\(\int \frac {(a+b x+c x^2)^{3/2}}{x^4 (d+e x)} \, dx\) [27]

Optimal result

Integrand size = 25, antiderivative size = 279 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x^4 (d+e x)} \, dx=-\frac {a \sqrt {a+b x+c x^2}}{3 d x^3}-\frac {(7 b d-6 a e) \sqrt {a+b x+c x^2}}{12 d^2 x^2}-\frac {\left (3 b^2 d^2+32 a c d^2-30 a b d e+24 a^2 e^2\right ) \sqrt {a+b x+c x^2}}{24 a d^3 x}+\frac {(b d-2 a e) \left (b^2 d^2+8 a b d e-4 a \left (3 c d^2+2 a e^2\right )\right ) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{16 a^{3/2} d^4}+\frac {\left (c d^2-b d e+a e^2\right )^{3/2} \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{d^4} \] Output:

-1/3*a*(c*x^2+b*x+a)^(1/2)/d/x^3-1/12*(-6*a*e+7*b*d)*(c*x^2+b*x+a)^(1/2)/d 
^2/x^2-1/24*(24*a^2*e^2-30*a*b*d*e+32*a*c*d^2+3*b^2*d^2)*(c*x^2+b*x+a)^(1/ 
2)/a/d^3/x+1/16*(-2*a*e+b*d)*(b^2*d^2+8*a*b*d*e-4*a*(2*a*e^2+3*c*d^2))*arc 
tanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/a^(3/2)/d^4+(a*e^2-b*d*e+c 
*d^2)^(3/2)*arctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d)*x)/(a*e^2-b*d*e+c*d^2)^(1/ 
2)/(c*x^2+b*x+a)^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 2.50 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x^4 (d+e x)} \, dx=\frac {-\sqrt {a} \left (d \sqrt {a+x (b+c x)} \left (3 b^2 d^2 x^2+2 a d x (7 b d+16 c d x-15 b e x)+4 a^2 \left (2 d^2-3 d e x+6 e^2 x^2\right )\right )+48 a \left (-c d^2+e (b d-a e)\right )^{3/2} x^3 \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+x (b+c x)}}{\sqrt {-c d^2+e (b d-a e)}}\right )\right )-48 a^3 e^3 x^3 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )+3 d \left (b^3 d^2+6 a b^2 d e+24 a^2 c d e-12 a b \left (c d^2+2 a e^2\right )\right ) x^3 \text {arctanh}\left (\frac {-\sqrt {c} x+\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{24 a^{3/2} d^4 x^3} \] Input:

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

Output:

(-(Sqrt[a]*(d*Sqrt[a + x*(b + c*x)]*(3*b^2*d^2*x^2 + 2*a*d*x*(7*b*d + 16*c 
*d*x - 15*b*e*x) + 4*a^2*(2*d^2 - 3*d*e*x + 6*e^2*x^2)) + 48*a*(-(c*d^2) + 
 e*(b*d - a*e))^(3/2)*x^3*ArcTan[(Sqrt[c]*(d + e*x) - e*Sqrt[a + x*(b + c* 
x)])/Sqrt[-(c*d^2) + e*(b*d - a*e)]])) - 48*a^3*e^3*x^3*ArcTanh[(Sqrt[c]*x 
 - Sqrt[a + x*(b + c*x)])/Sqrt[a]] + 3*d*(b^3*d^2 + 6*a*b^2*d*e + 24*a^2*c 
*d*e - 12*a*b*(c*d^2 + 2*a*e^2))*x^3*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + x*(b 
 + c*x)])/Sqrt[a]])/(24*a^(3/2)*d^4*x^3)
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(835\) vs. \(2(279)=558\).

Time = 1.29 (sec) , antiderivative size = 835, normalized size of antiderivative = 2.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1289, 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.

