\(\int \frac {c+d x}{x^3 (a+b x^2)^3} \, dx\) [228]

Optimal result

Integrand size = 18, antiderivative size = 122 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^3} \, dx=-\frac {c}{2 a^3 x^2}-\frac {d}{a^3 x}-\frac {b (c+d x)}{4 a^2 \left (a+b x^2\right )^2}-\frac {b (8 c+7 d x)}{8 a^3 \left (a+b x^2\right )}-\frac {15 \sqrt {b} d \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2}}-\frac {3 b c \log (x)}{a^4}+\frac {3 b c \log \left (a+b x^2\right )}{2 a^4} \] Output:

-1/2*c/a^3/x^2-d/a^3/x-1/4*b*(d*x+c)/a^2/(b*x^2+a)^2-1/8*b*(7*d*x+8*c)/a^3 
/(b*x^2+a)-15/8*b^(1/2)*d*arctan(b^(1/2)*x/a^(1/2))/a^(7/2)-3*b*c*ln(x)/a^ 
4+3/2*b*c*ln(b*x^2+a)/a^4
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.87 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^3} \, dx=-\frac {\frac {4 a c}{x^2}+\frac {8 a d}{x}+\frac {2 a^2 b (c+d x)}{\left (a+b x^2\right )^2}+\frac {a b (8 c+7 d x)}{a+b x^2}+15 \sqrt {a} \sqrt {b} d \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )+24 b c \log (x)-12 b c \log \left (a+b x^2\right )}{8 a^4} \] Input:

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

Output:

-1/8*((4*a*c)/x^2 + (8*a*d)/x + (2*a^2*b*(c + d*x))/(a + b*x^2)^2 + (a*b*( 
8*c + 7*d*x))/(a + b*x^2) + 15*Sqrt[a]*Sqrt[b]*d*ArcTan[(Sqrt[b]*x)/Sqrt[a 
]] + 24*b*c*Log[x] - 12*b*c*Log[a + b*x^2])/a^4
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, 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.

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

\(\Big \downarrow \) 532

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2336

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2333

\(\displaystyle \frac {\frac {\int \left (-\frac {24 b c}{a^2 x}+\frac {8 c}{a x^3}-\frac {3 b (5 a d-8 b c x)}{a^2 \left (b x^2+a\right )}+\frac {8 d}{a x^2}\right )dx}{2 a}-\frac {b (8 c+7 d x)}{2 a^2 \left (a+b x^2\right )}}{4 a}-\frac {b (c+d x)}{4 a^2 \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {-\frac {15 \sqrt {b} d \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}+\frac {12 b c \log \left (a+b x^2\right )}{a^2}-\frac {24 b c \log (x)}{a^2}-\frac {4 c}{a x^2}-\frac {8 d}{a x}}{2 a}-\frac {b (8 c+7 d x)}{2 a^2 \left (a+b x^2\right )}}{4 a}-\frac {b (c+d x)}{4 a^2 \left (a+b x^2\right )^2}\)

Input:

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

Output:

-1/4*(b*(c + d*x))/(a^2*(a + b*x^2)^2) + (-1/2*(b*(8*c + 7*d*x))/(a^2*(a + 
 b*x^2)) + ((-4*c)/(a*x^2) - (8*d)/(a*x) - (15*Sqrt[b]*d*ArcTan[(Sqrt[b]*x 
)/Sqrt[a]])/a^(3/2) - (24*b*c*Log[x])/a^2 + (12*b*c*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 = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.85

method	result	size
default	\(-\frac {c}{2 a^{3} x^{2}}-\frac {d}{a^{3} x}-\frac {3 b c \ln \left (x \right )}{a^{4}}-\frac {b \left (\frac {\frac {7}{8} a b d \,x^{3}+a b c \,x^{2}+\frac {9}{8} a^{2} d x +\frac {5}{4} a^{2} c}{\left (b \,x^{2}+a \right )^{2}}-\frac {3 c \ln \left (b \,x^{2}+a \right )}{2}+\frac {15 a d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{4}}\)	\(104\)
risch	\(\frac {-\frac {15 d \,b^{2} x^{5}}{8 a^{3}}-\frac {3 b^{2} c \,x^{4}}{2 a^{3}}-\frac {25 d b \,x^{3}}{8 a^{2}}-\frac {9 b c \,x^{2}}{4 a^{2}}-\frac {d x}{a}-\frac {c}{2 a}}{x^{2} \left (b \,x^{2}+a \right )^{2}}+\frac {15 \ln \left (\left (-5 d a b +24 \sqrt {-a b}\, b c \right ) x +5 \sqrt {-a b}\, a d +24 a b c \right ) \sqrt {-a b}\, d}{16 a^{4}}+\frac {3 \ln \left (\left (-5 d a b +24 \sqrt {-a b}\, b c \right ) x +5 \sqrt {-a b}\, a d +24 a b c \right ) b c}{2 a^{4}}-\frac {15 \ln \left (\left (-5 d a b -24 \sqrt {-a b}\, b c \right ) x -5 \sqrt {-a b}\, a d +24 a b c \right ) \sqrt {-a b}\, d}{16 a^{4}}+\frac {3 \ln \left (\left (-5 d a b -24 \sqrt {-a b}\, b c \right ) x -5 \sqrt {-a b}\, a d +24 a b c \right ) b c}{2 a^{4}}-\frac {3 b c \ln \left (x \right )}{a^{4}}\)	\(260\)

