\(\int \frac {x^2 (c+d x)^2 (A+B x+C x^2)}{(a+b x^2)^{3/2}} \, dx\) [141]

Optimal result

Integrand size = 32, antiderivative size = 301 \[ \int \frac {x^2 (c+d x)^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {a (b c (B c+2 A d)-a d (2 c C+B d))}{b^3 \sqrt {a+b x^2}}-\frac {\left (A b \left (b c^2-a d^2\right )+a \left (a C d^2-b c (c C+2 B d)\right )\right ) x}{b^3 \sqrt {a+b x^2}}+\frac {(b c (B c+2 A d)-2 a d (2 c C+B d)) \sqrt {a+b x^2}}{b^3}-\frac {\left (7 a C d^2-4 b \left (c^2 C+2 B c d+A d^2\right )\right ) x \sqrt {a+b x^2}}{8 b^3}+\frac {C d^2 x^3 \sqrt {a+b x^2}}{4 b^2}+\frac {d (2 c C+B d) \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {\left (4 A b \left (2 b c^2-3 a d^2\right )+3 a \left (5 a C d^2-4 b c (c C+2 B d)\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{7/2}} \] Output:

a*(b*c*(2*A*d+B*c)-a*d*(B*d+2*C*c))/b^3/(b*x^2+a)^(1/2)-(A*b*(-a*d^2+b*c^2 
)+a*(a*C*d^2-b*c*(2*B*d+C*c)))*x/b^3/(b*x^2+a)^(1/2)+(b*c*(2*A*d+B*c)-2*a* 
d*(B*d+2*C*c))*(b*x^2+a)^(1/2)/b^3-1/8*(7*a*C*d^2-4*b*(A*d^2+2*B*c*d+C*c^2 
))*x*(b*x^2+a)^(1/2)/b^3+1/4*C*d^2*x^3*(b*x^2+a)^(1/2)/b^2+1/3*d*(B*d+2*C* 
c)*(b*x^2+a)^(3/2)/b^3+1/8*(4*A*b*(-3*a*d^2+2*b*c^2)+3*a*(5*a*C*d^2-4*b*c* 
(2*B*d+C*c)))*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 1.80 (sec) , antiderivative size = 281, normalized size of antiderivative = 0.93 \[ \int \frac {x^2 (c+d x)^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {b} \left (-a^2 d (128 c C+64 B d+45 C d x)+a b \left (12 A d (8 c+3 d x)+C x \left (36 c^2-64 c d x-15 d^2 x^2\right )+8 B \left (6 c^2+9 c d x-4 d^2 x^2\right )\right )+2 b^2 x \left (6 A \left (-2 c^2+4 c d x+d^2 x^2\right )+x \left (4 B \left (3 c^2+3 c d x+d^2 x^2\right )+C x \left (6 c^2+8 c d x+3 d^2 x^2\right )\right )\right )\right )}{\sqrt {a+b x^2}}+72 a b \left (c^2 C+2 B c d+A d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a}-\sqrt {a+b x^2}}\right )+6 \left (8 A b^2 c^2+15 a^2 C d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{24 b^{7/2}} \] Input:

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

Output:

((Sqrt[b]*(-(a^2*d*(128*c*C + 64*B*d + 45*C*d*x)) + a*b*(12*A*d*(8*c + 3*d 
*x) + C*x*(36*c^2 - 64*c*d*x - 15*d^2*x^2) + 8*B*(6*c^2 + 9*c*d*x - 4*d^2* 
x^2)) + 2*b^2*x*(6*A*(-2*c^2 + 4*c*d*x + d^2*x^2) + x*(4*B*(3*c^2 + 3*c*d* 
x + d^2*x^2) + C*x*(6*c^2 + 8*c*d*x + 3*d^2*x^2)))))/Sqrt[a + b*x^2] + 72* 
a*b*(c^2*C + 2*B*c*d + A*d^2)*ArcTanh[(Sqrt[b]*x)/(Sqrt[a] - Sqrt[a + b*x^ 
2])] + 6*(8*A*b^2*c^2 + 15*a^2*C*d^2)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt 
[a + b*x^2])])/(24*b^(7/2))
 

Rubi [A] (verified)

Time = 1.65 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2176, 25, 2185, 2185, 27, 676, 224, 219}

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.

