\(\int \frac {(a+b x^2)^{3/2} (A+B x+C x^2)}{x^4 (c+d x)^2} \, dx\) [92]

Optimal result

Integrand size = 32, antiderivative size = 466 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^4 (c+d x)^2} \, dx=-\frac {\left (3 b c^2 \left (c^2 C-B c d+A d^2\right )+a d^2 \left (3 c^2 C-3 B c d+4 A d^2\right )\right ) \sqrt {a+b x^2}}{3 c^2 d^4 x^3}+\frac {\left (2 b c^2 \left (c^2 C-B c d+A d^2\right )+a d^2 \left (2 c^2 C-3 B c d+4 A d^2\right )\right ) \sqrt {a+b x^2}}{2 c^3 d^3 x^2}-\frac {\left (3 a d^2 \left (2 c^2 C-3 B c d+4 A d^2\right )+b c^2 \left (3 c^2 C-3 B c d+7 A d^2\right )\right ) \sqrt {a+b x^2}}{3 c^4 d^2 x}+\frac {\left (b c^2+a d^2\right ) \left (c^2 C-B c d+A d^2\right ) \sqrt {a+b x^2}}{c d^4 x^3 (c+d x)}+\frac {b^{3/2} C \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{d^2}-\frac {\sqrt {b c^2+a d^2} \left (a d^2 \left (2 c^2 C-3 B c d+4 A d^2\right )-b \left (c^4 C-A c^2 d^2\right )\right ) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{c^5 d^2}-\frac {\sqrt {a} \left (3 b c^2 (B c-2 A d)-2 a d \left (2 c^2 C-3 B c d+4 A d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 c^5} \] Output:

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

Mathematica [A] (verified)

Time = 2.49 (sec) , antiderivative size = 408, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^4 (c+d x)^2} \, dx=\frac {1}{6} \left (-\frac {\sqrt {a+b x^2} \left (2 b c^2 x^2 (3 c (c C-B d) x+A d (4 c+7 d x))+a d \left (2 A \left (c^3-2 c^2 d x+6 c d^2 x^2+12 d^3 x^3\right )+3 c x \left (2 c C x (c+2 d x)+B \left (c^2-3 c d x-6 d^2 x^2\right )\right )\right )\right )}{c^4 d x^3 (c+d x)}-\frac {12 \sqrt {-b c^2-a d^2} \left (a d^2 \left (-2 c^2 C+3 B c d-4 A d^2\right )+b \left (c^4 C-A c^2 d^2\right )\right ) \arctan \left (\frac {\sqrt {-b c^2-a d^2} x}{\sqrt {a} (c+d x)-c \sqrt {a+b x^2}}\right )}{c^5 d^2}+\frac {12 b^{3/2} C \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{d^2}+\frac {3 \sqrt {a} \left (-3 b c^2 (B c-2 A d)+2 a d \left (2 c^2 C-3 B c d+4 A d^2\right )\right ) \log (x)}{c^5}+\frac {3 \sqrt {a} \left (3 b c^2 (B c-2 A d)-2 a d \left (2 c^2 C-3 B c d+4 A d^2\right )\right ) \log \left (-\sqrt {a}+\sqrt {a+b x^2}\right )}{c^5}\right ) \] Input:

Integrate[((a + b*x^2)^(3/2)*(A + B*x + C*x^2))/(x^4*(c + d*x)^2),x]
 

Output:

(-((Sqrt[a + b*x^2]*(2*b*c^2*x^2*(3*c*(c*C - B*d)*x + A*d*(4*c + 7*d*x)) + 
 a*d*(2*A*(c^3 - 2*c^2*d*x + 6*c*d^2*x^2 + 12*d^3*x^3) + 3*c*x*(2*c*C*x*(c 
 + 2*d*x) + B*(c^2 - 3*c*d*x - 6*d^2*x^2)))))/(c^4*d*x^3*(c + d*x))) - (12 
*Sqrt[-(b*c^2) - a*d^2]*(a*d^2*(-2*c^2*C + 3*B*c*d - 4*A*d^2) + b*(c^4*C - 
 A*c^2*d^2))*ArcTan[(Sqrt[-(b*c^2) - a*d^2]*x)/(Sqrt[a]*(c + d*x) - c*Sqrt 
[a + b*x^2])])/(c^5*d^2) + (12*b^(3/2)*C*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + S 
qrt[a + b*x^2])])/d^2 + (3*Sqrt[a]*(-3*b*c^2*(B*c - 2*A*d) + 2*a*d*(2*c^2* 
C - 3*B*c*d + 4*A*d^2))*Log[x])/c^5 + (3*Sqrt[a]*(3*b*c^2*(B*c - 2*A*d) - 
2*a*d*(2*c^2*C - 3*B*c*d + 4*A*d^2))*Log[-Sqrt[a] + Sqrt[a + b*x^2]])/c^5) 
/6
 

Rubi [A] (verified)

Time = 2.07 (sec) , antiderivative size = 799, normalized size of antiderivative = 1.71, 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.