Input:

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

Output:

-1/2*c/a^3/x^2-d/a^3/x-3*b*c*ln(x)/a^4-b/a^4*((7/8*a*b*d*x^3+a*b*c*x^2+9/8 
*a^2*d*x+5/4*a^2*c)/(b*x^2+a)^2-3/2*c*ln(b*x^2+a)+15/8*a*d/(a*b)^(1/2)*arc 
tan(b*x/(a*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 428, normalized size of antiderivative = 3.51 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^3} \, dx=\left [-\frac {30 \, a b^{2} d x^{5} + 24 \, a b^{2} c x^{4} + 50 \, a^{2} b d x^{3} + 36 \, a^{2} b c x^{2} + 16 \, a^{3} d x + 8 \, a^{3} c - 15 \, {\left (a b^{2} d x^{6} + 2 \, a^{2} b d x^{4} + a^{3} d x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) - 24 \, {\left (b^{3} c x^{6} + 2 \, a b^{2} c x^{4} + a^{2} b c x^{2}\right )} \log \left (b x^{2} + a\right ) + 48 \, {\left (b^{3} c x^{6} + 2 \, a b^{2} c x^{4} + a^{2} b c x^{2}\right )} \log \left (x\right )}{16 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}, -\frac {15 \, a b^{2} d x^{5} + 12 \, a b^{2} c x^{4} + 25 \, a^{2} b d x^{3} + 18 \, a^{2} b c x^{2} + 8 \, a^{3} d x + 4 \, a^{3} c + 15 \, {\left (a b^{2} d x^{6} + 2 \, a^{2} b d x^{4} + a^{3} d x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) - 12 \, {\left (b^{3} c x^{6} + 2 \, a b^{2} c x^{4} + a^{2} b c x^{2}\right )} \log \left (b x^{2} + a\right ) + 24 \, {\left (b^{3} c x^{6} + 2 \, a b^{2} c x^{4} + a^{2} b c x^{2}\right )} \log \left (x\right )}{8 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}\right ] \] Input:

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

Output:

[-1/16*(30*a*b^2*d*x^5 + 24*a*b^2*c*x^4 + 50*a^2*b*d*x^3 + 36*a^2*b*c*x^2 
+ 16*a^3*d*x + 8*a^3*c - 15*(a*b^2*d*x^6 + 2*a^2*b*d*x^4 + a^3*d*x^2)*sqrt 
(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) - 24*(b^3*c*x^6 + 2 
*a*b^2*c*x^4 + a^2*b*c*x^2)*log(b*x^2 + a) + 48*(b^3*c*x^6 + 2*a*b^2*c*x^4 
 + a^2*b*c*x^2)*log(x))/(a^4*b^2*x^6 + 2*a^5*b*x^4 + a^6*x^2), -1/8*(15*a* 
b^2*d*x^5 + 12*a*b^2*c*x^4 + 25*a^2*b*d*x^3 + 18*a^2*b*c*x^2 + 8*a^3*d*x + 
 4*a^3*c + 15*(a*b^2*d*x^6 + 2*a^2*b*d*x^4 + a^3*d*x^2)*sqrt(b/a)*arctan(x 
*sqrt(b/a)) - 12*(b^3*c*x^6 + 2*a*b^2*c*x^4 + a^2*b*c*x^2)*log(b*x^2 + a) 
+ 24*(b^3*c*x^6 + 2*a*b^2*c*x^4 + a^2*b*c*x^2)*log(x))/(a^4*b^2*x^6 + 2*a^ 
5*b*x^4 + a^6*x^2)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 466 vs. \(2 (117) = 234\).

