\(\int \frac {(c+d x)^2 (A+B x+C x^2)}{x^6 \sqrt {a+b x^2}} \, dx\) [113]

Optimal result

Integrand size = 32, antiderivative size = 267 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2\right )}{x^6 \sqrt {a+b x^2}} \, dx=-\frac {A c^2 \sqrt {a+b x^2}}{5 a x^5}-\frac {c (B c+2 A d) \sqrt {a+b x^2}}{4 a x^4}-\frac {\left (5 a c (c C+2 B d)-A \left (4 b c^2-5 a d^2\right )\right ) \sqrt {a+b x^2}}{15 a^2 x^3}+\frac {(3 b c (B c+2 A d)-4 a d (2 c C+B d)) \sqrt {a+b x^2}}{8 a^2 x^2}-\frac {\left (2 A b \left (4 b c^2-5 a d^2\right )+5 a \left (3 a C d^2-2 b c (c C+2 B d)\right )\right ) \sqrt {a+b x^2}}{15 a^3 x}-\frac {b (3 b c (B c+2 A d)-4 a d (2 c C+B d)) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{5/2}} \] Output:

-1/5*A*c^2*(b*x^2+a)^(1/2)/a/x^5-1/4*c*(2*A*d+B*c)*(b*x^2+a)^(1/2)/a/x^4-1 
/15*(5*a*c*(2*B*d+C*c)-A*(-5*a*d^2+4*b*c^2))*(b*x^2+a)^(1/2)/a^2/x^3+1/8*( 
3*b*c*(2*A*d+B*c)-4*a*d*(B*d+2*C*c))*(b*x^2+a)^(1/2)/a^2/x^2-1/15*(2*A*b*( 
-5*a*d^2+4*b*c^2)+5*a*(3*a*C*d^2-2*b*c*(2*B*d+C*c)))*(b*x^2+a)^(1/2)/a^3/x 
-1/8*b*(3*b*c*(2*A*d+B*c)-4*a*d*(B*d+2*C*c))*arctanh((b*x^2+a)^(1/2)/a^(1/ 
2))/a^(5/2)
 

Mathematica [A] (verified)

Time = 1.85 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.81 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2\right )}{x^6 \sqrt {a+b x^2}} \, dx=\frac {\frac {\sqrt {a+b x^2} \left (-64 A b^2 c^2 x^4+a b x^2 \left (5 c x (9 B c+16 c C x+32 B d x)+A \left (32 c^2+90 c d x+80 d^2 x^2\right )\right )-2 a^2 \left (2 A \left (6 c^2+15 c d x+10 d^2 x^2\right )+5 x \left (4 C x \left (c^2+3 c d x+3 d^2 x^2\right )+B \left (3 c^2+8 c d x+6 d^2 x^2\right )\right )\right )\right )}{x^5}+30 \sqrt {a} b (-3 b c (B c+2 A d)+4 a d (2 c C+B d)) \text {arctanh}\left (\frac {-\sqrt {b} x+\sqrt {a+b x^2}}{\sqrt {a}}\right )}{120 a^3} \] Input:

Integrate[((c + d*x)^2*(A + B*x + C*x^2))/(x^6*Sqrt[a + b*x^2]),x]
 

Output:

((Sqrt[a + b*x^2]*(-64*A*b^2*c^2*x^4 + a*b*x^2*(5*c*x*(9*B*c + 16*c*C*x + 
32*B*d*x) + A*(32*c^2 + 90*c*d*x + 80*d^2*x^2)) - 2*a^2*(2*A*(6*c^2 + 15*c 
*d*x + 10*d^2*x^2) + 5*x*(4*C*x*(c^2 + 3*c*d*x + 3*d^2*x^2) + B*(3*c^2 + 8 
*c*d*x + 6*d^2*x^2)))))/x^5 + 30*Sqrt[a]*b*(-3*b*c*(B*c + 2*A*d) + 4*a*d*( 
2*c*C + B*d))*ArcTanh[(-(Sqrt[b]*x) + Sqrt[a + b*x^2])/Sqrt[a]])/(120*a^3)
 

Rubi [A] (verified)

Time = 1.77 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2338, 25, 2338, 25, 2338, 27, 539, 27, 534, 243, 73, 221}

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.

