\(\int \frac {x^4 (c+d x)^2}{(a+b x^2)^2} \, dx\) [178]

Optimal result

Integrand size = 20, antiderivative size = 134 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=\frac {\left (b c^2-2 a d^2\right ) x}{b^3}+\frac {c d x^2}{b^2}+\frac {d^2 x^3}{3 b^2}-\frac {a \left (2 a c d-\left (b c^2-a d^2\right ) x\right )}{2 b^3 \left (a+b x^2\right )}-\frac {\sqrt {a} \left (3 b c^2-5 a d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}}-\frac {2 a c d \log \left (a+b x^2\right )}{b^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.98 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=\frac {\left (b c^2-2 a d^2\right ) x}{b^3}+\frac {c d x^2}{b^2}+\frac {d^2 x^3}{3 b^2}+\frac {a \left (b c^2 x-a d (2 c+d x)\right )}{2 b^3 \left (a+b x^2\right )}+\frac {\sqrt {a} \left (-3 b c^2+5 a d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}}-\frac {2 a c d \log \left (a+b x^2\right )}{b^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {531, 2160, 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.

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

\(\Big \downarrow \) 531

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

\(\Big \downarrow \) 2160

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(a*x*(c + d*x)^2)/(2*b^2*(a + b*x^2)) - (-((a*(2*b*c^2 - 5*a*d^2)*x)/b^2) 
- (2*a*c*d*x^2)/b - (2*a*d^2*x^3)/(3*b) + (a^(3/2)*(3*b*c^2 - 5*a*d^2)*Arc 
Tan[(Sqrt[b]*x)/Sqrt[a]])/b^(5/2) + (4*a^2*c*d*Log[a + b*x^2])/b^2)/(2*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 2160

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {-\frac {1}{3} b \,d^{2} x^{3}-b c d \,x^{2}+2 a \,d^{2} x -c^{2} b x}{b^{3}}+\frac {a \left (\frac {\left (-\frac {a \,d^{2}}{2}+\frac {b \,c^{2}}{2}\right ) x -a c d}{b \,x^{2}+a}-2 d c \ln \left (b \,x^{2}+a \right )+\frac {\left (5 a \,d^{2}-3 b \,c^{2}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{3}}\)	\(117\)
risch	\(\frac {d^{2} x^{3}}{3 b^{2}}+\frac {c d \,x^{2}}{b^{2}}-\frac {2 a \,d^{2} x}{b^{3}}+\frac {c^{2} x}{b^{2}}+\frac {\left (-\frac {1}{2} a^{2} d^{2}+\frac {1}{2} a b \,c^{2}\right ) x -a^{2} c d}{b^{3} \left (b \,x^{2}+a \right )}-\frac {2 \ln \left (5 a^{2} d^{2}-3 a b \,c^{2}-\sqrt {-a b \left (5 a \,d^{2}-3 b \,c^{2}\right )^{2}}\, x \right ) a c d}{b^{3}}+\frac {\ln \left (5 a^{2} d^{2}-3 a b \,c^{2}-\sqrt {-a b \left (5 a \,d^{2}-3 b \,c^{2}\right )^{2}}\, x \right ) \sqrt {-a b \left (5 a \,d^{2}-3 b \,c^{2}\right )^{2}}}{4 b^{4}}-\frac {2 \ln \left (5 a^{2} d^{2}-3 a b \,c^{2}+\sqrt {-a b \left (5 a \,d^{2}-3 b \,c^{2}\right )^{2}}\, x \right ) a c d}{b^{3}}-\frac {\ln \left (5 a^{2} d^{2}-3 a b \,c^{2}+\sqrt {-a b \left (5 a \,d^{2}-3 b \,c^{2}\right )^{2}}\, x \right ) \sqrt {-a b \left (5 a \,d^{2}-3 b \,c^{2}\right )^{2}}}{4 b^{4}}\)	\(309\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.82 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=\left [\frac {4 \, b^{2} d^{2} x^{5} + 12 \, b^{2} c d x^{4} + 12 \, a b c d x^{2} - 12 \, a^{2} c d + 4 \, {\left (3 \, b^{2} c^{2} - 5 \, a b d^{2}\right )} x^{3} - 3 \, {\left (3 \, a b c^{2} - 5 \, a^{2} d^{2} + {\left (3 \, b^{2} c^{2} - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 6 \, {\left (3 \, a b c^{2} - 5 \, a^{2} d^{2}\right )} x - 24 \, {\left (a b c d x^{2} + a^{2} c d\right )} \log \left (b x^{2} + a\right )}{12 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, \frac {2 \, b^{2} d^{2} x^{5} + 6 \, b^{2} c d x^{4} + 6 \, a b c d x^{2} - 6 \, a^{2} c d + 2 \, {\left (3 \, b^{2} c^{2} - 5 \, a b d^{2}\right )} x^{3} - 3 \, {\left (3 \, a b c^{2} - 5 \, a^{2} d^{2} + {\left (3 \, b^{2} c^{2} - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 3 \, {\left (3 \, a b c^{2} - 5 \, a^{2} d^{2}\right )} x - 12 \, {\left (a b c d x^{2} + a^{2} c d\right )} \log \left (b x^{2} + a\right )}{6 \, {\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \] Input:

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

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (128) = 256\).