Time = 1.33 (sec) , antiderivative size = 466, normalized size of antiderivative = 3.82 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^3} \, dx=\left (\frac {3 b c}{2 a^{4}} - \frac {15 d \sqrt {- a^{9} b}}{16 a^{8}}\right ) \log {\left (x + \frac {- 2048 a^{8} c \left (\frac {3 b c}{2 a^{4}} - \frac {15 d \sqrt {- a^{9} b}}{16 a^{8}}\right )^{2} - 400 a^{5} d^{2} \cdot \left (\frac {3 b c}{2 a^{4}} - \frac {15 d \sqrt {- a^{9} b}}{16 a^{8}}\right ) - 3072 a^{4} b c^{2} \cdot \left (\frac {3 b c}{2 a^{4}} - \frac {15 d \sqrt {- a^{9} b}}{16 a^{8}}\right ) - 1200 a b c d^{2} + 9216 b^{2} c^{3}}{375 a b d^{3} + 8640 b^{2} c^{2} d} \right )} + \left (\frac {3 b c}{2 a^{4}} + \frac {15 d \sqrt {- a^{9} b}}{16 a^{8}}\right ) \log {\left (x + \frac {- 2048 a^{8} c \left (\frac {3 b c}{2 a^{4}} + \frac {15 d \sqrt {- a^{9} b}}{16 a^{8}}\right )^{2} - 400 a^{5} d^{2} \cdot \left (\frac {3 b c}{2 a^{4}} + \frac {15 d \sqrt {- a^{9} b}}{16 a^{8}}\right ) - 3072 a^{4} b c^{2} \cdot \left (\frac {3 b c}{2 a^{4}} + \frac {15 d \sqrt {- a^{9} b}}{16 a^{8}}\right ) - 1200 a b c d^{2} + 9216 b^{2} c^{3}}{375 a b d^{3} + 8640 b^{2} c^{2} d} \right )} + \frac {- 4 a^{2} c - 8 a^{2} d x - 18 a b c x^{2} - 25 a b d x^{3} - 12 b^{2} c x^{4} - 15 b^{2} d x^{5}}{8 a^{5} x^{2} + 16 a^{4} b x^{4} + 8 a^{3} b^{2} x^{6}} - \frac {3 b c \log {\left (x \right )}}{a^{4}} \] Input:

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

Output:

(3*b*c/(2*a**4) - 15*d*sqrt(-a**9*b)/(16*a**8))*log(x + (-2048*a**8*c*(3*b 
*c/(2*a**4) - 15*d*sqrt(-a**9*b)/(16*a**8))**2 - 400*a**5*d**2*(3*b*c/(2*a 
**4) - 15*d*sqrt(-a**9*b)/(16*a**8)) - 3072*a**4*b*c**2*(3*b*c/(2*a**4) - 
15*d*sqrt(-a**9*b)/(16*a**8)) - 1200*a*b*c*d**2 + 9216*b**2*c**3)/(375*a*b 
*d**3 + 8640*b**2*c**2*d)) + (3*b*c/(2*a**4) + 15*d*sqrt(-a**9*b)/(16*a**8 
))*log(x + (-2048*a**8*c*(3*b*c/(2*a**4) + 15*d*sqrt(-a**9*b)/(16*a**8))** 
2 - 400*a**5*d**2*(3*b*c/(2*a**4) + 15*d*sqrt(-a**9*b)/(16*a**8)) - 3072*a 
**4*b*c**2*(3*b*c/(2*a**4) + 15*d*sqrt(-a**9*b)/(16*a**8)) - 1200*a*b*c*d* 
*2 + 9216*b**2*c**3)/(375*a*b*d**3 + 8640*b**2*c**2*d)) + (-4*a**2*c - 8*a 
**2*d*x - 18*a*b*c*x**2 - 25*a*b*d*x**3 - 12*b**2*c*x**4 - 15*b**2*d*x**5) 
/(8*a**5*x**2 + 16*a**4*b*x**4 + 8*a**3*b**2*x**6) - 3*b*c*log(x)/a**4
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^3} \, dx=-\frac {15 \, b^{2} d x^{5} + 12 \, b^{2} c x^{4} + 25 \, a b d x^{3} + 18 \, a b c x^{2} + 8 \, a^{2} d x + 4 \, a^{2} c}{8 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )}} - \frac {15 \, b d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} + \frac {3 \, b c \log \left (b x^{2} + a\right )}{2 \, a^{4}} - \frac {3 \, b c \log \left (x\right )}{a^{4}} \] Input:

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

Output:

-1/8*(15*b^2*d*x^5 + 12*b^2*c*x^4 + 25*a*b*d*x^3 + 18*a*b*c*x^2 + 8*a^2*d* 
x + 4*a^2*c)/(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2) - 15/8*b*d*arctan(b*x/s 
qrt(a*b))/(sqrt(a*b)*a^3) + 3/2*b*c*log(b*x^2 + a)/a^4 - 3*b*c*log(x)/a^4
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.91 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^3} \, dx=-\frac {15 \, b d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} + \frac {3 \, b c \log \left (b x^{2} + a\right )}{2 \, a^{4}} - \frac {3 \, b c \log \left ({\left | x \right |}\right )}{a^{4}} - \frac {15 \, b^{2} d x^{5} + 12 \, b^{2} c x^{4} + 25 \, a b d x^{3} + 18 \, a b c x^{2} + 8 \, a^{2} d x + 4 \, a^{2} c}{8 \, {\left (b x^{3} + a x\right )}^{2} a^{3}} \] Input:

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

Output:

-15/8*b*d*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) + 3/2*b*c*log(b*x^2 + a)/a 
^4 - 3*b*c*log(abs(x))/a^4 - 1/8*(15*b^2*d*x^5 + 12*b^2*c*x^4 + 25*a*b*d*x 
^3 + 18*a*b*c*x^2 + 8*a^2*d*x + 4*a^2*c)/((b*x^3 + a*x)^2*a^3)
 

Mupad [B] (verification not implemented)

Time = 7.56 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.84 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^3} \, dx=\frac {3\,\ln \left (5\,a\,d\,\sqrt {-a^9\,b}+24\,a^5\,b\,c-5\,a^5\,b\,d\,x+24\,b\,c\,x\,\sqrt {-a^9\,b}\right )\,\left (5\,d\,\sqrt {-a^9\,b}+8\,a^4\,b\,c\right )}{16\,a^8}-\frac {3\,\ln \left (5\,a\,d\,\sqrt {-a^9\,b}-24\,a^5\,b\,c+5\,a^5\,b\,d\,x+24\,b\,c\,x\,\sqrt {-a^9\,b}\right )\,\left (5\,d\,\sqrt {-a^9\,b}-8\,a^4\,b\,c\right )}{16\,a^8}-\frac {\frac {c}{2\,a}+\frac {d\,x}{a}+\frac {3\,b^2\,c\,x^4}{2\,a^3}+\frac {15\,b^2\,d\,x^5}{8\,a^3}+\frac {9\,b\,c\,x^2}{4\,a^2}+\frac {25\,b\,d\,x^3}{8\,a^2}}{a^2\,x^2+2\,a\,b\,x^4+b^2\,x^6}-\frac {3\,b\,c\,\ln \left (x\right )}{a^4} \] Input:

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

Output:

(3*log(5*a*d*(-a^9*b)^(1/2) + 24*a^5*b*c - 5*a^5*b*d*x + 24*b*c*x*(-a^9*b) 
^(1/2))*(5*d*(-a^9*b)^(1/2) + 8*a^4*b*c))/(16*a^8) - (3*log(5*a*d*(-a^9*b) 
^(1/2) - 24*a^5*b*c + 5*a^5*b*d*x + 24*b*c*x*(-a^9*b)^(1/2))*(5*d*(-a^9*b) 
^(1/2) - 8*a^4*b*c))/(16*a^8) - (c/(2*a) + (d*x)/a + (3*b^2*c*x^4)/(2*a^3) 
 + (15*b^2*d*x^5)/(8*a^3) + (9*b*c*x^2)/(4*a^2) + (25*b*d*x^3)/(8*a^2))/(a 
^2*x^2 + b^2*x^6 + 2*a*b*x^4) - (3*b*c*log(x))/a^4
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.99 \[ \int \frac {c+d x}{x^3 \left (a+b x^2\right )^3} \, dx=\frac {-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} d \,x^{2}-30 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b d \,x^{4}-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} d \,x^{6}+12 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b c \,x^{2}+24 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{2} c \,x^{4}+12 \,\mathrm {log}\left (b \,x^{2}+a \right ) b^{3} c \,x^{6}-24 \,\mathrm {log}\left (x \right ) a^{2} b c \,x^{2}-48 \,\mathrm {log}\left (x \right ) a \,b^{2} c \,x^{4}-24 \,\mathrm {log}\left (x \right ) b^{3} c \,x^{6}-4 a^{3} c -8 a^{3} d x -12 a^{2} b c \,x^{2}-25 a^{2} b d \,x^{3}-15 a \,b^{2} d \,x^{5}+6 b^{3} c \,x^{6}}{8 a^{4} x^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

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

Output:

( - 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*d*x**2 - 30*sqrt 
(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b*d*x**4 - 15*sqrt(b)*sqrt(a)* 
atan((b*x)/(sqrt(b)*sqrt(a)))*b**2*d*x**6 + 12*log(a + b*x**2)*a**2*b*c*x* 
*2 + 24*log(a + b*x**2)*a*b**2*c*x**4 + 12*log(a + b*x**2)*b**3*c*x**6 - 2 
4*log(x)*a**2*b*c*x**2 - 48*log(x)*a*b**2*c*x**4 - 24*log(x)*b**3*c*x**6 - 
 4*a**3*c - 8*a**3*d*x - 12*a**2*b*c*x**2 - 25*a**2*b*d*x**3 - 15*a*b**2*d 
*x**5 + 6*b**3*c*x**6)/(8*a**4*x**2*(a**2 + 2*a*b*x**2 + b**2*x**4))