\(\int \frac {\sqrt {a+b x^2} (A+B x+C x^2)}{x^5 (c+d x)} \, dx\) [37]

Optimal result

Integrand size = 32, antiderivative size = 318 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^5 (c+d x)} \, dx=-\frac {A \sqrt {a+b x^2}}{4 c x^4}-\frac {(B c-A d) \sqrt {a+b x^2}}{3 c^2 x^3}-\frac {\left (4 a c (c C-B d)+A \left (b c^2+4 a d^2\right )\right ) \sqrt {a+b x^2}}{8 a c^3 x^2}-\frac {\left (b c^2 (B c-A d)-3 a d \left (c^2 C-B c d+A d^2\right )\right ) \sqrt {a+b x^2}}{3 a c^4 x}+\frac {d \sqrt {b c^2+a d^2} \left (c^2 C-B c d+A d^2\right ) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{c^5}-\frac {\left (4 a c (c C-B d) \left (b c^2+2 a d^2\right )-A \left (b^2 c^4-4 a b c^2 d^2-8 a^2 d^4\right )\right ) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{3/2} c^5} \] Output:

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

Mathematica [A] (verified)

Time = 2.26 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^5 (c+d x)} \, dx=\frac {\frac {c \sqrt {a+b x^2} \left (b c^2 x^2 (-3 A c-8 B c x+8 A d x)-2 a \left (A \left (3 c^3-4 c^2 d x+6 c d^2 x^2-12 d^3 x^3\right )+2 c x \left (3 c C x (c-2 d x)+B \left (2 c^2-3 c d x+6 d^2 x^2\right )\right )\right )\right )}{a x^4}-48 d \sqrt {-b c^2-a d^2} \left (c^2 C-B c d+A d^2\right ) \arctan \left (\frac {\sqrt {b} (c+d x)-d \sqrt {a+b x^2}}{\sqrt {-b c^2-a d^2}}\right )+48 \sqrt {a} A d^4 \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {6 c \left (A b c \left (b c^2-4 a d^2\right )-4 a (c C-B d) \left (b c^2+2 a d^2\right )\right ) \text {arctanh}\left (\frac {-\sqrt {b} x+\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}}{24 c^5} \] Input:

Integrate[(Sqrt[a + b*x^2]*(A + B*x + C*x^2))/(x^5*(c + d*x)),x]
 

Output:

((c*Sqrt[a + b*x^2]*(b*c^2*x^2*(-3*A*c - 8*B*c*x + 8*A*d*x) - 2*a*(A*(3*c^ 
3 - 4*c^2*d*x + 6*c*d^2*x^2 - 12*d^3*x^3) + 2*c*x*(3*c*C*x*(c - 2*d*x) + B 
*(2*c^2 - 3*c*d*x + 6*d^2*x^2)))))/(a*x^4) - 48*d*Sqrt[-(b*c^2) - a*d^2]*( 
c^2*C - B*c*d + A*d^2)*ArcTan[(Sqrt[b]*(c + d*x) - d*Sqrt[a + b*x^2])/Sqrt 
[-(b*c^2) - a*d^2]] + 48*Sqrt[a]*A*d^4*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2 
])/Sqrt[a]] + (6*c*(A*b*c*(b*c^2 - 4*a*d^2) - 4*a*(c*C - B*d)*(b*c^2 + 2*a 
*d^2))*ArcTanh[(-(Sqrt[b]*x) + Sqrt[a + b*x^2])/Sqrt[a]])/a^(3/2))/(24*c^5 
)
 

Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.11, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2353, 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.