\(\displaystyle \int \frac {(c+d x)^2 \left (A+B x+C x^2\right )}{x^6 \sqrt {a+b x^2}} \, dx\)

\(\Big \downarrow \) 2338

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2338

\(\displaystyle \frac {-\frac {\int -\frac {20 a^2 C d^2 x^2-5 a (3 b c (B c+2 A d)-4 a d (2 c C+B d)) x+4 a \left (5 a c (c C+2 B d)-A \left (4 b c^2-5 a d^2\right )\right )}{x^4 \sqrt {b x^2+a}}dx}{4 a}-\frac {5 c \sqrt {a+b x^2} (2 A d+B c)}{4 x^4}}{5 a}-\frac {A c^2 \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {20 a^2 C d^2 x^2-5 a (3 b c (B c+2 A d)-4 a d (2 c C+B d)) x+4 a \left (5 a c (c C+2 B d)-A \left (4 b c^2-5 a d^2\right )\right )}{x^4 \sqrt {b x^2+a}}dx}{4 a}-\frac {5 c \sqrt {a+b x^2} (2 A d+B c)}{4 x^4}}{5 a}-\frac {A c^2 \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 2338

\(\displaystyle \frac {\frac {-\frac {\int \frac {a \left (15 a (3 b c (B c+2 A d)-4 a d (2 c C+B d))-4 \left (2 A b \left (4 b c^2-5 a d^2\right )+5 a \left (3 a C d^2-2 b c (c C+2 B d)\right )\right ) x\right )}{x^3 \sqrt {b x^2+a}}dx}{3 a}-\frac {4 \sqrt {a+b x^2} \left (5 a c (2 B d+c C)-A \left (4 b c^2-5 a d^2\right )\right )}{3 x^3}}{4 a}-\frac {5 c \sqrt {a+b x^2} (2 A d+B c)}{4 x^4}}{5 a}-\frac {A c^2 \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {1}{3} \int \frac {15 a (3 b c (B c+2 A d)-4 a d (2 c C+B d))-4 \left (2 A b \left (4 b c^2-5 a d^2\right )+5 a \left (3 a C d^2-2 b c (c C+2 B d)\right )\right ) x}{x^3 \sqrt {b x^2+a}}dx-\frac {4 \sqrt {a+b x^2} \left (5 a c (2 B d+c C)-A \left (4 b c^2-5 a d^2\right )\right )}{3 x^3}}{4 a}-\frac {5 c \sqrt {a+b x^2} (2 A d+B c)}{4 x^4}}{5 a}-\frac {A c^2 \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 539

\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {\int \frac {a \left (8 \left (2 A b \left (4 b c^2-5 a d^2\right )+5 a \left (3 a C d^2-2 b c (c C+2 B d)\right )\right )+15 b (3 b c (B c+2 A d)-4 a d (2 c C+B d)) x\right )}{x^2 \sqrt {b x^2+a}}dx}{2 a}+\frac {15 \sqrt {a+b x^2} (3 b c (2 A d+B c)-4 a d (B d+2 c C))}{2 x^2}\right )-\frac {4 \sqrt {a+b x^2} \left (5 a c (2 B d+c C)-A \left (4 b c^2-5 a d^2\right )\right )}{3 x^3}}{4 a}-\frac {5 c \sqrt {a+b x^2} (2 A d+B c)}{4 x^4}}{5 a}-\frac {A c^2 \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {1}{2} \int \frac {8 \left (2 A b \left (4 b c^2-5 a d^2\right )+5 a \left (3 a C d^2-2 b c (c C+2 B d)\right )\right )+15 b (3 b c (B c+2 A d)-4 a d (2 c C+B d)) x}{x^2 \sqrt {b x^2+a}}dx+\frac {15 \sqrt {a+b x^2} (3 b c (2 A d+B c)-4 a d (B d+2 c C))}{2 x^2}\right )-\frac {4 \sqrt {a+b x^2} \left (5 a c (2 B d+c C)-A \left (4 b c^2-5 a d^2\right )\right )}{3 x^3}}{4 a}-\frac {5 c \sqrt {a+b x^2} (2 A d+B c)}{4 x^4}}{5 a}-\frac {A c^2 \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 534

