3.1154 \(\int \frac {a+b \tan ^{-1}(c x)}{x^3 (d+e x^2)} \, dx\)

Optimal. Leaf size=409 \[ -\frac {e \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}+\frac {e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 d^2}+\frac {e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 d^2}-\frac {a+b \tan ^{-1}(c x)}{2 d x^2}-\frac {a e \log (x)}{d^2}-\frac {b c^2 \tan ^{-1}(c x)}{2 d}-\frac {i b e \text {Li}_2(-i c x)}{2 d^2}+\frac {i b e \text {Li}_2(i c x)}{2 d^2}+\frac {i b e \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{2 d^2}-\frac {i b e \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 d^2}-\frac {i b e \text {Li}_2\left (1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{4 d^2}-\frac {b c}{2 d x} \]

[Out]

-1/2*b*c/d/x-1/2*b*c^2*arctan(c*x)/d+1/2*(-a-b*arctan(c*x))/d/x^2-a*e*ln(x)/d^2-e*(a+b*arctan(c*x))*ln(2/(1-I*
c*x))/d^2+1/2*e*(a+b*arctan(c*x))*ln(2*c*((-d)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2)))/d^2+1/2*e*
(a+b*arctan(c*x))*ln(2*c*((-d)^(1/2)+x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)+I*e^(1/2)))/d^2-1/2*I*b*e*polylog(2,-I
*c*x)/d^2+1/2*I*b*e*polylog(2,I*c*x)/d^2+1/2*I*b*e*polylog(2,1-2/(1-I*c*x))/d^2-1/4*I*b*e*polylog(2,1-2*c*((-d
)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2)))/d^2-1/4*I*b*e*polylog(2,1-2*c*((-d)^(1/2)+x*e^(1/2))/(1
-I*c*x)/(c*(-d)^(1/2)+I*e^(1/2)))/d^2

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Rubi [A]  time = 0.48, antiderivative size = 409, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 12, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {4918, 4852, 325, 203, 4928, 4848, 2391, 4980, 4856, 2402, 2315, 2447} \[ -\frac {i b e \text {PolyLog}(2,-i c x)}{2 d^2}+\frac {i b e \text {PolyLog}(2,i c x)}{2 d^2}+\frac {i b e \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 d^2}-\frac {i b e \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{4 d^2}-\frac {i b e \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{4 d^2}-\frac {e \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}+\frac {e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 d^2}+\frac {e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 d^2}-\frac {a+b \tan ^{-1}(c x)}{2 d x^2}-\frac {a e \log (x)}{d^2}-\frac {b c^2 \tan ^{-1}(c x)}{2 d}-\frac {b c}{2 d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c)/(2*d*x) - (b*c^2*ArcTan[c*x])/(2*d) - (a + b*ArcTan[c*x])/(2*d*x^2) - (a*e*Log[x])/d^2 - (e*(a + b*ArcT
an[c*x])*Log[2/(1 - I*c*x)])/d^2 + (e*(a + b*ArcTan[c*x])*Log[(2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sq
rt[e])*(1 - I*c*x))])/(2*d^2) + (e*(a + b*ArcTan[c*x])*Log[(2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[
e])*(1 - I*c*x))])/(2*d^2) - ((I/2)*b*e*PolyLog[2, (-I)*c*x])/d^2 + ((I/2)*b*e*PolyLog[2, I*c*x])/d^2 + ((I/2)
*b*e*PolyLog[2, 1 - 2/(1 - I*c*x)])/d^2 - ((I/4)*b*e*PolyLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d]
- I*Sqrt[e])*(1 - I*c*x))])/d^2 - ((I/4)*b*e*PolyLog[2, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt
[e])*(1 - I*c*x))])/d^2

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 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 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

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]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 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 4856

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])*Log[2/(1 -
I*c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d
+ e*x))/((c*d + I*e)*(1 - I*c*x))]/(1 + c^2*x^2), x], x] + Simp[((a + b*ArcTan[c*x])*Log[(2*c*(d + e*x))/((c*d
+ I*e)*(1 - I*c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]

Rule 4918

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTan[c*x])^p)/(d + e*
x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rule 4928

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[a
+ b*ArcTan[c*x], x^m/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IntegerQ[m] &&  !(EqQ[m, 1] && NeQ[a,
0])

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])