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

\(\Big \downarrow \) 1289

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^{3/2} \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {c x^2+b x+a}}\right ) e^3}{d^4}+\frac {b \left (b^2-12 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {c x^2+b x+a}}\right ) e^3}{16 c^{3/2} d^4}-\frac {\left (b^2+2 c x b+8 a c\right ) \sqrt {c x^2+b x+a} e^3}{8 c d^4}-\frac {\left (c x^2+b x+a\right )^{3/2} e^2}{d^3 x}-\frac {3 \sqrt {a} b \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {c x^2+b x+a}}\right ) e^2}{2 d^3}+\frac {3 \left (b^2+4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {c x^2+b x+a}}\right ) e^2}{8 \sqrt {c} d^3}+\frac {3 (3 b+2 c x) \sqrt {c x^2+b x+a} e^2}{4 d^3}+\frac {\left (c x^2+b x+a\right )^{3/2} e}{2 d^2 x^2}+\frac {3 \left (b^2+4 a c\right ) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {c x^2+b x+a}}\right ) e}{8 \sqrt {a} d^2}-\frac {3 b \sqrt {c} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {c x^2+b x+a}}\right ) e}{2 d^2}+\frac {3 (b-2 c x) \sqrt {c x^2+b x+a} e}{4 d^2 x}+\frac {\left (8 c^2 d^2+b^2 e^2-2 c e (5 b d-4 a e)-2 c e (2 c d-b e) x\right ) \sqrt {c x^2+b x+a} e}{8 c d^4}-\frac {\left (c x^2+b x+a\right )^{3/2}}{3 d x^3}+\frac {b \left (b^2-12 a c\right ) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {c x^2+b x+a}}\right )}{16 a^{3/2} d}-\frac {(2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {c x^2+b x+a}}\right )}{16 c^{3/2} d^4}+\frac {c^{3/2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {c x^2+b x+a}}\right )}{d}+\frac {\left (c d^2-b e d+a e^2\right )^{3/2} \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b e d+a e^2} \sqrt {c x^2+b x+a}}\right )}{d^4}-\frac {\left (2 a b+\left (b^2+8 a c\right ) x\right ) \sqrt {c x^2+b x+a}}{8 a d x^2}\)

Input:

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

Output:

(3*e*(b - 2*c*x)*Sqrt[a + b*x + c*x^2])/(4*d^2*x) + (3*e^2*(3*b + 2*c*x)*S 
qrt[a + b*x + c*x^2])/(4*d^3) - (e^3*(b^2 + 8*a*c + 2*b*c*x)*Sqrt[a + b*x 
+ c*x^2])/(8*c*d^4) - ((2*a*b + (b^2 + 8*a*c)*x)*Sqrt[a + b*x + c*x^2])/(8 
*a*d*x^2) + (e*(8*c^2*d^2 + b^2*e^2 - 2*c*e*(5*b*d - 4*a*e) - 2*c*e*(2*c*d 
 - b*e)*x)*Sqrt[a + b*x + c*x^2])/(8*c*d^4) - (a + b*x + c*x^2)^(3/2)/(3*d 
*x^3) + (e*(a + b*x + c*x^2)^(3/2))/(2*d^2*x^2) - (e^2*(a + b*x + c*x^2)^( 
3/2))/(d^3*x) + (b*(b^2 - 12*a*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + 
b*x + c*x^2])])/(16*a^(3/2)*d) + (3*(b^2 + 4*a*c)*e*ArcTanh[(2*a + b*x)/(2 
*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(8*Sqrt[a]*d^2) - (3*Sqrt[a]*b*e^2*ArcTa 
nh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(2*d^3) + (a^(3/2)*e^3* 
ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/d^4 + (c^(3/2)*Arc 
Tanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/d - (3*b*Sqrt[c]*e*Ar 
cTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*d^2) + (3*(b^2 + 
4*a*c)*e^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*Sqrt 
[c]*d^3) + (b*(b^2 - 12*a*c)*e^3*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b 
*x + c*x^2])])/(16*c^(3/2)*d^4) - ((2*c*d - b*e)*(8*c^2*d^2 - b^2*e^2 - 4* 
c*e*(2*b*d - 3*a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2]) 
])/(16*c^(3/2)*d^4) + ((c*d^2 - b*d*e + a*e^2)^(3/2)*ArcTanh[(b*d - 2*a*e 
+ (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])]) 
/d^4
 