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

\(\Big \downarrow \) 2353

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

\(\Big \downarrow \) 2009

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

Input:

Int[((a + b*x^2)^(3/2)*(A + B*x + C*x^2))/(x^4*(c + d*x)^2),x]
 

Output:

(3*b*(B*c - 2*A*d)*Sqrt[a + b*x^2])/(2*c^3) - (a*d*(2*c^2*C - 3*B*c*d + 4* 
A*d^2)*Sqrt[a + b*x^2])/c^5 - (A*b*Sqrt[a + b*x^2])/(c^2*x) + (3*b*(c^2*C 
- 2*B*c*d + 3*A*d^2)*x*Sqrt[a + b*x^2])/(2*c^4) - (3*b*(c^2*C - B*c*d + A* 
d^2)*(2*c - d*x)*Sqrt[a + b*x^2])/(2*c^4*d) + ((2*c^2*C - 3*B*c*d + 4*A*d^ 
2)*(2*(b*c^2 + a*d^2) - b*c*d*x)*Sqrt[a + b*x^2])/(2*c^5*d) - (A*(a + b*x^ 
2)^(3/2))/(3*c^2*x^3) - ((B*c - 2*A*d)*(a + b*x^2)^(3/2))/(2*c^3*x^2) - (( 
c^2*C - 2*B*c*d + 3*A*d^2)*(a + b*x^2)^(3/2))/(c^4*x) - (d*(c^2*C - B*c*d 
+ A*d^2)*(a + b*x^2)^(3/2))/(c^4*(c + d*x)) + (A*b^(3/2)*ArcTanh[(Sqrt[b]* 
x)/Sqrt[a + b*x^2]])/c^2 + (3*Sqrt[b]*(2*b*c^2 + a*d^2)*(c^2*C - B*c*d + A 
*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*c^4*d^2) + (3*a*Sqrt[b]*(c^ 
2*C - 2*B*c*d + 3*A*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*c^4) - ( 
Sqrt[b]*(2*b*c^2 + 3*a*d^2)*(2*c^2*C - 3*B*c*d + 4*A*d^2)*ArcTanh[(Sqrt[b] 
*x)/Sqrt[a + b*x^2]])/(2*c^4*d^2) + (3*b*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^3*d^2) - ((b*c^2 + a*d^2)^(3/2)*(2*c^2*C - 3*B*c*d + 4*A*d^2)*ArcTanh[(a 
*d - b*c*x)/(Sqrt[b*c^2 + a*d^2]*Sqrt[a + b*x^2])])/(c^5*d^2) - (3*Sqrt[a] 
*b*(B*c - 2*A*d)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(2*c^3) + (a^(3/2)*d*(2 
*c^2*C - 3*B*c*d + 4*A*d^2)*ArcTanh[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 = 0.42 (sec) , antiderivative size = 692, normalized size of antiderivative = 1.48

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (18 A a \,d^{2} x^{2}+8 A b \,c^{2} x^{2}-12 B a c d \,x^{2}+6 C a \,c^{2} x^{2}-6 A a c d x +3 B a \,c^{2} x +2 A \,c^{2} a \right )}{6 c^{4} x^{3}}+\frac {\frac {2 \left (A \,a^{2} d^{6}+2 A a b \,c^{2} d^{4}+A \,b^{2} c^{4} d^{2}-B \,a^{2} c \,d^{5}-2 B a b \,c^{3} d^{3}-c^{5} B \,b^{2} d +C \,a^{2} c^{2} d^{4}+2 C a b \,c^{4} d^{2}+c^{6} C \,b^{2}\right ) \left (-\frac {d^{2} \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{\left (a \,d^{2}+b \,c^{2}\right ) \left (x +\frac {c}{d}\right )}-\frac {b c d \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{\left (a \,d^{2}+b \,c^{2}\right ) \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}\right )}{d^{4}}-\frac {2 \left (4 A \,a^{2} d^{6}+4 A a b \,c^{2} d^{4}-3 B \,a^{2} c \,d^{5}-2 B a b \,c^{3} d^{3}+c^{5} B \,b^{2} d +2 C \,a^{2} c^{2} d^{4}-2 c^{6} C \,b^{2}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{3} c \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}+\frac {2 C \,b^{\frac {3}{2}} c^{4} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{d^{2}}+\frac {\sqrt {a}\, \left (8 A a \,d^{3}+6 A b \,c^{2} d -6 B a c \,d^{2}-3 B b \,c^{3}+4 C a \,c^{2} d \right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{c}}{2 c^{4}}\)	\(692\)
default	\(\text {Expression too large to display}\)	\(1769\)