Rubi steps

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

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Mathematica [C]  time = 0.30, size = 507, normalized size = 1.24 \[ \frac {\frac {-a-b \tan ^{-1}(c x)}{2 x^2}-\frac {b c \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-c^2 x^2\right )}{2 x}}{d}-\frac {e \left (-\frac {a \log \left (d+e x^2\right )}{2 d}+\frac {a \log (x)}{d}-\frac {i b \left (\text {Li}_2\left (-\frac {\sqrt {e} (1-i c x)}{i c \sqrt {-d}-\sqrt {e}}\right )+\log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )\right )}{4 d}-\frac {i b \left (\text {Li}_2\left (\frac {\sqrt {e} (1-i c x)}{i \sqrt {-d} c+\sqrt {e}}\right )+\log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )\right )}{4 d}+\frac {i b \left (\text {Li}_2\left (-\frac {\sqrt {e} (i c x+1)}{i c \sqrt {-d}-\sqrt {e}}\right )+\log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )\right )}{4 d}+\frac {i b \left (\text {Li}_2\left (\frac {\sqrt {e} (i c x+1)}{i \sqrt {-d} c+\sqrt {e}}\right )+\log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )\right )}{4 d}+\frac {i b \text {Li}_2(-i c x)}{2 d}-\frac {i b \text {Li}_2(i c x)}{2 d}\right )}{d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-a - b*ArcTan[c*x])/(2*x^2) - (b*c*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)])/(2*x))/d - (e*((a*Log[x])/d
- (a*Log[d + e*x^2])/(2*d) + ((I/2)*b*PolyLog[2, (-I)*c*x])/d - ((I/2)*b*PolyLog[2, I*c*x])/d - ((I/4)*b*(Log[
1 - I*c*x]*Log[(c*(Sqrt[-d] - Sqrt[e]*x))/(c*Sqrt[-d] + I*Sqrt[e])] + PolyLog[2, -((Sqrt[e]*(1 - I*c*x))/(I*c*
Sqrt[-d] - Sqrt[e]))]))/d - ((I/4)*b*(Log[1 - I*c*x]*Log[(c*(Sqrt[-d] + Sqrt[e]*x))/(c*Sqrt[-d] - I*Sqrt[e])]
+ PolyLog[2, (Sqrt[e]*(1 - I*c*x))/(I*c*Sqrt[-d] + Sqrt[e])]))/d + ((I/4)*b*(Log[1 + I*c*x]*Log[(c*(Sqrt[-d] +
Sqrt[e]*x))/(c*Sqrt[-d] + I*Sqrt[e])] + PolyLog[2, -((Sqrt[e]*(1 + I*c*x))/(I*c*Sqrt[-d] - Sqrt[e]))]))/d + (
(I/4)*b*(Log[1 + I*c*x]*Log[(c*(Sqrt[-d] - Sqrt[e]*x))/(c*Sqrt[-d] - I*Sqrt[e])] + PolyLog[2, (Sqrt[e]*(1 + I*
c*x))/(I*c*Sqrt[-d] + Sqrt[e])]))/d))/d

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arctan \left (c x\right ) + a}{e x^{5} + d x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x^3/(e*x^2+d),x, algorithm="fricas")

[Out]

