\(\int \frac {x^2}{(c+d x)^2 (a+b x^2)^3} \, dx\) [258]

Optimal result

Integrand size = 20, antiderivative size = 323 \[ \int \frac {x^2}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx=-\frac {c^2 d^3}{\left (b c^2+a d^2\right )^3 (c+d x)}-\frac {2 a c d+\left (b c^2-a d^2\right ) x}{4 \left (b c^2+a d^2\right )^2 \left (a+b x^2\right )^2}+\frac {8 a c d \left (b^2 c^4-a^2 d^4\right )+\left (b c^2+a d^2\right ) \left (b^2 c^4-12 a b c^2 d^2+3 a^2 d^4\right ) x}{8 a \left (b c^2+a d^2\right )^4 \left (a+b x^2\right )}+\frac {\left (b^3 c^6+13 a b^2 c^4 d^2-33 a^2 b c^2 d^4+3 a^3 d^6\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} \sqrt {b} \left (b c^2+a d^2\right )^4}+\frac {2 c d^3 \left (2 b c^2-a d^2\right ) \log (c+d x)}{\left (b c^2+a d^2\right )^4}-\frac {c d^3 \left (2 b c^2-a d^2\right ) \log \left (a+b x^2\right )}{\left (b c^2+a d^2\right )^4} \] Output:

-c^2*d^3/(a*d^2+b*c^2)^3/(d*x+c)-1/4*(2*a*c*d+(-a*d^2+b*c^2)*x)/(a*d^2+b*c 
^2)^2/(b*x^2+a)^2+1/8*(8*a*c*d*(-a^2*d^4+b^2*c^4)+(a*d^2+b*c^2)*(3*a^2*d^4 
-12*a*b*c^2*d^2+b^2*c^4)*x)/a/(a*d^2+b*c^2)^4/(b*x^2+a)+1/8*(3*a^3*d^6-33* 
a^2*b*c^2*d^4+13*a*b^2*c^4*d^2+b^3*c^6)*arctan(b^(1/2)*x/a^(1/2))/a^(3/2)/ 
b^(1/2)/(a*d^2+b*c^2)^4+2*c*d^3*(-a*d^2+2*b*c^2)*ln(d*x+c)/(a*d^2+b*c^2)^4 
-c*d^3*(-a*d^2+2*b*c^2)*ln(b*x^2+a)/(a*d^2+b*c^2)^4
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.82 \[ \int \frac {x^2}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx=\frac {-\frac {8 c^2 d^3 \left (b c^2+a d^2\right )}{c+d x}-\frac {2 \left (b c^2+a d^2\right )^2 \left (b c^2 x+a d (2 c-d x)\right )}{\left (a+b x^2\right )^2}+\frac {\left (b c^2+a d^2\right ) \left (b^2 c^4 x+4 a b c^2 d (2 c-3 d x)+a^2 d^3 (-8 c+3 d x)\right )}{a \left (a+b x^2\right )}+\frac {\left (b^3 c^6+13 a b^2 c^4 d^2-33 a^2 b c^2 d^4+3 a^3 d^6\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b}}+16 \left (2 b c^3 d^3-a c d^5\right ) \log (c+d x)-8 \left (2 b c^3 d^3-a c d^5\right ) \log \left (a+b x^2\right )}{8 \left (b c^2+a d^2\right )^4} \] Input:

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

Output:

((-8*c^2*d^3*(b*c^2 + a*d^2))/(c + d*x) - (2*(b*c^2 + a*d^2)^2*(b*c^2*x + 
a*d*(2*c - d*x)))/(a + b*x^2)^2 + ((b*c^2 + a*d^2)*(b^2*c^4*x + 4*a*b*c^2* 
d*(2*c - 3*d*x) + a^2*d^3*(-8*c + 3*d*x)))/(a*(a + b*x^2)) + ((b^3*c^6 + 1 
3*a*b^2*c^4*d^2 - 33*a^2*b*c^2*d^4 + 3*a^3*d^6)*ArcTan[(Sqrt[b]*x)/Sqrt[a] 
])/(a^(3/2)*Sqrt[b]) + 16*(2*b*c^3*d^3 - a*c*d^5)*Log[c + d*x] - 8*(2*b*c^ 
3*d^3 - a*c*d^5)*Log[a + b*x^2])/(8*(b*c^2 + a*d^2)^4)
 

Rubi [A] (verified)

Time = 2.20 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {601, 25, 2178, 25, 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.

