\(\int \frac {(c+d x)^2}{x^4 (a+b x^2)^2} \, dx\) [186]

Optimal result

Integrand size = 20, antiderivative size = 148 \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^2} \, dx=-\frac {c^2}{3 a^2 x^3}-\frac {c d}{a^2 x^2}+\frac {2 b c^2-a d^2}{a^3 x}-\frac {b \left (2 a c d-\left (b c^2-a d^2\right ) x\right )}{2 a^3 \left (a+b x^2\right )}+\frac {\sqrt {b} \left (5 b c^2-3 a d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2}}-\frac {4 b c d \log (x)}{a^3}+\frac {2 b c d \log \left (a+b x^2\right )}{a^3} \] Output:

-1/3*c^2/a^2/x^3-c*d/a^2/x^2+(-a*d^2+2*b*c^2)/a^3/x-1/2*b*(2*a*c*d-(-a*d^2 
+b*c^2)*x)/a^3/(b*x^2+a)+1/2*b^(1/2)*(-3*a*d^2+5*b*c^2)*arctan(b^(1/2)*x/a 
^(1/2))/a^(7/2)-4*b*c*d*ln(x)/a^3+2*b*c*d*ln(b*x^2+a)/a^3
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.98 \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^2} \, dx=-\frac {c^2}{3 a^2 x^3}-\frac {c d}{a^2 x^2}+\frac {2 b c^2-a d^2}{a^3 x}+\frac {b \left (b c^2 x-a d (2 c+d x)\right )}{2 a^3 \left (a+b x^2\right )}+\frac {\sqrt {b} \left (5 b c^2-3 a d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2}}-\frac {4 b c d \log (x)}{a^3}+\frac {2 b c d \log \left (a+b x^2\right )}{a^3} \] Input:

Integrate[(c + d*x)^2/(x^4*(a + b*x^2)^2),x]
 

Output:

-1/3*c^2/(a^2*x^3) - (c*d)/(a^2*x^2) + (2*b*c^2 - a*d^2)/(a^3*x) + (b*(b*c 
^2*x - a*d*(2*c + d*x)))/(2*a^3*(a + b*x^2)) + (Sqrt[b]*(5*b*c^2 - 3*a*d^2 
)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7/2)) - (4*b*c*d*Log[x])/a^3 + (2*b*c 
*d*Log[a + b*x^2])/a^3
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {532, 25, 2333, 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.

\(\displaystyle \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^2} \, dx\)

\(\Big \downarrow \) 532

\(\displaystyle -\frac {\int -\frac {\frac {b \left (b c^2-a d^2\right ) x^4}{a^2}-\frac {4 b c d x^3}{a}-2 \left (\frac {b c^2}{a}-d^2\right ) x^2+4 c d x+2 c^2}{x^4 \left (b x^2+a\right )}dx}{2 a}-\frac {b \left (2 a c d-x \left (b c^2-a d^2\right )\right )}{2 a^3 \left (a+b x^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\frac {b \left (b c^2-a d^2\right ) x^4}{a^2}-\frac {4 b c d x^3}{a}-2 \left (\frac {b c^2}{a}-d^2\right ) x^2+4 c d x+2 c^2}{x^4 \left (b x^2+a\right )}dx}{2 a}-\frac {b \left (2 a c d-x \left (b c^2-a d^2\right )\right )}{2 a^3 \left (a+b x^2\right )}\)

\(\Big \downarrow \) 2333

\(\displaystyle \frac {\int \left (\frac {2 c^2}{a x^4}-\frac {8 b d c}{a^2 x}+\frac {4 d c}{a x^3}-\frac {b \left (-5 b c^2-8 b d x c+3 a d^2\right )}{a^2 \left (b x^2+a\right )}+\frac {2 \left (a d^2-2 b c^2\right )}{a^2 x^2}\right )dx}{2 a}-\frac {b \left (2 a c d-x \left (b c^2-a d^2\right )\right )}{2 a^3 \left (a+b x^2\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (5 b c^2-3 a d^2\right )}{a^{5/2}}+\frac {2 \left (2 b c^2-a d^2\right )}{a^2 x}+\frac {4 b c d \log \left (a+b x^2\right )}{a^2}-\frac {8 b c d \log (x)}{a^2}-\frac {2 c^2}{3 a x^3}-\frac {2 c d}{a x^2}}{2 a}-\frac {b \left (2 a c d-x \left (b c^2-a d^2\right )\right )}{2 a^3 \left (a+b x^2\right )}\)

