)*ln(c^2*x^2+1)

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Rubi [A]  time = 0.23, antiderivative size = 186, normalized size of antiderivative = 1.00, 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} \[ -\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {b c \left (15 c^4 d^2-42 c^2 d e+35 e^2\right )}{210 x^2}+\frac {1}{210} b c^3 \left (15 c^4 d^2-42 c^2 d e+35 e^2\right ) \log \left (c^2 x^2+1\right )-\frac {1}{105} b c^3 \log (x) \left (15 c^4 d^2-42 c^2 d e+35 e^2\right )+\frac {b c d \left (5 c^2 d-14 e\right )}{140 x^4}-\frac {b c d^2}{42 x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*d^2)/(42*x^6) + (b*c*d*(5*c^2*d - 14*e))/(140*x^4) - (b*c*(15*c^4*d^2 - 42*c^2*d*e + 35*e^2))/(210*x^2)
- (d^2*(a + b*ArcTan[c*x]))/(7*x^7) - (2*d*e*(a + b*ArcTan[c*x]))/(5*x^5) - (e^2*(a + b*ArcTan[c*x]))/(3*x^3)
- (b*c^3*(15*c^4*d^2 - 42*c^2*d*e + 35*e^2)*Log[x])/105 + (b*c^3*(15*c^4*d^2 - 42*c^2*d*e + 35*e^2)*Log[1 + c^
2*x^2])/210

Rule 12

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

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 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]))

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 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]))

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right )^2 \left (a+b \tan ^{-1}(c x)\right )}{x^8} \, dx &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-(b c) \int \frac {-15 d^2-42 d e x^2-35 e^2 x^4}{105 x^7 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {1}{105} (b c) \int \frac {-15 d^2-42 d e x^2-35 e^2 x^4}{x^7 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {1}{210} (b c) \operatorname {Subst}\left (\int \frac {-15 d^2-42 d e x-35 e^2 x^2}{x^4 \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {1}{210} (b c) \operatorname {Subst}\left (\int \left (-\frac {15 d^2}{x^4}+\frac {3 d \left (5 c^2 d-14 e\right )}{x^3}+\frac {-15 c^4 d^2+42 c^2 d e-35 e^2}{x^2}+\frac {15 c^6 d^2-42 c^4 d e+35 c^2 e^2}{x}+\frac {-15 c^8 d^2+42 c^6 d e-35 c^4 e^2}{1+c^2 x}\right ) \, dx,x,x^2\right )\\ &=-\frac {b c d^2}{42 x^6}+\frac {b c d \left (5 c^2 d-14 e\right )}{140 x^4}-\frac {b c \left (15 c^4 d^2-42 c^2 d e+35 e^2\right )}{210 x^2}-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {1}{105} b c^3 \left (15 c^4 d^2-42 c^2 d e+35 e^2\right ) \log (x)+\frac {1}{210} b c^3 \left (15 c^4 d^2-42 c^2 d e+35 e^2\right ) \log \left (1+c^2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 177, normalized size = 0.95 \[ \frac {1}{420} \left (-\frac {60 d^2 \left (a+b \tan ^{-1}(c x)\right )}{x^7}-\frac {168 d e \left (a+b \tan ^{-1}(c x)\right )}{x^5}-\frac {140 e^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3}-70 b c e^2 \left (-c^2 \log \left (c^2 x^2+1\right )+2 c^2 \log (x)+\frac {1}{x^2}\right )-42 b c d e \left (-4 c^4 \log (x)-\frac {2 c^2}{x^2}+2 c^4 \log \left (c^2 x^2+1\right )+\frac {1}{x^4}\right )-5 b c d^2 \left (12 c^6 \log (x)-6 c^6 \log \left (c^2 x^2+1\right )+\frac {6 c^4 x^4-3 c^2 x^2+2}{x^6}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-60*d^2*(a + b*ArcTan[c*x]))/x^7 - (168*d*e*(a + b*ArcTan[c*x]))/x^5 - (140*e^2*(a + b*ArcTan[c*x]))/x^3 - 7
0*b*c*e^2*(x^(-2) + 2*c^2*Log[x] - c^2*Log[1 + c^2*x^2]) - 42*b*c*d*e*(x^(-4) - (2*c^2)/x^2 - 4*c^4*Log[x] + 2
*c^4*Log[1 + c^2*x^2]) - 5*b*c*d^2*((2 - 3*c^2*x^2 + 6*c^4*x^4)/x^6 + 12*c^6*Log[x] - 6*c^6*Log[1 + c^2*x^2]))
/420