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

\(\Big \downarrow \) 601

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2178

\(\displaystyle \frac {\frac {x \left (3 a^2 d^4-12 a b c^2 d^2+b^2 c^4\right )+8 a c d \left (b c^2-a d^2\right )}{2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )^3}-\frac {\int -\frac {\frac {a b \left (b^2 c^4+12 a b d^2 c^2-5 a^2 d^4\right ) c^2}{\left (b c^2+a d^2\right )^3}+\frac {2 a b d \left (b c^2-5 a d^2\right ) x c}{\left (b c^2+a d^2\right )^2}+\frac {a b d^2 \left (b^2 c^4-12 a b d^2 c^2+3 a^2 d^4\right ) x^2}{\left (b c^2+a d^2\right )^3}}{(c+d x)^2 \left (b x^2+a\right )}dx}{2 a b}}{4 a}-\frac {x \left (b c^2-a d^2\right )+2 a c d}{4 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\frac {a b \left (b^2 c^4+12 a b d^2 c^2-5 a^2 d^4\right ) c^2}{\left (b c^2+a d^2\right )^3}+\frac {2 a b d \left (b c^2-5 a d^2\right ) x c}{\left (b c^2+a d^2\right )^2}+\frac {a b d^2 \left (b^2 c^4-12 a b d^2 c^2+3 a^2 d^4\right ) x^2}{\left (b c^2+a d^2\right )^3}}{(c+d x)^2 \left (b x^2+a\right )}dx}{2 a b}+\frac {x \left (3 a^2 d^4-12 a b c^2 d^2+b^2 c^4\right )+8 a c d \left (b c^2-a d^2\right )}{2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )^3}}{4 a}-\frac {x \left (b c^2-a d^2\right )+2 a c d}{4 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )^2}\)

\(\Big \downarrow \) 2160

\(\displaystyle \frac {\frac {\int \left (-\frac {16 a^2 b c \left (a d^2-2 b c^2\right ) d^4}{\left (b c^2+a d^2\right )^4 (c+d x)}+\frac {8 a^2 b c^2 d^4}{\left (b c^2+a d^2\right )^3 (c+d x)^2}+\frac {a b \left (b^3 c^6+13 a b^2 d^2 c^4-33 a^2 b d^4 c^2-16 a b d^3 \left (2 b c^2-a d^2\right ) x c+3 a^3 d^6\right )}{\left (b c^2+a d^2\right )^4 \left (b x^2+a\right )}\right )dx}{2 a b}+\frac {x \left (3 a^2 d^4-12 a b c^2 d^2+b^2 c^4\right )+8 a c d \left (b c^2-a d^2\right )}{2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )^3}}{4 a}-\frac {x \left (b c^2-a d^2\right )+2 a c d}{4 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {x \left (3 a^2 d^4-12 a b c^2 d^2+b^2 c^4\right )+8 a c d \left (b c^2-a d^2\right )}{2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )^3}+\frac {-\frac {8 a^2 b c d^3 \left (2 b c^2-a d^2\right ) \log \left (a+b x^2\right )}{\left (a d^2+b c^2\right )^4}-\frac {8 a^2 b c^2 d^3}{(c+d x) \left (a d^2+b c^2\right )^3}+\frac {16 a^2 b c d^3 \left (2 b c^2-a d^2\right ) \log (c+d x)}{\left (a d^2+b c^2\right )^4}+\frac {\sqrt {a} \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (3 a^3 d^6-33 a^2 b c^2 d^4+13 a b^2 c^4 d^2+b^3 c^6\right )}{\left (a d^2+b c^2\right )^4}}{2 a b}}{4 a}-\frac {x \left (b c^2-a d^2\right )+2 a c d}{4 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )^2}\)

Input:

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

Output:

-1/4*(2*a*c*d + (b*c^2 - a*d^2)*x)/((b*c^2 + a*d^2)^2*(a + b*x^2)^2) + ((8 
*a*c*d*(b*c^2 - a*d^2) + (b^2*c^4 - 12*a*b*c^2*d^2 + 3*a^2*d^4)*x)/(2*(b*c 
^2 + a*d^2)^3*(a + b*x^2)) + ((-8*a^2*b*c^2*d^3)/((b*c^2 + a*d^2)^3*(c + d 
*x)) + (Sqrt[a]*Sqrt[b]*(b^3*c^6 + 13*a*b^2*c^4*d^2 - 33*a^2*b*c^2*d^4 + 3 
*a^3*d^6)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(b*c^2 + a*d^2)^4 + (16*a^2*b*c*d^3 
*(2*b*c^2 - a*d^2)*Log[c + d*x])/(b*c^2 + a*d^2)^4 - (8*a^2*b*c*d^3*(2*b*c 
^2 - a*d^2)*Log[a + b*x^2])/(b*c^2 + a*d^2)^4)/(2*a*b))/(4*a)
 

Defintions of rubi rules used

rule 25

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

rule 601

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[(c 
+ d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* 
(2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] 
 && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
 

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]
 

rule 2178

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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {\frac {\frac {b \left (3 a^{3} d^{6}-9 a^{2} b \,c^{2} d^{4}-11 a \,b^{2} c^{4} d^{2}+b^{3} c^{6}\right ) x^{3}}{8 a}+\left (-b c \,d^{5} a^{2}+c^{5} d \,b^{3}\right ) x^{2}+\left (\frac {5}{8} a^{3} d^{6}-\frac {7}{8} a^{2} b \,c^{2} d^{4}-\frac {13}{8} a \,b^{2} c^{4} d^{2}-\frac {1}{8} b^{3} c^{6}\right ) x -\frac {a c d \left (3 a^{2} d^{4}+2 b \,c^{2} d^{2} a -b^{2} c^{4}\right )}{2}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\frac {\left (16 b c \,d^{5} a^{2}-32 a \,b^{2} c^{3} d^{3}\right ) \ln \left (b \,x^{2}+a \right )}{2 b}+\frac {\left (3 a^{3} d^{6}-33 a^{2} b \,c^{2} d^{4}+13 a \,b^{2} c^{4} d^{2}+b^{3} c^{6}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{8 a}}{\left (a \,d^{2}+b \,c^{2}\right )^{4}}-\frac {c^{2} d^{3}}{\left (a \,d^{2}+b \,c^{2}\right )^{3} \left (d x +c \right )}-\frac {2 c \,d^{3} \left (a \,d^{2}-2 b \,c^{2}\right ) \ln \left (d x +c \right )}{\left (a \,d^{2}+b \,c^{2}\right )^{4}}\)	\(337\)
risch	\(\text {Expression too large to display}\)	\(5276\)

Input:

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

Output:

1/(a*d^2+b*c^2)^4*((1/8*b*(3*a^3*d^6-9*a^2*b*c^2*d^4-11*a*b^2*c^4*d^2+b^3* 
c^6)/a*x^3+(-a^2*b*c*d^5+b^3*c^5*d)*x^2+(5/8*a^3*d^6-7/8*a^2*b*c^2*d^4-13/ 
8*a*b^2*c^4*d^2-1/8*b^3*c^6)*x-1/2*a*c*d*(3*a^2*d^4+2*a*b*c^2*d^2-b^2*c^4) 
)/(b*x^2+a)^2+1/8/a*(1/2*(16*a^2*b*c*d^5-32*a*b^2*c^3*d^3)/b*ln(b*x^2+a)+( 
3*a^3*d^6-33*a^2*b*c^2*d^4+13*a*b^2*c^4*d^2+b^3*c^6)/(a*b)^(1/2)*arctan(b* 
x/(a*b)^(1/2))))-c^2*d^3/(a*d^2+b*c^2)^3/(d*x+c)-2*c*d^3*(a*d^2-2*b*c^2)/( 
a*d^2+b*c^2)^4*ln(d*x+c)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1356 vs. \(2 (309) = 618\).