Input:

Int[(c + d*x)^2/(x^4*(a + b*x^2)^2),x]
 

Output:

-1/2*(b*(2*a*c*d - (b*c^2 - a*d^2)*x))/(a^3*(a + b*x^2)) + ((-2*c^2)/(3*a* 
x^3) - (2*c*d)/(a*x^2) + (2*(2*b*c^2 - a*d^2))/(a^2*x) + (Sqrt[b]*(5*b*c^2 
 - 3*a*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/a^(5/2) - (8*b*c*d*Log[x])/a^2 + 
(4*b*c*d*Log[a + b*x^2])/a^2)/(2*a)
 

Defintions of rubi rules used

rule 25

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

rule 532

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[x^m 
*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), 
x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, 
 -1] && IntegerQ[2*p]
 

rule 2009

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

rule 2333

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] 
&& PolyQ[Pq, x] && IGtQ[p, -2]
 

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {c^{2}}{3 a^{2} x^{3}}-\frac {a \,d^{2}-2 b \,c^{2}}{a^{3} x}-\frac {c d}{a^{2} x^{2}}-\frac {4 b c d \ln \left (x \right )}{a^{3}}-\frac {b \left (\frac {\left (\frac {a \,d^{2}}{2}-\frac {b \,c^{2}}{2}\right ) x +a c d}{b \,x^{2}+a}-2 d c \ln \left (b \,x^{2}+a \right )+\frac {\left (3 a \,d^{2}-5 b \,c^{2}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}\)	\(131\)
risch	\(\frac {-\frac {b \left (3 a \,d^{2}-5 b \,c^{2}\right ) x^{4}}{2 a^{3}}-\frac {2 b c d \,x^{3}}{a^{2}}-\frac {\left (3 a \,d^{2}-5 b \,c^{2}\right ) x^{2}}{3 a^{2}}-\frac {c d x}{a}-\frac {c^{2}}{3 a}}{\left (b \,x^{2}+a \right ) x^{3}}+\frac {2 \ln \left (\left (-9 a^{2} b \,d^{4}+30 a \,b^{2} c^{2} d^{2}-25 b^{3} c^{4}+24 \sqrt {-a b \left (3 a \,d^{2}-5 b \,c^{2}\right )^{2}}\, b c d \right ) x +72 a^{2} b c \,d^{3}-120 a \,b^{2} c^{3} d +3 \sqrt {-a b \left (3 a \,d^{2}-5 b \,c^{2}\right )^{2}}\, a \,d^{2}-5 \sqrt {-a b \left (3 a \,d^{2}-5 b \,c^{2}\right )^{2}}\, b \,c^{2}\right ) b c d}{a^{3}}+\frac {\ln \left (\left (-9 a^{2} b \,d^{4}+30 a \,b^{2} c^{2} d^{2}-25 b^{3} c^{4}+24 \sqrt {-a b \left (3 a \,d^{2}-5 b \,c^{2}\right )^{2}}\, b c d \right ) x +72 a^{2} b c \,d^{3}-120 a \,b^{2} c^{3} d +3 \sqrt {-a b \left (3 a \,d^{2}-5 b \,c^{2}\right )^{2}}\, a \,d^{2}-5 \sqrt {-a b \left (3 a \,d^{2}-5 b \,c^{2}\right )^{2}}\, b \,c^{2}\right ) \sqrt {-a b \left (3 a \,d^{2}-5 b \,c^{2}\right )^{2}}}{4 a^{4}}+\frac {2 \ln \left (\left (-9 a^{2} b \,d^{4}+30 a \,b^{2} c^{2} d^{2}-25 b^{3} c^{4}-24 \sqrt {-a b \left (3 a \,d^{2}-5 b \,c^{2}\right )^{2}}\, b c d \right ) x +72 a^{2} b c \,d^{3}-120 a \,b^{2} c^{3} d -3 \sqrt {-a b \left (3 a \,d^{2}-5 b \,c^{2}\right )^{2}}\, a \,d^{2}+5 \sqrt {-a b \left (3 a \,d^{2}-5 b \,c^{2}\right )^{2}}\, b \,c^{2}\right ) b c d}{a^{3}}-\frac {\ln \left (\left (-9 a^{2} b \,d^{4}+30 a \,b^{2} c^{2} d^{2}-25 b^{3} c^{4}-24 \sqrt {-a b \left (3 a \,d^{2}-5 b \,c^{2}\right )^{2}}\, b c d \right ) x +72 a^{2} b c \,d^{3}-120 a \,b^{2} c^{3} d -3 \sqrt {-a b \left (3 a \,d^{2}-5 b \,c^{2}\right )^{2}}\, a \,d^{2}+5 \sqrt {-a b \left (3 a \,d^{2}-5 b \,c^{2}\right )^{2}}\, b \,c^{2}\right ) \sqrt {-a b \left (3 a \,d^{2}-5 b \,c^{2}\right )^{2}}}{4 a^{4}}-\frac {4 b c d \ln \left (x \right )}{a^{3}}\)	\(700\)