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

\(\Big \downarrow \) 2353

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

\(\Big \downarrow \) 2009

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

Input:

Int[(Sqrt[a + b*x^2]*(A + B*x + C*x^2))/(x^5*(c + d*x)),x]
 

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2353

Int[(Px_)*((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2) 
^(p_), x_Symbol] :> Int[ExpandIntegrand[Px*(e*x)^m*(c + d*x)^n*(a + b*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && PolyQ[Px, x] && (Integer 
Q[p] || (IntegerQ[2*p] && IntegerQ[m] && ILtQ[n, 0]))
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.33

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-24 A a \,d^{3} x^{3}-8 A b \,c^{2} d \,x^{3}+24 B a c \,d^{2} x^{3}+8 B b \,c^{3} x^{3}-24 C a \,c^{2} d \,x^{3}+12 A a c \,d^{2} x^{2}+3 A b \,c^{3} x^{2}-12 B a \,c^{2} d \,x^{2}+12 C a \,c^{3} x^{2}-8 A a \,c^{2} d x +8 B a \,c^{3} x +6 A a \,c^{3}\right )}{24 a \,c^{4} x^{4}}-\frac {\frac {\left (8 A \,a^{2} d^{4}+4 A a b \,c^{2} d^{2}-A \,b^{2} c^{4}-8 B \,a^{2} c \,d^{3}-4 B a b \,c^{3} d +8 C \,a^{2} c^{2} d^{2}+4 C a b \,c^{4}\right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{c \sqrt {a}}-\frac {8 a \left (A a \,d^{4}+A b \,c^{2} d^{2}-B a c \,d^{3}-c^{3} B b d +C a \,c^{2} d^{2}+c^{4} C b \right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{c \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}}{8 c^{4} a}\)	\(424\)
default	\(\frac {A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )}{4 a}\right )}{c}+\frac {\left (A d -B c \right ) \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 c^{2} a \,x^{3}}+\frac {\left (A \,d^{2}-B c d +C \,c^{2}\right ) \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )}{c^{3}}+\frac {\left (A \,d^{2}-B c d +C \,c^{2}\right ) d^{2} \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{c^{5}}-\frac {\left (A \,d^{2}-B c d +C \,c^{2}\right ) d \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{a x}+\frac {2 b \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{a}\right )}{c^{4}}-\frac {\left (A \,d^{2}-B c d +C \,c^{2}\right ) d^{2} \left (\sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}-\frac {\sqrt {b}\, c \ln \left (\frac {-\frac {b c}{d}+b \left (x +\frac {c}{d}\right )}{\sqrt {b}}+\sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\right )}{d}-\frac {\left (a \,d^{2}+b \,c^{2}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}\right )}{c^{5}}\)	\(625\)

Input:

int((b*x^2+a)^(1/2)*(C*x^2+B*x+A)/x^5/(d*x+c),x,method=_RETURNVERBOSE)
 

Output:

-1/24*(b*x^2+a)^(1/2)*(-24*A*a*d^3*x^3-8*A*b*c^2*d*x^3+24*B*a*c*d^2*x^3+8* 
B*b*c^3*x^3-24*C*a*c^2*d*x^3+12*A*a*c*d^2*x^2+3*A*b*c^3*x^2-12*B*a*c^2*d*x 
^2+12*C*a*c^3*x^2-8*A*a*c^2*d*x+8*B*a*c^3*x+6*A*a*c^3)/a/c^4/x^4-1/8/c^4/a 
*((8*A*a^2*d^4+4*A*a*b*c^2*d^2-A*b^2*c^4-8*B*a^2*c*d^3-4*B*a*b*c^3*d+8*C*a 
^2*c^2*d^2+4*C*a*b*c^4)/c/a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)-8* 
a*(A*a*d^4+A*b*c^2*d^2-B*a*c*d^3-B*b*c^3*d+C*a*c^2*d^2+C*b*c^4)/c/((a*d^2+ 
b*c^2)/d^2)^(1/2)*ln((2*(a*d^2+b*c^2)/d^2-2*b*c/d*(x+c/d)+2*((a*d^2+b*c^2) 
/d^2)^(1/2)*(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))/(x+c/d) 
))
 

Fricas [A] (verification not implemented)

