\(\int \frac {1}{(d+e x)^2 (a+c x^2)} \, dx\) [503]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 17, antiderivative size = 123 \[ \int \frac {1}{(d+e x)^2 \left (a+c x^2\right )} \, dx=-\frac {e}{\left (c d^2+a e^2\right ) (d+e x)}+\frac {\sqrt {c} \left (c d^2-a e^2\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} \left (c d^2+a e^2\right )^2}+\frac {2 c d e \log (d+e x)}{\left (c d^2+a e^2\right )^2}-\frac {c d e \log \left (a+c x^2\right )}{\left (c d^2+a e^2\right )^2} \]

[Out]

-e/(a*e^2+c*d^2)/(e*x+d)+2*c*d*e*ln(e*x+d)/(a*e^2+c*d^2)^2-c*d*e*ln(c*x^2+a)/(a*e^2+c*d^2)^2+(-a*e^2+c*d^2)*ar
ctan(x*c^(1/2)/a^(1/2))*c^(1/2)/(a*e^2+c*d^2)^2/a^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {724, 815, 649, 211, 266} \[ \int \frac {1}{(d+e x)^2 \left (a+c x^2\right )} \, dx=\frac {\sqrt {c} \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (c d^2-a e^2\right )}{\sqrt {a} \left (a e^2+c d^2\right )^2}-\frac {c d e \log \left (a+c x^2\right )}{\left (a e^2+c d^2\right )^2}-\frac {e}{(d+e x) \left (a e^2+c d^2\right )}+\frac {2 c d e \log (d+e x)}{\left (a e^2+c d^2\right )^2} \]

[In]

Int[1/((d + e*x)^2*(a + c*x^2)),x]

[Out]

-(e/((c*d^2 + a*e^2)*(d + e*x))) + (Sqrt[c]*(c*d^2 - a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*(c*d^2 + a*e
^2)^2) + (2*c*d*e*Log[d + e*x])/(c*d^2 + a*e^2)^2 - (c*d*e*Log[a + c*x^2])/(c*d^2 + a*e^2)^2

Rule 211

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 724

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

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps \begin{align*} \text {integral}& = -\frac {e}{\left (c d^2+a e^2\right ) (d+e x)}+\frac {c \int \frac {d-e x}{(d+e x) \left (a+c x^2\right )} \, dx}{c d^2+a e^2} \\ & = -\frac {e}{\left (c d^2+a e^2\right ) (d+e x)}+\frac {c \int \left (\frac {2 d e^2}{\left (c d^2+a e^2\right ) (d+e x)}+\frac {c d^2-a e^2-2 c d e x}{\left (c d^2+a e^2\right ) \left (a+c x^2\right )}\right ) \, dx}{c d^2+a e^2} \\ & = -\frac {e}{\left (c d^2+a e^2\right ) (d+e x)}+\frac {2 c d e \log (d+e x)}{\left (c d^2+a e^2\right )^2}+\frac {c \int \frac {c d^2-a e^2-2 c d e x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^2} \\ & = -\frac {e}{\left (c d^2+a e^2\right ) (d+e x)}+\frac {2 c d e \log (d+e x)}{\left (c d^2+a e^2\right )^2}-\frac {\left (2 c^2 d e\right ) \int \frac {x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^2}+\frac {\left (c \left (c d^2-a e^2\right )\right ) \int \frac {1}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^2} \\ & = -\frac {e}{\left (c d^2+a e^2\right ) (d+e x)}+\frac {\sqrt {c} \left (c d^2-a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} \left (c d^2+a e^2\right )^2}+\frac {2 c d e \log (d+e x)}{\left (c d^2+a e^2\right )^2}-\frac {c d e \log \left (a+c x^2\right )}{\left (c d^2+a e^2\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(d+e x)^2 \left (a+c x^2\right )} \, dx=\frac {\sqrt {c} \left (c d^2-a e^2\right ) (d+e x) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )-\sqrt {a} e \left (c d^2+a e^2-2 c d (d+e x) \log (d+e x)+c d (d+e x) \log \left (a+c x^2\right )\right )}{\sqrt {a} \left (c d^2+a e^2\right )^2 (d+e x)} \]

