\(\int \frac {x^4 (c+d x)^2}{(a+b x^2)^3} \, dx\) [231]

Optimal result

Integrand size = 20, antiderivative size = 138 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^3} \, dx=\frac {d^2 x}{b^3}-\frac {a \left (2 a c d-\left (b c^2-a d^2\right ) x\right )}{4 b^3 \left (a+b x^2\right )^2}+\frac {16 a c d-\left (5 b c^2-9 a d^2\right ) x}{8 b^3 \left (a+b x^2\right )}+\frac {3 \left (b c^2-5 a d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}+\frac {c d \log \left (a+b x^2\right )}{b^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.96 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^3} \, dx=\frac {d^2 x}{b^3}+\frac {a \left (b c^2 x-a d (2 c+d x)\right )}{4 b^3 \left (a+b x^2\right )^2}+\frac {-5 b c^2 x+a d (16 c+9 d x)}{8 b^3 \left (a+b x^2\right )}+\frac {3 \left (b c^2-5 a d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}+\frac {c d \log \left (a+b x^2\right )}{b^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {531, 2176, 2341, 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.

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

\(\Big \downarrow \) 531

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

\(\Big \downarrow \) 2176

\(\displaystyle \frac {a x (c+d x)^2}{4 b^2 \left (a+b x^2\right )^2}-\frac {-\frac {\int \frac {8 d^2 x^2 a^2+\left (3 c^2-\frac {7 a d^2}{b}\right ) a^2+16 c d x a^2}{b x^2+a}dx}{2 a b}-\frac {a (c+d x) (7 a d-5 b c x)}{2 b^2 \left (a+b x^2\right )}}{4 a b}\)

\(\Big \downarrow \) 2341

\(\displaystyle \frac {a x (c+d x)^2}{4 b^2 \left (a+b x^2\right )^2}-\frac {-\frac {\int \left (\frac {8 a^2 d^2}{b}+\frac {3 \left (b c^2-5 a d^2\right ) a^2+16 b c d x a^2}{b \left (b x^2+a\right )}\right )dx}{2 a b}-\frac {a (c+d x) (7 a d-5 b c x)}{2 b^2 \left (a+b x^2\right )}}{4 a b}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 531

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb 
ol] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomi 
alRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a 
 + b*x^2, x], x, 1]}, Simp[(c + d*x)^n*(a*f - b*e*x)*((a + b*x^2)^(p + 1)/( 
2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(c + d*x)^(n - 1)*(a + b 
*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(c + d*x)*Qx - a*d*f*n + b*c*e*(2*p 
 + 3) + b*d*e*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGt 
Q[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && GtQ[n, 1] && IntegerQ[2*p]
 

rule 2009

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

rule 2176

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a + b*x^2, x], R = Coeff[PolynomialRema 
inder[Pq, a + b*x^2, x], x, 0], S = Coeff[PolynomialRemainder[Pq, a + b*x^2 
, x], x, 1]}, Simp[(d + e*x)^m*(a + b*x^2)^(p + 1)*((a*S - b*R*x)/(2*a*b*(p 
 + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^(p 
 + 1)*ExpandToSum[2*a*b*(p + 1)*(d + e*x)*Qx - a*e*S*m + b*d*R*(2*p + 3) + 
b*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x 
] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && R 
ationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
 

