\(\int \frac {x^3 (a+b x+c x^2)^{3/2}}{d-f x^2} \, dx\) [51]

Optimal result

Integrand size = 28, antiderivative size = 535 \[ \int \frac {x^3 \left (a+b x+c x^2\right )^{3/2}}{d-f x^2} \, dx=-\frac {\left (1920 c^4 d^2+2560 a c^3 d f+45 b^4 f^2-300 a b^2 c f^2+48 c^2 f \left (5 b^2 d+8 a^2 f\right )\right ) \sqrt {a+b x+c x^2}}{1920 c^3 f^3}-\frac {b \left (560 c^2 d-15 b^2 f+84 a c f\right ) x \sqrt {a+b x+c x^2}}{960 c^2 f^2}-\frac {\left (80 c^2 d+3 b^2 f+96 a c f\right ) x^2 \sqrt {a+b x+c x^2}}{240 c f^2}-\frac {11 b x^3 \sqrt {a+b x+c x^2}}{40 f}-\frac {c x^4 \sqrt {a+b x+c x^2}}{5 f}-\frac {b \left (384 c^4 d^2+192 a c^3 d f-3 b^4 f^2+24 a b^2 c f^2-16 c^2 f \left (b^2 d+3 a^2 f\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{7/2} f^3}-\frac {d \left (c d-b \sqrt {d} \sqrt {f}+a f\right )^{3/2} \text {arctanh}\left (\frac {b \sqrt {d}-2 a \sqrt {f}+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x}{2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f^{7/2}}+\frac {d \left (c d+b \sqrt {d} \sqrt {f}+a f\right )^{3/2} \text {arctanh}\left (\frac {b \sqrt {d}+2 a \sqrt {f}+\left (2 c \sqrt {d}+b \sqrt {f}\right ) x}{2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f^{7/2}} \] Output:

-1/1920*(1920*c^4*d^2+2560*a*c^3*d*f+45*b^4*f^2-300*a*b^2*c*f^2+48*c^2*f*( 
8*a^2*f+5*b^2*d))*(c*x^2+b*x+a)^(1/2)/c^3/f^3-1/960*b*(84*a*c*f-15*b^2*f+5 
60*c^2*d)*x*(c*x^2+b*x+a)^(1/2)/c^2/f^2-1/240*(96*a*c*f+3*b^2*f+80*c^2*d)* 
x^2*(c*x^2+b*x+a)^(1/2)/c/f^2-11/40*b*x^3*(c*x^2+b*x+a)^(1/2)/f-1/5*c*x^4* 
(c*x^2+b*x+a)^(1/2)/f-1/256*b*(384*c^4*d^2+192*a*c^3*d*f-3*b^4*f^2+24*a*b^ 
2*c*f^2-16*c^2*f*(3*a^2*f+b^2*d))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x 
+a)^(1/2))/c^(7/2)/f^3-1/2*d*(c*d-b*d^(1/2)*f^(1/2)+a*f)^(3/2)*arctanh(1/2 
*(b*d^(1/2)-2*a*f^(1/2)+(2*c*d^(1/2)-b*f^(1/2))*x)/(c*d-b*d^(1/2)*f^(1/2)+ 
a*f)^(1/2)/(c*x^2+b*x+a)^(1/2))/f^(7/2)+1/2*d*(c*d+b*d^(1/2)*f^(1/2)+a*f)^ 
(3/2)*arctanh(1/2*(b*d^(1/2)+2*a*f^(1/2)+(2*c*d^(1/2)+b*f^(1/2))*x)/(c*d+b 
*d^(1/2)*f^(1/2)+a*f)^(1/2)/(c*x^2+b*x+a)^(1/2))/f^(7/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 3.30 (sec) , antiderivative size = 734, normalized size of antiderivative = 1.37 \[ \int \frac {x^3 \left (a+b x+c x^2\right )^{3/2}}{d-f x^2} \, dx=\frac {-2 \sqrt {c} \sqrt {a+x (b+c x)} \left (45 b^4 f^2-30 b^2 c f^2 (10 a+b x)+16 c^3 f \left (160 a d+70 b d x+48 a f x^2+33 b f x^3\right )+128 c^4 \left (15 d^2+5 d f x^2+3 f^2 x^4\right )+24 c^2 f \left (16 a^2 f+7 a b f x+b^2 \left (10 d+f x^2\right )\right )\right )-15 b \left (-384 c^4 d^2-192 a c^3 d f+3 b^4 f^2-24 a b^2 c f^2+16 c^2 f \left (b^2 d+3 a^2 f\right )\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )-1920 c^{7/2} d \text {RootSum}\left [b^2 d-a^2 f-4 b \sqrt {c} d \text {$\#$1}+4 c d \text {$\#$1}^2+2 a f \text {$\#$1}^2-f \text {$\#$1}^4\&,\frac {2 b^2 c d^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a c^2 d^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )+a b^2 d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 a^2 c d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a^3 f^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-4 b c^{3/2} d^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-4 a b \sqrt {c} d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+c^2 d^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+b^2 d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+2 a c d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+a^2 f^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{b \sqrt {c} d-2 c d \text {$\#$1}-a f \text {$\#$1}+f \text {$\#$1}^3}\&\right ]}{3840 c^{7/2} f^3} \] Input:

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

Output:

(-2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(45*b^4*f^2 - 30*b^2*c*f^2*(10*a + b*x) 
+ 16*c^3*f*(160*a*d + 70*b*d*x + 48*a*f*x^2 + 33*b*f*x^3) + 128*c^4*(15*d^ 
2 + 5*d*f*x^2 + 3*f^2*x^4) + 24*c^2*f*(16*a^2*f + 7*a*b*f*x + b^2*(10*d + 
f*x^2))) - 15*b*(-384*c^4*d^2 - 192*a*c^3*d*f + 3*b^4*f^2 - 24*a*b^2*c*f^2 
 + 16*c^2*f*(b^2*d + 3*a^2*f))*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c 
*x)]] - 1920*c^(7/2)*d*RootSum[b^2*d - a^2*f - 4*b*Sqrt[c]*d*#1 + 4*c*d*#1 
^2 + 2*a*f*#1^2 - f*#1^4 & , (2*b^2*c*d^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x 
+ c*x^2] - #1] - a*c^2*d^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] 
+ a*b^2*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*a^2*c*d*f*L 
og[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a^3*f^2*Log[-(Sqrt[c]*x) + 
 Sqrt[a + b*x + c*x^2] - #1] - 4*b*c^(3/2)*d^2*Log[-(Sqrt[c]*x) + Sqrt[a + 
 b*x + c*x^2] - #1]*#1 - 4*a*b*Sqrt[c]*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x 
 + c*x^2] - #1]*#1 + c^2*d^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1 
]*#1^2 + b^2*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 + 2*a 
*c*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 + a^2*f^2*Log[- 
(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2)/(b*Sqrt[c]*d - 2*c*d*#1 - 
a*f*#1 + f*#1^3) & ])/(3840*c^(7/2)*f^3)
 

Rubi [A] (verified)

Time = 1.64 (sec) , antiderivative size = 501, normalized size of antiderivative = 0.94, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {7276, 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.

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

\(\Big \downarrow \) 7276

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 b \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{7/2} f}-\frac {b d \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (12 a c f+b^2 (-f)+24 c^2 d\right )}{16 c^{3/2} f^3}-\frac {d \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2} \text {arctanh}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 f^{7/2}}+\frac {d \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2} \text {arctanh}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 f^{7/2}}-\frac {3 b \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^3 f}-\frac {d \sqrt {a+b x+c x^2} \left (8 a c f+b^2 f+2 b c f x+8 c^2 d\right )}{8 c f^3}+\frac {b (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2 f}-\frac {d \left (a+b x+c x^2\right )^{3/2}}{3 f^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{5 c f}\)

Input:

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

Output:

(-3*b*(b^2 - 4*a*c)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(128*c^3*f) - (d*(8 
*c^2*d + b^2*f + 8*a*c*f + 2*b*c*f*x)*Sqrt[a + b*x + c*x^2])/(8*c*f^3) - ( 
d*(a + b*x + c*x^2)^(3/2))/(3*f^2) + (b*(b + 2*c*x)*(a + b*x + c*x^2)^(3/2 
))/(16*c^2*f) - (a + b*x + c*x^2)^(5/2)/(5*c*f) + (3*b*(b^2 - 4*a*c)^2*Arc 
Tanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(256*c^(7/2)*f) - (b* 
d*(24*c^2*d - b^2*f + 12*a*c*f)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b* 
x + c*x^2])])/(16*c^(3/2)*f^3) - (d*(c*d - b*Sqrt[d]*Sqrt[f] + a*f)^(3/2)* 
ArcTanh[(b*Sqrt[d] - 2*a*Sqrt[f] + (2*c*Sqrt[d] - b*Sqrt[f])*x)/(2*Sqrt[c* 
d - b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a + b*x + c*x^2])])/(2*f^(7/2)) + (d*(c* 
d + b*Sqrt[d]*Sqrt[f] + a*f)^(3/2)*ArcTanh[(b*Sqrt[d] + 2*a*Sqrt[f] + (2*c 
*Sqrt[d] + b*Sqrt[f])*x)/(2*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a + b 
*x + c*x^2])])/(2*f^(7/2))
 

Defintions of rubi rules used

rule 2009

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

rule 7276

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE 
xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ 
[n, 0]
 

Maple [A] (verified)

Time = 2.53 (sec) , antiderivative size = 758, normalized size of antiderivative = 1.42

method	result	size
risch	\(-\frac {\left (384 c^{4} f^{2} x^{4}+528 b \,f^{2} c^{3} x^{3}+768 a \,c^{3} f^{2} x^{2}+24 b^{2} c^{2} f^{2} x^{2}+640 c^{4} d f \,x^{2}+168 a b \,c^{2} f^{2} x -30 b^{3} c \,f^{2} x +1120 b \,c^{3} d f x +384 a^{2} c^{2} f^{2}-300 a \,b^{2} c \,f^{2}+2560 a \,c^{3} d f +45 b^{4} f^{2}+240 b^{2} c^{2} d f +1920 c^{4} d^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{1920 c^{3} f^{3}}+\frac {\frac {b \left (48 a^{2} c^{2} f^{2}-24 a \,b^{2} c \,f^{2}-192 a \,c^{3} d f +3 b^{4} f^{2}+16 b^{2} c^{2} d f -384 c^{4} d^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+\frac {128 c^{3} d \left (\sqrt {d f}\, a^{2} f^{2}+2 \sqrt {d f}\, a c d f +\sqrt {d f}\, b^{2} d f +\sqrt {d f}\, c^{2} d^{2}-2 d \,f^{2} a b -2 d^{2} f b c \right ) \ln \left (\frac {\frac {-2 b \sqrt {d f}+2 a f +2 c d}{f}+\frac {\left (-2 c \sqrt {d f}+f b \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {-b \sqrt {d f}+a f +c d}{f}}\, \sqrt {c \left (x +\frac {\sqrt {d f}}{f}\right )^{2}+\frac {\left (-2 c \sqrt {d f}+f b \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+\frac {-b \sqrt {d f}+a f +c d}{f}}}{x +\frac {\sqrt {d f}}{f}}\right )}{\sqrt {d f}\, f \sqrt {\frac {-b \sqrt {d f}+a f +c d}{f}}}+\frac {128 c^{3} d \left (\sqrt {d f}\, a^{2} f^{2}+2 \sqrt {d f}\, a c d f +\sqrt {d f}\, b^{2} d f +\sqrt {d f}\, c^{2} d^{2}+2 d \,f^{2} a b +2 d^{2} f b c \right ) \ln \left (\frac {\frac {2 b \sqrt {d f}+2 a f +2 c d}{f}+\frac {\left (2 c \sqrt {d f}+f b \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {b \sqrt {d f}+a f +c d}{f}}\, \sqrt {c \left (x -\frac {\sqrt {d f}}{f}\right )^{2}+\frac {\left (2 c \sqrt {d f}+f b \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+\frac {b \sqrt {d f}+a f +c d}{f}}}{x -\frac {\sqrt {d f}}{f}}\right )}{\sqrt {d f}\, f \sqrt {\frac {b \sqrt {d f}+a f +c d}{f}}}}{256 c^{3} f^{3}}\)	\(758\)
default	\(\text {Expression too large to display}\)	\(1607\)