[In]

Integrate[1/((d + e*x)^2*(a + c*x^2)),x]

[Out]

(Sqrt[c]*(c*d^2 - a*e^2)*(d + e*x)*ArcTan[(Sqrt[c]*x)/Sqrt[a]] - Sqrt[a]*e*(c*d^2 + a*e^2 - 2*c*d*(d + e*x)*Lo
g[d + e*x] + c*d*(d + e*x)*Log[a + c*x^2]))/(Sqrt[a]*(c*d^2 + a*e^2)^2*(d + e*x))

Maple [A] (verified)

Time = 2.31 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.85

method result size
default \(-\frac {c \left (d e \ln \left (c \,x^{2}+a \right )+\frac {\left (e^{2} a -c \,d^{2}\right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}\right )}{\left (e^{2} a +c \,d^{2}\right )^{2}}-\frac {e}{\left (e^{2} a +c \,d^{2}\right ) \left (e x +d \right )}+\frac {2 c d e \ln \left (e x +d \right )}{\left (e^{2} a +c \,d^{2}\right )^{2}}\) \(104\)
risch \(-\frac {e}{\left (e^{2} a +c \,d^{2}\right ) \left (e x +d \right )}+\frac {2 c d e \ln \left (e x +d \right )}{a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (e^{4} a^{3}+2 d^{2} e^{2} a^{2} c +d^{4} c^{2} a \right ) \textit {\_Z}^{2}+4 a c d e \textit {\_Z} +c \right )}{\sum }\textit {\_R} \ln \left (\left (\left (3 e^{6} a^{3}+5 d^{2} e^{4} a^{2} c +d^{4} e^{2} c^{2} a -c^{3} d^{6}\right ) \textit {\_R}^{2}+\left (4 d \,e^{3} a c +4 d^{3} e \,c^{2}\right ) \textit {\_R} +2 e^{2} c \right ) x +\left (4 d \,e^{5} a^{3}+8 a^{2} c \,d^{3} e^{3}+4 a \,c^{2} d^{5} e \right ) \textit {\_R}^{2}+\left (a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \textit {\_R} -2 c d e \right )\right )}{2}\) \(255\)

[In]

int(1/(e*x+d)^2/(c*x^2+a),x,method=_RETURNVERBOSE)

[Out]

-c/(a*e^2+c*d^2)^2*(d*e*ln(c*x^2+a)+(a*e^2-c*d^2)/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2)))-e/(a*e^2+c*d^2)/(e*x+d)
+2*c*d*e*ln(e*x+d)/(a*e^2+c*d^2)^2

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.85 \[ \int \frac {1}{(d+e x)^2 \left (a+c x^2\right )} \, dx=\left [-\frac {2 \, c d^{2} e + 2 \, a e^{3} + {\left (c d^{3} - a d e^{2} + {\left (c d^{2} e - a e^{3}\right )} x\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{2} - 2 \, a x \sqrt {-\frac {c}{a}} - a}{c x^{2} + a}\right ) + 2 \, {\left (c d e^{2} x + c d^{2} e\right )} \log \left (c x^{2} + a\right ) - 4 \, {\left (c d e^{2} x + c d^{2} e\right )} \log \left (e x + d\right )}{2 \, {\left (c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\right )}}, -\frac {c d^{2} e + a e^{3} - {\left (c d^{3} - a d e^{2} + {\left (c d^{2} e - a e^{3}\right )} x\right )} \sqrt {\frac {c}{a}} \arctan \left (x \sqrt {\frac {c}{a}}\right ) + {\left (c d e^{2} x + c d^{2} e\right )} \log \left (c x^{2} + a\right ) - 2 \, {\left (c d e^{2} x + c d^{2} e\right )} \log \left (e x + d\right )}{c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x}\right ] \]