Time = 6.43 (sec) , antiderivative size = 1514, normalized size of antiderivative = 4.76 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^5 (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((b*x^2+a)^(1/2)*(C*x^2+B*x+A)/x^5/(d*x+c),x, algorithm="fricas")
 

Output:

[1/48*(24*(C*a^2*c^2*d - B*a^2*c*d^2 + A*a^2*d^3)*sqrt(b*c^2 + a*d^2)*x^4* 
log((2*a*b*c*d*x - a*b*c^2 - 2*a^2*d^2 - (2*b^2*c^2 + a*b*d^2)*x^2 + 2*sqr 
t(b*c^2 + a*d^2)*(b*c*x - a*d)*sqrt(b*x^2 + a))/(d^2*x^2 + 2*c*d*x + c^2)) 
 - 3*(4*B*a*b*c^3*d + 8*B*a^2*c*d^3 - 8*A*a^2*d^4 - (4*C*a*b - A*b^2)*c^4 
- 4*(2*C*a^2 + A*a*b)*c^2*d^2)*sqrt(a)*x^4*log(-(b*x^2 - 2*sqrt(b*x^2 + a) 
*sqrt(a) + 2*a)/x^2) - 2*(6*A*a^2*c^4 + 8*(B*a*b*c^4 + 3*B*a^2*c^2*d^2 - 3 
*A*a^2*c*d^3 - (3*C*a^2 + A*a*b)*c^3*d)*x^3 - 3*(4*B*a^2*c^3*d - 4*A*a^2*c 
^2*d^2 - (4*C*a^2 + A*a*b)*c^4)*x^2 + 8*(B*a^2*c^4 - A*a^2*c^3*d)*x)*sqrt( 
b*x^2 + a))/(a^2*c^5*x^4), 1/48*(48*(C*a^2*c^2*d - B*a^2*c*d^2 + A*a^2*d^3 
)*sqrt(-b*c^2 - a*d^2)*x^4*arctan(sqrt(-b*c^2 - a*d^2)*(b*c*x - a*d)*sqrt( 
b*x^2 + a)/(a*b*c^2 + a^2*d^2 + (b^2*c^2 + a*b*d^2)*x^2)) - 3*(4*B*a*b*c^3 
*d + 8*B*a^2*c*d^3 - 8*A*a^2*d^4 - (4*C*a*b - A*b^2)*c^4 - 4*(2*C*a^2 + A* 
a*b)*c^2*d^2)*sqrt(a)*x^4*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x 
^2) - 2*(6*A*a^2*c^4 + 8*(B*a*b*c^4 + 3*B*a^2*c^2*d^2 - 3*A*a^2*c*d^3 - (3 
*C*a^2 + A*a*b)*c^3*d)*x^3 - 3*(4*B*a^2*c^3*d - 4*A*a^2*c^2*d^2 - (4*C*a^2 
 + A*a*b)*c^4)*x^2 + 8*(B*a^2*c^4 - A*a^2*c^3*d)*x)*sqrt(b*x^2 + a))/(a^2* 
c^5*x^4), -1/24*(3*(4*B*a*b*c^3*d + 8*B*a^2*c*d^3 - 8*A*a^2*d^4 - (4*C*a*b 
 - A*b^2)*c^4 - 4*(2*C*a^2 + A*a*b)*c^2*d^2)*sqrt(-a)*x^4*arctan(sqrt(b*x^ 
2 + a)*sqrt(-a)/a) - 12*(C*a^2*c^2*d - B*a^2*c*d^2 + A*a^2*d^3)*sqrt(b*c^2 
 + a*d^2)*x^4*log((2*a*b*c*d*x - a*b*c^2 - 2*a^2*d^2 - (2*b^2*c^2 + a*b...
 