\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {1}{2} \left (15 b (3 b c (2 A d+B c)-4 a d (B d+2 c C)) \int \frac {1}{x \sqrt {b x^2+a}}dx-\frac {8 \sqrt {a+b x^2} \left (2 A b \left (4 b c^2-5 a d^2\right )+5 a \left (3 a C d^2-2 b c (2 B d+c C)\right )\right )}{a x}\right )+\frac {15 \sqrt {a+b x^2} (3 b c (2 A d+B c)-4 a d (B d+2 c C))}{2 x^2}\right )-\frac {4 \sqrt {a+b x^2} \left (5 a c (2 B d+c C)-A \left (4 b c^2-5 a d^2\right )\right )}{3 x^3}}{4 a}-\frac {5 c \sqrt {a+b x^2} (2 A d+B c)}{4 x^4}}{5 a}-\frac {A c^2 \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {1}{2} \left (\frac {15}{2} b (3 b c (2 A d+B c)-4 a d (B d+2 c C)) \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2-\frac {8 \sqrt {a+b x^2} \left (2 A b \left (4 b c^2-5 a d^2\right )+5 a \left (3 a C d^2-2 b c (2 B d+c C)\right )\right )}{a x}\right )+\frac {15 \sqrt {a+b x^2} (3 b c (2 A d+B c)-4 a d (B d+2 c C))}{2 x^2}\right )-\frac {4 \sqrt {a+b x^2} \left (5 a c (2 B d+c C)-A \left (4 b c^2-5 a d^2\right )\right )}{3 x^3}}{4 a}-\frac {5 c \sqrt {a+b x^2} (2 A d+B c)}{4 x^4}}{5 a}-\frac {A c^2 \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {1}{2} \left (15 (3 b c (2 A d+B c)-4 a d (B d+2 c C)) \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}-\frac {8 \sqrt {a+b x^2} \left (2 A b \left (4 b c^2-5 a d^2\right )+5 a \left (3 a C d^2-2 b c (2 B d+c C)\right )\right )}{a x}\right )+\frac {15 \sqrt {a+b x^2} (3 b c (2 A d+B c)-4 a d (B d+2 c C))}{2 x^2}\right )-\frac {4 \sqrt {a+b x^2} \left (5 a c (2 B d+c C)-A \left (4 b c^2-5 a d^2\right )\right )}{3 x^3}}{4 a}-\frac {5 c \sqrt {a+b x^2} (2 A d+B c)}{4 x^4}}{5 a}-\frac {A c^2 \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {1}{3} \left (\frac {1}{2} \left (-\frac {15 b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) (3 b c (2 A d+B c)-4 a d (B d+2 c C))}{\sqrt {a}}-\frac {8 \sqrt {a+b x^2} \left (2 A b \left (4 b c^2-5 a d^2\right )+5 a \left (3 a C d^2-2 b c (2 B d+c C)\right )\right )}{a x}\right )+\frac {15 \sqrt {a+b x^2} (3 b c (2 A d+B c)-4 a d (B d+2 c C))}{2 x^2}\right )-\frac {4 \sqrt {a+b x^2} \left (5 a c (2 B d+c C)-A \left (4 b c^2-5 a d^2\right )\right )}{3 x^3}}{4 a}-\frac {5 c \sqrt {a+b x^2} (2 A d+B c)}{4 x^4}}{5 a}-\frac {A c^2 \sqrt {a+b x^2}}{5 a x^5}\)

Input:

Int[((c + d*x)^2*(A + B*x + C*x^2))/(x^6*Sqrt[a + b*x^2]),x]
 

Output:

-1/5*(A*c^2*Sqrt[a + b*x^2])/(a*x^5) + ((-5*c*(B*c + 2*A*d)*Sqrt[a + b*x^2 
])/(4*x^4) + ((-4*(5*a*c*(c*C + 2*B*d) - A*(4*b*c^2 - 5*a*d^2))*Sqrt[a + b 
*x^2])/(3*x^3) + ((15*(3*b*c*(B*c + 2*A*d) - 4*a*d*(2*c*C + B*d))*Sqrt[a + 
 b*x^2])/(2*x^2) + ((-8*(2*A*b*(4*b*c^2 - 5*a*d^2) + 5*a*(3*a*C*d^2 - 2*b* 
c*(c*C + 2*B*d)))*Sqrt[a + b*x^2])/(a*x) - (15*b*(3*b*c*(B*c + 2*A*d) - 4* 
a*d*(2*c*C + B*d))*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a])/2)/3)/(4*a)) 
/(5*a)
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 539