[In]

integrate(1/(e*x+d)^2/(c*x^2+a),x, algorithm="fricas")

[Out]

[-1/2*(2*c*d^2*e + 2*a*e^3 + (c*d^3 - a*d*e^2 + (c*d^2*e - a*e^3)*x)*sqrt(-c/a)*log((c*x^2 - 2*a*x*sqrt(-c/a)
- a)/(c*x^2 + a)) + 2*(c*d*e^2*x + c*d^2*e)*log(c*x^2 + a) - 4*(c*d*e^2*x + c*d^2*e)*log(e*x + d))/(c^2*d^5 +
2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^2*e^3 + a^2*e^5)*x), -(c*d^2*e + a*e^3 - (c*d^3 - a*d*e^2 + (
c*d^2*e - a*e^3)*x)*sqrt(c/a)*arctan(x*sqrt(c/a)) + (c*d*e^2*x + c*d^2*e)*log(c*x^2 + a) - 2*(c*d*e^2*x + c*d^
2*e)*log(e*x + d))/(c^2*d^5 + 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^2*e^3 + a^2*e^5)*x)]

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x)^2 \left (a+c x^2\right )} \, dx=\text {Timed out} \]

[In]

integrate(1/(e*x+d)**2/(c*x**2+a),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.36 \[ \int \frac {1}{(d+e x)^2 \left (a+c x^2\right )} \, dx=-\frac {c d e \log \left (c x^{2} + a\right )}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}} + \frac {2 \, c d e \log \left (e x + d\right )}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}} + \frac {{\left (c^{2} d^{2} - a c e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a c}} - \frac {e}{c d^{3} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x} \]

[In]

integrate(1/(e*x+d)^2/(c*x^2+a),x, algorithm="maxima")

[Out]

-c*d*e*log(c*x^2 + a)/(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4) + 2*c*d*e*log(e*x + d)/(c^2*d^4 + 2*a*c*d^2*e^2 + a^
2*e^4) + (c^2*d^2 - a*c*e^2)*arctan(c*x/sqrt(a*c))/((c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(a*c)) - e/(c*d^3
+ a*d*e^2 + (c*d^2*e + a*e^3)*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.57 \[ \int \frac {1}{(d+e x)^2 \left (a+c x^2\right )} \, dx=-\frac {c d e \log \left (c - \frac {2 \, c d}{e x + d} + \frac {c d^{2}}{{\left (e x + d\right )}^{2}} + \frac {a e^{2}}{{\left (e x + d\right )}^{2}}\right )}{c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}} - \frac {e^{3}}{{\left (c d^{2} e^{2} + a e^{4}\right )} {\left (e x + d\right )}} + \frac {{\left (c^{2} d^{2} e^{2} - a c e^{4}\right )} \arctan \left (\frac {c d - \frac {c d^{2}}{e x + d} - \frac {a e^{2}}{e x + d}}{\sqrt {a c} e}\right )}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {a c} e^{2}} \]

[In]

integrate(1/(e*x+d)^2/(c*x^2+a),x, algorithm="giac")

[Out]

-c*d*e*log(c - 2*c*d/(e*x + d) + c*d^2/(e*x + d)^2 + a*e^2/(e*x + d)^2)/(c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4) -
e^3/((c*d^2*e^2 + a*e^4)*(e*x + d)) + (c^2*d^2*e^2 - a*c*e^4)*arctan((c*d - c*d^2/(e*x + d) - a*e^2/(e*x + d))
/(sqrt(a*c)*e))/((c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(a*c)*e^2)

Mupad [B] (verification not implemented)

