\(\int \frac {(d x)^{5/2}}{(a x+b x^2+c x^3)^{5/2}} \, dx\) [2]

Optimal result

Integrand size = 26, antiderivative size = 129 \[ \int \frac {(d x)^{5/2}}{\left (a x+b x^2+c x^3\right )^{5/2}} \, dx=\frac {2 (d x)^{5/2} \left (b^2-2 a c+b c x\right )}{3 a \left (b^2-4 a c\right ) \left (a x+b x^2+c x^3\right )^{3/2}}-\frac {2 (d x)^{3/2} \left (b \left (b^2-12 a c\right )-16 a c^2 x\right ) \left (a d+b d x+c d x^2\right )}{3 a \left (b^2-4 a c\right )^2 \left (a x+b x^2+c x^3\right )^{3/2}} \] Output:

2/3*(d*x)^(5/2)*(b*c*x-2*a*c+b^2)/a/(-4*a*c+b^2)/(c*x^3+b*x^2+a*x)^(3/2)-2 
/3*(d*x)^(3/2)*(b*(-12*a*c+b^2)-16*a*c^2*x)*(c*d*x^2+b*d*x+a*d)/a/(-4*a*c+ 
b^2)^2/(c*x^3+b*x^2+a*x)^(3/2)
 

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.52 \[ \int \frac {(d x)^{5/2}}{\left (a x+b x^2+c x^3\right )^{5/2}} \, dx=\frac {2 d (d x)^{3/2} (b+2 c x) \left (-b^2+8 b c x+4 c \left (3 a+2 c x^2\right )\right )}{3 \left (b^2-4 a c\right )^2 (x (a+x (b+c x)))^{3/2}} \] Input:

Integrate[(d*x)^(5/2)/(a*x + b*x^2 + c*x^3)^(5/2),x]
 

Output:

(2*d*(d*x)^(3/2)*(b + 2*c*x)*(-b^2 + 8*b*c*x + 4*c*(3*a + 2*c*x^2)))/(3*(b 
^2 - 4*a*c)^2*(x*(a + x*(b + c*x)))^(3/2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2467, 30, 1089, 1088}

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.

\(\displaystyle \int \frac {(d x)^{5/2}}{\left (a x+b x^2+c x^3\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 30

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

\(\Big \downarrow \) 1089

\(\displaystyle \frac {d^2 \sqrt {d x} \sqrt {a+b x+c x^2} \left (-\frac {8 c \int \frac {1}{\left (c x^2+b x+a\right )^{3/2}}dx}{3 \left (b^2-4 a c\right )}-\frac {2 (b+2 c x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\right )}{\sqrt {a x+b x^2+c x^3}}\)

\(\Big \downarrow \) 1088

\(\displaystyle \frac {d^2 \sqrt {d x} \sqrt {a+b x+c x^2} \left (\frac {16 c (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 (b+2 c x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\right )}{\sqrt {a x+b x^2+c x^3}}\)

Input:

Int[(d*x)^(5/2)/(a*x + b*x^2 + c*x^3)^(5/2),x]
 

Output:

(d^2*Sqrt[d*x]*Sqrt[a + b*x + c*x^2]*((-2*(b + 2*c*x))/(3*(b^2 - 4*a*c)*(a 
 + b*x + c*x^2)^(3/2)) + (16*c*(b + 2*c*x))/(3*(b^2 - 4*a*c)^2*Sqrt[a + b* 
x + c*x^2])))/Sqrt[a*x + b*x^2 + c*x^3]
 

Defintions of rubi rules used

rule 30

Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I 
ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) 
Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & 
&  !IntegerQ[p]
 

rule 1088

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 
2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c}, x] && 
 NeQ[b^2 - 4*a*c, 0]
 

rule 1089

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 
 3)/((p + 1)*(b^2 - 4*a*c)))   Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre 
eQ[{a, b, c}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.75

method	result	size
gosper	\(\frac {2 \left (c \,x^{2}+b x +a \right ) \left (16 c^{3} x^{3}+24 b \,c^{2} x^{2}+24 a \,c^{2} x +6 b^{2} c x +12 a b c -b^{3}\right ) \left (d x \right )^{\frac {5}{2}}}{3 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \left (c \,x^{3}+b \,x^{2}+x a \right )^{\frac {5}{2}}}\)	\(97\)
orering	\(\frac {2 \left (c \,x^{2}+b x +a \right ) \left (16 c^{3} x^{3}+24 b \,c^{2} x^{2}+24 a \,c^{2} x +6 b^{2} c x +12 a b c -b^{3}\right ) \left (d x \right )^{\frac {5}{2}}}{3 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \left (c \,x^{3}+b \,x^{2}+x a \right )^{\frac {5}{2}}}\)	\(97\)
default	\(\frac {16 d^{2} \sqrt {d x}\, \sqrt {x \left (c \,x^{2}+b x +a \right )}\, \left (16 c^{3} x^{3}+24 b \,c^{2} x^{2}+24 a \,c^{2} x +6 b^{2} c x +12 a b c -b^{3}\right ) c^{2} \sqrt {-\frac {d \left (2 c x +\sqrt {-4 a c +b^{2}}+b \right ) \left (-2 c x +\sqrt {-4 a c +b^{2}}-b \right )}{c}}}{3 x \left (2 c x +\sqrt {-4 a c +b^{2}}+b \right )^{2} \left (b -\sqrt {-4 a c +b^{2}}+2 c x \right )^{2} \left (4 a c -b^{2}\right )^{2} \sqrt {d \left (c \,x^{2}+b x +a \right )}}\)	\(179\)