Defintions of rubi rules used

rule 1289

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

rule 2009

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

Maple [A] (verified)

Time = 1.55 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.44

method	result	size
risch	\(-\frac {\sqrt {c \,x^{2}+b x +a}\, \left (24 a^{2} e^{2} x^{2}-30 a b d e \,x^{2}+32 a \,d^{2} x^{2} c +3 b^{2} d^{2} x^{2}-12 a^{2} d e x +14 a b \,d^{2} x +8 a^{2} d^{2}\right )}{24 a \,d^{3} x^{3}}-\frac {-\frac {\left (16 e^{3} a^{3}-24 a^{2} b d \,e^{2}+24 a^{2} c \,d^{2} e +6 a \,b^{2} d^{2} e -12 a b c \,d^{3}+b^{3} d^{3}\right ) \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{d \sqrt {a}}+\frac {16 a \left (a^{2} e^{4}-2 d \,e^{3} a b +2 a c \,d^{2} e^{2}+d^{2} e^{2} b^{2}-2 b c \,d^{3} e +c^{2} d^{4}\right ) \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{d e \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}}{16 d^{3} a}\)	\(402\)
default	\(\text {Expression too large to display}\)	\(2431\)

Input:

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

Output:

-1/24*(c*x^2+b*x+a)^(1/2)*(24*a^2*e^2*x^2-30*a*b*d*e*x^2+32*a*c*d^2*x^2+3* 
b^2*d^2*x^2-12*a^2*d*e*x+14*a*b*d^2*x+8*a^2*d^2)/a/d^3/x^3-1/16/d^3/a*(-1/ 
d*(16*a^3*e^3-24*a^2*b*d*e^2+24*a^2*c*d^2*e+6*a*b^2*d^2*e-12*a*b*c*d^3+b^3 
*d^3)/a^(1/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+16*a*(a^2*e^4- 
2*a*b*d*e^3+2*a*c*d^2*e^2+b^2*d^2*e^2-2*b*c*d^3*e+c^2*d^4)/d/e/((a*e^2-b*d 
*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2 
*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+(a*e^2 
-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e)))
 

Fricas [A] (verification not implemented)

Time = 1.48 (sec) , antiderivative size = 1591, normalized size of antiderivative = 5.70 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x^4 (d+e x)} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/96*(48*(a^2*c*d^2 - a^2*b*d*e + a^3*e^2)*sqrt(c*d^2 - b*d*e + a*e^2)*x^ 
3*log((8*a*b*d*e - 8*a^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e 
+ (b^2 + 4*a*c)*e^2)*x^2 - 4*sqrt(c*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x 
+ a)*(b*d - 2*a*e + (2*c*d - b*e)*x) - 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 
 4*a*c)*d*e)*x)/(e^2*x^2 + 2*d*e*x + d^2)) - 3*(24*a^2*b*d*e^2 - 16*a^3*e^ 
3 - (b^3 - 12*a*b*c)*d^3 - 6*(a*b^2 + 4*a^2*c)*d^2*e)*sqrt(a)*x^3*log(-(8* 
a*b*x + (b^2 + 4*a*c)*x^2 + 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 
8*a^2)/x^2) - 4*(8*a^3*d^3 - (30*a^2*b*d^2*e - 24*a^3*d*e^2 - (3*a*b^2 + 3 
2*a^2*c)*d^3)*x^2 + 2*(7*a^2*b*d^3 - 6*a^3*d^2*e)*x)*sqrt(c*x^2 + b*x + a) 
)/(a^2*d^4*x^3), 1/96*(96*(a^2*c*d^2 - a^2*b*d*e + a^3*e^2)*sqrt(-c*d^2 + 
b*d*e - a*e^2)*x^3*arctan(-1/2*sqrt(-c*d^2 + b*d*e - a*e^2)*sqrt(c*x^2 + b 
*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x)/(a*c*d^2 - a*b*d*e + a^2*e^2 + (c^ 
2*d^2 - b*c*d*e + a*c*e^2)*x^2 + (b*c*d^2 - b^2*d*e + a*b*e^2)*x)) - 3*(24 
*a^2*b*d*e^2 - 16*a^3*e^3 - (b^3 - 12*a*b*c)*d^3 - 6*(a*b^2 + 4*a^2*c)*d^2 
*e)*sqrt(a)*x^3*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 + 4*sqrt(c*x^2 + b*x + a 
)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) - 4*(8*a^3*d^3 - (30*a^2*b*d^2*e - 24* 
a^3*d*e^2 - (3*a*b^2 + 32*a^2*c)*d^3)*x^2 + 2*(7*a^2*b*d^3 - 6*a^3*d^2*e)* 
x)*sqrt(c*x^2 + b*x + a))/(a^2*d^4*x^3), 1/48*(3*(24*a^2*b*d*e^2 - 16*a^3* 
e^3 - (b^3 - 12*a*b*c)*d^3 - 6*(a*b^2 + 4*a^2*c)*d^2*e)*sqrt(-a)*x^3*arcta 
n(1/2*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2...
 