Time = 5.88 (sec) , antiderivative size = 2736, normalized size of antiderivative = 8.47 \[ \int \frac {x^2}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/16*(8*a^3*b^3*c^6*d - 32*a^4*b^2*c^4*d^3 - 40*a^5*b*c^2*d^5 + 2*(a*b^5* 
c^6*d - 19*a^2*b^4*c^4*d^3 - 17*a^3*b^3*c^2*d^5 + 3*a^4*b^2*d^7)*x^4 + 2*( 
a*b^5*c^7 - 3*a^2*b^4*c^5*d^2 - 9*a^3*b^3*c^3*d^4 - 5*a^4*b^2*c*d^6)*x^3 + 
 2*(7*a^2*b^4*c^6*d - 29*a^3*b^3*c^4*d^3 - 31*a^4*b^2*c^2*d^5 + 5*a^5*b*d^ 
7)*x^2 - (a^2*b^3*c^7 + 13*a^3*b^2*c^5*d^2 - 33*a^4*b*c^3*d^4 + 3*a^5*c*d^ 
6 + (b^5*c^6*d + 13*a*b^4*c^4*d^3 - 33*a^2*b^3*c^2*d^5 + 3*a^3*b^2*d^7)*x^ 
5 + (b^5*c^7 + 13*a*b^4*c^5*d^2 - 33*a^2*b^3*c^3*d^4 + 3*a^3*b^2*c*d^6)*x^ 
4 + 2*(a*b^4*c^6*d + 13*a^2*b^3*c^4*d^3 - 33*a^3*b^2*c^2*d^5 + 3*a^4*b*d^7 
)*x^3 + 2*(a*b^4*c^7 + 13*a^2*b^3*c^5*d^2 - 33*a^3*b^2*c^3*d^4 + 3*a^4*b*c 
*d^6)*x^2 + (a^2*b^3*c^6*d + 13*a^3*b^2*c^4*d^3 - 33*a^4*b*c^2*d^5 + 3*a^5 
*d^7)*x)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 2*(a^2 
*b^4*c^7 + 9*a^3*b^3*c^5*d^2 + 15*a^4*b^2*c^3*d^4 + 7*a^5*b*c*d^6)*x - 16* 
(2*a^4*b^2*c^4*d^3 - a^5*b*c^2*d^5 + (2*a^2*b^4*c^3*d^4 - a^3*b^3*c*d^6)*x 
^5 + (2*a^2*b^4*c^4*d^3 - a^3*b^3*c^2*d^5)*x^4 + 2*(2*a^3*b^3*c^3*d^4 - a^ 
4*b^2*c*d^6)*x^3 + 2*(2*a^3*b^3*c^4*d^3 - a^4*b^2*c^2*d^5)*x^2 + (2*a^4*b^ 
2*c^3*d^4 - a^5*b*c*d^6)*x)*log(b*x^2 + a) + 32*(2*a^4*b^2*c^4*d^3 - a^5*b 
*c^2*d^5 + (2*a^2*b^4*c^3*d^4 - a^3*b^3*c*d^6)*x^5 + (2*a^2*b^4*c^4*d^3 - 
a^3*b^3*c^2*d^5)*x^4 + 2*(2*a^3*b^3*c^3*d^4 - a^4*b^2*c*d^6)*x^3 + 2*(2*a^ 
3*b^3*c^4*d^3 - a^4*b^2*c^2*d^5)*x^2 + (2*a^4*b^2*c^3*d^4 - a^5*b*c*d^6)*x 
)*log(d*x + c))/(a^4*b^5*c^9 + 4*a^5*b^4*c^7*d^2 + 6*a^6*b^3*c^5*d^4 + ...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 741 vs. \(2 (309) = 618\).