Input:

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

Output:

-1/1920*(384*c^4*f^2*x^4+528*b*c^3*f^2*x^3+768*a*c^3*f^2*x^2+24*b^2*c^2*f^ 
2*x^2+640*c^4*d*f*x^2+168*a*b*c^2*f^2*x-30*b^3*c*f^2*x+1120*b*c^3*d*f*x+38 
4*a^2*c^2*f^2-300*a*b^2*c*f^2+2560*a*c^3*d*f+45*b^4*f^2+240*b^2*c^2*d*f+19 
20*c^4*d^2)/c^3*(c*x^2+b*x+a)^(1/2)/f^3+1/256/c^3/f^3*(b*(48*a^2*c^2*f^2-2 
4*a*b^2*c*f^2-192*a*c^3*d*f+3*b^4*f^2+16*b^2*c^2*d*f-384*c^4*d^2)*ln((1/2* 
b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)+128*c^3*d*((d*f)^(1/2)*a^2*f^2 
+2*(d*f)^(1/2)*a*c*d*f+(d*f)^(1/2)*b^2*d*f+(d*f)^(1/2)*c^2*d^2-2*d*f^2*a*b 
-2*d^2*f*b*c)/(d*f)^(1/2)/f/(1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*ln((2/f*( 
-b*(d*f)^(1/2)+a*f+c*d)+1/f*(-2*c*(d*f)^(1/2)+f*b)*(x+(d*f)^(1/2)/f)+2*(1/ 
f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2)*(c*(x+(d*f)^(1/2)/f)^2+1/f*(-2*c*(d*f)^( 
1/2)+f*b)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+a*f+c*d))^(1/2))/(x+(d*f)^ 
(1/2)/f))+128*c^3*d*((d*f)^(1/2)*a^2*f^2+2*(d*f)^(1/2)*a*c*d*f+(d*f)^(1/2) 
*b^2*d*f+(d*f)^(1/2)*c^2*d^2+2*d*f^2*a*b+2*d^2*f*b*c)/(d*f)^(1/2)/f/((b*(d 
*f)^(1/2)+a*f+c*d)/f)^(1/2)*ln((2*(b*(d*f)^(1/2)+a*f+c*d)/f+(2*c*(d*f)^(1/ 
2)+f*b)/f*(x-(d*f)^(1/2)/f)+2*((b*(d*f)^(1/2)+a*f+c*d)/f)^(1/2)*(c*(x-(d*f 
)^(1/2)/f)^2+(2*c*(d*f)^(1/2)+f*b)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)+a*f+ 
c*d)/f)^(1/2))/(x-(d*f)^(1/2)/f)))
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

\[ \int \frac {x^3 \left (a+b x+c x^2\right )^{3/2}}{d-f x^2} \, dx=- \int \frac {a x^{3} \sqrt {a + b x + c x^{2}}}{- d + f x^{2}}\, dx - \int \frac {b x^{4} \sqrt {a + b x + c x^{2}}}{- d + f x^{2}}\, dx - \int \frac {c x^{5} \sqrt {a + b x + c x^{2}}}{- d + f x^{2}}\, dx \] Input:

integrate(x**3*(c*x**2+b*x+a)**(3/2)/(-f*x**2+d),x)
 

Output:

-Integral(a*x**3*sqrt(a + b*x + c*x**2)/(-d + f*x**2), x) - Integral(b*x** 
4*sqrt(a + b*x + c*x**2)/(-d + f*x**2), x) - Integral(c*x**5*sqrt(a + b*x 
+ c*x**2)/(-d + f*x**2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3 \left (a+b x+c x^2\right )^{3/2}}{d-f x^2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(((c*sqrt(4*d*f))/(2*f^2)>0)', se 
e `assume?
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [F(-1)]

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

int((x^3*(a + b*x + c*x^2)^(3/2))/(d - f*x^2),x)
 

Output:

int((x^3*(a + b*x + c*x^2)^(3/2))/(d - f*x^2), x)
 

Reduce [F]

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

int(x^3*(c*x^2+b*x+a)^(3/2)/(-f*x^2+d),x)
 

Output:

int(x^3*(c*x^2+b*x+a)^(3/2)/(-f*x^2+d),x)