Time = 0.57 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.21 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=x \left (- \frac {2 a d^{2}}{b^{3}} + \frac {c^{2}}{b^{2}}\right ) + \left (- \frac {2 a c d}{b^{3}} - \frac {\sqrt {- a b^{7}} \cdot \left (5 a d^{2} - 3 b c^{2}\right )}{4 b^{7}}\right ) \log {\left (x + \frac {8 a c d + 4 b^{3} \left (- \frac {2 a c d}{b^{3}} - \frac {\sqrt {- a b^{7}} \cdot \left (5 a d^{2} - 3 b c^{2}\right )}{4 b^{7}}\right )}{5 a d^{2} - 3 b c^{2}} \right )} + \left (- \frac {2 a c d}{b^{3}} + \frac {\sqrt {- a b^{7}} \cdot \left (5 a d^{2} - 3 b c^{2}\right )}{4 b^{7}}\right ) \log {\left (x + \frac {8 a c d + 4 b^{3} \left (- \frac {2 a c d}{b^{3}} + \frac {\sqrt {- a b^{7}} \cdot \left (5 a d^{2} - 3 b c^{2}\right )}{4 b^{7}}\right )}{5 a d^{2} - 3 b c^{2}} \right )} + \frac {- 2 a^{2} c d + x \left (- a^{2} d^{2} + a b c^{2}\right )}{2 a b^{3} + 2 b^{4} x^{2}} + \frac {c d x^{2}}{b^{2}} + \frac {d^{2} x^{3}}{3 b^{2}} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.99 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=-\frac {2 \, a c d \log \left (b x^{2} + a\right )}{b^{3}} - \frac {2 \, a^{2} c d - {\left (a b c^{2} - a^{2} d^{2}\right )} x}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} - \frac {{\left (3 \, a b c^{2} - 5 \, a^{2} d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} + \frac {b d^{2} x^{3} + 3 \, b c d x^{2} + 3 \, {\left (b c^{2} - 2 \, a d^{2}\right )} x}{3 \, b^{3}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.02 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=-\frac {2 \, a c d \log \left (b x^{2} + a\right )}{b^{3}} - \frac {{\left (3 \, a b c^{2} - 5 \, a^{2} d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} - \frac {2 \, a^{2} c d - {\left (a b c^{2} - a^{2} d^{2}\right )} x}{2 \, {\left (b x^{2} + a\right )} b^{3}} + \frac {b^{4} d^{2} x^{3} + 3 \, b^{4} c d x^{2} + 3 \, b^{4} c^{2} x - 6 \, a b^{3} d^{2} x}{3 \, b^{6}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 7.00 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.19 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=x\,\left (\frac {c^2}{b^2}-\frac {2\,a\,d^2}{b^3}\right )-\frac {x\,\left (\frac {a^2\,d^2}{2}-\frac {a\,b\,c^2}{2}\right )+a^2\,c\,d}{b^4\,x^2+a\,b^3}+\frac {d^2\,x^3}{3\,b^2}+\frac {c\,d\,x^2}{b^2}+\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,x\,\left (5\,a\,d^2-3\,b\,c^2\right )}{5\,a^2\,d^2-3\,a\,b\,c^2}\right )\,\left (5\,a\,d^2-3\,b\,c^2\right )}{2\,b^{7/2}}-\frac {2\,a\,c\,d\,\ln \left (b\,x^2+a\right )}{b^3} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.68 \[ \int \frac {x^4 (c+d x)^2}{\left (a+b x^2\right )^2} \, dx=\frac {15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} d^{2}-9 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,c^{2}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,d^{2} x^{2}-9 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} c^{2} x^{2}-12 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b c d -12 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{2} c d \,x^{2}-15 a^{2} b \,d^{2} x +9 a \,b^{2} c^{2} x +12 a \,b^{2} c d \,x^{2}-10 a \,b^{2} d^{2} x^{3}+6 b^{3} c^{2} x^{3}+6 b^{3} c d \,x^{4}+2 b^{3} d^{2} x^{5}}{6 b^{4} \left (b \,x^{2}+a \right )} \] Input:

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

Output:

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