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[c*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] + Simp[1/(a*(m + 1)) 
   Int[x^(m + 1)*(a + b*x^2)^p*(a*d*(m + 1) - b*c*(m + 2*p + 3)*x), x], x] 
/; FreeQ[{a, b, c, d, p}, x] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 2338

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ 
Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( 
m + 1))   Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( 
m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.94

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-80 A b \,d^{2} x^{4} a +64 A \,b^{2} c^{2} x^{4}-160 B b c d \,x^{4} a +120 C \,a^{2} d^{2} x^{4}-80 C b \,c^{2} x^{4} a -90 A b c d \,x^{3} a +60 B \,a^{2} d^{2} x^{3}-45 B b \,c^{2} x^{3} a +120 C \,a^{2} c d \,x^{3}+40 A \,a^{2} d^{2} x^{2}-32 A b \,c^{2} x^{2} a +80 B \,a^{2} c d \,x^{2}+40 C \,a^{2} c^{2} x^{2}+60 A \,a^{2} c d x +30 B \,a^{2} c^{2} x +24 A \,a^{2} c^{2}\right )}{120 a^{3} x^{5}}-\frac {\left (6 A b c d -4 a B \,d^{2}+3 b B \,c^{2}-8 C a c d \right ) b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{8 a^{\frac {5}{2}}}\)	\(250\)
default	\(\left (A \,d^{2}+2 B c d +C \,c^{2}\right ) \left (-\frac {\sqrt {b \,x^{2}+a}}{3 a \,x^{3}}+\frac {2 b \sqrt {b \,x^{2}+a}}{3 a^{2} x}\right )+A \,c^{2} \left (-\frac {\sqrt {b \,x^{2}+a}}{5 a \,x^{5}}-\frac {4 b \left (-\frac {\sqrt {b \,x^{2}+a}}{3 a \,x^{3}}+\frac {2 b \sqrt {b \,x^{2}+a}}{3 a^{2} x}\right )}{5 a}\right )-\frac {C \,d^{2} \sqrt {b \,x^{2}+a}}{a x}+c \left (2 A d +B c \right ) \left (-\frac {\sqrt {b \,x^{2}+a}}{4 a \,x^{4}}-\frac {3 b \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )+d \left (B d +2 C c \right ) \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )\)	\(279\)

Input:

int((d*x+c)^2*(C*x^2+B*x+A)/x^6/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/120*(b*x^2+a)^(1/2)*(-80*A*a*b*d^2*x^4+64*A*b^2*c^2*x^4-160*B*a*b*c*d*x 
^4+120*C*a^2*d^2*x^4-80*C*a*b*c^2*x^4-90*A*a*b*c*d*x^3+60*B*a^2*d^2*x^3-45 
*B*a*b*c^2*x^3+120*C*a^2*c*d*x^3+40*A*a^2*d^2*x^2-32*A*a*b*c^2*x^2+80*B*a^ 
2*c*d*x^2+40*C*a^2*c^2*x^2+60*A*a^2*c*d*x+30*B*a^2*c^2*x+24*A*a^2*c^2)/a^3 
/x^5-1/8*(6*A*b*c*d-4*B*a*d^2+3*B*b*c^2-8*C*a*c*d)/a^(5/2)*b*ln((2*a+2*a^( 
1/2)*(b*x^2+a)^(1/2))/x)
 

