\(\int \frac {(c+d x)^2}{x^4 (a+b x^2)^3} \, dx\) [239]

Optimal result

Integrand size = 20, antiderivative size = 187 \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^3} \, dx=-\frac {c^2}{3 a^3 x^3}-\frac {c d}{a^3 x^2}+\frac {3 b c^2-a d^2}{a^4 x}-\frac {b \left (2 a c d-\left (b c^2-a d^2\right ) x\right )}{4 a^3 \left (a+b x^2\right )^2}-\frac {b \left (16 a c d-\left (11 b c^2-7 a d^2\right ) x\right )}{8 a^4 \left (a+b x^2\right )}+\frac {5 \sqrt {b} \left (7 b c^2-3 a d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2}}-\frac {6 b c d \log (x)}{a^4}+\frac {3 b c d \log \left (a+b x^2\right )}{a^4} \] Output:

-1/3*c^2/a^3/x^3-c*d/a^3/x^2+(-a*d^2+3*b*c^2)/a^4/x-1/4*b*(2*a*c*d-(-a*d^2 
+b*c^2)*x)/a^3/(b*x^2+a)^2-1/8*b*(16*a*c*d-(-7*a*d^2+11*b*c^2)*x)/a^4/(b*x 
^2+a)+5/8*b^(1/2)*(-3*a*d^2+7*b*c^2)*arctan(b^(1/2)*x/a^(1/2))/a^(9/2)-6*b 
*c*d*ln(x)/a^4+3*b*c*d*ln(b*x^2+a)/a^4
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.97 \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^3} \, dx=-\frac {c^2}{3 a^3 x^3}-\frac {c d}{a^3 x^2}+\frac {3 b c^2-a d^2}{a^4 x}+\frac {b \left (b c^2 x-a d (2 c+d x)\right )}{4 a^3 \left (a+b x^2\right )^2}+\frac {b \left (11 b c^2 x-a d (16 c+7 d x)\right )}{8 a^4 \left (a+b x^2\right )}+\frac {5 \sqrt {b} \left (7 b c^2-3 a d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2}}-\frac {6 b c d \log (x)}{a^4}+\frac {3 b c d \log \left (a+b x^2\right )}{a^4} \] Input:

Integrate[(c + d*x)^2/(x^4*(a + b*x^2)^3),x]
 

Output:

-1/3*c^2/(a^3*x^3) - (c*d)/(a^3*x^2) + (3*b*c^2 - a*d^2)/(a^4*x) + (b*(b*c 
^2*x - a*d*(2*c + d*x)))/(4*a^3*(a + b*x^2)^2) + (b*(11*b*c^2*x - a*d*(16* 
c + 7*d*x)))/(8*a^4*(a + b*x^2)) + (5*Sqrt[b]*(7*b*c^2 - 3*a*d^2)*ArcTan[( 
Sqrt[b]*x)/Sqrt[a]])/(8*a^(9/2)) - (6*b*c*d*Log[x])/a^4 + (3*b*c*d*Log[a + 
 b*x^2])/a^4
 

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {532, 25, 2336, 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.

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

\(\Big \downarrow \) 532

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2336

\(\displaystyle \frac {-\frac {\int -\frac {\frac {b \left (11 b c^2-7 a d^2\right ) x^4}{a^2}-\frac {32 b c d x^3}{a}-8 \left (\frac {2 b c^2}{a}-d^2\right ) x^2+16 c d x+8 c^2}{x^4 \left (b x^2+a\right )}dx}{2 a}-\frac {b \left (16 a c d-x \left (11 b c^2-7 a d^2\right )\right )}{2 a^3 \left (a+b x^2\right )}}{4 a}-\frac {b \left (2 a c d-x \left (b c^2-a d^2\right )\right )}{4 a^3 \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\frac {b \left (11 b c^2-7 a d^2\right ) x^4}{a^2}-\frac {32 b c d x^3}{a}-8 \left (\frac {2 b c^2}{a}-d^2\right ) x^2+16 c d x+8 c^2}{x^4 \left (b x^2+a\right )}dx}{2 a}-\frac {b \left (16 a c d-x \left (11 b c^2-7 a d^2\right )\right )}{2 a^3 \left (a+b x^2\right )}}{4 a}-\frac {b \left (2 a c d-x \left (b c^2-a d^2\right )\right )}{4 a^3 \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 2333

