\(\int \frac {(c+d x) \sqrt {a+b x^2} (A+B x+C x^2)}{x^6} \, dx\) [12]

Optimal result

Integrand size = 30, antiderivative size = 170 \[ \int \frac {(c+d x) \sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^6} \, dx=-\frac {(4 a C d-b (B c+A d)) \sqrt {a+b x^2}}{8 a x^2}-\frac {A c \left (a+b x^2\right )^{3/2}}{5 a x^5}-\frac {(B c+A d) \left (a+b x^2\right )^{3/2}}{4 a x^4}+\frac {(2 A b c-5 a (c C+B d)) \left (a+b x^2\right )^{3/2}}{15 a^2 x^3}-\frac {b (4 a C d-b (B c+A d)) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.64 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.13 \[ \int \frac {(c+d x) \sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^6} \, dx=-\frac {\sqrt {a+b x^2} \left (-16 A b^2 c x^4+a b x^2 (A (8 c+15 d x)+5 x (3 B c+8 c C x+8 B d x))+a^2 \left (6 A (4 c+5 d x)+10 x \left (3 B c+4 c C x+4 B d x+6 C d x^2\right )\right )\right )}{120 a^2 x^5}+\frac {b C d \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {b^2 (B c+A d) \text {arctanh}\left (\frac {-\sqrt {b} x+\sqrt {a+b x^2}}{\sqrt {a}}\right )}{4 a^{3/2}} \] Input:

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

Output:

-1/120*(Sqrt[a + b*x^2]*(-16*A*b^2*c*x^4 + a*b*x^2*(A*(8*c + 15*d*x) + 5*x 
*(3*B*c + 8*c*C*x + 8*B*d*x)) + a^2*(6*A*(4*c + 5*d*x) + 10*x*(3*B*c + 4*c 
*C*x + 4*B*d*x + 6*C*d*x^2))))/(a^2*x^5) + (b*C*d*ArcTanh[(Sqrt[b]*x - Sqr 
t[a + b*x^2])/Sqrt[a]])/Sqrt[a] + (b^2*(B*c + A*d)*ArcTanh[(-(Sqrt[b]*x) + 
 Sqrt[a + b*x^2])/Sqrt[a]])/(4*a^(3/2))
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2338, 25, 2338, 27, 534, 243, 51, 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.

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

\(\Big \downarrow \) 2338

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2338

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 534

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-1/5*(A*c*(a + b*x^2)^(3/2))/(a*x^5) + ((-5*(B*c + A*d)*(a + b*x^2)^(3/2)) 
/(4*x^4) + ((4*(2*A*b*c - 5*a*(c*C + B*d))*(a + b*x^2)^(3/2))/(3*a*x^3) - 
(5*(b*B*c + A*b*d - 4*a*C*d)*(-(Sqrt[a + b*x^2]/x^2) - (b*ArcTanh[Sqrt[a + 
 b*x^2]/Sqrt[a]])/Sqrt[a]))/2)/4)/(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 51

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

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 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 = 0.20 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.01

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-16 A \,b^{2} c \,x^{4}+40 B a b d \,x^{4}+40 C a b c \,x^{4}+15 A b d \,x^{3} a +15 B b c \,x^{3} a +60 C \,a^{2} d \,x^{3}+8 A a b c \,x^{2}+40 B \,a^{2} d \,x^{2}+40 C \,a^{2} c \,x^{2}+30 A \,a^{2} d x +30 B \,a^{2} c x +24 A \,a^{2} c \right )}{120 x^{5} a^{2}}+\frac {\left (A b d +B b c -4 a C d \right ) b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{8 a^{\frac {3}{2}}}\)	\(171\)
default	\(\left (A d +B c \right ) \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )}{4 a}\right )-\frac {\left (B d +C c \right ) \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 a \,x^{3}}+A c \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 a \,x^{5}}+\frac {2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 a^{2} x^{3}}\right )+d C \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )\)	\(224\)

Input:

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

Output:

-1/120*(b*x^2+a)^(1/2)*(-16*A*b^2*c*x^4+40*B*a*b*d*x^4+40*C*a*b*c*x^4+15*A 
*a*b*d*x^3+15*B*a*b*c*x^3+60*C*a^2*d*x^3+8*A*a*b*c*x^2+40*B*a^2*d*x^2+40*C 
*a^2*c*x^2+30*A*a^2*d*x+30*B*a^2*c*x+24*A*a^2*c)/x^5/a^2+1/8*(A*b*d+B*b*c- 
4*C*a*d)/a^(3/2)*b*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.06 \[ \int \frac {(c+d x) \sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^6} \, dx=\left [\frac {15 \, {\left (B b^{2} c - {\left (4 \, C a b - A b^{2}\right )} d\right )} \sqrt {a} x^{5} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (8 \, {\left (5 \, B a b d + {\left (5 \, C a b - 2 \, A b^{2}\right )} c\right )} x^{4} + 24 \, A a^{2} c + 15 \, {\left (B a b c + {\left (4 \, C a^{2} + A a b\right )} d\right )} x^{3} + 8 \, {\left (5 \, B a^{2} d + {\left (5 \, C a^{2} + A a b\right )} c\right )} x^{2} + 30 \, {\left (B a^{2} c + A a^{2} d\right )} x\right )} \sqrt {b x^{2} + a}}{240 \, a^{2} x^{5}}, -\frac {15 \, {\left (B b^{2} c - {\left (4 \, C a b - A b^{2}\right )} d\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + {\left (8 \, {\left (5 \, B a b d + {\left (5 \, C a b - 2 \, A b^{2}\right )} c\right )} x^{4} + 24 \, A a^{2} c + 15 \, {\left (B a b c + {\left (4 \, C a^{2} + A a b\right )} d\right )} x^{3} + 8 \, {\left (5 \, B a^{2} d + {\left (5 \, C a^{2} + A a b\right )} c\right )} x^{2} + 30 \, {\left (B a^{2} c + A a^{2} d\right )} x\right )} \sqrt {b x^{2} + a}}{120 \, a^{2} x^{5}}\right ] \] Input:

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

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 440 vs. \(2 (153) = 306\).

Time = 6.29 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.59 \[ \int \frac {(c+d x) \sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^6} \, dx=- \frac {A a d}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {A \sqrt {b} c \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {3 A \sqrt {b} d}{8 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {A b^{\frac {3}{2}} c \sqrt {\frac {a}{b x^{2}} + 1}}{15 a x^{2}} - \frac {A b^{\frac {3}{2}} d}{8 a x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {2 A b^{\frac {5}{2}} c \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{2}} + \frac {A b^{2} d \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 a^{\frac {3}{2}}} - \frac {B a c}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 B \sqrt {b} c}{8 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {B \sqrt {b} d \sqrt {\frac {a}{b x^{2}} + 1}}{3 x^{2}} - \frac {B b^{\frac {3}{2}} c}{8 a x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {B b^{\frac {3}{2}} d \sqrt {\frac {a}{b x^{2}} + 1}}{3 a} + \frac {B b^{2} c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 a^{\frac {3}{2}}} - \frac {C \sqrt {b} c \sqrt {\frac {a}{b x^{2}} + 1}}{3 x^{2}} - \frac {C \sqrt {b} d \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} - \frac {C b^{\frac {3}{2}} c \sqrt {\frac {a}{b x^{2}} + 1}}{3 a} - \frac {C b d \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 \sqrt {a}} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.38 \[ \int \frac {(c+d x) \sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^6} \, dx=-\frac {C b d \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, \sqrt {a}} + \frac {\sqrt {b x^{2} + a} C b d}{2 \, a} + \frac {{\left (B c + A d\right )} b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {3}{2}}} - \frac {\sqrt {b x^{2} + a} {\left (B c + A d\right )} b^{2}}{8 \, a^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C d}{2 \, a x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C c}{3 \, a x^{3}} + \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b c}{15 \, a^{2} x^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B d}{3 \, a x^{3}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} {\left (B c + A d\right )} b}{8 \, a^{2} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A c}{5 \, a x^{5}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} {\left (B c + A d\right )}}{4 \, a x^{4}} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 751 vs. \(2 (146) = 292\).