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

\(\Big \downarrow \) 2176

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\frac {\int \frac {(c+d x) \left (-a b (c C-4 B d) x^2 d^3+a (4 A b c-7 a C c-8 a B d) d^3-a \left (15 a C d^2+b \left (C c^2-4 B d c-12 A d^2\right )\right ) x d^2\right )}{\sqrt {b x^2+a}}dx}{4 b d^3}+\frac {a C \sqrt {a+b x^2} (c+d x)^3}{4 b d}}{a b}+\frac {(c+d x)^2 \left (a B-b x \left (A-\frac {a C}{b}\right )\right )}{b^2 \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\frac {\frac {\int \frac {a b d^4 (c+d x) \left (d (12 A b c-19 a C c-32 a B d)-\left (45 a C d^2+2 b \left (C c^2-4 B d c-18 A d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx}{3 b d^2}-\frac {1}{3} a d^2 \sqrt {a+b x^2} (c+d x)^2 (c C-4 B d)}{4 b d^3}+\frac {a C \sqrt {a+b x^2} (c+d x)^3}{4 b d}}{a b}+\frac {(c+d x)^2 \left (a B-b x \left (A-\frac {a C}{b}\right )\right )}{b^2 \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {1}{3} a d^2 \int \frac {(c+d x) \left (d (12 A b c-19 a C c-32 a B d)-\left (45 a C d^2+2 b \left (C c^2-4 B d c-18 A d^2\right )\right ) x\right )}{\sqrt {b x^2+a}}dx-\frac {1}{3} a d^2 \sqrt {a+b x^2} (c+d x)^2 (c C-4 B d)}{4 b d^3}+\frac {a C \sqrt {a+b x^2} (c+d x)^3}{4 b d}}{a b}+\frac {(c+d x)^2 \left (a B-b x \left (A-\frac {a C}{b}\right )\right )}{b^2 \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {\frac {\frac {1}{3} a d^2 \left (\frac {3 d \left (4 A b \left (2 b c^2-3 a d^2\right )+3 a \left (5 a C d^2-4 b c (2 B d+c C)\right )\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}-\frac {2 \sqrt {a+b x^2} \left (16 a d^2 (B d+2 c C)+b c \left (-24 A d^2-4 B c d+c^2 C\right )\right )}{b}-\frac {d x \sqrt {a+b x^2} \left (45 a C d^2+b \left (-36 A d^2-8 B c d+2 c^2 C\right )\right )}{2 b}\right )-\frac {1}{3} a d^2 \sqrt {a+b x^2} (c+d x)^2 (c C-4 B d)}{4 b d^3}+\frac {a C \sqrt {a+b x^2} (c+d x)^3}{4 b d}}{a b}+\frac {(c+d x)^2 \left (a B-b x \left (A-\frac {a C}{b}\right )\right )}{b^2 \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\frac {1}{3} a d^2 \left (\frac {3 d \left (4 A b \left (2 b c^2-3 a d^2\right )+3 a \left (5 a C d^2-4 b c (2 B d+c C)\right )\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}-\frac {2 \sqrt {a+b x^2} \left (16 a d^2 (B d+2 c C)+b c \left (-24 A d^2-4 B c d+c^2 C\right )\right )}{b}-\frac {d x \sqrt {a+b x^2} \left (45 a C d^2+b \left (-36 A d^2-8 B c d+2 c^2 C\right )\right )}{2 b}\right )-\frac {1}{3} a d^2 \sqrt {a+b x^2} (c+d x)^2 (c C-4 B d)}{4 b d^3}+\frac {a C \sqrt {a+b x^2} (c+d x)^3}{4 b d}}{a b}+\frac {(c+d x)^2 \left (a B-b x \left (A-\frac {a C}{b}\right )\right )}{b^2 \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {1}{3} a d^2 \left (\frac {3 d \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (4 A b \left (2 b c^2-3 a d^2\right )+3 a \left (5 a C d^2-4 b c (2 B d+c C)\right )\right )}{2 b^{3/2}}-\frac {2 \sqrt {a+b x^2} \left (16 a d^2 (B d+2 c C)+b c \left (-24 A d^2-4 B c d+c^2 C\right )\right )}{b}-\frac {d x \sqrt {a+b x^2} \left (45 a C d^2+b \left (-36 A d^2-8 B c d+2 c^2 C\right )\right )}{2 b}\right )-\frac {1}{3} a d^2 \sqrt {a+b x^2} (c+d x)^2 (c C-4 B d)}{4 b d^3}+\frac {a C \sqrt {a+b x^2} (c+d x)^3}{4 b d}}{a b}+\frac {(c+d x)^2 \left (a B-b x \left (A-\frac {a C}{b}\right )\right )}{b^2 \sqrt {a+b x^2}}\)