\(\displaystyle \frac {\frac {\int \left (\frac {8 c^2}{a x^4}-\frac {48 b d c}{a^2 x}+\frac {16 d c}{a x^3}-\frac {b \left (-35 b c^2-48 b d x c+15 a d^2\right )}{a^2 \left (b x^2+a\right )}+\frac {8 \left (a d^2-3 b c^2\right )}{a^2 x^2}\right )dx}{2 a}-\frac {b \left (16 a c d-x \left (11 b c^2-7 a d^2\right )\right )}{2 a^3 \left (a+b x^2\right )}}{4 a}-\frac {b \left (2 a c d-x \left (b c^2-a d^2\right )\right )}{4 a^3 \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\frac {5 \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (7 b c^2-3 a d^2\right )}{a^{5/2}}+\frac {8 \left (3 b c^2-a d^2\right )}{a^2 x}+\frac {24 b c d \log \left (a+b x^2\right )}{a^2}-\frac {48 b c d \log (x)}{a^2}-\frac {8 c^2}{3 a x^3}-\frac {8 c d}{a x^2}}{2 a}-\frac {b \left (16 a c d-x \left (11 b c^2-7 a d^2\right )\right )}{2 a^3 \left (a+b x^2\right )}}{4 a}-\frac {b \left (2 a c d-x \left (b c^2-a d^2\right )\right )}{4 a^3 \left (a+b x^2\right )^2}\)

Input:

Int[(c + d*x)^2/(x^4*(a + b*x^2)^3),x]
 

Output:

-1/4*(b*(2*a*c*d - (b*c^2 - a*d^2)*x))/(a^3*(a + b*x^2)^2) + (-1/2*(b*(16* 
a*c*d - (11*b*c^2 - 7*a*d^2)*x))/(a^3*(a + b*x^2)) + ((-8*c^2)/(3*a*x^3) - 
 (8*c*d)/(a*x^2) + (8*(3*b*c^2 - a*d^2))/(a^2*x) + (5*Sqrt[b]*(7*b*c^2 - 3 
*a*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/a^(5/2) - (48*b*c*d*Log[x])/a^2 + (24 
*b*c*d*Log[a + b*x^2])/a^2)/(2*a))/(4*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]
 

rule 2336

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema 
inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) 
^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* 
b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex 
pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F 
reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.88

method	result	size
default	\(-\frac {c^{2}}{3 a^{3} x^{3}}-\frac {a \,d^{2}-3 b \,c^{2}}{a^{4} x}-\frac {c d}{a^{3} x^{2}}-\frac {6 b c d \ln \left (x \right )}{a^{4}}-\frac {b \left (\frac {\left (\frac {7}{8} a b \,d^{2}-\frac {11}{8} b^{2} c^{2}\right ) x^{3}+2 a b c d \,x^{2}+\frac {a \left (9 a \,d^{2}-13 b \,c^{2}\right ) x}{8}+\frac {5 a^{2} c d}{2}}{\left (b \,x^{2}+a \right )^{2}}-3 d c \ln \left (b \,x^{2}+a \right )+\frac {\left (15 a \,d^{2}-35 b \,c^{2}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{4}}\)	\(165\)
risch	\(\frac {-\frac {5 b^{2} \left (3 a \,d^{2}-7 b \,c^{2}\right ) x^{6}}{8 a^{4}}-\frac {3 c d \,b^{2} x^{5}}{a^{3}}-\frac {25 b \left (3 a \,d^{2}-7 b \,c^{2}\right ) x^{4}}{24 a^{3}}-\frac {9 b c d \,x^{3}}{2 a^{2}}-\frac {\left (3 a \,d^{2}-7 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 )^{2} x^{3}}+\frac {3 \ln \left (\left (-45 a^{2} b \,d^{4}+210 a \,b^{2} c^{2} d^{2}-245 b^{3} c^{4}+144 \sqrt {-a b \left (3 a \,d^{2}-7 b \,c^{2}\right )^{2}}\, b c d \right ) x +432 a^{2} b c \,d^{3}-1008 a \,b^{2} c^{3} d +15 \sqrt {-a b \left (3 a \,d^{2}-7 b \,c^{2}\right )^{2}}\, a \,d^{2}-35 \sqrt {-a b \left (3 a \,d^{2}-7 b \,c^{2}\right )^{2}}\, b \,c^{2}\right ) b c d}{a^{4}}+\frac {5 \ln \left (\left (-45 a^{2} b \,d^{4}+210 a \,b^{2} c^{2} d^{2}-245 b^{3} c^{4}+144 \sqrt {-a b \left (3 a \,d^{2}-7 b \,c^{2}\right )^{2}}\, b c d \right ) x +432 a^{2} b c \,d^{3}-1008 a \,b^{2} c^{3} d +15 \sqrt {-a b \left (3 a \,d^{2}-7 b \,c^{2}\right )^{2}}\, a \,d^{2}-35 \sqrt {-a b \left (3 a \,d^{2}-7 b \,c^{2}\right )^{2}}\, b \,c^{2}\right ) \sqrt {-a b \left (3 a \,d^{2}-7 b \,c^{2}\right )^{2}}}{16 a^{5}}+\frac {3 \ln \left (\left (-45 a^{2} b \,d^{4}+210 a \,b^{2} c^{2} d^{2}-245 b^{3} c^{4}-144 \sqrt {-a b \left (3 a \,d^{2}-7 b \,c^{2}\right )^{2}}\, b c d \right ) x +432 a^{2} b c \,d^{3}-1008 a \,b^{2} c^{3} d -15 \sqrt {-a b \left (3 a \,d^{2}-7 b \,c^{2}\right )^{2}}\, a \,d^{2}+35 \sqrt {-a b \left (3 a \,d^{2}-7 b \,c^{2}\right )^{2}}\, b \,c^{2}\right ) b c d}{a^{4}}-\frac {5 \ln \left (\left (-45 a^{2} b \,d^{4}+210 a \,b^{2} c^{2} d^{2}-245 b^{3} c^{4}-144 \sqrt {-a b \left (3 a \,d^{2}-7 b \,c^{2}\right )^{2}}\, b c d \right ) x +432 a^{2} b c \,d^{3}-1008 a \,b^{2} c^{3} d -15 \sqrt {-a b \left (3 a \,d^{2}-7 b \,c^{2}\right )^{2}}\, a \,d^{2}+35 \sqrt {-a b \left (3 a \,d^{2}-7 b \,c^{2}\right )^{2}}\, b \,c^{2}\right ) \sqrt {-a b \left (3 a \,d^{2}-7 b \,c^{2}\right )^{2}}}{16 a^{5}}-\frac {6 b c d \ln \left (x \right )}{a^{4}}\)	\(737\)