Time = 0.14 (sec) , antiderivative size = 741, normalized size of antiderivative = 2.29 \[ \int \frac {x^2}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx=-\frac {{\left (2 \, b c^{3} d^{3} - a c d^{5}\right )} \log \left (b x^{2} + a\right )}{b^{4} c^{8} + 4 \, a b^{3} c^{6} d^{2} + 6 \, a^{2} b^{2} c^{4} d^{4} + 4 \, a^{3} b c^{2} d^{6} + a^{4} d^{8}} + \frac {2 \, {\left (2 \, b c^{3} d^{3} - a c d^{5}\right )} \log \left (d x + c\right )}{b^{4} c^{8} + 4 \, a b^{3} c^{6} d^{2} + 6 \, a^{2} b^{2} c^{4} d^{4} + 4 \, a^{3} b c^{2} d^{6} + a^{4} d^{8}} + \frac {{\left (b^{3} c^{6} + 13 \, a b^{2} c^{4} d^{2} - 33 \, a^{2} b c^{2} d^{4} + 3 \, a^{3} d^{6}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, {\left (a b^{4} c^{8} + 4 \, a^{2} b^{3} c^{6} d^{2} + 6 \, a^{3} b^{2} c^{4} d^{4} + 4 \, a^{4} b c^{2} d^{6} + a^{5} d^{8}\right )} \sqrt {a b}} + \frac {4 \, a^{2} b c^{4} d - 20 \, a^{3} c^{2} d^{3} + {\left (b^{3} c^{4} d - 20 \, a b^{2} c^{2} d^{3} + 3 \, a^{2} b d^{5}\right )} x^{4} + {\left (b^{3} c^{5} - 4 \, a b^{2} c^{3} d^{2} - 5 \, a^{2} b c d^{4}\right )} x^{3} + {\left (7 \, a b^{2} c^{4} d - 36 \, a^{2} b c^{2} d^{3} + 5 \, a^{3} d^{5}\right )} x^{2} - {\left (a b^{2} c^{5} + 8 \, a^{2} b c^{3} d^{2} + 7 \, a^{3} c d^{4}\right )} x}{8 \, {\left (a^{3} b^{3} c^{7} + 3 \, a^{4} b^{2} c^{5} d^{2} + 3 \, a^{5} b c^{3} d^{4} + a^{6} c d^{6} + {\left (a b^{5} c^{6} d + 3 \, a^{2} b^{4} c^{4} d^{3} + 3 \, a^{3} b^{3} c^{2} d^{5} + a^{4} b^{2} d^{7}\right )} x^{5} + {\left (a b^{5} c^{7} + 3 \, a^{2} b^{4} c^{5} d^{2} + 3 \, a^{3} b^{3} c^{3} d^{4} + a^{4} b^{2} c d^{6}\right )} x^{4} + 2 \, {\left (a^{2} b^{4} c^{6} d + 3 \, a^{3} b^{3} c^{4} d^{3} + 3 \, a^{4} b^{2} c^{2} d^{5} + a^{5} b d^{7}\right )} x^{3} + 2 \, {\left (a^{2} b^{4} c^{7} + 3 \, a^{3} b^{3} c^{5} d^{2} + 3 \, a^{4} b^{2} c^{3} d^{4} + a^{5} b c d^{6}\right )} x^{2} + {\left (a^{3} b^{3} c^{6} d + 3 \, a^{4} b^{2} c^{4} d^{3} + 3 \, a^{5} b c^{2} d^{5} + a^{6} d^{7}\right )} x\right )}} \] Input:

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

Output:

-(2*b*c^3*d^3 - a*c*d^5)*log(b*x^2 + a)/(b^4*c^8 + 4*a*b^3*c^6*d^2 + 6*a^2 
*b^2*c^4*d^4 + 4*a^3*b*c^2*d^6 + a^4*d^8) + 2*(2*b*c^3*d^3 - a*c*d^5)*log( 
d*x + c)/(b^4*c^8 + 4*a*b^3*c^6*d^2 + 6*a^2*b^2*c^4*d^4 + 4*a^3*b*c^2*d^6 
+ a^4*d^8) + 1/8*(b^3*c^6 + 13*a*b^2*c^4*d^2 - 33*a^2*b*c^2*d^4 + 3*a^3*d^ 
6)*arctan(b*x/sqrt(a*b))/((a*b^4*c^8 + 4*a^2*b^3*c^6*d^2 + 6*a^3*b^2*c^4*d 
^4 + 4*a^4*b*c^2*d^6 + a^5*d^8)*sqrt(a*b)) + 1/8*(4*a^2*b*c^4*d - 20*a^3*c 
^2*d^3 + (b^3*c^4*d - 20*a*b^2*c^2*d^3 + 3*a^2*b*d^5)*x^4 + (b^3*c^5 - 4*a 
*b^2*c^3*d^2 - 5*a^2*b*c*d^4)*x^3 + (7*a*b^2*c^4*d - 36*a^2*b*c^2*d^3 + 5* 
a^3*d^5)*x^2 - (a*b^2*c^5 + 8*a^2*b*c^3*d^2 + 7*a^3*c*d^4)*x)/(a^3*b^3*c^7 
 + 3*a^4*b^2*c^5*d^2 + 3*a^5*b*c^3*d^4 + a^6*c*d^6 + (a*b^5*c^6*d + 3*a^2* 
b^4*c^4*d^3 + 3*a^3*b^3*c^2*d^5 + a^4*b^2*d^7)*x^5 + (a*b^5*c^7 + 3*a^2*b^ 
4*c^5*d^2 + 3*a^3*b^3*c^3*d^4 + a^4*b^2*c*d^6)*x^4 + 2*(a^2*b^4*c^6*d + 3* 
a^3*b^3*c^4*d^3 + 3*a^4*b^2*c^2*d^5 + a^5*b*d^7)*x^3 + 2*(a^2*b^4*c^7 + 3* 
a^3*b^3*c^5*d^2 + 3*a^4*b^2*c^3*d^4 + a^5*b*c*d^6)*x^2 + (a^3*b^3*c^6*d + 
3*a^4*b^2*c^4*d^3 + 3*a^5*b*c^2*d^5 + a^6*d^7)*x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 600, normalized size of antiderivative = 1.86 \[ \int \frac {x^2}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx=-\frac {c^{2} d^{9}}{{\left (b^{3} c^{6} d^{6} + 3 \, a b^{2} c^{4} d^{8} + 3 \, a^{2} b c^{2} d^{10} + a^{3} d^{12}\right )} {\left (d x + c\right )}} - \frac {{\left (2 \, b c^{3} d^{3} - a c d^{5}\right )} \log \left (b - \frac {2 \, b c}{d x + c} + \frac {b c^{2}}{{\left (d x + c\right )}^{2}} + \frac {a d^{2}}{{\left (d x + c\right )}^{2}}\right )}{b^{4} c^{8} + 4 \, a b^{3} c^{6} d^{2} + 6 \, a^{2} b^{2} c^{4} d^{4} + 4 \, a^{3} b c^{2} d^{6} + a^{4} d^{8}} + \frac {{\left (b^{3} c^{6} d^{2} + 13 \, a b^{2} c^{4} d^{4} - 33 \, a^{2} b c^{2} d^{6} + 3 \, a^{3} d^{8}\right )} \arctan \left (\frac {b c - \frac {b c^{2}}{d x + c} - \frac {a d^{2}}{d x + c}}{\sqrt {a b} d}\right )}{8 \, {\left (a b^{4} c^{8} + 4 \, a^{2} b^{3} c^{6} d^{2} + 6 \, a^{3} b^{2} c^{4} d^{4} + 4 \, a^{4} b c^{2} d^{6} + a^{5} d^{8}\right )} \sqrt {a b} d^{2}} + \frac {b^{4} c^{5} d - 22 \, a b^{3} c^{3} d^{3} + 17 \, a^{2} b^{2} c d^{5} - \frac {3 \, b^{4} c^{6} d^{2} - 77 \, a b^{3} c^{4} d^{4} + 77 \, a^{2} b^{2} c^{2} d^{6} - 3 \, a^{3} b d^{8}}{{\left (d x + c\right )} d} + \frac {3 \, b^{4} c^{7} d^{3} - 89 \, a b^{3} c^{5} d^{5} + 85 \, a^{2} b^{2} c^{3} d^{7} + 17 \, a^{3} b c d^{9}}{{\left (d x + c\right )}^{2} d^{2}} - \frac {b^{4} c^{8} d^{4} - 34 \, a b^{3} c^{6} d^{6} + 20 \, a^{2} b^{2} c^{4} d^{8} + 50 \, a^{3} b c^{2} d^{10} - 5 \, a^{4} d^{12}}{{\left (d x + c\right )}^{3} d^{3}}}{8 \, {\left (b c^{2} + a d^{2}\right )}^{4} a {\left (b - \frac {2 \, b c}{d x + c} + \frac {b c^{2}}{{\left (d x + c\right )}^{2}} + \frac {a d^{2}}{{\left (d x + c\right )}^{2}}\right )}^{2}} \] Input:

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

Output:

-c^2*d^9/((b^3*c^6*d^6 + 3*a*b^2*c^4*d^8 + 3*a^2*b*c^2*d^10 + a^3*d^12)*(d 
*x + c)) - (2*b*c^3*d^3 - a*c*d^5)*log(b - 2*b*c/(d*x + c) + b*c^2/(d*x + 
c)^2 + a*d^2/(d*x + c)^2)/(b^4*c^8 + 4*a*b^3*c^6*d^2 + 6*a^2*b^2*c^4*d^4 + 
 4*a^3*b*c^2*d^6 + a^4*d^8) + 1/8*(b^3*c^6*d^2 + 13*a*b^2*c^4*d^4 - 33*a^2 
*b*c^2*d^6 + 3*a^3*d^8)*arctan((b*c - b*c^2/(d*x + c) - a*d^2/(d*x + c))/( 
sqrt(a*b)*d))/((a*b^4*c^8 + 4*a^2*b^3*c^6*d^2 + 6*a^3*b^2*c^4*d^4 + 4*a^4* 
b*c^2*d^6 + a^5*d^8)*sqrt(a*b)*d^2) + 1/8*(b^4*c^5*d - 22*a*b^3*c^3*d^3 + 
17*a^2*b^2*c*d^5 - (3*b^4*c^6*d^2 - 77*a*b^3*c^4*d^4 + 77*a^2*b^2*c^2*d^6 
- 3*a^3*b*d^8)/((d*x + c)*d) + (3*b^4*c^7*d^3 - 89*a*b^3*c^5*d^5 + 85*a^2* 
b^2*c^3*d^7 + 17*a^3*b*c*d^9)/((d*x + c)^2*d^2) - (b^4*c^8*d^4 - 34*a*b^3* 
c^6*d^6 + 20*a^2*b^2*c^4*d^8 + 50*a^3*b*c^2*d^10 - 5*a^4*d^12)/((d*x + c)^ 
3*d^3))/((b*c^2 + a*d^2)^4*a*(b - 2*b*c/(d*x + c) + b*c^2/(d*x + c)^2 + a* 
d^2/(d*x + c)^2)^2)
 

Mupad [B] (verification not implemented)

Time = 8.45 (sec) , antiderivative size = 1357, normalized size of antiderivative = 4.20 \[ \int \frac {x^2}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