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fricas [A]  time = 0.45, size = 194, normalized size = 1.04 \[ \frac {2 \, {\left (15 \, b c^{7} d^{2} - 42 \, b c^{5} d e + 35 \, b c^{3} e^{2}\right )} x^{7} \log \left (c^{2} x^{2} + 1\right ) - 4 \, {\left (15 \, b c^{7} d^{2} - 42 \, b c^{5} d e + 35 \, b c^{3} e^{2}\right )} x^{7} \log \relax (x) - 140 \, a e^{2} x^{4} - 2 \, {\left (15 \, b c^{5} d^{2} - 42 \, b c^{3} d e + 35 \, b c e^{2}\right )} x^{5} - 10 \, b c d^{2} x - 168 \, a d e x^{2} + 3 \, {\left (5 \, b c^{3} d^{2} - 14 \, b c d e\right )} x^{3} - 60 \, a d^{2} - 4 \, {\left (35 \, b e^{2} x^{4} + 42 \, b d e x^{2} + 15 \, b d^{2}\right )} \arctan \left (c x\right )}{420 \, x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arctan(c*x))/x^8,x, algorithm="fricas")

[Out]

1/420*(2*(15*b*c^7*d^2 - 42*b*c^5*d*e + 35*b*c^3*e^2)*x^7*log(c^2*x^2 + 1) - 4*(15*b*c^7*d^2 - 42*b*c^5*d*e +
35*b*c^3*e^2)*x^7*log(x) - 140*a*e^2*x^4 - 2*(15*b*c^5*d^2 - 42*b*c^3*d*e + 35*b*c*e^2)*x^5 - 10*b*c*d^2*x - 1
68*a*d*e*x^2 + 3*(5*b*c^3*d^2 - 14*b*c*d*e)*x^3 - 60*a*d^2 - 4*(35*b*e^2*x^4 + 42*b*d*e*x^2 + 15*b*d^2)*arctan
(c*x))/x^7

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arctan(c*x))/x^8,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.05, size = 224, normalized size = 1.20 \[ -\frac {a \,d^{2}}{7 x^{7}}-\frac {a \,e^{2}}{3 x^{3}}-\frac {2 a e d}{5 x^{5}}-\frac {b \arctan \left (c x \right ) d^{2}}{7 x^{7}}-\frac {b \arctan \left (c x \right ) e^{2}}{3 x^{3}}-\frac {2 b \arctan \left (c x \right ) e d}{5 x^{5}}-\frac {c^{5} b \,d^{2}}{14 x^{2}}+\frac {c^{3} b e d}{5 x^{2}}-\frac {c b \,e^{2}}{6 x^{2}}-\frac {c^{7} b \,d^{2} \ln \left (c x \right )}{7}+\frac {2 c^{5} b \ln \left (c x \right ) d e}{5}-\frac {c^{3} b \ln \left (c x \right ) e^{2}}{3}-\frac {b c \,d^{2}}{42 x^{6}}+\frac {c^{3} b \,d^{2}}{28 x^{4}}-\frac {c b e d}{10 x^{4}}+\frac {c^{7} b \ln \left (c^{2} x^{2}+1\right ) d^{2}}{14}-\frac {c^{5} b \ln \left (c^{2} x^{2}+1\right ) e d}{5}+\frac {c^{3} b \ln \left (c^{2} x^{2}+1\right ) e^{2}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^2*(a+b*arctan(c*x))/x^8,x)

[Out]

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

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maxima [A]  time = 0.32, size = 197, normalized size = 1.06 \[ \frac {1}{84} \, {\left ({\left (6 \, c^{6} \log \left (c^{2} x^{2} + 1\right ) - 6 \, c^{6} \log \left (x^{2}\right ) - \frac {6 \, c^{4} x^{4} - 3 \, c^{2} x^{2} + 2}{x^{6}}\right )} c - \frac {12 \, \arctan \left (c x\right )}{x^{7}}\right )} b d^{2} - \frac {1}{10} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d e + \frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b e^{2} - \frac {a e^{2}}{3 \, x^{3}} - \frac {2 \, a d e}{5 \, x^{5}} - \frac {a d^{2}}{7 \, x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(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^
2 - 1/10*((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*e + 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*e^2 - 1/3*a*e^2/x^3 - 2/5*a*d*e/x^5 - 1
/7*a*d^2/x^7