Input:

int((d*x+c)^2/x^4/(b*x^2+a)^3,x,method=_RETURNVERBOSE)
 

Output:

-1/3*c^2/a^3/x^3-(a*d^2-3*b*c^2)/a^4/x-c*d/a^3/x^2-6*b*c*d*ln(x)/a^4-b/a^4 
*(((7/8*a*b*d^2-11/8*b^2*c^2)*x^3+2*a*b*c*d*x^2+1/8*a*(9*a*d^2-13*b*c^2)*x 
+5/2*a^2*c*d)/(b*x^2+a)^2-3*d*c*ln(b*x^2+a)+1/8*(15*a*d^2-35*b*c^2)/(a*b)^ 
(1/2)*arctan(b*x/(a*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 632, normalized size of antiderivative = 3.38 \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^3} \, dx=\left [-\frac {144 \, a b^{2} c d x^{5} + 216 \, a^{2} b c d x^{3} - 30 \, {\left (7 \, b^{3} c^{2} - 3 \, a b^{2} d^{2}\right )} x^{6} + 48 \, a^{3} c d x + 16 \, a^{3} c^{2} - 50 \, {\left (7 \, a b^{2} c^{2} - 3 \, a^{2} b d^{2}\right )} x^{4} - 16 \, {\left (7 \, a^{2} b c^{2} - 3 \, a^{3} d^{2}\right )} x^{2} + 15 \, {\left ({\left (7 \, b^{3} c^{2} - 3 \, a b^{2} d^{2}\right )} x^{7} + 2 \, {\left (7 \, a b^{2} c^{2} - 3 \, a^{2} b d^{2}\right )} x^{5} + {\left (7 \, a^{2} b c^{2} - 3 \, a^{3} d^{2}\right )} x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) - 144 \, {\left (b^{3} c d x^{7} + 2 \, a b^{2} c d x^{5} + a^{2} b c d x^{3}\right )} \log \left (b x^{2} + a\right ) + 288 \, {\left (b^{3} c d x^{7} + 2 \, a b^{2} c d x^{5} + a^{2} b c d x^{3}\right )} \log \left (x\right )}{48 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}}, -\frac {72 \, a b^{2} c d x^{5} + 108 \, a^{2} b c d x^{3} - 15 \, {\left (7 \, b^{3} c^{2} - 3 \, a b^{2} d^{2}\right )} x^{6} + 24 \, a^{3} c d x + 8 \, a^{3} c^{2} - 25 \, {\left (7 \, a b^{2} c^{2} - 3 \, a^{2} b d^{2}\right )} x^{4} - 8 \, {\left (7 \, a^{2} b c^{2} - 3 \, a^{3} d^{2}\right )} x^{2} - 15 \, {\left ({\left (7 \, b^{3} c^{2} - 3 \, a b^{2} d^{2}\right )} x^{7} + 2 \, {\left (7 \, a b^{2} c^{2} - 3 \, a^{2} b d^{2}\right )} x^{5} + {\left (7 \, a^{2} b c^{2} - 3 \, a^{3} d^{2}\right )} x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) - 72 \, {\left (b^{3} c d x^{7} + 2 \, a b^{2} c d x^{5} + a^{2} b c d x^{3}\right )} \log \left (b x^{2} + a\right ) + 144 \, {\left (b^{3} c d x^{7} + 2 \, a b^{2} c d x^{5} + a^{2} b c d x^{3}\right )} \log \left (x\right )}{24 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}}\right ] \] Input:

integrate((d*x+c)^2/x^4/(b*x^2+a)^3,x, algorithm="fricas")
 

Output:

[-1/48*(144*a*b^2*c*d*x^5 + 216*a^2*b*c*d*x^3 - 30*(7*b^3*c^2 - 3*a*b^2*d^ 
2)*x^6 + 48*a^3*c*d*x + 16*a^3*c^2 - 50*(7*a*b^2*c^2 - 3*a^2*b*d^2)*x^4 - 
16*(7*a^2*b*c^2 - 3*a^3*d^2)*x^2 + 15*((7*b^3*c^2 - 3*a*b^2*d^2)*x^7 + 2*( 
7*a*b^2*c^2 - 3*a^2*b*d^2)*x^5 + (7*a^2*b*c^2 - 3*a^3*d^2)*x^3)*sqrt(-b/a) 
*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) - 144*(b^3*c*d*x^7 + 2*a* 
b^2*c*d*x^5 + a^2*b*c*d*x^3)*log(b*x^2 + a) + 288*(b^3*c*d*x^7 + 2*a*b^2*c 
*d*x^5 + a^2*b*c*d*x^3)*log(x))/(a^4*b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3), -1/ 
24*(72*a*b^2*c*d*x^5 + 108*a^2*b*c*d*x^3 - 15*(7*b^3*c^2 - 3*a*b^2*d^2)*x^ 
6 + 24*a^3*c*d*x + 8*a^3*c^2 - 25*(7*a*b^2*c^2 - 3*a^2*b*d^2)*x^4 - 8*(7*a 
^2*b*c^2 - 3*a^3*d^2)*x^2 - 15*((7*b^3*c^2 - 3*a*b^2*d^2)*x^7 + 2*(7*a*b^2 
*c^2 - 3*a^2*b*d^2)*x^5 + (7*a^2*b*c^2 - 3*a^3*d^2)*x^3)*sqrt(b/a)*arctan( 
x*sqrt(b/a)) - 72*(b^3*c*d*x^7 + 2*a*b^2*c*d*x^5 + a^2*b*c*d*x^3)*log(b*x^ 
2 + a) + 144*(b^3*c*d*x^7 + 2*a*b^2*c*d*x^5 + a^2*b*c*d*x^3)*log(x))/(a^4* 
b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)**2/x**4/(b*x**2+a)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.08 \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^3} \, dx=-\frac {72 \, a b^{2} c d x^{5} + 108 \, a^{2} b c d x^{3} - 15 \, {\left (7 \, b^{3} c^{2} - 3 \, a b^{2} d^{2}\right )} x^{6} + 24 \, a^{3} c d x + 8 \, a^{3} c^{2} - 25 \, {\left (7 \, a b^{2} c^{2} - 3 \, a^{2} b d^{2}\right )} x^{4} - 8 \, {\left (7 \, a^{2} b c^{2} - 3 \, a^{3} d^{2}\right )} x^{2}}{24 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}} + \frac {3 \, b c d \log \left (b x^{2} + a\right )}{a^{4}} - \frac {6 \, b c d \log \left (x\right )}{a^{4}} + \frac {5 \, {\left (7 \, b^{2} c^{2} - 3 \, a b d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} \] Input:

integrate((d*x+c)^2/x^4/(b*x^2+a)^3,x, algorithm="maxima")
 

Output:

-1/24*(72*a*b^2*c*d*x^5 + 108*a^2*b*c*d*x^3 - 15*(7*b^3*c^2 - 3*a*b^2*d^2) 
*x^6 + 24*a^3*c*d*x + 8*a^3*c^2 - 25*(7*a*b^2*c^2 - 3*a^2*b*d^2)*x^4 - 8*( 
7*a^2*b*c^2 - 3*a^3*d^2)*x^2)/(a^4*b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3) + 3*b* 
c*d*log(b*x^2 + a)/a^4 - 6*b*c*d*log(x)/a^4 + 5/8*(7*b^2*c^2 - 3*a*b*d^2)* 
arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^4)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.01 \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^3} \, dx=\frac {3 \, b c d \log \left (b x^{2} + a\right )}{a^{4}} - \frac {6 \, b c d \log \left ({\left | x \right |}\right )}{a^{4}} + \frac {5 \, {\left (7 \, b^{2} c^{2} - 3 \, a b d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} - \frac {72 \, a b^{2} c d x^{5} + 108 \, a^{2} b c d x^{3} - 15 \, {\left (7 \, b^{3} c^{2} - 3 \, a b^{2} d^{2}\right )} x^{6} + 24 \, a^{3} c d x + 8 \, a^{3} c^{2} - 25 \, {\left (7 \, a b^{2} c^{2} - 3 \, a^{2} b d^{2}\right )} x^{4} - 8 \, {\left (7 \, a^{2} b c^{2} - 3 \, a^{3} d^{2}\right )} x^{2}}{24 \, {\left (b x^{2} + a\right )}^{2} a^{4} x^{3}} \] Input:

integrate((d*x+c)^2/x^4/(b*x^2+a)^3,x, algorithm="giac")
 

Output:

3*b*c*d*log(b*x^2 + a)/a^4 - 6*b*c*d*log(abs(x))/a^4 + 5/8*(7*b^2*c^2 - 3* 
a*b*d^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^4) - 1/24*(72*a*b^2*c*d*x^5 + 
108*a^2*b*c*d*x^3 - 15*(7*b^3*c^2 - 3*a*b^2*d^2)*x^6 + 24*a^3*c*d*x + 8*a^ 
3*c^2 - 25*(7*a*b^2*c^2 - 3*a^2*b*d^2)*x^4 - 8*(7*a^2*b*c^2 - 3*a^3*d^2)*x 
^2)/((b*x^2 + a)^2*a^4*x^3)
 

Mupad [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.01 \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^3} \, dx=\frac {\ln \left (15\,a\,d^2\,\sqrt {-a^9\,b}-35\,b\,c^2\,\sqrt {-a^9\,b}-35\,a^4\,b^2\,c^2\,x-144\,a^5\,b\,c\,d+15\,a^5\,b\,d^2\,x+144\,b\,c\,d\,x\,\sqrt {-a^9\,b}\right )\,\left (35\,b\,c^2\,\sqrt {-a^9\,b}-15\,a\,d^2\,\sqrt {-a^9\,b}+48\,a^5\,b\,c\,d\right )}{16\,a^9}-\frac {\frac {c^2}{3\,a}+\frac {x^2\,\left (3\,a\,d^2-7\,b\,c^2\right )}{3\,a^2}+\frac {5\,b^2\,x^6\,\left (3\,a\,d^2-7\,b\,c^2\right )}{8\,a^4}+\frac {c\,d\,x}{a}+\frac {25\,b\,x^4\,\left (3\,a\,d^2-7\,b\,c^2\right )}{24\,a^3}+\frac {9\,b\,c\,d\,x^3}{2\,a^2}+\frac {3\,b^2\,c\,d\,x^5}{a^3}}{a^2\,x^3+2\,a\,b\,x^5+b^2\,x^7}+\frac {\ln \left (15\,a\,d^2\,\sqrt {-a^9\,b}-35\,b\,c^2\,\sqrt {-a^9\,b}+35\,a^4\,b^2\,c^2\,x+144\,a^5\,b\,c\,d-15\,a^5\,b\,d^2\,x+144\,b\,c\,d\,x\,\sqrt {-a^9\,b}\right )\,\left (15\,a\,d^2\,\sqrt {-a^9\,b}-35\,b\,c^2\,\sqrt {-a^9\,b}+48\,a^5\,b\,c\,d\right )}{16\,a^9}-\frac {6\,b\,c\,d\,\ln \left (x\right )}{a^4} \] Input:

int((c + d*x)^2/(x^4*(a + b*x^2)^3),x)
 