Input:

int((b*x^2+a)^(3/2)*(C*x^2+B*x+A)/x^4/(d*x+c)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/6*(b*x^2+a)^(1/2)*(18*A*a*d^2*x^2+8*A*b*c^2*x^2-12*B*a*c*d*x^2+6*C*a*c^ 
2*x^2-6*A*a*c*d*x+3*B*a*c^2*x+2*A*a*c^2)/c^4/x^3+1/2/c^4*(2*(A*a^2*d^6+2*A 
*a*b*c^2*d^4+A*b^2*c^4*d^2-B*a^2*c*d^5-2*B*a*b*c^3*d^3-B*b^2*c^5*d+C*a^2*c 
^2*d^4+2*C*a*b*c^4*d^2+C*b^2*c^6)/d^4*(-1/(a*d^2+b*c^2)*d^2/(x+c/d)*(b*(x+ 
c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)-b*c*d/(a*d^2+b*c^2)/((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)))-2/d^3*(4*A*a^2*d^6+4*A*a*b*c^2*d^4-3*B*a^2*c*d^5-2*B*a*b*c^3*d^3+B*b^ 
2*c^5*d+2*C*a^2*c^2*d^4-2*C*b^2*c^6)/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))+2*C*b^(3/2)*c^4/d^2*ln(b^ 
(1/2)*x+(b*x^2+a)^(1/2))+a^(1/2)/c*(8*A*a*d^3+6*A*b*c^2*d-6*B*a*c*d^2-3*B* 
b*c^3+4*C*a*c^2*d)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^4 (c+d x)^2} \, dx=\text {Timed out} \] Input:

integrate((b*x^2+a)^(3/2)*(C*x^2+B*x+A)/x^4/(d*x+c)^2,x, algorithm="fricas 
")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^4 (c+d x)^2} \, dx=\int \frac {\left (a + b x^{2}\right )^{\frac {3}{2}} \left (A + B x + C x^{2}\right )}{x^{4} \left (c + d x\right )^{2}}\, dx \] Input:

integrate((b*x**2+a)**(3/2)*(C*x**2+B*x+A)/x**4/(d*x+c)**2,x)
 

Output:

Integral((a + b*x**2)**(3/2)*(A + B*x + C*x**2)/(x**4*(c + d*x)**2), x)
 

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^4 (c+d x)^2} \, dx=\int { \frac {{\left (C x^{2} + B x + A\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}}}{{\left (d x + c\right )}^{2} x^{4}} \,d x } \] Input:

integrate((b*x^2+a)^(3/2)*(C*x^2+B*x+A)/x^4/(d*x+c)^2,x, algorithm="maxima 
")
 

Output:

integrate((C*x^2 + B*x + A)*(b*x^2 + a)^(3/2)/((d*x + c)^2*x^4), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^4 (c+d x)^2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((b*x^2+a)^(3/2)*(C*x^2+B*x+A)/x^4/(d*x+c)^2,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^4 (c+d x)^2} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{3/2}\,\left (C\,x^2+B\,x+A\right )}{x^4\,{\left (c+d\,x\right )}^2} \,d x \] Input:

int(((a + b*x^2)^(3/2)*(A + B*x + C*x^2))/(x^4*(c + d*x)^2),x)
 

Output:

int(((a + b*x^2)^(3/2)*(A + B*x + C*x^2))/(x^4*(c + d*x)^2), x)
 

Reduce [B] (verification not implemented)

Time = 11.50 (sec) , antiderivative size = 1779, normalized size of antiderivative = 3.82 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^4 (c+d x)^2} \, dx =\text {Too large to display} \] Input:

int((b*x^2+a)^(3/2)*(C*x^2+B*x+A)/x^4/(d*x+c)^2,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*c*d**4*x**3 + 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**5*x**4 + 12*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**3*d 
**2*x**3 + 12*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**2*d**3*x**4 - 36*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**2*d**3*x**3 - 36* 
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**4*x**4 + 24*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**4*d**2*x**3 + 24*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**3*x 
**4 - 12*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)*b*c**6*x**3 - 12*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)*b*c**5*d*x**4 - 48*sqrt(a*d**2 + b*c 
**2)*log(c + d*x)*a**2*c*d**4*x**3 - 48*sqrt(a*d**2 + b*c**2)*log(c + d*x) 
*a**2*d**5*x**4 - 12*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a*b*c**3*d**2*x**3 
 - 12*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a*b*c**2*d**3*x**4 + 36*sqrt(a*d* 
*2 + b*c**2)*log(c + d*x)*a*b*c**2*d**3*x**3 + 36*sqrt(a*d**2 + b*c**2)*lo 
g(c + d*x)*a*b*c*d**4*x**4 - 24*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a*c**4* 
d**2*x**3 - 24*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a*c**3*d**3*x**4 + 12...