Sympy [F]

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

integrate((c*x**2+b*x+a)**(3/2)/x**4/(e*x+d),x)
 

Output:

Integral((a + b*x + c*x**2)**(3/2)/(x**4*(d + e*x)), x)
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 916 vs. \(2 (249) = 498\).

Time = 0.34 (sec) , antiderivative size = 916, normalized size of antiderivative = 3.28 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x^4 (d+e x)} \, dx =\text {Too large to display} \] Input:

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

Output:

2*(c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + 2*a*c*d^2*e^2 - 2*a*b*d*e^3 + a^2 
*e^4)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + b*x + a))*e + sqrt(c)*d)/sqrt(-c* 
d^2 + b*d*e - a*e^2))/(sqrt(-c*d^2 + b*d*e - a*e^2)*d^4) - 1/8*(b^3*d^3 - 
12*a*b*c*d^3 + 6*a*b^2*d^2*e + 24*a^2*c*d^2*e - 24*a^2*b*d*e^2 + 16*a^3*e^ 
3)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt(-a)*a*d^4) 
+ 1/24*(3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*b^3*d^2 + 60*(sqrt(c)*x - 
sqrt(c*x^2 + b*x + a))^5*a*b*c*d^2 - 30*(sqrt(c)*x - sqrt(c*x^2 + b*x + a) 
)^5*a*b^2*d*e - 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a^2*c*d*e + 24*(s 
qrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a^2*b*e^2 + 48*(sqrt(c)*x - sqrt(c*x^2 
 + b*x + a))^4*a*b^2*sqrt(c)*d^2 + 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^ 
4*a^2*c^(3/2)*d^2 - 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a^2*b*sqrt(c) 
*d*e + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a^3*sqrt(c)*e^2 + 8*(sqrt( 
c)*x - sqrt(c*x^2 + b*x + a))^3*a*b^3*d^2 + 48*(sqrt(c)*x - sqrt(c*x^2 + b 
*x + a))^3*a^2*b^2*d*e - 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^3*b*e^ 
2 - 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^3*c^(3/2)*d^2 + 144*(sqrt(c 
)*x - sqrt(c*x^2 + b*x + a))^2*a^3*b*sqrt(c)*d*e - 96*(sqrt(c)*x - sqrt(c* 
x^2 + b*x + a))^2*a^4*sqrt(c)*e^2 - 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))* 
a^2*b^3*d^2 + 36*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^3*b*c*d^2 - 18*(sqr 
t(c)*x - sqrt(c*x^2 + b*x + a))*a^3*b^2*d*e + 24*(sqrt(c)*x - sqrt(c*x^2 + 
 b*x + a))*a^4*c*d*e + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^4*b*e^2...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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