Output:

(log(15*a*d^2*(-a^9*b)^(1/2) - 35*b*c^2*(-a^9*b)^(1/2) - 35*a^4*b^2*c^2*x 
- 144*a^5*b*c*d + 15*a^5*b*d^2*x + 144*b*c*d*x*(-a^9*b)^(1/2))*(35*b*c^2*( 
-a^9*b)^(1/2) - 15*a*d^2*(-a^9*b)^(1/2) + 48*a^5*b*c*d))/(16*a^9) - (c^2/( 
3*a) + (x^2*(3*a*d^2 - 7*b*c^2))/(3*a^2) + (5*b^2*x^6*(3*a*d^2 - 7*b*c^2)) 
/(8*a^4) + (c*d*x)/a + (25*b*x^4*(3*a*d^2 - 7*b*c^2))/(24*a^3) + (9*b*c*d* 
x^3)/(2*a^2) + (3*b^2*c*d*x^5)/a^3)/(a^2*x^3 + b^2*x^7 + 2*a*b*x^5) + (log 
(15*a*d^2*(-a^9*b)^(1/2) - 35*b*c^2*(-a^9*b)^(1/2) + 35*a^4*b^2*c^2*x + 14 
4*a^5*b*c*d - 15*a^5*b*d^2*x + 144*b*c*d*x*(-a^9*b)^(1/2))*(15*a*d^2*(-a^9 
*b)^(1/2) - 35*b*c^2*(-a^9*b)^(1/2) + 48*a^5*b*c*d))/(16*a^9) - (6*b*c*d*l 
og(x))/a^4
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.18 \[ \int \frac {(c+d x)^2}{x^4 \left (a+b x^2\right )^3} \, dx=\frac {-45 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} d^{2} x^{3}+105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,c^{2} x^{3}-90 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,d^{2} x^{5}+210 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} c^{2} x^{5}-45 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} d^{2} x^{7}+105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} c^{2} x^{7}+72 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3} b c d \,x^{3}+144 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b^{2} c d \,x^{5}+72 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{3} c d \,x^{7}-144 \,\mathrm {log}\left (x \right ) a^{3} b c d \,x^{3}-288 \,\mathrm {log}\left (x \right ) a^{2} b^{2} c d \,x^{5}-144 \,\mathrm {log}\left (x \right ) a \,b^{3} c d \,x^{7}-8 a^{4} c^{2}-24 a^{4} c d x -24 a^{4} d^{2} x^{2}+56 a^{3} b \,c^{2} x^{2}-72 a^{3} b c d \,x^{3}-75 a^{3} b \,d^{2} x^{4}+175 a^{2} b^{2} c^{2} x^{4}-45 a^{2} b^{2} d^{2} x^{6}+105 a \,b^{3} c^{2} x^{6}+36 a \,b^{3} c d \,x^{7}}{24 a^{5} x^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int((d*x+c)^2/x^4/(b*x^2+a)^3,x)
 

Output:

( - 45*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*d**2*x**3 + 105* 
sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b*c**2*x**3 - 90*sqrt(b 
)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b*d**2*x**5 + 210*sqrt(b)*sqr 
t(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**2*c**2*x**5 - 45*sqrt(b)*sqrt(a)*a 
tan((b*x)/(sqrt(b)*sqrt(a)))*a*b**2*d**2*x**7 + 105*sqrt(b)*sqrt(a)*atan(( 
b*x)/(sqrt(b)*sqrt(a)))*b**3*c**2*x**7 + 72*log(a + b*x**2)*a**3*b*c*d*x** 
3 + 144*log(a + b*x**2)*a**2*b**2*c*d*x**5 + 72*log(a + b*x**2)*a*b**3*c*d 
*x**7 - 144*log(x)*a**3*b*c*d*x**3 - 288*log(x)*a**2*b**2*c*d*x**5 - 144*l 
og(x)*a*b**3*c*d*x**7 - 8*a**4*c**2 - 24*a**4*c*d*x - 24*a**4*d**2*x**2 + 
56*a**3*b*c**2*x**2 - 72*a**3*b*c*d*x**3 - 75*a**3*b*d**2*x**4 + 175*a**2* 
b**2*c**2*x**4 - 45*a**2*b**2*d**2*x**6 + 105*a*b**3*c**2*x**6 + 36*a*b**3 
*c*d*x**7)/(24*a**5*x**3*(a**2 + 2*a*b*x**2 + b**2*x**4))