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 485, normalized size of antiderivative = 1.82 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2\right )}{x^6 \sqrt {a+b x^2}} \, dx=\left [-\frac {15 \, {\left (3 \, B b^{2} c^{2} - 4 \, B a b d^{2} - 2 \, {\left (4 \, C a b - 3 \, A b^{2}\right )} c d\right )} \sqrt {a} x^{5} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (24 \, A a^{2} c^{2} - 8 \, {\left (20 \, B a b c d + 2 \, {\left (5 \, C a b - 4 \, A b^{2}\right )} c^{2} - 5 \, {\left (3 \, C a^{2} - 2 \, A a b\right )} d^{2}\right )} x^{4} - 15 \, {\left (3 \, B a b c^{2} - 4 \, B a^{2} d^{2} - 2 \, {\left (4 \, C a^{2} - 3 \, A a b\right )} c d\right )} x^{3} + 8 \, {\left (10 \, B a^{2} c d + 5 \, A a^{2} d^{2} + {\left (5 \, C a^{2} - 4 \, A a b\right )} c^{2}\right )} x^{2} + 30 \, {\left (B a^{2} c^{2} + 2 \, A a^{2} c d\right )} x\right )} \sqrt {b x^{2} + a}}{240 \, a^{3} x^{5}}, \frac {15 \, {\left (3 \, B b^{2} c^{2} - 4 \, B a b d^{2} - 2 \, {\left (4 \, C a b - 3 \, A b^{2}\right )} c d\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) - {\left (24 \, A a^{2} c^{2} - 8 \, {\left (20 \, B a b c d + 2 \, {\left (5 \, C a b - 4 \, A b^{2}\right )} c^{2} - 5 \, {\left (3 \, C a^{2} - 2 \, A a b\right )} d^{2}\right )} x^{4} - 15 \, {\left (3 \, B a b c^{2} - 4 \, B a^{2} d^{2} - 2 \, {\left (4 \, C a^{2} - 3 \, A a b\right )} c d\right )} x^{3} + 8 \, {\left (10 \, B a^{2} c d + 5 \, A a^{2} d^{2} + {\left (5 \, C a^{2} - 4 \, A a b\right )} c^{2}\right )} x^{2} + 30 \, {\left (B a^{2} c^{2} + 2 \, A a^{2} c d\right )} x\right )} \sqrt {b x^{2} + a}}{120 \, a^{3} x^{5}}\right ] \] Input:

integrate((d*x+c)^2*(C*x^2+B*x+A)/x^6/(b*x^2+a)^(1/2),x, algorithm="fricas 
")
 

Output:

[-1/240*(15*(3*B*b^2*c^2 - 4*B*a*b*d^2 - 2*(4*C*a*b - 3*A*b^2)*c*d)*sqrt(a 
)*x^5*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(24*A*a^2*c^ 
2 - 8*(20*B*a*b*c*d + 2*(5*C*a*b - 4*A*b^2)*c^2 - 5*(3*C*a^2 - 2*A*a*b)*d^ 
2)*x^4 - 15*(3*B*a*b*c^2 - 4*B*a^2*d^2 - 2*(4*C*a^2 - 3*A*a*b)*c*d)*x^3 + 
8*(10*B*a^2*c*d + 5*A*a^2*d^2 + (5*C*a^2 - 4*A*a*b)*c^2)*x^2 + 30*(B*a^2*c 
^2 + 2*A*a^2*c*d)*x)*sqrt(b*x^2 + a))/(a^3*x^5), 1/120*(15*(3*B*b^2*c^2 - 
4*B*a*b*d^2 - 2*(4*C*a*b - 3*A*b^2)*c*d)*sqrt(-a)*x^5*arctan(sqrt(b*x^2 + 
a)*sqrt(-a)/a) - (24*A*a^2*c^2 - 8*(20*B*a*b*c*d + 2*(5*C*a*b - 4*A*b^2)*c 
^2 - 5*(3*C*a^2 - 2*A*a*b)*d^2)*x^4 - 15*(3*B*a*b*c^2 - 4*B*a^2*d^2 - 2*(4 
*C*a^2 - 3*A*a*b)*c*d)*x^3 + 8*(10*B*a^2*c*d + 5*A*a^2*d^2 + (5*C*a^2 - 4* 
A*a*b)*c^2)*x^2 + 30*(B*a^2*c^2 + 2*A*a^2*c*d)*x)*sqrt(b*x^2 + a))/(a^3*x^ 
5)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 860 vs. \(2 (255) = 510\).

Time = 8.62 (sec) , antiderivative size = 860, normalized size of antiderivative = 3.22 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2\right )}{x^6 \sqrt {a+b x^2}} \, dx =\text {Too large to display} \] Input:

integrate((d*x+c)**2*(C*x**2+B*x+A)/x**6/(b*x**2+a)**(1/2),x)
 

Output:

-3*A*a**4*b**(9/2)*c**2*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4* 
b**5*x**6 + 15*a**3*b**6*x**8) - 2*A*a**3*b**(11/2)*c**2*x**2*sqrt(a/(b*x* 
*2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 3*A 
*a**2*b**(13/2)*c**2*x**4*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a** 
4*b**5*x**6 + 15*a**3*b**6*x**8) - 12*A*a*b**(15/2)*c**2*x**6*sqrt(a/(b*x* 
*2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 8*A 
*b**(17/2)*c**2*x**8*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b** 
5*x**6 + 15*a**3*b**6*x**8) - A*c*d/(2*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) 
+ A*sqrt(b)*c*d/(4*a*x**3*sqrt(a/(b*x**2) + 1)) - A*sqrt(b)*d**2*sqrt(a/(b 
*x**2) + 1)/(3*a*x**2) + 3*A*b**(3/2)*c*d/(4*a**2*x*sqrt(a/(b*x**2) + 1)) 
+ 2*A*b**(3/2)*d**2*sqrt(a/(b*x**2) + 1)/(3*a**2) - 3*A*b**2*c*d*asinh(sqr 
t(a)/(sqrt(b)*x))/(4*a**(5/2)) - B*c**2/(4*sqrt(b)*x**5*sqrt(a/(b*x**2) + 
1)) + B*sqrt(b)*c**2/(8*a*x**3*sqrt(a/(b*x**2) + 1)) - 2*B*sqrt(b)*c*d*sqr 
t(a/(b*x**2) + 1)/(3*a*x**2) - B*sqrt(b)*d**2*sqrt(a/(b*x**2) + 1)/(2*a*x) 
 + 3*B*b**(3/2)*c**2/(8*a**2*x*sqrt(a/(b*x**2) + 1)) + 4*B*b**(3/2)*c*d*sq 
rt(a/(b*x**2) + 1)/(3*a**2) + B*b*d**2*asinh(sqrt(a)/(sqrt(b)*x))/(2*a**(3 
/2)) - 3*B*b**2*c**2*asinh(sqrt(a)/(sqrt(b)*x))/(8*a**(5/2)) - C*sqrt(b)*c 
**2*sqrt(a/(b*x**2) + 1)/(3*a*x**2) - C*sqrt(b)*c*d*sqrt(a/(b*x**2) + 1)/( 
a*x) - C*sqrt(b)*d**2*sqrt(a/(b*x**2) + 1)/a + 2*C*b**(3/2)*c**2*sqrt(a/(b 
*x**2) + 1)/(3*a**2) + C*b*c*d*asinh(sqrt(a)/(sqrt(b)*x))/a**(3/2)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.13 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2\right )}{x^6 \sqrt {a+b x^2}} \, dx=-\frac {8 \, \sqrt {b x^{2} + a} A b^{2} c^{2}}{15 \, a^{3} x} - \frac {\sqrt {b x^{2} + a} C d^{2}}{a x} + \frac {{\left (2 \, C c d + B d^{2}\right )} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {3}{2}}} - \frac {3 \, {\left (B c^{2} + 2 \, A c d\right )} b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {5}{2}}} + \frac {4 \, \sqrt {b x^{2} + a} A b c^{2}}{15 \, a^{2} x^{3}} + \frac {2 \, {\left (C c^{2} + 2 \, B c d + A d^{2}\right )} \sqrt {b x^{2} + a} b}{3 \, a^{2} x} - \frac {{\left (2 \, C c d + B d^{2}\right )} \sqrt {b x^{2} + a}}{2 \, a x^{2}} + \frac {3 \, {\left (B c^{2} + 2 \, A c d\right )} \sqrt {b x^{2} + a} b}{8 \, a^{2} x^{2}} - \frac {\sqrt {b x^{2} + a} A c^{2}}{5 \, a x^{5}} - \frac {{\left (C c^{2} + 2 \, B c d + A d^{2}\right )} \sqrt {b x^{2} + a}}{3 \, a x^{3}} - \frac {{\left (B c^{2} + 2 \, A c d\right )} \sqrt {b x^{2} + a}}{4 \, a x^{4}} \] Input:

integrate((d*x+c)^2*(C*x^2+B*x+A)/x^6/(b*x^2+a)^(1/2),x, algorithm="maxima 
")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1071 vs. \(2 (239) = 478\).