Input:

int((d*x+c)^2/x^4/(b*x^2+a)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/3*c^2/a^2/x^3-(a*d^2-2*b*c^2)/a^3/x-c*d/a^2/x^2-4*b*c*d*ln(x)/a^3-b/a^3 
*(((1/2*a*d^2-1/2*b*c^2)*x+a*c*d)/(b*x^2+a)-2*d*c*ln(b*x^2+a)+1/2*(3*a*d^2 
-5*b*c^2)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 420, normalized size of antiderivative = 2.84 \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^2} \, dx=\left [-\frac {24 \, a b c d x^{3} + 12 \, a^{2} c d x - 6 \, {\left (5 \, b^{2} c^{2} - 3 \, a b d^{2}\right )} x^{4} + 4 \, a^{2} c^{2} - 4 \, {\left (5 \, a b c^{2} - 3 \, a^{2} d^{2}\right )} x^{2} + 3 \, {\left ({\left (5 \, b^{2} c^{2} - 3 \, a b d^{2}\right )} x^{5} + {\left (5 \, a b c^{2} - 3 \, a^{2} d^{2}\right )} x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) - 24 \, {\left (b^{2} c d x^{5} + a b c d x^{3}\right )} \log \left (b x^{2} + a\right ) + 48 \, {\left (b^{2} c d x^{5} + a b c d x^{3}\right )} \log \left (x\right )}{12 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}, -\frac {12 \, a b c d x^{3} + 6 \, a^{2} c d x - 3 \, {\left (5 \, b^{2} c^{2} - 3 \, a b d^{2}\right )} x^{4} + 2 \, a^{2} c^{2} - 2 \, {\left (5 \, a b c^{2} - 3 \, a^{2} d^{2}\right )} x^{2} - 3 \, {\left ({\left (5 \, b^{2} c^{2} - 3 \, a b d^{2}\right )} x^{5} + {\left (5 \, a b c^{2} - 3 \, a^{2} d^{2}\right )} x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) - 12 \, {\left (b^{2} c d x^{5} + a b c d x^{3}\right )} \log \left (b x^{2} + a\right ) + 24 \, {\left (b^{2} c d x^{5} + a b c d x^{3}\right )} \log \left (x\right )}{6 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}\right ] \] Input:

integrate((d*x+c)^2/x^4/(b*x^2+a)^2,x, algorithm="fricas")
 

Output:

[-1/12*(24*a*b*c*d*x^3 + 12*a^2*c*d*x - 6*(5*b^2*c^2 - 3*a*b*d^2)*x^4 + 4* 
a^2*c^2 - 4*(5*a*b*c^2 - 3*a^2*d^2)*x^2 + 3*((5*b^2*c^2 - 3*a*b*d^2)*x^5 + 
 (5*a*b*c^2 - 3*a^2*d^2)*x^3)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a 
)/(b*x^2 + a)) - 24*(b^2*c*d*x^5 + a*b*c*d*x^3)*log(b*x^2 + a) + 48*(b^2*c 
*d*x^5 + a*b*c*d*x^3)*log(x))/(a^3*b*x^5 + a^4*x^3), -1/6*(12*a*b*c*d*x^3 
+ 6*a^2*c*d*x - 3*(5*b^2*c^2 - 3*a*b*d^2)*x^4 + 2*a^2*c^2 - 2*(5*a*b*c^2 - 
 3*a^2*d^2)*x^2 - 3*((5*b^2*c^2 - 3*a*b*d^2)*x^5 + (5*a*b*c^2 - 3*a^2*d^2) 
*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)) - 12*(b^2*c*d*x^5 + a*b*c*d*x^3)*log(b 
*x^2 + a) + 24*(b^2*c*d*x^5 + a*b*c*d*x^3)*log(x))/(a^3*b*x^5 + a^4*x^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)**2/x**4/(b*x**2+a)**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.01 \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {2 \, b c d \log \left (b x^{2} + a\right )}{a^{3}} - \frac {4 \, b c d \log \left (x\right )}{a^{3}} - \frac {12 \, a b c d x^{3} + 6 \, a^{2} c d x - 3 \, {\left (5 \, b^{2} c^{2} - 3 \, a b d^{2}\right )} x^{4} + 2 \, a^{2} c^{2} - 2 \, {\left (5 \, a b c^{2} - 3 \, a^{2} d^{2}\right )} x^{2}}{6 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}} + \frac {{\left (5 \, b^{2} c^{2} - 3 \, a b d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} \] Input:

integrate((d*x+c)^2/x^4/(b*x^2+a)^2,x, algorithm="maxima")
 

Output:

2*b*c*d*log(b*x^2 + a)/a^3 - 4*b*c*d*log(x)/a^3 - 1/6*(12*a*b*c*d*x^3 + 6* 
a^2*c*d*x - 3*(5*b^2*c^2 - 3*a*b*d^2)*x^4 + 2*a^2*c^2 - 2*(5*a*b*c^2 - 3*a 
^2*d^2)*x^2)/(a^3*b*x^5 + a^4*x^3) + 1/2*(5*b^2*c^2 - 3*a*b*d^2)*arctan(b* 
x/sqrt(a*b))/(sqrt(a*b)*a^3)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00 \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {2 \, b c d \log \left (b x^{2} + a\right )}{a^{3}} - \frac {4 \, b c d \log \left ({\left | x \right |}\right )}{a^{3}} + \frac {{\left (5 \, b^{2} c^{2} - 3 \, a b d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} - \frac {12 \, a b c d x^{3} + 6 \, a^{2} c d x - 3 \, {\left (5 \, b^{2} c^{2} - 3 \, a b d^{2}\right )} x^{4} + 2 \, a^{2} c^{2} - 2 \, {\left (5 \, a b c^{2} - 3 \, a^{2} d^{2}\right )} x^{2}}{6 \, {\left (b x^{2} + a\right )} a^{3} x^{3}} \] Input:

integrate((d*x+c)^2/x^4/(b*x^2+a)^2,x, algorithm="giac")
 

Output:

2*b*c*d*log(b*x^2 + a)/a^3 - 4*b*c*d*log(abs(x))/a^3 + 1/2*(5*b^2*c^2 - 3* 
a*b*d^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) - 1/6*(12*a*b*c*d*x^3 + 6*a 
^2*c*d*x - 3*(5*b^2*c^2 - 3*a*b*d^2)*x^4 + 2*a^2*c^2 - 2*(5*a*b*c^2 - 3*a^ 
2*d^2)*x^2)/((b*x^2 + a)*a^3*x^3)
 