Time = 0.22 (sec) , antiderivative size = 751, normalized size of antiderivative = 4.42 \[ \int \frac {(c+d x) \sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^6} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/4*(B*b^2*c - 4*C*a*b*d + A*b^2*d)*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a)) 
/sqrt(-a))/(sqrt(-a)*a) + 1/60*(15*(sqrt(b)*x - sqrt(b*x^2 + a))^9*B*b^2*c 
 + 60*(sqrt(b)*x - sqrt(b*x^2 + a))^9*C*a*b*d + 15*(sqrt(b)*x - sqrt(b*x^2 
 + a))^9*A*b^2*d + 120*(sqrt(b)*x - sqrt(b*x^2 + a))^8*C*a*b^(3/2)*c + 120 
*(sqrt(b)*x - sqrt(b*x^2 + a))^8*B*a*b^(3/2)*d + 90*(sqrt(b)*x - sqrt(b*x^ 
2 + a))^7*B*a*b^2*c - 120*(sqrt(b)*x - sqrt(b*x^2 + a))^7*C*a^2*b*d + 90*( 
sqrt(b)*x - sqrt(b*x^2 + a))^7*A*a*b^2*d - 240*(sqrt(b)*x - sqrt(b*x^2 + a 
))^6*C*a^2*b^(3/2)*c + 240*(sqrt(b)*x - sqrt(b*x^2 + a))^6*A*a*b^(5/2)*c - 
 240*(sqrt(b)*x - sqrt(b*x^2 + a))^6*B*a^2*b^(3/2)*d + 160*(sqrt(b)*x - sq 
rt(b*x^2 + a))^4*C*a^3*b^(3/2)*c + 80*(sqrt(b)*x - sqrt(b*x^2 + a))^4*A*a^ 
2*b^(5/2)*c + 160*(sqrt(b)*x - sqrt(b*x^2 + a))^4*B*a^3*b^(3/2)*d - 90*(sq 
rt(b)*x - sqrt(b*x^2 + a))^3*B*a^3*b^2*c + 120*(sqrt(b)*x - sqrt(b*x^2 + a 
))^3*C*a^4*b*d - 90*(sqrt(b)*x - sqrt(b*x^2 + a))^3*A*a^3*b^2*d - 80*(sqrt 
(b)*x - sqrt(b*x^2 + a))^2*C*a^4*b^(3/2)*c + 80*(sqrt(b)*x - sqrt(b*x^2 + 
a))^2*A*a^3*b^(5/2)*c - 80*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^4*b^(3/2)*d 
 - 15*(sqrt(b)*x - sqrt(b*x^2 + a))*B*a^4*b^2*c - 60*(sqrt(b)*x - sqrt(b*x 
^2 + a))*C*a^5*b*d - 15*(sqrt(b)*x - sqrt(b*x^2 + a))*A*a^4*b^2*d + 40*C*a 
^5*b^(3/2)*c - 16*A*a^4*b^(5/2)*c + 40*B*a^5*b^(3/2)*d)/(((sqrt(b)*x - sqr 
t(b*x^2 + a))^2 - a)^5*a)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 453, normalized size of antiderivative = 2.66 \[ \int \frac {(c+d x) \sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^6} \, dx=\frac {-24 \sqrt {b \,x^{2}+a}\, a^{3} c -30 \sqrt {b \,x^{2}+a}\, a^{3} d x -8 \sqrt {b \,x^{2}+a}\, a^{2} b c \,x^{2}-30 \sqrt {b \,x^{2}+a}\, a^{2} b c x -15 \sqrt {b \,x^{2}+a}\, a^{2} b d \,x^{3}-40 \sqrt {b \,x^{2}+a}\, a^{2} b d \,x^{2}-40 \sqrt {b \,x^{2}+a}\, a^{2} c^{2} x^{2}-60 \sqrt {b \,x^{2}+a}\, a^{2} c d \,x^{3}+16 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{4}-15 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{3}-40 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{4}-40 \sqrt {b \,x^{2}+a}\, a b \,c^{2} x^{4}-15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} d \,x^{5}+60 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b c d \,x^{5}-15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} c \,x^{5}+15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} d \,x^{5}-60 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b c d \,x^{5}+15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} c \,x^{5}-16 \sqrt {b}\, a \,b^{2} c \,x^{5}-8 \sqrt {b}\, a \,b^{2} d \,x^{5}-8 \sqrt {b}\, a b \,c^{2} x^{5}}{120 a^{2} x^{5}} \] Input:

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

Output:

( - 24*sqrt(a + b*x**2)*a**3*c - 30*sqrt(a + b*x**2)*a**3*d*x - 8*sqrt(a + 
 b*x**2)*a**2*b*c*x**2 - 30*sqrt(a + b*x**2)*a**2*b*c*x - 15*sqrt(a + b*x* 
*2)*a**2*b*d*x**3 - 40*sqrt(a + b*x**2)*a**2*b*d*x**2 - 40*sqrt(a + b*x**2 
)*a**2*c**2*x**2 - 60*sqrt(a + b*x**2)*a**2*c*d*x**3 + 16*sqrt(a + b*x**2) 
*a*b**2*c*x**4 - 15*sqrt(a + b*x**2)*a*b**2*c*x**3 - 40*sqrt(a + b*x**2)*a 
*b**2*d*x**4 - 40*sqrt(a + b*x**2)*a*b*c**2*x**4 - 15*sqrt(a)*log((sqrt(a 
+ b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b**2*d*x**5 + 60*sqrt(a)*log(( 
sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b*c*d*x**5 - 15*sqrt(a) 
*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*b**3*c*x**5 + 15*sq 
rt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b**2*d*x**5 
- 60*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b*c*d 
*x**5 + 15*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b 
**3*c*x**5 - 16*sqrt(b)*a*b**2*c*x**5 - 8*sqrt(b)*a*b**2*d*x**5 - 8*sqrt(b 
)*a*b*c**2*x**5)/(120*a**2*x**5)