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mupad [B]  time = 0.63, size = 232, normalized size = 1.25 \[ -\frac {60\,a\,d^2+60\,b\,d^2\,\mathrm {atan}\left (c\,x\right )+140\,a\,e^2\,x^4-15\,b\,c^3\,d^2\,x^3+30\,b\,c^5\,d^2\,x^5+10\,b\,c\,d^2\,x+168\,a\,d\,e\,x^2+70\,b\,c\,e^2\,x^5+140\,b\,e^2\,x^4\,\mathrm {atan}\left (c\,x\right )+60\,b\,c^7\,d^2\,x^7\,\ln \relax (x)+140\,b\,c^3\,e^2\,x^7\,\ln \relax (x)-84\,b\,c^3\,d\,e\,x^5+42\,b\,c\,d\,e\,x^3-30\,b\,c^7\,d^2\,x^7\,\ln \left (c^2\,x^2+1\right )-70\,b\,c^3\,e^2\,x^7\,\ln \left (c^2\,x^2+1\right )+168\,b\,d\,e\,x^2\,\mathrm {atan}\left (c\,x\right )-168\,b\,c^5\,d\,e\,x^7\,\ln \relax (x)+84\,b\,c^5\,d\,e\,x^7\,\ln \left (c^2\,x^2+1\right )}{420\,x^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + e*x^2)^2)/x^8,x)

[Out]

-(60*a*d^2 + 60*b*d^2*atan(c*x) + 140*a*e^2*x^4 - 15*b*c^3*d^2*x^3 + 30*b*c^5*d^2*x^5 + 10*b*c*d^2*x + 168*a*d
*e*x^2 + 70*b*c*e^2*x^5 + 140*b*e^2*x^4*atan(c*x) + 60*b*c^7*d^2*x^7*log(x) + 140*b*c^3*e^2*x^7*log(x) - 84*b*
c^3*d*e*x^5 + 42*b*c*d*e*x^3 - 30*b*c^7*d^2*x^7*log(c^2*x^2 + 1) - 70*b*c^3*e^2*x^7*log(c^2*x^2 + 1) + 168*b*d
*e*x^2*atan(c*x) - 168*b*c^5*d*e*x^7*log(x) + 84*b*c^5*d*e*x^7*log(c^2*x^2 + 1))/(420*x^7)

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sympy [A]  time = 3.98, size = 289, normalized size = 1.55 \[ \begin {cases} - \frac {a d^{2}}{7 x^{7}} - \frac {2 a d e}{5 x^{5}} - \frac {a e^{2}}{3 x^{3}} - \frac {b c^{7} d^{2} \log {\relax (x )}}{7} + \frac {b c^{7} d^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{14} - \frac {b c^{5} d^{2}}{14 x^{2}} + \frac {2 b c^{5} d e \log {\relax (x )}}{5} - \frac {b c^{5} d e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{5} + \frac {b c^{3} d^{2}}{28 x^{4}} + \frac {b c^{3} d e}{5 x^{2}} - \frac {b c^{3} e^{2} \log {\relax (x )}}{3} + \frac {b c^{3} e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{6} - \frac {b c d^{2}}{42 x^{6}} - \frac {b c d e}{10 x^{4}} - \frac {b c e^{2}}{6 x^{2}} - \frac {b d^{2} \operatorname {atan}{\left (c x \right )}}{7 x^{7}} - \frac {2 b d e \operatorname {atan}{\left (c x \right )}}{5 x^{5}} - \frac {b e^{2} \operatorname {atan}{\left (c x \right )}}{3 x^{3}} & \text {for}\: c \neq 0 \\a \left (- \frac {d^{2}}{7 x^{7}} - \frac {2 d e}{5 x^{5}} - \frac {e^{2}}{3 x^{3}}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**2*(a+b*atan(c*x))/x**8,x)

[Out]

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

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