rule 2341

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {d^{2} x}{b^{3}}-\frac {\frac {\left (-\frac {9}{8} a b \,d^{2}+\frac {5}{8} b^{2} c^{2}\right ) x^{3}-2 a b c d \,x^{2}-\frac {a \left (7 a \,d^{2}-3 b \,c^{2}\right ) x}{8}-\frac {3 a^{2} c d}{2}}{\left (b \,x^{2}+a \right )^{2}}-d c \ln \left (b \,x^{2}+a \right )+\frac {\left (15 a \,d^{2}-3 b \,c^{2}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{b^{3}}\)	\(121\)
risch	\(\frac {d^{2} x}{b^{3}}+\frac {\left (\frac {9}{8} a b \,d^{2}-\frac {5}{8} b^{2} c^{2}\right ) x^{3}+2 a b c d \,x^{2}+\frac {a \left (7 a \,d^{2}-3 b \,c^{2}\right ) x}{8}+\frac {3 a^{2} c d}{2}}{b^{3} \left (b \,x^{2}+a \right )^{2}}+\frac {\ln \left (-5 a^{2} d^{2}+a b \,c^{2}-\sqrt {-a b \left (5 a \,d^{2}-b \,c^{2}\right )^{2}}\, x \right ) c d}{b^{3}}+\frac {3 \ln \left (-5 a^{2} d^{2}+a b \,c^{2}-\sqrt {-a b \left (5 a \,d^{2}-b \,c^{2}\right )^{2}}\, x \right ) \sqrt {-a b \left (5 a \,d^{2}-b \,c^{2}\right )^{2}}}{16 b^{4} a}+\frac {\ln \left (-5 a^{2} d^{2}+a b \,c^{2}+\sqrt {-a b \left (5 a \,d^{2}-b \,c^{2}\right )^{2}}\, x \right ) c d}{b^{3}}-\frac {3 \ln \left (-5 a^{2} d^{2}+a b \,c^{2}+\sqrt {-a b \left (5 a \,d^{2}-b \,c^{2}\right )^{2}}\, x \right ) \sqrt {-a b \left (5 a \,d^{2}-b \,c^{2}\right )^{2}}}{16 b^{4} a}\)	\(305\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 486, normalized size of antiderivative = 3.52 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^3} \, dx=\left [\frac {16 \, a b^{3} d^{2} x^{5} + 32 \, a^{2} b^{2} c d x^{2} + 24 \, a^{3} b c d - 10 \, {\left (a b^{3} c^{2} - 5 \, a^{2} b^{2} d^{2}\right )} x^{3} + 3 \, {\left (a^{2} b c^{2} - 5 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 5 \, a b^{2} d^{2}\right )} x^{4} + 2 \, {\left (a b^{2} c^{2} - 5 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 6 \, {\left (a^{2} b^{2} c^{2} - 5 \, a^{3} b d^{2}\right )} x + 16 \, {\left (a b^{3} c d x^{4} + 2 \, a^{2} b^{2} c d x^{2} + a^{3} b c d\right )} \log \left (b x^{2} + a\right )}{16 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}, \frac {8 \, a b^{3} d^{2} x^{5} + 16 \, a^{2} b^{2} c d x^{2} + 12 \, a^{3} b c d - 5 \, {\left (a b^{3} c^{2} - 5 \, a^{2} b^{2} d^{2}\right )} x^{3} + 3 \, {\left (a^{2} b c^{2} - 5 \, a^{3} d^{2} + {\left (b^{3} c^{2} - 5 \, a b^{2} d^{2}\right )} x^{4} + 2 \, {\left (a b^{2} c^{2} - 5 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - 3 \, {\left (a^{2} b^{2} c^{2} - 5 \, a^{3} b d^{2}\right )} x + 8 \, {\left (a b^{3} c d x^{4} + 2 \, a^{2} b^{2} c d x^{2} + a^{3} b c d\right )} \log \left (b x^{2} + a\right )}{8 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}\right ] \] Input:

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

Output:

[1/16*(16*a*b^3*d^2*x^5 + 32*a^2*b^2*c*d*x^2 + 24*a^3*b*c*d - 10*(a*b^3*c^ 
2 - 5*a^2*b^2*d^2)*x^3 + 3*(a^2*b*c^2 - 5*a^3*d^2 + (b^3*c^2 - 5*a*b^2*d^2 
)*x^4 + 2*(a*b^2*c^2 - 5*a^2*b*d^2)*x^2)*sqrt(-a*b)*log((b*x^2 + 2*sqrt(-a 
*b)*x - a)/(b*x^2 + a)) - 6*(a^2*b^2*c^2 - 5*a^3*b*d^2)*x + 16*(a*b^3*c*d* 
x^4 + 2*a^2*b^2*c*d*x^2 + a^3*b*c*d)*log(b*x^2 + a))/(a*b^6*x^4 + 2*a^2*b^ 
5*x^2 + a^3*b^4), 1/8*(8*a*b^3*d^2*x^5 + 16*a^2*b^2*c*d*x^2 + 12*a^3*b*c*d 
 - 5*(a*b^3*c^2 - 5*a^2*b^2*d^2)*x^3 + 3*(a^2*b*c^2 - 5*a^3*d^2 + (b^3*c^2 
 - 5*a*b^2*d^2)*x^4 + 2*(a*b^2*c^2 - 5*a^2*b*d^2)*x^2)*sqrt(a*b)*arctan(sq 
rt(a*b)*x/a) - 3*(a^2*b^2*c^2 - 5*a^3*b*d^2)*x + 8*(a*b^3*c*d*x^4 + 2*a^2* 
b^2*c*d*x^2 + a^3*b*c*d)*log(b*x^2 + a))/(a*b^6*x^4 + 2*a^2*b^5*x^2 + a^3* 
b^4)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (131) = 262\).

