3.1147 \(\int \frac{(d+e x^2)^3 (a+b \tan ^{-1}(c x))}{x^7} \, dx\)

Optimal. Leaf size=228 \[ \frac{1}{2} i b e^3 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b e^3 \text{PolyLog}(2,i c x)-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac{3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+a e^3 \log (x)+\frac{3 b c^3 d^2 e}{4 x}+\frac{3}{4} b c^4 d^2 e \tan ^{-1}(c x)+\frac{b c^3 d^3}{18 x^3}-\frac{b c^5 d^3}{6 x}-\frac{1}{6} b c^6 d^3 \tan ^{-1}(c x)-\frac{3}{2} b c^2 d e^2 \tan ^{-1}(c x)-\frac{b c d^2 e}{4 x^3}-\frac{b c d^3}{30 x^5}-\frac{3 b c d e^2}{2 x} \]

[Out]

-(b*c*d^3)/(30*x^5) + (b*c^3*d^3)/(18*x^3) - (b*c*d^2*e)/(4*x^3) - (b*c^5*d^3)/(6*x) + (3*b*c^3*d^2*e)/(4*x) -
(3*b*c*d*e^2)/(2*x) - (b*c^6*d^3*ArcTan[c*x])/6 + (3*b*c^4*d^2*e*ArcTan[c*x])/4 - (3*b*c^2*d*e^2*ArcTan[c*x])
/2 - (d^3*(a + b*ArcTan[c*x]))/(6*x^6) - (3*d^2*e*(a + b*ArcTan[c*x]))/(4*x^4) - (3*d*e^2*(a + b*ArcTan[c*x]))
/(2*x^2) + a*e^3*Log[x] + (I/2)*b*e^3*PolyLog[2, (-I)*c*x] - (I/2)*b*e^3*PolyLog[2, I*c*x]

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Rubi [A]  time = 0.230883, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4980, 4852, 325, 203, 4848, 2391} \[ \frac{1}{2} i b e^3 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b e^3 \text{PolyLog}(2,i c x)-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac{3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+a e^3 \log (x)+\frac{3 b c^3 d^2 e}{4 x}+\frac{3}{4} b c^4 d^2 e \tan ^{-1}(c x)+\frac{b c^3 d^3}{18 x^3}-\frac{b c^5 d^3}{6 x}-\frac{1}{6} b c^6 d^3 \tan ^{-1}(c x)-\frac{3}{2} b c^2 d e^2 \tan ^{-1}(c x)-\frac{b c d^2 e}{4 x^3}-\frac{b c d^3}{30 x^5}-\frac{3 b c d e^2}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^7,x]

[Out]

-(b*c*d^3)/(30*x^5) + (b*c^3*d^3)/(18*x^3) - (b*c*d^2*e)/(4*x^3) - (b*c^5*d^3)/(6*x) + (3*b*c^3*d^2*e)/(4*x) -
(3*b*c*d*e^2)/(2*x) - (b*c^6*d^3*ArcTan[c*x])/6 + (3*b*c^4*d^2*e*ArcTan[c*x])/4 - (3*b*c^2*d*e^2*ArcTan[c*x])
/2 - (d^3*(a + b*ArcTan[c*x]))/(6*x^6) - (3*d^2*e*(a + b*ArcTan[c*x]))/(4*x^4) - (3*d*e^2*(a + b*ArcTan[c*x]))
/(2*x^2) + a*e^3*Log[x] + (I/2)*b*e^3*PolyLog[2, (-I)*c*x] - (I/2)*b*e^3*PolyLog[2, I*c*x]

Rule 4980

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With
[{u = ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b,
c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && ((EqQ[p, 1] && GtQ[q, 0]) || IntegerQ[m])