Input:

int((d*x)^(5/2)/(c*x^3+b*x^2+a*x)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/3*(c*x^2+b*x+a)*(16*c^3*x^3+24*b*c^2*x^2+24*a*c^2*x+6*b^2*c*x+12*a*b*c-b 
^3)*(d*x)^(5/2)/(16*a^2*c^2-8*a*b^2*c+b^4)/(c*x^3+b*x^2+a*x)^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.70 \[ \int \frac {(d x)^{5/2}}{\left (a x+b x^2+c x^3\right )^{5/2}} \, dx=\frac {2 \, {\left (16 \, c^{3} d^{2} x^{3} + 24 \, b c^{2} d^{2} x^{2} + 6 \, {\left (b^{2} c + 4 \, a c^{2}\right )} d^{2} x - {\left (b^{3} - 12 \, a b c\right )} d^{2}\right )} \sqrt {c x^{3} + b x^{2} + a x} \sqrt {d x}}{3 \, {\left ({\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{5} + 2 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{4} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{3} + 2 \, {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x^{2} + {\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2}\right )} x\right )}} \] Input:

integrate((d*x)^(5/2)/(c*x^3+b*x^2+a*x)^(5/2),x, algorithm="fricas")
 

Output:

2/3*(16*c^3*d^2*x^3 + 24*b*c^2*d^2*x^2 + 6*(b^2*c + 4*a*c^2)*d^2*x - (b^3 
- 12*a*b*c)*d^2)*sqrt(c*x^3 + b*x^2 + a*x)*sqrt(d*x)/((b^4*c^2 - 8*a*b^2*c 
^3 + 16*a^2*c^4)*x^5 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^4 + (b^6 - 
 6*a*b^4*c + 32*a^3*c^3)*x^3 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x^2 
+ (a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*x)
 

Sympy [F]

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

integrate((d*x)**(5/2)/(c*x**3+b*x**2+a*x)**(5/2),x)
 

Output:

Integral((d*x)**(5/2)/(x*(a + b*x + c*x**2))**(5/2), x)
 

Maxima [F]

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

integrate((d*x)^(5/2)/(c*x^3+b*x^2+a*x)^(5/2),x, algorithm="maxima")
 

Output:

integrate((d*x)^(5/2)/(c*x^3 + b*x^2 + a*x)^(5/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (118) = 236\).

Time = 0.18 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.09 \[ \int \frac {(d x)^{5/2}}{\left (a x+b x^2+c x^3\right )^{5/2}} \, dx=\frac {2}{3} \, {\left (\frac {{\left (2 \, {\left (4 \, {\left (\frac {2 \, c^{3} d x}{b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}} + \frac {3 \, b c^{2} d}{b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}}\right )} d x + \frac {3 \, {\left (b^{2} c d^{2} + 4 \, a c^{2} d^{2}\right )}}{b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}}\right )} d x - \frac {b^{3} d^{3} - 12 \, a b c d^{3}}{b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}}\right )} d^{6}}{{\left (c d^{3} x^{2} + b d^{3} x + a d^{3}\right )}^{\frac {3}{2}}} + \frac {\sqrt {a d^{3}} b^{3} - 12 \, \sqrt {a d^{3}} a b c}{a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2}}\right )} d \] Input:

integrate((d*x)^(5/2)/(c*x^3+b*x^2+a*x)^(5/2),x, algorithm="giac")
 

Output:

2/3*((2*(4*(2*c^3*d*x/(b^4*d^3 - 8*a*b^2*c*d^3 + 16*a^2*c^2*d^3) + 3*b*c^2 
*d/(b^4*d^3 - 8*a*b^2*c*d^3 + 16*a^2*c^2*d^3))*d*x + 3*(b^2*c*d^2 + 4*a*c^ 
2*d^2)/(b^4*d^3 - 8*a*b^2*c*d^3 + 16*a^2*c^2*d^3))*d*x - (b^3*d^3 - 12*a*b 
*c*d^3)/(b^4*d^3 - 8*a*b^2*c*d^3 + 16*a^2*c^2*d^3))*d^6/(c*d^3*x^2 + b*d^3 
*x + a*d^3)^(3/2) + (sqrt(a*d^3)*b^3 - 12*sqrt(a*d^3)*a*b*c)/(a^2*b^4 - 8* 
a^3*b^2*c + 16*a^4*c^2))*d
 

Mupad [B] (verification not implemented)