Time = 1.26 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.20 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^3} \, dx=\left (\frac {c d}{b^{3}} - \frac {3 \sqrt {- a b^{7}} \cdot \left (5 a d^{2} - b c^{2}\right )}{16 a b^{7}}\right ) \log {\left (x + \frac {- 16 a b^{3} \left (\frac {c d}{b^{3}} - \frac {3 \sqrt {- a b^{7}} \cdot \left (5 a d^{2} - b c^{2}\right )}{16 a b^{7}}\right ) + 16 a c d}{15 a d^{2} - 3 b c^{2}} \right )} + \left (\frac {c d}{b^{3}} + \frac {3 \sqrt {- a b^{7}} \cdot \left (5 a d^{2} - b c^{2}\right )}{16 a b^{7}}\right ) \log {\left (x + \frac {- 16 a b^{3} \left (\frac {c d}{b^{3}} + \frac {3 \sqrt {- a b^{7}} \cdot \left (5 a d^{2} - b c^{2}\right )}{16 a b^{7}}\right ) + 16 a c d}{15 a d^{2} - 3 b c^{2}} \right )} + \frac {12 a^{2} c d + 16 a b c d x^{2} + x^{3} \cdot \left (9 a b d^{2} - 5 b^{2} c^{2}\right ) + x \left (7 a^{2} d^{2} - 3 a b c^{2}\right )}{8 a^{2} b^{3} + 16 a b^{4} x^{2} + 8 b^{5} x^{4}} + \frac {d^{2} x}{b^{3}} \] Input:

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

Output:

(c*d/b**3 - 3*sqrt(-a*b**7)*(5*a*d**2 - b*c**2)/(16*a*b**7))*log(x + (-16* 
a*b**3*(c*d/b**3 - 3*sqrt(-a*b**7)*(5*a*d**2 - b*c**2)/(16*a*b**7)) + 16*a 
*c*d)/(15*a*d**2 - 3*b*c**2)) + (c*d/b**3 + 3*sqrt(-a*b**7)*(5*a*d**2 - b* 
c**2)/(16*a*b**7))*log(x + (-16*a*b**3*(c*d/b**3 + 3*sqrt(-a*b**7)*(5*a*d* 
*2 - b*c**2)/(16*a*b**7)) + 16*a*c*d)/(15*a*d**2 - 3*b*c**2)) + (12*a**2*c 
*d + 16*a*b*c*d*x**2 + x**3*(9*a*b*d**2 - 5*b**2*c**2) + x*(7*a**2*d**2 - 
3*a*b*c**2))/(8*a**2*b**3 + 16*a*b**4*x**2 + 8*b**5*x**4) + d**2*x/b**3
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.01 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^3} \, dx=\frac {16 \, a b c d x^{2} + 12 \, a^{2} c d - {\left (5 \, b^{2} c^{2} - 9 \, a b d^{2}\right )} x^{3} - {\left (3 \, a b c^{2} - 7 \, a^{2} d^{2}\right )} x}{8 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} + \frac {d^{2} x}{b^{3}} + \frac {c d \log \left (b x^{2} + a\right )}{b^{3}} + \frac {3 \, {\left (b c^{2} - 5 \, a d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} \] Input:

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

Output:

1/8*(16*a*b*c*d*x^2 + 12*a^2*c*d - (5*b^2*c^2 - 9*a*b*d^2)*x^3 - (3*a*b*c^ 
2 - 7*a^2*d^2)*x)/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3) + d^2*x/b^3 + c*d*log( 
b*x^2 + a)/b^3 + 3/8*(b*c^2 - 5*a*d^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^ 
3)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.91 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^3} \, dx=\frac {d^{2} x}{b^{3}} + \frac {c d \log \left (b x^{2} + a\right )}{b^{3}} + \frac {3 \, {\left (b c^{2} - 5 \, a d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} + \frac {16 \, a b c d x^{2} + 12 \, a^{2} c d - {\left (5 \, b^{2} c^{2} - 9 \, a b d^{2}\right )} x^{3} - {\left (3 \, a b c^{2} - 7 \, a^{2} d^{2}\right )} x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{3}} \] Input:

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

Output:

d^2*x/b^3 + c*d*log(b*x^2 + a)/b^3 + 3/8*(b*c^2 - 5*a*d^2)*arctan(b*x/sqrt 
(a*b))/(sqrt(a*b)*b^3) + 1/8*(16*a*b*c*d*x^2 + 12*a^2*c*d - (5*b^2*c^2 - 9 
*a*b*d^2)*x^3 - (3*a*b*c^2 - 7*a^2*d^2)*x)/((b*x^2 + a)^2*b^3)
 

Mupad [B] (verification not implemented)

Time = 7.32 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.99 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^3} \, dx=\frac {x\,\left (\frac {7\,a^2\,d^2}{8}-\frac {3\,a\,b\,c^2}{8}\right )-x^3\,\left (\frac {5\,b^2\,c^2}{8}-\frac {9\,a\,b\,d^2}{8}\right )+\frac {3\,a^2\,c\,d}{2}+2\,a\,b\,c\,d\,x^2}{a^2\,b^3+2\,a\,b^4\,x^2+b^5\,x^4}+\frac {d^2\,x}{b^3}-\frac {3\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (5\,a\,d^2-b\,c^2\right )}{8\,\sqrt {a}\,b^{7/2}}+\frac {c\,d\,\ln \left (b\,x^2+a\right )}{b^3} \] Input:

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

Output:

(x*((7*a^2*d^2)/8 - (3*a*b*c^2)/8) - x^3*((5*b^2*c^2)/8 - (9*a*b*d^2)/8) + 
 (3*a^2*c*d)/2 + 2*a*b*c*d*x^2)/(a^2*b^3 + b^5*x^4 + 2*a*b^4*x^2) + (d^2*x 
)/b^3 - (3*atan((b^(1/2)*x)/a^(1/2))*(5*a*d^2 - b*c^2))/(8*a^(1/2)*b^(7/2) 
) + (c*d*log(a + b*x^2))/b^3
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.35 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^3} \, dx=\frac {-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} d^{2}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,c^{2}-30 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,d^{2} x^{2}+6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} c^{2} x^{2}-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} d^{2} x^{4}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} c^{2} x^{4}+8 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3} b c d +16 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b^{2} c d \,x^{2}+8 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{3} c d \,x^{4}+4 a^{3} b c d +15 a^{3} b \,d^{2} x -3 a^{2} b^{2} c^{2} x +25 a^{2} b^{2} d^{2} x^{3}-5 a \,b^{3} c^{2} x^{3}-8 a \,b^{3} c d \,x^{4}+8 a \,b^{3} d^{2} x^{5}}{8 a \,b^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

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

Output:

( - 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*d**2 + 3*sqrt(b) 
*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b*c**2 - 30*sqrt(b)*sqrt(a)*at 
an((b*x)/(sqrt(b)*sqrt(a)))*a**2*b*d**2*x**2 + 6*sqrt(b)*sqrt(a)*atan((b*x 
)/(sqrt(b)*sqrt(a)))*a*b**2*c**2*x**2 - 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqr 
t(b)*sqrt(a)))*a*b**2*d**2*x**4 + 3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sq 
rt(a)))*b**3*c**2*x**4 + 8*log(a + b*x**2)*a**3*b*c*d + 16*log(a + b*x**2) 
*a**2*b**2*c*d*x**2 + 8*log(a + b*x**2)*a*b**3*c*d*x**4 + 4*a**3*b*c*d + 1 
5*a**3*b*d**2*x - 3*a**2*b**2*c**2*x + 25*a**2*b**2*d**2*x**3 - 5*a*b**3*c 
**2*x**3 - 8*a*b**3*c*d*x**4 + 8*a*b**3*d**2*x**5)/(8*a*b**4*(a**2 + 2*a*b 
*x**2 + b**2*x**4))