\(\int \frac {x}{(c+d x)^2 (a+b x^2)} \, dx\) [153]

Optimal result

Integrand size = 18, antiderivative size = 133 \[ \int \frac {x}{(c+d x)^2 \left (a+b x^2\right )} \, dx=\frac {c}{\left (b c^2+a d^2\right ) (c+d x)}+\frac {2 \sqrt {a} \sqrt {b} c d \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\left (b c^2+a d^2\right )^2}-\frac {\left (b c^2-a d^2\right ) \log (c+d x)}{\left (b c^2+a d^2\right )^2}+\frac {\left (b c^2-a d^2\right ) \log \left (a+b x^2\right )}{2 \left (b c^2+a d^2\right )^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.80 \[ \int \frac {x}{(c+d x)^2 \left (a+b x^2\right )} \, dx=\frac {\frac {2 c \left (b c^2+a d^2\right )}{c+d x}+4 \sqrt {a} \sqrt {b} c d \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )+2 \left (-b c^2+a d^2\right ) \log (c+d x)+\left (b c^2-a d^2\right ) \log \left (a+b x^2\right )}{2 \left (b c^2+a d^2\right )^2} \] Input:

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

Output:

((2*c*(b*c^2 + a*d^2))/(c + d*x) + 4*Sqrt[a]*Sqrt[b]*c*d*ArcTan[(Sqrt[b]*x 
)/Sqrt[a]] + 2*(-(b*c^2) + a*d^2)*Log[c + d*x] + (b*c^2 - a*d^2)*Log[a + b 
*x^2])/(2*(b*c^2 + a*d^2)^2)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {594, 25, 657, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x}{\left (a+b x^2\right ) (c+d x)^2} \, dx\)

\(\Big \downarrow \) 594

\(\displaystyle \frac {c}{(c+d x) \left (a d^2+b c^2\right )}-\frac {\int -\frac {a d+b c x}{(c+d x) \left (b x^2+a\right )}dx}{a d^2+b c^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {a d+b c x}{(c+d x) \left (b x^2+a\right )}dx}{a d^2+b c^2}+\frac {c}{(c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 657

\(\displaystyle \frac {\int \left (\frac {d \left (a d^2-b c^2\right )}{\left (b c^2+a d^2\right ) (c+d x)}+\frac {b \left (2 a c d+\left (b c^2-a d^2\right ) x\right )}{\left (b c^2+a d^2\right ) \left (b x^2+a\right )}\right )dx}{a d^2+b c^2}+\frac {c}{(c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {2 \sqrt {a} \sqrt {b} c d \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a d^2+b c^2}+\frac {\left (b c^2-a d^2\right ) \log \left (a+b x^2\right )}{2 \left (a d^2+b c^2\right )}-\frac {\left (b c^2-a d^2\right ) \log (c+d x)}{a d^2+b c^2}}{a d^2+b c^2}+\frac {c}{(c+d x) \left (a d^2+b c^2\right )}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 594

Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))) 
, x] + Simp[1/((n + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^(n + 1)*(a + b*x^2) 
^p*(a*d*(n + 1) + b*c*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] 
 && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 657

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {b \left (\frac {\left (-a \,d^{2}+b \,c^{2}\right ) \ln \left (b \,x^{2}+a \right )}{2 b}+\frac {2 a c d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{\left (a \,d^{2}+b \,c^{2}\right )^{2}}+\frac {\left (a \,d^{2}-b \,c^{2}\right ) \ln \left (d x +c \right )}{\left (a \,d^{2}+b \,c^{2}\right )^{2}}+\frac {c}{\left (a \,d^{2}+b \,c^{2}\right ) \left (d x +c \right )}\)	\(116\)
risch	\(\frac {c}{\left (a \,d^{2}+b \,c^{2}\right ) \left (d x +c \right )}+\frac {\ln \left (d x +c \right ) a \,d^{2}}{a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}}-\frac {\ln \left (d x +c \right ) b \,c^{2}}{a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right ) \textit {\_Z}^{2}+\left (2 a \,d^{2}-2 b \,c^{2}\right ) \textit {\_Z} \right )}{\sum }\textit {\_R} \ln \left (\left (\left (3 a^{3} d^{6}+5 a^{2} b \,c^{2} d^{4}+a \,b^{2} c^{4} d^{2}-b^{3} c^{6}\right ) \textit {\_R}^{2}+\left (3 a^{2} d^{4}+2 b \,c^{2} d^{2} a -b^{2} c^{4}\right ) \textit {\_R} +2 b \,c^{2}\right ) x +\left (4 a^{3} c \,d^{5}+8 a^{2} b \,c^{3} d^{3}+4 a \,b^{2} c^{5} d \right ) \textit {\_R}^{2}+2 a c d \right )\right )}{2}\)	\(280\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 370, normalized size of antiderivative = 2.78 \[ \int \frac {x}{(c+d x)^2 \left (a+b x^2\right )} \, dx=\left [\frac {2 \, b c^{3} + 2 \, a c d^{2} + 2 \, {\left (c d^{2} x + c^{2} d\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + {\left (b c^{3} - a c d^{2} + {\left (b c^{2} d - a d^{3}\right )} x\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (b c^{3} - a c d^{2} + {\left (b c^{2} d - a d^{3}\right )} x\right )} \log \left (d x + c\right )}{2 \, {\left (b^{2} c^{5} + 2 \, a b c^{3} d^{2} + a^{2} c d^{4} + {\left (b^{2} c^{4} d + 2 \, a b c^{2} d^{3} + a^{2} d^{5}\right )} x\right )}}, \frac {2 \, b c^{3} + 2 \, a c d^{2} + 4 \, {\left (c d^{2} x + c^{2} d\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (b c^{3} - a c d^{2} + {\left (b c^{2} d - a d^{3}\right )} x\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (b c^{3} - a c d^{2} + {\left (b c^{2} d - a d^{3}\right )} x\right )} \log \left (d x + c\right )}{2 \, {\left (b^{2} c^{5} + 2 \, a b c^{3} d^{2} + a^{2} c d^{4} + {\left (b^{2} c^{4} d + 2 \, a b c^{2} d^{3} + a^{2} d^{5}\right )} x\right )}}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.31 \[ \int \frac {x}{(c+d x)^2 \left (a+b x^2\right )} \, dx=\frac {2 \, a b c d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}\right )} \sqrt {a b}} + \frac {{\left (b c^{2} - a d^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}\right )}} - \frac {{\left (b c^{2} - a d^{2}\right )} \log \left (d x + c\right )}{b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}} + \frac {c}{b c^{3} + a c d^{2} + {\left (b c^{2} d + a d^{3}\right )} x} \] Input:

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

Output:

2*a*b*c*d*arctan(b*x/sqrt(a*b))/((b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4)*sqrt( 
a*b)) + 1/2*(b*c^2 - a*d^2)*log(b*x^2 + a)/(b^2*c^4 + 2*a*b*c^2*d^2 + a^2* 
d^4) - (b*c^2 - a*d^2)*log(d*x + c)/(b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4) + 
c/(b*c^3 + a*c*d^2 + (b*c^2*d + a*d^3)*x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.46 \[ \int \frac {x}{(c+d x)^2 \left (a+b x^2\right )} \, dx=\frac {\frac {4 \, a b c d^{2} \arctan \left (\frac {b c - \frac {b c^{2}}{d x + c} - \frac {a d^{2}}{d x + c}}{\sqrt {a b} d}\right )}{{\left (b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}\right )} \sqrt {a b}} + \frac {2 \, c d^{3}}{{\left (b c^{2} d^{2} + a d^{4}\right )} {\left (d x + c\right )}} + \frac {{\left (b c^{2} d - a d^{3}\right )} \log \left (b - \frac {2 \, b c}{d x + c} + \frac {b c^{2}}{{\left (d x + c\right )}^{2}} + \frac {a d^{2}}{{\left (d x + c\right )}^{2}}\right )}{b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}}}{2 \, d} \] Input:

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

Output:

1/2*(4*a*b*c*d^2*arctan((b*c - b*c^2/(d*x + c) - a*d^2/(d*x + c))/(sqrt(a* 
b)*d))/((b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4)*sqrt(a*b)) + 2*c*d^3/((b*c^2*d 
^2 + a*d^4)*(d*x + c)) + (b*c^2*d - a*d^3)*log(b - 2*b*c/(d*x + c) + b*c^2 
/(d*x + c)^2 + a*d^2/(d*x + c)^2)/(b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4))/d
 

Mupad [B] (verification not implemented)

Time = 8.43 (sec) , antiderivative size = 454, normalized size of antiderivative = 3.41 \[ \int \frac {x}{(c+d x)^2 \left (a+b x^2\right )} \, dx=\frac {\ln \left (9\,b^5\,c^8\,x+9\,a^4\,d^8\,\sqrt {-a\,b}+9\,b^4\,c^8\,\sqrt {-a\,b}-10\,c^4\,d^4\,{\left (-a\,b\right )}^{5/2}-4\,a^2\,c^2\,d^6\,{\left (-a\,b\right )}^{3/2}-4\,b^2\,c^6\,d^2\,{\left (-a\,b\right )}^{3/2}+9\,a^4\,b\,d^8\,x+4\,a\,b^4\,c^6\,d^2\,x-10\,a^2\,b^3\,c^4\,d^4\,x+4\,a^3\,b^2\,c^2\,d^6\,x\right )\,\left (b\,c^2-a\,d^2+2\,c\,d\,\sqrt {-a\,b}\right )}{2\,\left (a^2\,d^4+2\,a\,b\,c^2\,d^2+b^2\,c^4\right )}-\frac {\ln \left (9\,b^5\,c^8\,x-9\,a^4\,d^8\,\sqrt {-a\,b}-9\,b^4\,c^8\,\sqrt {-a\,b}+10\,c^4\,d^4\,{\left (-a\,b\right )}^{5/2}+4\,a^2\,c^2\,d^6\,{\left (-a\,b\right )}^{3/2}+4\,b^2\,c^6\,d^2\,{\left (-a\,b\right )}^{3/2}+9\,a^4\,b\,d^8\,x+4\,a\,b^4\,c^6\,d^2\,x-10\,a^2\,b^3\,c^4\,d^4\,x+4\,a^3\,b^2\,c^2\,d^6\,x\right )\,\left (a\,d^2-b\,c^2+2\,c\,d\,\sqrt {-a\,b}\right )}{2\,\left (a^2\,d^4+2\,a\,b\,c^2\,d^2+b^2\,c^4\right )}+\frac {\ln \left (c+d\,x\right )\,\left (a\,d^2-b\,c^2\right )}{a^2\,d^4+2\,a\,b\,c^2\,d^2+b^2\,c^4}+\frac {c}{\left (b\,c^2+a\,d^2\right )\,\left (c+d\,x\right )} \] Input:

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

Output:

(log(9*b^5*c^8*x + 9*a^4*d^8*(-a*b)^(1/2) + 9*b^4*c^8*(-a*b)^(1/2) - 10*c^ 
4*d^4*(-a*b)^(5/2) - 4*a^2*c^2*d^6*(-a*b)^(3/2) - 4*b^2*c^6*d^2*(-a*b)^(3/ 
2) + 9*a^4*b*d^8*x + 4*a*b^4*c^6*d^2*x - 10*a^2*b^3*c^4*d^4*x + 4*a^3*b^2* 
c^2*d^6*x)*(b*c^2 - a*d^2 + 2*c*d*(-a*b)^(1/2)))/(2*(a^2*d^4 + b^2*c^4 + 2 
*a*b*c^2*d^2)) - (log(9*b^5*c^8*x - 9*a^4*d^8*(-a*b)^(1/2) - 9*b^4*c^8*(-a 
*b)^(1/2) + 10*c^4*d^4*(-a*b)^(5/2) + 4*a^2*c^2*d^6*(-a*b)^(3/2) + 4*b^2*c 
^6*d^2*(-a*b)^(3/2) + 9*a^4*b*d^8*x + 4*a*b^4*c^6*d^2*x - 10*a^2*b^3*c^4*d 
^4*x + 4*a^3*b^2*c^2*d^6*x)*(a*d^2 - b*c^2 + 2*c*d*(-a*b)^(1/2)))/(2*(a^2* 
d^4 + b^2*c^4 + 2*a*b*c^2*d^2)) + (log(c + d*x)*(a*d^2 - b*c^2))/(a^2*d^4 
+ b^2*c^4 + 2*a*b*c^2*d^2) + c/((a*d^2 + b*c^2)*(c + d*x))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.74 \[ \int \frac {x}{(c+d x)^2 \left (a+b x^2\right )} \, dx=\frac {4 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) c^{2} d +4 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) c \,d^{2} x -\mathrm {log}\left (b \,x^{2}+a \right ) a c \,d^{2}-\mathrm {log}\left (b \,x^{2}+a \right ) a \,d^{3} x +\mathrm {log}\left (b \,x^{2}+a \right ) b \,c^{3}+\mathrm {log}\left (b \,x^{2}+a \right ) b \,c^{2} d x +2 \,\mathrm {log}\left (d x +c \right ) a c \,d^{2}+2 \,\mathrm {log}\left (d x +c \right ) a \,d^{3} x -2 \,\mathrm {log}\left (d x +c \right ) b \,c^{3}-2 \,\mathrm {log}\left (d x +c \right ) b \,c^{2} d x -2 a \,d^{3} x -2 b \,c^{2} d x}{2 a^{2} d^{5} x +4 a b \,c^{2} d^{3} x +2 b^{2} c^{4} d x +2 a^{2} c \,d^{4}+4 a b \,c^{3} d^{2}+2 b^{2} c^{5}} \] Input:

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

Output:

(4*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*c**2*d + 4*sqrt(b)*sqrt(a 
)*atan((b*x)/(sqrt(b)*sqrt(a)))*c*d**2*x - log(a + b*x**2)*a*c*d**2 - log( 
a + b*x**2)*a*d**3*x + log(a + b*x**2)*b*c**3 + log(a + b*x**2)*b*c**2*d*x 
 + 2*log(c + d*x)*a*c*d**2 + 2*log(c + d*x)*a*d**3*x - 2*log(c + d*x)*b*c* 
*3 - 2*log(c + d*x)*b*c**2*d*x - 2*a*d**3*x - 2*b*c**2*d*x)/(2*(a**2*c*d** 
4 + a**2*d**5*x + 2*a*b*c**3*d**2 + 2*a*b*c**2*d**3*x + b**2*c**5 + b**2*c 
**4*d*x))