Time = 10.05 (sec) , antiderivative size = 452, normalized size of antiderivative = 3.67 \[ \int \frac {1}{(d+e x)^2 \left (a+c x^2\right )} \, dx=\frac {\ln \left (a^5\,e^8\,\sqrt {-a\,c}-c^3\,d^8\,{\left (-a\,c\right )}^{3/2}-36\,a^3\,d^2\,e^6\,{\left (-a\,c\right )}^{3/2}+a\,c^5\,d^8\,x+a^5\,c\,e^8\,x+70\,a\,d^4\,e^4\,{\left (-a\,c\right )}^{5/2}+36\,c\,d^6\,e^2\,{\left (-a\,c\right )}^{5/2}+36\,a^2\,c^4\,d^6\,e^2\,x+70\,a^3\,c^3\,d^4\,e^4\,x+36\,a^4\,c^2\,d^2\,e^6\,x\right )\,\left (c\,\left (\frac {d^2\,\sqrt {-a\,c}}{2}-a\,d\,e\right )-\frac {a\,e^2\,\sqrt {-a\,c}}{2}\right )}{a^3\,e^4+2\,a^2\,c\,d^2\,e^2+a\,c^2\,d^4}+\frac {\ln \left (c^3\,d^8\,{\left (-a\,c\right )}^{3/2}-a^5\,e^8\,\sqrt {-a\,c}+36\,a^3\,d^2\,e^6\,{\left (-a\,c\right )}^{3/2}+a\,c^5\,d^8\,x+a^5\,c\,e^8\,x-70\,a\,d^4\,e^4\,{\left (-a\,c\right )}^{5/2}-36\,c\,d^6\,e^2\,{\left (-a\,c\right )}^{5/2}+36\,a^2\,c^4\,d^6\,e^2\,x+70\,a^3\,c^3\,d^4\,e^4\,x+36\,a^4\,c^2\,d^2\,e^6\,x\right )\,\left (a\,\left (\frac {e^2\,\sqrt {-a\,c}}{2}-c\,d\,e\right )-\frac {c\,d^2\,\sqrt {-a\,c}}{2}\right )}{a^3\,e^4+2\,a^2\,c\,d^2\,e^2+a\,c^2\,d^4}-\frac {e}{\left (c\,d^2+a\,e^2\right )\,\left (d+e\,x\right )}+\frac {2\,c\,d\,e\,\ln \left (d+e\,x\right )}{{\left (c\,d^2+a\,e^2\right )}^2} \]

[In]

int(1/((a + c*x^2)*(d + e*x)^2),x)

[Out]

(log(a^5*e^8*(-a*c)^(1/2) - c^3*d^8*(-a*c)^(3/2) - 36*a^3*d^2*e^6*(-a*c)^(3/2) + a*c^5*d^8*x + a^5*c*e^8*x + 7
0*a*d^4*e^4*(-a*c)^(5/2) + 36*c*d^6*e^2*(-a*c)^(5/2) + 36*a^2*c^4*d^6*e^2*x + 70*a^3*c^3*d^4*e^4*x + 36*a^4*c^
2*d^2*e^6*x)*(c*((d^2*(-a*c)^(1/2))/2 - a*d*e) - (a*e^2*(-a*c)^(1/2))/2))/(a^3*e^4 + a*c^2*d^4 + 2*a^2*c*d^2*e
^2) + (log(c^3*d^8*(-a*c)^(3/2) - a^5*e^8*(-a*c)^(1/2) + 36*a^3*d^2*e^6*(-a*c)^(3/2) + a*c^5*d^8*x + a^5*c*e^8
*x - 70*a*d^4*e^4*(-a*c)^(5/2) - 36*c*d^6*e^2*(-a*c)^(5/2) + 36*a^2*c^4*d^6*e^2*x + 70*a^3*c^3*d^4*e^4*x + 36*
a^4*c^2*d^2*e^6*x)*(a*((e^2*(-a*c)^(1/2))/2 - c*d*e) - (c*d^2*(-a*c)^(1/2))/2))/(a^3*e^4 + a*c^2*d^4 + 2*a^2*c
*d^2*e^2) - e/((a*e^2 + c*d^2)*(d + e*x)) + (2*c*d*e*log(d + e*x))/(a*e^2 + c*d^2)^2