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^3 \left (a+b \tan ^{-1}(c x)\right )}{x^7} \, dx &=\int \left (\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^7}+\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x^5}+\frac{3 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}\right ) \, dx\\ &=d^3 \int \frac{a+b \tan ^{-1}(c x)}{x^7} \, dx+\left (3 d^2 e\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^5} \, dx+\left (3 d e^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^3} \, dx+e^3 \int \frac{a+b \tan ^{-1}(c x)}{x} \, dx\\ &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+a e^3 \log (x)+\frac{1}{6} \left (b c d^3\right ) \int \frac{1}{x^6 \left (1+c^2 x^2\right )} \, dx+\frac{1}{4} \left (3 b c d^2 e\right ) \int \frac{1}{x^4 \left (1+c^2 x^2\right )} \, dx+\frac{1}{2} \left (3 b c d e^2\right ) \int \frac{1}{x^2 \left (1+c^2 x^2\right )} \, dx+\frac{1}{2} \left (i b e^3\right ) \int \frac{\log (1-i c x)}{x} \, dx-\frac{1}{2} \left (i b e^3\right ) \int \frac{\log (1+i c x)}{x} \, dx\\ &=-\frac{b c d^3}{30 x^5}-\frac{b c d^2 e}{4 x^3}-\frac{3 b c d e^2}{2 x}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+a e^3 \log (x)+\frac{1}{2} i b e^3 \text{Li}_2(-i c x)-\frac{1}{2} i b e^3 \text{Li}_2(i c x)-\frac{1}{6} \left (b c^3 d^3\right ) \int \frac{1}{x^4 \left (1+c^2 x^2\right )} \, dx-\frac{1}{4} \left (3 b c^3 d^2 e\right ) \int \frac{1}{x^2 \left (1+c^2 x^2\right )} \, dx-\frac{1}{2} \left (3 b c^3 d e^2\right ) \int \frac{1}{1+c^2 x^2} \, dx\\ &=-\frac{b c d^3}{30 x^5}+\frac{b c^3 d^3}{18 x^3}-\frac{b c d^2 e}{4 x^3}+\frac{3 b c^3 d^2 e}{4 x}-\frac{3 b c d e^2}{2 x}-\frac{3}{2} b c^2 d e^2 \tan ^{-1}(c x)-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+a e^3 \log (x)+\frac{1}{2} i b e^3 \text{Li}_2(-i c x)-\frac{1}{2} i b e^3 \text{Li}_2(i c x)+\frac{1}{6} \left (b c^5 d^3\right ) \int \frac{1}{x^2 \left (1+c^2 x^2\right )} \, dx+\frac{1}{4} \left (3 b c^5 d^2 e\right ) \int \frac{1}{1+c^2 x^2} \, dx\\ &=-\frac{b c d^3}{30 x^5}+\frac{b c^3 d^3}{18 x^3}-\frac{b c d^2 e}{4 x^3}-\frac{b c^5 d^3}{6 x}+\frac{3 b c^3 d^2 e}{4 x}-\frac{3 b c d e^2}{2 x}+\frac{3}{4} b c^4 d^2 e \tan ^{-1}(c x)-\frac{3}{2} b c^2 d e^2 \tan ^{-1}(c x)-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+a e^3 \log (x)+\frac{1}{2} i b e^3 \text{Li}_2(-i c x)-\frac{1}{2} i b e^3 \text{Li}_2(i c x)-\frac{1}{6} \left (b c^7 d^3\right ) \int \frac{1}{1+c^2 x^2} \, dx\\ &=-\frac{b c d^3}{30 x^5}+\frac{b c^3 d^3}{18 x^3}-\frac{b c d^2 e}{4 x^3}-\frac{b c^5 d^3}{6 x}+\frac{3 b c^3 d^2 e}{4 x}-\frac{3 b c d e^2}{2 x}-\frac{1}{6} b c^6 d^3 \tan ^{-1}(c x)+\frac{3}{4} b c^4 d^2 e \tan ^{-1}(c x)-\frac{3}{2} b c^2 d e^2 \tan ^{-1}(c x)-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac{3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+a e^3 \log (x)+\frac{1}{2} i b e^3 \text{Li}_2(-i c x)-\frac{1}{2} i b e^3 \text{Li}_2(i c x)\\ \end{align*}

Mathematica [C]  time = 0.138102, size = 175, normalized size = 0.77 \[ \frac{1}{60} \left (-\frac{15 b c d^2 e \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-c^2 x^2\right )}{x^3}-\frac{2 b c d^3 \text{Hypergeometric2F1}\left (-\frac{5}{2},1,-\frac{3}{2},-c^2 x^2\right )}{x^5}-\frac{90 b c d e^2 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-c^2 x^2\right )}{x}+30 i b e^3 \text{PolyLog}(2,-i c x)-30 i b e^3 \text{PolyLog}(2,i c x)-\frac{45 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x^4}-\frac{10 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^6}-\frac{90 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+60 a e^3 \log (x)\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^7,x]

[Out]

((-10*d^3*(a + b*ArcTan[c*x]))/x^6 - (45*d^2*e*(a + b*ArcTan[c*x]))/x^4 - (90*d*e^2*(a + b*ArcTan[c*x]))/x^2 -
(2*b*c*d^3*Hypergeometric2F1[-5/2, 1, -3/2, -(c^2*x^2)])/x^5 - (15*b*c*d^2*e*Hypergeometric2F1[-3/2, 1, -1/2,
-(c^2*x^2)])/x^3 - (90*b*c*d*e^2*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)])/x + 60*a*e^3*Log[x] + (30*I)*b*
e^3*PolyLog[2, (-I)*c*x] - (30*I)*b*e^3*PolyLog[2, I*c*x])/60