Time = 0.18 (sec) , antiderivative size = 1071, normalized size of antiderivative = 4.01 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2\right )}{x^6 \sqrt {a+b x^2}} \, dx =\text {Too large to display} \] Input:

integrate((d*x+c)^2*(C*x^2+B*x+A)/x^6/(b*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

1/4*(3*B*b^2*c^2 - 8*C*a*b*c*d + 6*A*b^2*c*d - 4*B*a*b*d^2)*arctan(-(sqrt( 
b)*x - sqrt(b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) - 1/60*(45*(sqrt(b)*x - s 
qrt(b*x^2 + a))^9*B*b^2*c^2 - 120*(sqrt(b)*x - sqrt(b*x^2 + a))^9*C*a*b*c* 
d + 90*(sqrt(b)*x - sqrt(b*x^2 + a))^9*A*b^2*c*d - 60*(sqrt(b)*x - sqrt(b* 
x^2 + a))^9*B*a*b*d^2 - 120*(sqrt(b)*x - sqrt(b*x^2 + a))^8*C*a^2*sqrt(b)* 
d^2 - 210*(sqrt(b)*x - sqrt(b*x^2 + a))^7*B*a*b^2*c^2 + 240*(sqrt(b)*x - s 
qrt(b*x^2 + a))^7*C*a^2*b*c*d - 420*(sqrt(b)*x - sqrt(b*x^2 + a))^7*A*a*b^ 
2*c*d + 120*(sqrt(b)*x - sqrt(b*x^2 + a))^7*B*a^2*b*d^2 - 240*(sqrt(b)*x - 
 sqrt(b*x^2 + a))^6*C*a^2*b^(3/2)*c^2 - 480*(sqrt(b)*x - sqrt(b*x^2 + a))^ 
6*B*a^2*b^(3/2)*c*d + 480*(sqrt(b)*x - sqrt(b*x^2 + a))^6*C*a^3*sqrt(b)*d^ 
2 - 240*(sqrt(b)*x - sqrt(b*x^2 + a))^6*A*a^2*b^(3/2)*d^2 + 560*(sqrt(b)*x 
 - sqrt(b*x^2 + a))^4*C*a^3*b^(3/2)*c^2 - 640*(sqrt(b)*x - sqrt(b*x^2 + a) 
)^4*A*a^2*b^(5/2)*c^2 + 1120*(sqrt(b)*x - sqrt(b*x^2 + a))^4*B*a^3*b^(3/2) 
*c*d - 720*(sqrt(b)*x - sqrt(b*x^2 + a))^4*C*a^4*sqrt(b)*d^2 + 560*(sqrt(b 
)*x - sqrt(b*x^2 + a))^4*A*a^3*b^(3/2)*d^2 + 210*(sqrt(b)*x - sqrt(b*x^2 + 
 a))^3*B*a^3*b^2*c^2 - 240*(sqrt(b)*x - sqrt(b*x^2 + a))^3*C*a^4*b*c*d + 4 
20*(sqrt(b)*x - sqrt(b*x^2 + a))^3*A*a^3*b^2*c*d - 120*(sqrt(b)*x - sqrt(b 
*x^2 + a))^3*B*a^4*b*d^2 - 400*(sqrt(b)*x - sqrt(b*x^2 + a))^2*C*a^4*b^(3/ 
2)*c^2 + 320*(sqrt(b)*x - sqrt(b*x^2 + a))^2*A*a^3*b^(5/2)*c^2 - 800*(sqrt 
(b)*x - sqrt(b*x^2 + a))^2*B*a^4*b^(3/2)*c*d + 480*(sqrt(b)*x - sqrt(b*...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2\right )}{x^6 \sqrt {a+b x^2}} \, dx=\int \frac {{\left (c+d\,x\right )}^2\,\left (C\,x^2+B\,x+A\right )}{x^6\,\sqrt {b\,x^2+a}} \,d x \] Input:

int(((c + d*x)^2*(A + B*x + C*x^2))/(x^6*(a + b*x^2)^(1/2)),x)
 

Output:

int(((c + d*x)^2*(A + B*x + C*x^2))/(x^6*(a + b*x^2)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 661, normalized size of antiderivative = 2.48 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2\right )}{x^6 \sqrt {a+b x^2}} \, dx=\frac {-120 \sqrt {b \,x^{2}+a}\, a^{2} c \,d^{2} x^{4}-64 \sqrt {b \,x^{2}+a}\, a \,b^{2} c^{2} x^{4}+80 \sqrt {b \,x^{2}+a}\, a b \,c^{3} x^{4}+45 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} c^{2} x^{5}-45 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} c^{2} x^{5}-80 \sqrt {b}\, a^{2} b \,d^{2} x^{5}+72 \sqrt {b}\, a^{2} c \,d^{2} x^{5}+64 \sqrt {b}\, a \,b^{2} c^{2} x^{5}-80 \sqrt {b}\, a b \,c^{3} x^{5}-80 \sqrt {b \,x^{2}+a}\, a^{2} b c d \,x^{2}+160 \sqrt {b \,x^{2}+a}\, a \,b^{2} c d \,x^{4}-30 \sqrt {b \,x^{2}+a}\, a^{2} b \,c^{2} x +45 \sqrt {b \,x^{2}+a}\, a \,b^{2} c^{2} x^{3}-60 \sqrt {b \,x^{2}+a}\, a^{3} c d x +32 \sqrt {b \,x^{2}+a}\, a^{2} b \,c^{2} x^{2}+80 \sqrt {b \,x^{2}+a}\, a^{2} b \,d^{2} x^{4}-60 \sqrt {b \,x^{2}+a}\, a^{2} b \,d^{2} x^{3}-120 \sqrt {b \,x^{2}+a}\, a^{2} c^{2} d \,x^{3}-40 \sqrt {b \,x^{2}+a}\, a^{3} d^{2} x^{2}-40 \sqrt {b \,x^{2}+a}\, a^{2} c^{3} x^{2}-24 \sqrt {b \,x^{2}+a}\, a^{3} c^{2}+90 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} c d \,x^{5}-120 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,c^{2} d \,x^{5}-90 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} c d \,x^{5}+120 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,c^{2} d \,x^{5}-60 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} d^{2} x^{5}+60 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} d^{2} x^{5}-160 \sqrt {b}\, a \,b^{2} c d \,x^{5}+90 \sqrt {b \,x^{2}+a}\, a^{2} b c d \,x^{3}}{120 a^{3} x^{5}} \] Input:

int((d*x+c)^2*(C*x^2+B*x+A)/x^6/(b*x^2+a)^(1/2),x)
 

Output:

( - 24*sqrt(a + b*x**2)*a**3*c**2 - 60*sqrt(a + b*x**2)*a**3*c*d*x - 40*sq 
rt(a + b*x**2)*a**3*d**2*x**2 + 32*sqrt(a + b*x**2)*a**2*b*c**2*x**2 - 30* 
sqrt(a + b*x**2)*a**2*b*c**2*x + 90*sqrt(a + b*x**2)*a**2*b*c*d*x**3 - 80* 
sqrt(a + b*x**2)*a**2*b*c*d*x**2 + 80*sqrt(a + b*x**2)*a**2*b*d**2*x**4 - 
60*sqrt(a + b*x**2)*a**2*b*d**2*x**3 - 40*sqrt(a + b*x**2)*a**2*c**3*x**2 
- 120*sqrt(a + b*x**2)*a**2*c**2*d*x**3 - 120*sqrt(a + b*x**2)*a**2*c*d**2 
*x**4 - 64*sqrt(a + b*x**2)*a*b**2*c**2*x**4 + 45*sqrt(a + b*x**2)*a*b**2* 
c**2*x**3 + 160*sqrt(a + b*x**2)*a*b**2*c*d*x**4 + 80*sqrt(a + b*x**2)*a*b 
*c**3*x**4 + 90*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt( 
a))*a*b**2*c*d*x**5 - 60*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b) 
*x)/sqrt(a))*a*b**2*d**2*x**5 - 120*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a 
) + sqrt(b)*x)/sqrt(a))*a*b*c**2*d*x**5 + 45*sqrt(a)*log((sqrt(a + b*x**2) 
 - sqrt(a) + sqrt(b)*x)/sqrt(a))*b**3*c**2*x**5 - 90*sqrt(a)*log((sqrt(a + 
 b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b**2*c*d*x**5 + 60*sqrt(a)*log( 
(sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b**2*d**2*x**5 + 120*s 
qrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b*c**2*d*x* 
*5 - 45*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b**3 
*c**2*x**5 - 80*sqrt(b)*a**2*b*d**2*x**5 + 72*sqrt(b)*a**2*c*d**2*x**5 + 6 
4*sqrt(b)*a*b**2*c**2*x**5 - 160*sqrt(b)*a*b**2*c*d*x**5 - 80*sqrt(b)*a*b* 
c**3*x**5)/(120*a**3*x**5)