Input:

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

Output:

((a*B - b*(A - (a*C)/b)*x)*(c + d*x)^2)/(b^2*Sqrt[a + b*x^2]) + ((a*C*(c + 
 d*x)^3*Sqrt[a + b*x^2])/(4*b*d) + (-1/3*(a*d^2*(c*C - 4*B*d)*(c + d*x)^2* 
Sqrt[a + b*x^2]) + (a*d^2*((-2*(16*a*d^2*(2*c*C + B*d) + b*c*(c^2*C - 4*B* 
c*d - 24*A*d^2))*Sqrt[a + b*x^2])/b - (d*(45*a*C*d^2 + b*(2*c^2*C - 8*B*c* 
d - 36*A*d^2))*x*Sqrt[a + b*x^2])/(2*b) + (3*d*(4*A*b*(2*b*c^2 - 3*a*d^2) 
+ 3*a*(5*a*C*d^2 - 4*b*c*(c*C + 2*B*d)))*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^ 
2]])/(2*b^(3/2))))/3)/(4*b*d^3))/(a*b)
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

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 2185

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x 
)^(q - 2)*(a*e^2*(m + q - 1) - b*d^2*(m + q + 2*p + 1) - 2*b*d*e*(m + q + p 
)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d 
, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && 
True) &&  !(IGtQ[m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 
 1/2, 0]))
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 299, normalized size of antiderivative = 0.99

method	result	size
risch	\(\frac {\left (6 C b \,d^{2} x^{3}+8 B b \,d^{2} x^{2}+16 C b c d \,x^{2}+12 A \,d^{2} x b +24 B c d x b -21 C a \,d^{2} x +12 x b \,c^{2} C +48 A b c d -40 a B \,d^{2}+24 b B \,c^{2}-80 C a c d \right ) \sqrt {b \,x^{2}+a}}{24 b^{3}}-\frac {-\frac {8 a \left (2 A b c d -a B \,d^{2}+b B \,c^{2}-2 C a c d \right )}{\sqrt {b \,x^{2}+a}}+b \left (12 A a b \,d^{2}-8 A \,b^{2} c^{2}+24 B a c d b -15 a^{2} C \,d^{2}+12 a b \,c^{2} C \right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )-\frac {7 C \,a^{2} d^{2} x}{\sqrt {b \,x^{2}+a}}+\frac {4 A a b \,d^{2} x}{\sqrt {b \,x^{2}+a}}+\frac {4 C a b \,c^{2} x}{\sqrt {b \,x^{2}+a}}+\frac {8 B a b c d x}{\sqrt {b \,x^{2}+a}}}{8 b^{3}}\)	\(299\)
default	\(c \left (2 A d +B c \right ) \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )+d \left (B d +2 C c \right ) \left (\frac {x^{4}}{3 b \sqrt {b \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )}{3 b}\right )+\left (A \,d^{2}+2 B c d +C \,c^{2}\right ) \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )+A \,c^{2} \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )+C \,d^{2} \left (\frac {x^{5}}{4 b \sqrt {b \,x^{2}+a}}-\frac {5 a \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )}{4 b}\right )\)	\(317\)

Input:

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

Output:

1/24*(6*C*b*d^2*x^3+8*B*b*d^2*x^2+16*C*b*c*d*x^2+12*A*b*d^2*x+24*B*b*c*d*x 
-21*C*a*d^2*x+12*C*b*c^2*x+48*A*b*c*d-40*B*a*d^2+24*B*b*c^2-80*C*a*c*d)*(b 
*x^2+a)^(1/2)/b^3-1/8/b^3*(-8*a*(2*A*b*c*d-B*a*d^2+B*b*c^2-2*C*a*c*d)/(b*x 
^2+a)^(1/2)+b*(12*A*a*b*d^2-8*A*b^2*c^2+24*B*a*b*c*d-15*C*a^2*d^2+12*C*a*b 
*c^2)*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))-7*C*a 
^2*d^2*x/(b*x^2+a)^(1/2)+4*A*a*b*d^2*x/(b*x^2+a)^(1/2)+4*C*a*b*c^2*x/(b*x^ 
2+a)^(1/2)+8*B*a*b*c*d*x/(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 739, normalized size of antiderivative = 2.46 \[ \int \frac {x^2 (c+d x)^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/48*(3*(24*B*a^2*b*c*d + 4*(3*C*a^2*b - 2*A*a*b^2)*c^2 - 3*(5*C*a^3 - 4* 
A*a^2*b)*d^2 + (24*B*a*b^2*c*d + 4*(3*C*a*b^2 - 2*A*b^3)*c^2 - 3*(5*C*a^2* 
b - 4*A*a*b^2)*d^2)*x^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)* 
x - a) + 2*(6*C*b^3*d^2*x^5 + 48*B*a*b^2*c^2 - 64*B*a^2*b*d^2 + 8*(2*C*b^3 
*c*d + B*b^3*d^2)*x^4 + 3*(4*C*b^3*c^2 + 8*B*b^3*c*d - (5*C*a*b^2 - 4*A*b^ 
3)*d^2)*x^3 - 32*(4*C*a^2*b - 3*A*a*b^2)*c*d + 8*(3*B*b^3*c^2 - 4*B*a*b^2* 
d^2 - 2*(4*C*a*b^2 - 3*A*b^3)*c*d)*x^2 + 3*(24*B*a*b^2*c*d + 4*(3*C*a*b^2 
- 2*A*b^3)*c^2 - 3*(5*C*a^2*b - 4*A*a*b^2)*d^2)*x)*sqrt(b*x^2 + a))/(b^5*x 
^2 + a*b^4), 1/24*(3*(24*B*a^2*b*c*d + 4*(3*C*a^2*b - 2*A*a*b^2)*c^2 - 3*( 
5*C*a^3 - 4*A*a^2*b)*d^2 + (24*B*a*b^2*c*d + 4*(3*C*a*b^2 - 2*A*b^3)*c^2 - 
 3*(5*C*a^2*b - 4*A*a*b^2)*d^2)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 
 + a)) + (6*C*b^3*d^2*x^5 + 48*B*a*b^2*c^2 - 64*B*a^2*b*d^2 + 8*(2*C*b^3*c 
*d + B*b^3*d^2)*x^4 + 3*(4*C*b^3*c^2 + 8*B*b^3*c*d - (5*C*a*b^2 - 4*A*b^3) 
*d^2)*x^3 - 32*(4*C*a^2*b - 3*A*a*b^2)*c*d + 8*(3*B*b^3*c^2 - 4*B*a*b^2*d^ 
2 - 2*(4*C*a*b^2 - 3*A*b^3)*c*d)*x^2 + 3*(24*B*a*b^2*c*d + 4*(3*C*a*b^2 - 
2*A*b^3)*c^2 - 3*(5*C*a^2*b - 4*A*a*b^2)*d^2)*x)*sqrt(b*x^2 + a))/(b^5*x^2 
 + a*b^4)]
 

Sympy [F]

\[ \int \frac {x^2 (c+d x)^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {x^{2} \left (c + d x\right )^{2} \left (A + B x + C x^{2}\right )}{\left (a + b x^{2}\right )^{\frac {3}{2}}}\, dx \] Input:

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

Output:

Integral(x**2*(c + d*x)**2*(A + B*x + C*x**2)/(a + b*x**2)**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.19 \[ \int \frac {x^2 (c+d x)^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {C d^{2} x^{5}}{4 \, \sqrt {b x^{2} + a} b} - \frac {5 \, C a d^{2} x^{3}}{8 \, \sqrt {b x^{2} + a} b^{2}} + \frac {{\left (2 \, C c d + B d^{2}\right )} x^{4}}{3 \, \sqrt {b x^{2} + a} b} - \frac {A c^{2} x}{\sqrt {b x^{2} + a} b} - \frac {15 \, C a^{2} d^{2} x}{8 \, \sqrt {b x^{2} + a} b^{3}} + \frac {{\left (C c^{2} + 2 \, B c d + A d^{2}\right )} x^{3}}{2 \, \sqrt {b x^{2} + a} b} + \frac {A c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} + \frac {15 \, C a^{2} d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {7}{2}}} - \frac {4 \, {\left (2 \, C c d + B d^{2}\right )} a x^{2}}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {{\left (B c^{2} + 2 \, A c d\right )} x^{2}}{\sqrt {b x^{2} + a} b} + \frac {3 \, {\left (C c^{2} + 2 \, B c d + A d^{2}\right )} a x}{2 \, \sqrt {b x^{2} + a} b^{2}} - \frac {3 \, {\left (C c^{2} + 2 \, B c d + A d^{2}\right )} a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} - \frac {8 \, {\left (2 \, C c d + B d^{2}\right )} a^{2}}{3 \, \sqrt {b x^{2} + a} b^{3}} + \frac {2 \, {\left (B c^{2} + 2 \, A c d\right )} a}{\sqrt {b x^{2} + a} b^{2}} \] Input:

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

Output:

1/4*C*d^2*x^5/(sqrt(b*x^2 + a)*b) - 5/8*C*a*d^2*x^3/(sqrt(b*x^2 + a)*b^2) 
+ 1/3*(2*C*c*d + B*d^2)*x^4/(sqrt(b*x^2 + a)*b) - A*c^2*x/(sqrt(b*x^2 + a) 
*b) - 15/8*C*a^2*d^2*x/(sqrt(b*x^2 + a)*b^3) + 1/2*(C*c^2 + 2*B*c*d + A*d^ 
2)*x^3/(sqrt(b*x^2 + a)*b) + A*c^2*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 15/8*C 
*a^2*d^2*arcsinh(b*x/sqrt(a*b))/b^(7/2) - 4/3*(2*C*c*d + B*d^2)*a*x^2/(sqr 
t(b*x^2 + a)*b^2) + (B*c^2 + 2*A*c*d)*x^2/(sqrt(b*x^2 + a)*b) + 3/2*(C*c^2 
 + 2*B*c*d + A*d^2)*a*x/(sqrt(b*x^2 + a)*b^2) - 3/2*(C*c^2 + 2*B*c*d + A*d 
^2)*a*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 8/3*(2*C*c*d + B*d^2)*a^2/(sqrt(b*x 
^2 + a)*b^3) + 2*(B*c^2 + 2*A*c*d)*a/(sqrt(b*x^2 + a)*b^2)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.03 \[ \int \frac {x^2 (c+d x)^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left ({\left (2 \, {\left (\frac {3 \, C d^{2} x}{b} + \frac {4 \, {\left (2 \, C b^{5} c d + B b^{5} d^{2}\right )}}{b^{6}}\right )} x + \frac {3 \, {\left (4 \, C b^{5} c^{2} + 8 \, B b^{5} c d - 5 \, C a b^{4} d^{2} + 4 \, A b^{5} d^{2}\right )}}{b^{6}}\right )} x + \frac {8 \, {\left (3 \, B b^{5} c^{2} - 8 \, C a b^{4} c d + 6 \, A b^{5} c d - 4 \, B a b^{4} d^{2}\right )}}{b^{6}}\right )} x + \frac {3 \, {\left (12 \, C a b^{4} c^{2} - 8 \, A b^{5} c^{2} + 24 \, B a b^{4} c d - 15 \, C a^{2} b^{3} d^{2} + 12 \, A a b^{4} d^{2}\right )}}{b^{6}}\right )} x + \frac {16 \, {\left (3 \, B a b^{4} c^{2} - 8 \, C a^{2} b^{3} c d + 6 \, A a b^{4} c d - 4 \, B a^{2} b^{3} d^{2}\right )}}{b^{6}}}{24 \, \sqrt {b x^{2} + a}} + \frac {{\left (12 \, C a b c^{2} - 8 \, A b^{2} c^{2} + 24 \, B a b c d - 15 \, C a^{2} d^{2} + 12 \, A a b d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {7}{2}}} \] Input:

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

Output:

1/24*((((2*(3*C*d^2*x/b + 4*(2*C*b^5*c*d + B*b^5*d^2)/b^6)*x + 3*(4*C*b^5* 
c^2 + 8*B*b^5*c*d - 5*C*a*b^4*d^2 + 4*A*b^5*d^2)/b^6)*x + 8*(3*B*b^5*c^2 - 
 8*C*a*b^4*c*d + 6*A*b^5*c*d - 4*B*a*b^4*d^2)/b^6)*x + 3*(12*C*a*b^4*c^2 - 
 8*A*b^5*c^2 + 24*B*a*b^4*c*d - 15*C*a^2*b^3*d^2 + 12*A*a*b^4*d^2)/b^6)*x 
+ 16*(3*B*a*b^4*c^2 - 8*C*a^2*b^3*c*d + 6*A*a*b^4*c*d - 4*B*a^2*b^3*d^2)/b 
^6)/sqrt(b*x^2 + a) + 1/8*(12*C*a*b*c^2 - 8*A*b^2*c^2 + 24*B*a*b*c*d - 15* 
C*a^2*d^2 + 12*A*a*b*d^2)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^2 (c+d x)^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {x^{2} \left (d x +c \right )^{2} \left (C \,x^{2}+B x +A \right )}{\left (b \,x^{2}+a \right )^{\frac {3}{2}}}d x \] Input:

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

Output:

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