Time = 21.51 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.46 \[ \int \frac {(d x)^{5/2}}{\left (a x+b x^2+c x^3\right )^{5/2}} \, dx=\frac {\sqrt {c\,x^3+b\,x^2+a\,x}\,\left (\frac {16\,b\,d^2\,x^2\,\sqrt {d\,x}}{{\left (4\,a\,c-b^2\right )}^2}+\frac {32\,c\,d^2\,x^3\,\sqrt {d\,x}}{3\,{\left (4\,a\,c-b^2\right )}^2}+\frac {4\,d^2\,x\,\sqrt {d\,x}\,\left (b^2+4\,a\,c\right )}{c\,{\left (4\,a\,c-b^2\right )}^2}+\frac {2\,b\,d^2\,\sqrt {d\,x}\,\left (12\,a\,c-b^2\right )}{3\,c^2\,{\left (4\,a\,c-b^2\right )}^2}\right )}{x^5+\frac {x^3\,\left (b^2+2\,a\,c\right )}{c^2}+\frac {a^2\,x}{c^2}+\frac {2\,b\,x^4}{c}+\frac {2\,a\,b\,x^2}{c^2}} \] Input:

int((d*x)^(5/2)/(a*x + b*x^2 + c*x^3)^(5/2),x)
 

Output:

((a*x + b*x^2 + c*x^3)^(1/2)*((16*b*d^2*x^2*(d*x)^(1/2))/(4*a*c - b^2)^2 + 
 (32*c*d^2*x^3*(d*x)^(1/2))/(3*(4*a*c - b^2)^2) + (4*d^2*x*(d*x)^(1/2)*(4* 
a*c + b^2))/(c*(4*a*c - b^2)^2) + (2*b*d^2*(d*x)^(1/2)*(12*a*c - b^2))/(3* 
c^2*(4*a*c - b^2)^2)))/(x^5 + (x^3*(2*a*c + b^2))/c^2 + (a^2*x)/c^2 + (2*b 
*x^4)/c + (2*a*b*x^2)/c^2)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.52 \[ \int \frac {(d x)^{5/2}}{\left (a x+b x^2+c x^3\right )^{5/2}} \, dx=\frac {2 \sqrt {d}\, d^{2} \left (12 \sqrt {c \,x^{2}+b x +a}\, a b c +24 \sqrt {c \,x^{2}+b x +a}\, a \,c^{2} x -\sqrt {c \,x^{2}+b x +a}\, b^{3}+6 \sqrt {c \,x^{2}+b x +a}\, b^{2} c x +24 \sqrt {c \,x^{2}+b x +a}\, b \,c^{2} x^{2}+16 \sqrt {c \,x^{2}+b x +a}\, c^{3} x^{3}-16 \sqrt {c}\, a^{2} c -32 \sqrt {c}\, a b c x -32 \sqrt {c}\, a \,c^{2} x^{2}-16 \sqrt {c}\, b^{2} c \,x^{2}-32 \sqrt {c}\, b \,c^{2} x^{3}-16 \sqrt {c}\, c^{3} x^{4}\right )}{48 a^{2} c^{4} x^{4}-24 a \,b^{2} c^{3} x^{4}+3 b^{4} c^{2} x^{4}+96 a^{2} b \,c^{3} x^{3}-48 a \,b^{3} c^{2} x^{3}+6 b^{5} c \,x^{3}+96 a^{3} c^{3} x^{2}-18 a \,b^{4} c \,x^{2}+3 b^{6} x^{2}+96 a^{3} b \,c^{2} x -48 a^{2} b^{3} c x +6 a \,b^{5} x +48 a^{4} c^{2}-24 a^{3} b^{2} c +3 a^{2} b^{4}} \] Input:

int((d*x)^(5/2)/(c*x^3+b*x^2+a*x)^(5/2),x)
 

Output:

(2*sqrt(d)*d**2*(12*sqrt(a + b*x + c*x**2)*a*b*c + 24*sqrt(a + b*x + c*x** 
2)*a*c**2*x - sqrt(a + b*x + c*x**2)*b**3 + 6*sqrt(a + b*x + c*x**2)*b**2* 
c*x + 24*sqrt(a + b*x + c*x**2)*b*c**2*x**2 + 16*sqrt(a + b*x + c*x**2)*c* 
*3*x**3 - 16*sqrt(c)*a**2*c - 32*sqrt(c)*a*b*c*x - 32*sqrt(c)*a*c**2*x**2 
- 16*sqrt(c)*b**2*c*x**2 - 32*sqrt(c)*b*c**2*x**3 - 16*sqrt(c)*c**3*x**4)) 
/(3*(16*a**4*c**2 - 8*a**3*b**2*c + 32*a**3*b*c**2*x + 32*a**3*c**3*x**2 + 
 a**2*b**4 - 16*a**2*b**3*c*x + 32*a**2*b*c**3*x**3 + 16*a**2*c**4*x**4 + 
2*a*b**5*x - 6*a*b**4*c*x**2 - 16*a*b**3*c**2*x**3 - 8*a*b**2*c**3*x**4 + 
b**6*x**2 + 2*b**5*c*x**3 + b**4*c**2*x**4))