((x^2*(5*a^2*d^5 + 7*b^2*c^4*d - 36*a*b*c^2*d^3))/(8*(a*d^2 + b*c^2)*(a^2* 
d^4 + b^2*c^4 + 2*a*b*c^2*d^2)) - (x*(b*c^3 + 7*a*c*d^2))/(8*(a^2*d^4 + b^ 
2*c^4 + 2*a*b*c^2*d^2)) + (x^4*(3*a^2*b*d^5 + b^3*c^4*d - 20*a*b^2*c^2*d^3 
))/(8*a*(a^3*d^6 + b^3*c^6 + 3*a*b^2*c^4*d^2 + 3*a^2*b*c^2*d^4)) - (c*(5*a 
^2*c*d^3 - a*b*c^3*d))/(2*(a*d^2 + b*c^2)*(a^2*d^4 + b^2*c^4 + 2*a*b*c^2*d 
^2)) + (x^3*(b^2*c^3 - 5*a*b*c*d^2))/(8*a*(a^2*d^4 + b^2*c^4 + 2*a*b*c^2*d 
^2)))/(a^2*c + b^2*c*x^4 + b^2*d*x^5 + a^2*d*x + 2*a*b*c*x^2 + 2*a*b*d*x^3 
) - (log(b^7*c^16*(-a^3*b)^(3/2) - 9*a^11*d^16*(-a^3*b)^(1/2) + a^4*b^9*c^ 
16*x + 3828*a^3*c^4*d^12*(-a^3*b)^(5/2) + 2124*a^7*c^2*d^14*(-a^3*b)^(3/2) 
 - 8596*b^3*c^10*d^6*(-a^3*b)^(5/2) + 9*a^12*b*d^16*x + 156*a^2*b^5*c^12*d 
^4*(-a^3*b)^(3/2) + 28*a^5*b^8*c^14*d^2*x + 156*a^6*b^7*c^12*d^4*x + 8596* 
a^7*b^6*c^10*d^6*x + 8782*a^8*b^5*c^8*d^8*x - 5628*a^9*b^4*c^6*d^10*x - 38 
28*a^10*b^3*c^4*d^12*x + 2124*a^11*b^2*c^2*d^14*x - 8782*a*b^2*c^8*d^8*(-a 
^3*b)^(5/2) + 5628*a^2*b*c^6*d^10*(-a^3*b)^(5/2) + 28*a*b^6*c^14*d^2*(-a^3 
*b)^(3/2))*(b^2*(2*a^3*c^3*d^3 + (13*a*c^4*d^2*(-a^3*b)^(1/2))/16) - b*(a^ 
4*c*d^5 + (33*a^2*c^2*d^4*(-a^3*b)^(1/2))/16) + (3*a^3*d^6*(-a^3*b)^(1/2)) 
/16 + (b^3*c^6*(-a^3*b)^(1/2))/16))/(a^7*b*d^8 + a^3*b^5*c^8 + 4*a^4*b^4*c 
^6*d^2 + 6*a^5*b^3*c^4*d^4 + 4*a^6*b^2*c^2*d^6) + (log(9*a^11*d^16*(-a^3*b 
)^(1/2) - b^7*c^16*(-a^3*b)^(3/2) + a^4*b^9*c^16*x - 3828*a^3*c^4*d^12*(-a 
^3*b)^(5/2) - 2124*a^7*c^2*d^14*(-a^3*b)^(3/2) + 8596*b^3*c^10*d^6*(-a^...
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 2011, normalized size of antiderivative = 6.23 \[ \int \frac {x^2}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**5*c**2*d**6 + 3*sqrt(b 
)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**5*c*d**7*x - 33*sqrt(b)*sqrt(a) 
*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4*b*c**4*d**4 - 33*sqrt(b)*sqrt(a)*atan( 
(b*x)/(sqrt(b)*sqrt(a)))*a**4*b*c**3*d**5*x + 6*sqrt(b)*sqrt(a)*atan((b*x) 
/(sqrt(b)*sqrt(a)))*a**4*b*c**2*d**6*x**2 + 6*sqrt(b)*sqrt(a)*atan((b*x)/( 
sqrt(b)*sqrt(a)))*a**4*b*c*d**7*x**3 + 13*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt 
(b)*sqrt(a)))*a**3*b**2*c**6*d**2 + 13*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b) 
*sqrt(a)))*a**3*b**2*c**5*d**3*x - 66*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)* 
sqrt(a)))*a**3*b**2*c**4*d**4*x**2 - 66*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b 
)*sqrt(a)))*a**3*b**2*c**3*d**5*x**3 + 3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt( 
b)*sqrt(a)))*a**3*b**2*c**2*d**6*x**4 + 3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt 
(b)*sqrt(a)))*a**3*b**2*c*d**7*x**5 + sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)* 
sqrt(a)))*a**2*b**3*c**8 + sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a 
**2*b**3*c**7*d*x + 26*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2* 
b**3*c**6*d**2*x**2 + 26*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a** 
2*b**3*c**5*d**3*x**3 - 33*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a 
**2*b**3*c**4*d**4*x**4 - 33*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a))) 
*a**2*b**3*c**3*d**5*x**5 + 2*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)) 
)*a*b**4*c**8*x**2 + 2*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b** 
4*c**7*d*x**3 + 13*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**4...