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Maple [A]  time = 0.06, size = 272, normalized size = 1.2 \begin{align*} -{\frac{3\,ad{e}^{2}}{2\,{x}^{2}}}-{\frac{3\,a{d}^{2}e}{4\,{x}^{4}}}-{\frac{a{d}^{3}}{6\,{x}^{6}}}+a{e}^{3}\ln \left ( cx \right ) -{\frac{3\,\arctan \left ( cx \right ) bd{e}^{2}}{2\,{x}^{2}}}-{\frac{3\,b{d}^{2}\arctan \left ( cx \right ) e}{4\,{x}^{4}}}-{\frac{b{d}^{3}\arctan \left ( cx \right ) }{6\,{x}^{6}}}+b\arctan \left ( cx \right ){e}^{3}\ln \left ( cx \right ) -{\frac{i}{2}}b{e}^{3}\ln \left ( cx \right ) \ln \left ( 1-icx \right ) +{\frac{i}{2}}b{e}^{3}\ln \left ( cx \right ) \ln \left ( 1+icx \right ) -{\frac{i}{2}}b{e}^{3}{\it dilog} \left ( 1-icx \right ) +{\frac{i}{2}}b{e}^{3}{\it dilog} \left ( 1+icx \right ) -{\frac{b{c}^{6}{d}^{3}\arctan \left ( cx \right ) }{6}}+{\frac{3\,b{c}^{4}{d}^{2}e\arctan \left ( cx \right ) }{4}}-{\frac{3\,b{c}^{2}d{e}^{2}\arctan \left ( cx \right ) }{2}}-{\frac{b{c}^{5}{d}^{3}}{6\,x}}+{\frac{3\,b{c}^{3}{d}^{2}e}{4\,x}}-{\frac{3\,bcd{e}^{2}}{2\,x}}-{\frac{bc{d}^{3}}{30\,{x}^{5}}}+{\frac{b{c}^{3}{d}^{3}}{18\,{x}^{3}}}-{\frac{bc{d}^{2}e}{4\,{x}^{3}}} \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^7,x)

[Out]

-3/2*a*d*e^2/x^2-3/4*a*d^2*e/x^4-1/6*a*d^3/x^6+a*e^3*ln(c*x)-3/2*b*arctan(c*x)*d*e^2/x^2-3/4*b*arctan(c*x)*d^2
*e/x^4-1/6*b*arctan(c*x)*d^3/x^6+b*arctan(c*x)*e^3*ln(c*x)-1/2*I*b*e^3*ln(c*x)*ln(1-I*c*x)+1/2*I*b*e^3*ln(c*x)
*ln(1+I*c*x)-1/2*I*b*e^3*dilog(1-I*c*x)+1/2*I*b*e^3*dilog(1+I*c*x)-1/6*b*c^6*d^3*arctan(c*x)+3/4*b*c^4*d^2*e*a
rctan(c*x)-3/2*b*c^2*d*e^2*arctan(c*x)-1/6*b*c^5*d^3/x+3/4*b*c^3*d^2*e/x-3/2*b*c*d*e^2/x-1/30*b*c*d^3/x^5+1/18
*b*c^3*d^3/x^3-1/4*b*c*d^2*e/x^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{90} \,{\left ({\left (15 \, c^{5} \arctan \left (c x\right ) + \frac{15 \, c^{4} x^{4} - 5 \, c^{2} x^{2} + 3}{x^{5}}\right )} c + \frac{15 \, \arctan \left (c x\right )}{x^{6}}\right )} b d^{3} + \frac{1}{4} \,{\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac{3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac{3 \, \arctan \left (c x\right )}{x^{4}}\right )} b d^{2} e - \frac{3}{2} \,{\left ({\left (c \arctan \left (c x\right ) + \frac{1}{x}\right )} c + \frac{\arctan \left (c x\right )}{x^{2}}\right )} b d e^{2} + b e^{3} \int \frac{\arctan \left (c x\right )}{x}\,{d x} + a e^{3} \log \left (x\right ) - \frac{3 \, a d e^{2}}{2 \, x^{2}} - \frac{3 \, a d^{2} e}{4 \, x^{4}} - \frac{a d^{3}}{6 \, x^{6}} \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^7,x, algorithm="maxima")

[Out]

-1/90*((15*c^5*arctan(c*x) + (15*c^4*x^4 - 5*c^2*x^2 + 3)/x^5)*c + 15*arctan(c*x)/x^6)*b*d^3 + 1/4*((3*c^3*arc
tan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*b*d^2*e - 3/2*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^
2)*b*d*e^2 + b*e^3*integrate(arctan(c*x)/x, x) + a*e^3*log(x) - 3/2*a*d*e^2/x^2 - 3/4*a*d^2*e/x^4 - 1/6*a*d^3/
x^6

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a e^{3} x^{6} + 3 \, a d e^{2} x^{4} + 3 \, a d^{2} e x^{2} + a d^{3} +{\left (b e^{3} x^{6} + 3 \, b d e^{2} x^{4} + 3 \, b d^{2} e x^{2} + b d^{3}\right )} \arctan \left (c x\right )}{x^{7}}, x\right ) \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^7,x, algorithm="fricas")

[Out]

integral((a*e^3*x^6 + 3*a*d*e^2*x^4 + 3*a*d^2*e*x^2 + a*d^3 + (b*e^3*x^6 + 3*b*d*e^2*x^4 + 3*b*d^2*e*x^2 + b*d
^3)*arctan(c*x))/x^7, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{3}}{x^{7}}\, dx \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**7,x)

[Out]

Integral((a + b*atan(c*x))*(d + e*x**2)**3/x**7, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}^{3}{\left (b \arctan \left (c x\right ) + a\right )}}{x^{7}}\,{d 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^7,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^3*(b*arctan(c*x) + a)/x^7, x)