Mupad [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 328, normalized size of antiderivative = 2.22 \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {\ln \left (3\,a\,d^2\,\sqrt {-a^7\,b}-5\,b\,c^2\,\sqrt {-a^7\,b}-5\,a^3\,b^2\,c^2\,x-24\,a^4\,b\,c\,d+3\,a^4\,b\,d^2\,x+24\,b\,c\,d\,x\,\sqrt {-a^7\,b}\right )\,\left (5\,b\,c^2\,\sqrt {-a^7\,b}-3\,a\,d^2\,\sqrt {-a^7\,b}+8\,a^4\,b\,c\,d\right )}{4\,a^7}-\frac {\frac {c^2}{3\,a}+\frac {x^2\,\left (3\,a\,d^2-5\,b\,c^2\right )}{3\,a^2}+\frac {c\,d\,x}{a}+\frac {b\,x^4\,\left (3\,a\,d^2-5\,b\,c^2\right )}{2\,a^3}+\frac {2\,b\,c\,d\,x^3}{a^2}}{b\,x^5+a\,x^3}+\frac {\ln \left (3\,a\,d^2\,\sqrt {-a^7\,b}-5\,b\,c^2\,\sqrt {-a^7\,b}+5\,a^3\,b^2\,c^2\,x+24\,a^4\,b\,c\,d-3\,a^4\,b\,d^2\,x+24\,b\,c\,d\,x\,\sqrt {-a^7\,b}\right )\,\left (3\,a\,d^2\,\sqrt {-a^7\,b}-5\,b\,c^2\,\sqrt {-a^7\,b}+8\,a^4\,b\,c\,d\right )}{4\,a^7}-\frac {4\,b\,c\,d\,\ln \left (x\right )}{a^3} \] Input:

int((c + d*x)^2/(x^4*(a + b*x^2)^2),x)
 

Output:

(log(3*a*d^2*(-a^7*b)^(1/2) - 5*b*c^2*(-a^7*b)^(1/2) - 5*a^3*b^2*c^2*x - 2 
4*a^4*b*c*d + 3*a^4*b*d^2*x + 24*b*c*d*x*(-a^7*b)^(1/2))*(5*b*c^2*(-a^7*b) 
^(1/2) - 3*a*d^2*(-a^7*b)^(1/2) + 8*a^4*b*c*d))/(4*a^7) - (c^2/(3*a) + (x^ 
2*(3*a*d^2 - 5*b*c^2))/(3*a^2) + (c*d*x)/a + (b*x^4*(3*a*d^2 - 5*b*c^2))/( 
2*a^3) + (2*b*c*d*x^3)/a^2)/(a*x^3 + b*x^5) + (log(3*a*d^2*(-a^7*b)^(1/2) 
- 5*b*c^2*(-a^7*b)^(1/2) + 5*a^3*b^2*c^2*x + 24*a^4*b*c*d - 3*a^4*b*d^2*x 
+ 24*b*c*d*x*(-a^7*b)^(1/2))*(3*a*d^2*(-a^7*b)^(1/2) - 5*b*c^2*(-a^7*b)^(1 
/2) + 8*a^4*b*c*d))/(4*a^7) - (4*b*c*d*log(x))/a^3
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.77 \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {-9 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} d^{2} x^{3}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,c^{2} x^{3}-9 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,d^{2} x^{5}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} c^{2} x^{5}+12 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b c d \,x^{3}+12 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{2} c d \,x^{5}-24 \,\mathrm {log}\left (x \right ) a^{2} b c d \,x^{3}-24 \,\mathrm {log}\left (x \right ) a \,b^{2} c d \,x^{5}-2 a^{3} c^{2}-6 a^{3} c d x -6 a^{3} d^{2} x^{2}+10 a^{2} b \,c^{2} x^{2}-9 a^{2} b \,d^{2} x^{4}+15 a \,b^{2} c^{2} x^{4}+12 a \,b^{2} c d \,x^{5}}{6 a^{4} x^{3} \left (b \,x^{2}+a \right )} \] Input:

int((d*x+c)^2/x^4/(b*x^2+a)^2,x)
 

Output:

( - 9*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*d**2*x**3 + 15*sq 
rt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b*c**2*x**3 - 9*sqrt(b)*sqrt 
(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b*d**2*x**5 + 15*sqrt(b)*sqrt(a)*atan( 
(b*x)/(sqrt(b)*sqrt(a)))*b**2*c**2*x**5 + 12*log(a + b*x**2)*a**2*b*c*d*x* 
*3 + 12*log(a + b*x**2)*a*b**2*c*d*x**5 - 24*log(x)*a**2*b*c*d*x**3 - 24*l 
og(x)*a*b**2*c*d*x**5 - 2*a**3*c**2 - 6*a**3*c*d*x - 6*a**3*d**2*x**2 + 10 
*a**2*b*c**2*x**2 - 9*a**2*b*d**2*x**4 + 15*a*b**2*c**2*x**4 + 12*a*b**2*c 
*d*x**5)/(6*a**4*x**3*(a + b*x**2))