integral((b*arctan(c*x) + a)/(e*x^5 + d*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x^3/(e*x^2+d),x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 0.39, size = 801, normalized size = 1.96 \[ \frac {a e \ln \left (c^{2} e \,x^{2}+c^{2} d \right )}{2 d^{2}}-\frac {a}{2 d \,x^{2}}-\frac {a e \ln \left (c x \right )}{d^{2}}+\frac {b \arctan \left (c x \right ) e \ln \left (c^{2} e \,x^{2}+c^{2} d \right )}{2 d^{2}}-\frac {b \arctan \left (c x \right )}{2 d \,x^{2}}-\frac {b \arctan \left (c x \right ) e \ln \left (c x \right )}{d^{2}}-\frac {i b e \dilog \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )-c x +i}{\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )}\right )}{4 d^{2}}+\frac {i b e \ln \left (c x +i\right ) \ln \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )-c x -i}{\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )}\right )}{4 d^{2}}+\frac {i b e \ln \left (c x +i\right ) \ln \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )-c x -i}{\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )}\right )}{4 d^{2}}-\frac {i b e \ln \left (c x \right ) \ln \left (i c x +1\right )}{2 d^{2}}+\frac {i b e \dilog \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )-c x -i}{\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )}\right )}{4 d^{2}}-\frac {i b e \ln \left (c x +i\right ) \ln \left (c^{2} e \,x^{2}+c^{2} d \right )}{4 d^{2}}+\frac {i b e \ln \left (c x -i\right ) \ln \left (c^{2} e \,x^{2}+c^{2} d \right )}{4 d^{2}}-\frac {i b e \dilog \left (i c x +1\right )}{2 d^{2}}-\frac {i b e \ln \left (c x -i\right ) \ln \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )-c x +i}{\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )}\right )}{4 d^{2}}+\frac {i b e \dilog \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )-c x -i}{\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )}\right )}{4 d^{2}}-\frac {b c}{2 d x}-\frac {b \,c^{2} \arctan \left (c x \right )}{2 d}-\frac {i b e \ln \left (c x -i\right ) \ln \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )-c x +i}{\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )}\right )}{4 d^{2}}+\frac {i b e \dilog \left (-i c x +1\right )}{2 d^{2}}+\frac {i b e \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2 d^{2}}-\frac {i b e \dilog \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )-c x +i}{\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )}\right )}{4 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*e/d^2*ln(c^2*e*x^2+c^2*d)-1/2*a/d/x^2-a/d^2*e*ln(c*x)+1/2*b*arctan(c*x)*e/d^2*ln(c^2*e*x^2+c^2*d)-1/2*b*
arctan(c*x)/d/x^2-b*arctan(c*x)/d^2*e*ln(c*x)-1/2*I*b/d^2*e*ln(c*x)*ln(1+I*c*x)+1/4*I*b/d^2*e*dilog((RootOf(e*
_Z^2-2*I*_Z*e+c^2*d-e,index=1)-c*x-I)/RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=1))-1/4*I*b/d^2*e*ln(I+c*x)*ln(c^2*
e*x^2+c^2*d)-1/4*I*b/d^2*e*ln(c*x-I)*ln((RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=2)-c*x+I)/RootOf(e*_Z^2+2*I*_Z*e
+c^2*d-e,index=2))+1/4*I*b/d^2*e*ln(I+c*x)*ln((RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=2)-c*x-I)/RootOf(e*_Z^2-2*
I*_Z*e+c^2*d-e,index=2))+1/4*I*b/d^2*e*ln(c*x-I)*ln(c^2*e*x^2+c^2*d)-1/4*I*b/d^2*e*dilog((RootOf(e*_Z^2+2*I*_Z
*e+c^2*d-e,index=2)-c*x+I)/RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=2))-1/2*I*b/d^2*e*dilog(1+I*c*x)+1/4*I*b/d^2*e
*ln(I+c*x)*ln((RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=1)-c*x-I)/RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=1))+1/4*I*b
/d^2*e*dilog((RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=2)-c*x-I)/RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=2))-1/2*b*c/
d/x-1/2*b*c^2*arctan(c*x)/d+1/2*I*b/d^2*e*ln(c*x)*ln(1-I*c*x)-1/4*I*b/d^2*e*dilog((RootOf(e*_Z^2+2*I*_Z*e+c^2*
d-e,index=1)-c*x+I)/RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=1))-1/4*I*b/d^2*e*ln(c*x-I)*ln((RootOf(e*_Z^2+2*I*_Z*
e+c^2*d-e,index=1)-c*x+I)/RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=1))+1/2*I*b/d^2*e*dilog(1-I*c*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a {\left (\frac {e \log \left (e x^{2} + d\right )}{d^{2}} - \frac {2 \, e \log \relax (x)}{d^{2}} - \frac {1}{d x^{2}}\right )} + 2 \, b \int \frac {\arctan \left (c x\right )}{2 \, {\left (e x^{5} + d x^{3}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x^3/(e*x^2+d),x, algorithm="maxima")

[Out]

1/2*a*(e*log(e*x^2 + d)/d^2 - 2*e*log(x)/d^2 - 1/(d*x^2)) + 2*b*integrate(1/2*arctan(c*x)/(e*x^5 + d*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x^3\,\left (e\,x^2+d\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))/(x^3*(d + e*x^2)),x)

[Out]

int((a + b*atan(c*x))/(x^3*(d + e*x^2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {atan}{\left (c x \right )}}{x^{3} \left (d + e x^{2}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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