\(\int \frac {(b d+2 c d x)^{7/2}}{(a+b x+c x^2)^{3/2}} \, dx\) [276]

Optimal result

Integrand size = 28, antiderivative size = 162 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 d (b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}}+\frac {40}{3} c d^3 \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {20 \left (b^2-4 a c\right )^{5/4} d^{7/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 \sqrt {a+b x+c x^2}} \] Output:

-2*d*(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^(1/2)+40/3*c*d^3*(2*c*d*x+b*d)^(1/2 
)*(c*x^2+b*x+a)^(1/2)+20/3*(-4*a*c+b^2)^(5/4)*d^(7/2)*(-c*(c*x^2+b*x+a)/(- 
4*a*c+b^2))^(1/2)*EllipticF((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2) 
,I)/(c*x^2+b*x+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.19 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.75 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {2 d^3 \sqrt {d (b+2 c x)} \left (-3 b^2+8 b c x+4 c \left (5 a+2 c x^2\right )+10 \left (b^2-4 a c\right ) \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{3 \sqrt {a+x (b+c x)}} \] Input:

Integrate[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(3/2),x]
 

Output:

(2*d^3*Sqrt[d*(b + 2*c*x)]*(-3*b^2 + 8*b*c*x + 4*c*(5*a + 2*c*x^2) + 10*(b 
^2 - 4*a*c)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hypergeometric2F1[1 
/4, 1/2, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(3*Sqrt[a + x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1110, 1116, 1115, 1113, 762}

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.

\(\displaystyle \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1110

\(\displaystyle 10 c d^2 \int \frac {(b d+2 c x d)^{3/2}}{\sqrt {c x^2+b x+a}}dx-\frac {2 d (b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}}\)

\(\Big \downarrow \) 1116

\(\displaystyle 10 c d^2 \left (\frac {1}{3} d^2 \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {c x^2+b x+a}}dx+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )-\frac {2 d (b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}}\)

\(\Big \downarrow \) 1115

\(\displaystyle 10 c d^2 \left (\frac {d^2 \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{3 \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )-\frac {2 d (b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}}\)

\(\Big \downarrow \) 1113

\(\displaystyle 10 c d^2 \left (\frac {2 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{3 c \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )-\frac {2 d (b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}}\)

\(\Big \downarrow \) 762

\(\displaystyle 10 c d^2 \left (\frac {2 d^{3/2} \left (b^2-4 a c\right )^{5/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )-\frac {2 d (b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}}\)

Input:

Int[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(3/2),x]
 

Output:

(-2*d*(b*d + 2*c*d*x)^(5/2))/Sqrt[a + b*x + c*x^2] + 10*c*d^2*((4*d*Sqrt[b 
*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2])/3 + (2*(b^2 - 4*a*c)^(5/4)*d^(3/2)*Sq 
rt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSin[Sqrt[b*d + 2*c 
*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(3*c*Sqrt[a + b*x + c*x^2]))
 

Defintions of rubi rules used

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 1110

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), 
x] - Simp[d*e*((m - 1)/(b*(p + 1)))   Int[(d + e*x)^(m - 2)*(a + b*x + c*x^ 
2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0] && N 
eQ[m + 2*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]
 

rule 1113

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_ 
Symbol] :> Simp[(4/e)*Sqrt[-c/(b^2 - 4*a*c)]   Subst[Int[1/Sqrt[Simp[1 - b^ 
2*(x^4/(d^2*(b^2 - 4*a*c))), x]], x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, 
 c, d, e}, x] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]
 

rule 1115

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Sym 
bol] :> Simp[Sqrt[(-c)*((a + b*x + c*x^2)/(b^2 - 4*a*c))]/Sqrt[a + b*x + c* 
x^2]   Int[(d + e*x)^m/Sqrt[(-a)*(c/(b^2 - 4*a*c)) - b*c*(x/(b^2 - 4*a*c)) 
- c^2*(x^2/(b^2 - 4*a*c))], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c* 
d - b*e, 0] && EqQ[m^2, 1/4]
 

rule 1116

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[2*d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p 
 + 1))), x] + Simp[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1)))   Int[(d 
 + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p 
+ 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || OddQ[m])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(365\) vs. \(2(136)=272\).

Time = 7.96 (sec) , antiderivative size = 366, normalized size of antiderivative = 2.26

method	result	size
default	\(-\frac {2 \sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, d^{3} \left (20 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) a c -5 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) b^{2}-16 c^{3} x^{3}-24 b \,c^{2} x^{2}-40 a \,c^{2} x -2 b^{2} c x -20 a b c +3 b^{3}\right )}{3 \left (2 x^{3} c^{2}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right )}\)	\(366\)
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \sqrt {d \left (2 c x +b \right )}\, \left (\frac {2 \left (2 c^{2} d x +d b c \right ) \left (4 a c -b^{2}\right ) d^{3}}{c \sqrt {\left (\frac {a}{c}+\frac {b x}{c}+x^{2}\right ) \left (2 c^{2} d x +d b c \right )}}+\frac {16 c \,d^{3} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{3}+\frac {2 \left (-8 c \left (2 a c -b^{2}\right ) d^{4}+2 d^{4} \left (4 a c -b^{2}\right ) c -\frac {16 c \,d^{3} \left (a c d +\frac {1}{2} b^{2} d \right )}{3}\right ) \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right )}{\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}\, d}\)	\(572\)
risch	\(\frac {16 \sqrt {c \,x^{2}+b x +a}\, \left (2 c x +b \right ) c \,d^{4}}{3 \sqrt {d \left (2 c x +b \right )}}+\frac {\left (-\frac {\left (-48 a^{2} c^{2}+24 c a \,b^{2}-3 b^{4}\right ) \left (\frac {4 c^{2} d x +2 d b c}{\left (4 a c -b^{2}\right ) d c \sqrt {\left (\frac {a}{c}+\frac {b x}{c}+x^{2}\right ) \left (2 c^{2} d x +d b c \right )}}+\frac {4 c \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{\left (4 a c -b^{2}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right )}{3}-\frac {32 \left (4 a c -b^{2}\right ) c \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{3 \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right ) d^{4} \sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(921\)