Sympy [F]

\[ \int \frac {\sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^5 (c+d x)} \, dx=\int \frac {\sqrt {a + b x^{2}} \left (A + B x + C x^{2}\right )}{x^{5} \left (c + d x\right )}\, dx \] Input:

integrate((b*x**2+a)**(1/2)*(C*x**2+B*x+A)/x**5/(d*x+c),x)
 

Output:

Integral(sqrt(a + b*x**2)*(A + B*x + C*x**2)/(x**5*(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {\sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^5 (c+d x)} \, dx=\int { \frac {{\left (C x^{2} + B x + A\right )} \sqrt {b x^{2} + a}}{{\left (d x + c\right )} x^{5}} \,d x } \] Input:

integrate((b*x^2+a)^(1/2)*(C*x^2+B*x+A)/x^5/(d*x+c),x, algorithm="maxima")
 

Output:

integrate((C*x^2 + B*x + A)*sqrt(b*x^2 + a)/((d*x + c)*x^5), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1249 vs. \(2 (289) = 578\).

Time = 0.24 (sec) , antiderivative size = 1249, normalized size of antiderivative = 3.93 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^5 (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((b*x^2+a)^(1/2)*(C*x^2+B*x+A)/x^5/(d*x+c),x, algorithm="giac")
 

Output:

-2*(C*b*c^4*d - B*b*c^3*d^2 + C*a*c^2*d^3 + A*b*c^2*d^3 - B*a*c*d^4 + A*a* 
d^5)*arctan(-((sqrt(b)*x - sqrt(b*x^2 + a))*d + sqrt(b)*c)/sqrt(-b*c^2 - a 
*d^2))/(sqrt(-b*c^2 - a*d^2)*c^5) + 1/4*(4*C*a*b*c^4 - A*b^2*c^4 - 4*B*a*b 
*c^3*d + 8*C*a^2*c^2*d^2 + 4*A*a*b*c^2*d^2 - 8*B*a^2*c*d^3 + 8*A*a^2*d^4)* 
arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a*c^5) + 1/12*(1 
2*(sqrt(b)*x - sqrt(b*x^2 + a))^7*C*a*b*c^3 + 3*(sqrt(b)*x - sqrt(b*x^2 + 
a))^7*A*b^2*c^3 - 12*(sqrt(b)*x - sqrt(b*x^2 + a))^7*B*a*b*c^2*d + 12*(sqr 
t(b)*x - sqrt(b*x^2 + a))^7*A*a*b*c*d^2 + 24*(sqrt(b)*x - sqrt(b*x^2 + a)) 
^6*B*a*b^(3/2)*c^3 - 24*(sqrt(b)*x - sqrt(b*x^2 + a))^6*C*a^2*sqrt(b)*c^2* 
d - 24*(sqrt(b)*x - sqrt(b*x^2 + a))^6*A*a*b^(3/2)*c^2*d + 24*(sqrt(b)*x - 
 sqrt(b*x^2 + a))^6*B*a^2*sqrt(b)*c*d^2 - 24*(sqrt(b)*x - sqrt(b*x^2 + a)) 
^6*A*a^2*sqrt(b)*d^3 - 12*(sqrt(b)*x - sqrt(b*x^2 + a))^5*C*a^2*b*c^3 + 21 
*(sqrt(b)*x - sqrt(b*x^2 + a))^5*A*a*b^2*c^3 + 12*(sqrt(b)*x - sqrt(b*x^2 
+ a))^5*B*a^2*b*c^2*d - 12*(sqrt(b)*x - sqrt(b*x^2 + a))^5*A*a^2*b*c*d^2 - 
 24*(sqrt(b)*x - sqrt(b*x^2 + a))^4*B*a^2*b^(3/2)*c^3 + 72*(sqrt(b)*x - sq 
rt(b*x^2 + a))^4*C*a^3*sqrt(b)*c^2*d + 24*(sqrt(b)*x - sqrt(b*x^2 + a))^4* 
A*a^2*b^(3/2)*c^2*d - 72*(sqrt(b)*x - sqrt(b*x^2 + a))^4*B*a^3*sqrt(b)*c*d 
^2 + 72*(sqrt(b)*x - sqrt(b*x^2 + a))^4*A*a^3*sqrt(b)*d^3 - 12*(sqrt(b)*x 
- sqrt(b*x^2 + a))^3*C*a^3*b*c^3 + 21*(sqrt(b)*x - sqrt(b*x^2 + a))^3*A*a^ 
2*b^2*c^3 + 12*(sqrt(b)*x - sqrt(b*x^2 + a))^3*B*a^3*b*c^2*d - 12*(sqrt...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^5 (c+d x)} \, dx=\int \frac {\sqrt {b\,x^2+a}\,\left (C\,x^2+B\,x+A\right )}{x^5\,\left (c+d\,x\right )} \,d x \] Input:

int(((a + b*x^2)^(1/2)*(A + B*x + C*x^2))/(x^5*(c + d*x)),x)
 

Output:

int(((a + b*x^2)^(1/2)*(A + B*x + C*x^2))/(x^5*(c + d*x)), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 853, normalized size of antiderivative = 2.68 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x+C x^2\right )}{x^5 (c+d x)} \, dx =\text {Too large to display} \] Input:

int((b*x^2+a)^(1/2)*(C*x^2+B*x+A)/x^5/(d*x+c),x)
 

Output:

(48*sqrt(a*d**2 + b*c**2)*log( - sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - 
a*d + b*c*x)*a**2*d**3*x**4 - 48*sqrt(a*d**2 + b*c**2)*log( - sqrt(a + b*x 
**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a*b*c*d**2*x**4 + 48*sqrt(a*d**2 
 + b*c**2)*log( - sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a* 
c**3*d*x**4 - 48*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a**2*d**3*x**4 + 48*sq 
rt(a*d**2 + b*c**2)*log(c + d*x)*a*b*c*d**2*x**4 - 48*sqrt(a*d**2 + b*c**2 
)*log(c + d*x)*a*c**3*d*x**4 - 12*sqrt(a + b*x**2)*a**2*c**4 + 16*sqrt(a + 
 b*x**2)*a**2*c**3*d*x - 24*sqrt(a + b*x**2)*a**2*c**2*d**2*x**2 + 48*sqrt 
(a + b*x**2)*a**2*c*d**3*x**3 - 6*sqrt(a + b*x**2)*a*b*c**4*x**2 - 16*sqrt 
(a + b*x**2)*a*b*c**4*x + 16*sqrt(a + b*x**2)*a*b*c**3*d*x**3 + 24*sqrt(a 
+ b*x**2)*a*b*c**3*d*x**2 - 48*sqrt(a + b*x**2)*a*b*c**2*d**2*x**3 - 24*sq 
rt(a + b*x**2)*a*c**5*x**2 + 48*sqrt(a + b*x**2)*a*c**4*d*x**3 - 16*sqrt(a 
 + b*x**2)*b**2*c**4*x**3 + 24*sqrt(a)*log(sqrt(a + b*x**2) - sqrt(a))*a** 
2*d**4*x**4 + 12*sqrt(a)*log(sqrt(a + b*x**2) - sqrt(a))*a*b*c**2*d**2*x** 
4 - 24*sqrt(a)*log(sqrt(a + b*x**2) - sqrt(a))*a*b*c*d**3*x**4 + 24*sqrt(a 
)*log(sqrt(a + b*x**2) - sqrt(a))*a*c**3*d**2*x**4 - 3*sqrt(a)*log(sqrt(a 
+ b*x**2) - sqrt(a))*b**2*c**4*x**4 - 12*sqrt(a)*log(sqrt(a + b*x**2) - sq 
rt(a))*b**2*c**3*d*x**4 + 12*sqrt(a)*log(sqrt(a + b*x**2) - sqrt(a))*b*c** 
5*x**4 - 24*sqrt(a)*log(sqrt(a + b*x**2) + sqrt(a))*a**2*d**4*x**4 - 12*sq 
rt(a)*log(sqrt(a + b*x**2) + sqrt(a))*a*b*c**2*d**2*x**4 + 24*sqrt(a)*l...