Input:

int((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-2/3*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*d^3*(20*(-4*a*c+b^2)^(1/2)*(1 
/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2)*(-(2*c*x+b)/(-4*a* 
c+b^2)^(1/2))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/ 
2)*EllipticF(1/2*2^(1/2)*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b 
))^(1/2),2^(1/2))*a*c-5*(-4*a*c+b^2)^(1/2)*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(- 
4*a*c+b^2)^(1/2)+b))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x+ 
(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*2^(1/2)*(1/( 
-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2),2^(1/2))*b^2-16*c^3* 
x^3-24*b*c^2*x^2-40*a*c^2*x-2*b^2*c*x-20*a*b*c+3*b^3)/(2*c^2*x^3+3*b*c*x^2 
+2*a*c*x+b^2*x+a*b)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.06 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (5 \, \sqrt {2} {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{3} x^{2} + {\left (b^{3} - 4 \, a b c\right )} d^{3} x + {\left (a b^{2} - 4 \, a^{2} c\right )} d^{3}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) + {\left (8 \, c^{3} d^{3} x^{2} + 8 \, b c^{2} d^{3} x - {\left (3 \, b^{2} c - 20 \, a c^{2}\right )} d^{3}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}\right )}}{3 \, {\left (c^{2} x^{2} + b c x + a c\right )}} \] Input:

integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")
 

Output:

2/3*(5*sqrt(2)*((b^2*c - 4*a*c^2)*d^3*x^2 + (b^3 - 4*a*b*c)*d^3*x + (a*b^2 
 - 4*a^2*c)*d^3)*sqrt(c^2*d)*weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2 
*(2*c*x + b)/c) + (8*c^3*d^3*x^2 + 8*b*c^2*d^3*x - (3*b^2*c - 20*a*c^2)*d^ 
3)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/(c^2*x^2 + b*c*x + a*c)
 

Sympy [F]

\[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {\left (d \left (b + 2 c x\right )\right )^{\frac {7}{2}}}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((2*c*d*x+b*d)**(7/2)/(c*x**2+b*x+a)**(3/2),x)
 

Output:

Integral((d*(b + 2*c*x))**(7/2)/(a + b*x + c*x**2)**(3/2), x)
 

Maxima [F]

\[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {7}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate((2*c*d*x + b*d)^(7/2)/(c*x^2 + b*x + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {7}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")
 

Output:

integrate((2*c*d*x + b*d)^(7/2)/(c*x^2 + b*x + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \] Input:

int((b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(3/2),x)
 

Output:

int((b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\text {too large to display} \] Input:

int((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(3/2),x)
 

Output:

(2*sqrt(d)*d**3*( - 16*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**2*c + 1 
6*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b*c**2*x + 16*sqrt(b + 2*c*x)*s 
qrt(a + b*x + c*x**2)*a*c**3*x**2 + 3*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x** 
2)*b**4 - 8*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**3*c*x - 8*sqrt(b + 2 
*c*x)*sqrt(a + b*x + c*x**2)*b**2*c**2*x**2 - 160*int((sqrt(b + 2*c*x)*sqr 
t(a + b*x + c*x**2)*x**2)/(2*a**3*b*c + 4*a**3*c**2*x - a**2*b**3 + 2*a**2 
*b**2*c*x + 12*a**2*b*c**2*x**2 + 8*a**2*c**3*x**3 - 2*a*b**4*x - 4*a*b**3 
*c*x**2 + 4*a*b**2*c**2*x**3 + 10*a*b*c**3*x**4 + 4*a*c**4*x**5 - b**5*x** 
2 - 4*b**4*c*x**3 - 5*b**3*c**2*x**4 - 2*b**2*c**3*x**5),x)*a**4*c**5 + 16 
0*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(2*a**3*b*c + 4*a**3*c 
**2*x - a**2*b**3 + 2*a**2*b**2*c*x + 12*a**2*b*c**2*x**2 + 8*a**2*c**3*x* 
*3 - 2*a*b**4*x - 4*a*b**3*c*x**2 + 4*a*b**2*c**2*x**3 + 10*a*b*c**3*x**4 
+ 4*a*c**4*x**5 - b**5*x**2 - 4*b**4*c*x**3 - 5*b**3*c**2*x**4 - 2*b**2*c* 
*3*x**5),x)*a**3*b**2*c**4 - 160*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x** 
2)*x**2)/(2*a**3*b*c + 4*a**3*c**2*x - a**2*b**3 + 2*a**2*b**2*c*x + 12*a* 
*2*b*c**2*x**2 + 8*a**2*c**3*x**3 - 2*a*b**4*x - 4*a*b**3*c*x**2 + 4*a*b** 
2*c**2*x**3 + 10*a*b*c**3*x**4 + 4*a*c**4*x**5 - b**5*x**2 - 4*b**4*c*x**3 
 - 5*b**3*c**2*x**4 - 2*b**2*c**3*x**5),x)*a**3*b*c**5*x - 160*int((sqrt(b 
 + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(2*a**3*b*c + 4*a**3*c**2*x - a**2* 
b**3 + 2*a**2*b**2*c*x + 12*a**2*b*c**2*x**2 + 8*a**2*c**3*x**3 - 2*a*b...