\(\int \frac {(b d+2 c d x)^{11/2}}{(a+b x+c x^2)^{5/2}} \, dx\) [286]

Optimal result

Integrand size = 28, antiderivative size = 196 \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 d (b d+2 c d x)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {12 c d^3 (b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}}+80 c^2 d^5 \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {40 c \left (b^2-4 a c\right )^{5/4} d^{11/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{\sqrt {a+b x+c x^2}} \] Output:

-2/3*d*(2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^(3/2)-12*c*d^3*(2*c*d*x+b*d)^(5/2 
)/(c*x^2+b*x+a)^(1/2)+80*c^2*d^5*(2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^(1/2)+4 
0*c*(-4*a*c+b^2)^(5/4)*d^(11/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)*Elli 
pticF((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 = 10.17 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.91 \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 d^5 \sqrt {d (b+2 c x)} \left (-b^4-26 b^3 c x+6 b^2 c \left (-3 a+c x^2\right )+8 b c^2 x \left (21 a+8 c x^2\right )+8 c^2 \left (15 a^2+21 a c x^2+4 c^2 x^4\right )+60 c \left (b^2-4 a c\right ) (a+x (b+c x)) \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{3 (a+x (b+c x))^{3/2}} \] Input:

Integrate[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^(5/2),x]
 

Output:

(2*d^5*Sqrt[d*(b + 2*c*x)]*(-b^4 - 26*b^3*c*x + 6*b^2*c*(-3*a + c*x^2) + 8 
*b*c^2*x*(21*a + 8*c*x^2) + 8*c^2*(15*a^2 + 21*a*c*x^2 + 4*c^2*x^4) + 60*c 
*(b^2 - 4*a*c)*(a + x*(b + c*x))*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*(a + 
x*(b + c*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1110, 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.

\(\displaystyle \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1110

\(\displaystyle 6 c d^2 \int \frac {(b d+2 c x d)^{7/2}}{\left (c x^2+b x+a\right )^{3/2}}dx-\frac {2 d (b d+2 c d x)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1110

\(\displaystyle 6 c d^2 \left (10 c d^2 \int \frac {(b d+2 c x d)^{3/2}}{\sqrt {c x^2+b x+a}}dx-\frac {2 d (b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1116

\(\displaystyle 6 c d^2 \left (10 c d^2 \left (\frac {1}{3} d^2 \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {c x^2+b x+a}}dx+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )-\frac {2 d (b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1115

\(\displaystyle 6 c d^2 \left (10 c d^2 \left (\frac {d^2 \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{3 \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )-\frac {2 d (b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1113

\(\displaystyle 6 c d^2 \left (10 c d^2 \left (\frac {2 d \left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{3 c \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )-\frac {2 d (b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle 6 c d^2 \left (10 c d^2 \left (\frac {2 d^{3/2} \left (b^2-4 a c\right )^{5/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}\right )-\frac {2 d (b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}}\right )-\frac {2 d (b d+2 c d x)^{9/2}}{3 \left (a+b x+c x^2\right )^{3/2}}\)

Input:

Int[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^(5/2),x]
 

Output:

(-2*d*(b*d + 2*c*d*x)^(9/2))/(3*(a + b*x + c*x^2)^(3/2)) + 6*c*d^2*((-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)*Sqrt[-(( 
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. \(656\) vs. \(2(168)=336\).

Time = 12.65 (sec) , antiderivative size = 657, normalized size of antiderivative = 3.35

method	result	size
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \sqrt {d \left (2 c x +b \right )}\, \left (-\frac {2 d^{5} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{3 c^{2} \left (\frac {a}{c}+\frac {b x}{c}+x^{2}\right )^{2}}+\frac {52 \left (2 c^{2} d x +d b c \right ) \left (4 a c -b^{2}\right ) d^{5}}{3 \sqrt {\left (\frac {a}{c}+\frac {b x}{c}+x^{2}\right ) \left (2 c^{2} d x +d b c \right )}}+\frac {64 c^{2} d^{5} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{3}+\frac {2 \left (-16 c^{2} \left (8 a c -3 b^{2}\right ) d^{6}+\frac {52 d^{6} c^{2} \left (4 a c -b^{2}\right )}{3}-\frac {64 c^{2} d^{5} \left (a c d +\frac {1}{2} b^{2} d \right )}{3}\right ) \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right )}{\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}\, d}\)	\(657\)
default	\(-\frac {2 \left (120 \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 ) \sqrt {-4 a c +b^{2}}\, a \,c^{3} x^{2}-30 \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 ) \sqrt {-4 a c +b^{2}}\, b^{2} c^{2} x^{2}+120 \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 ) \sqrt {-4 a c +b^{2}}\, a b \,c^{2} x -30 \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 ) \sqrt {-4 a c +b^{2}}\, b^{3} c x -64 c^{5} x^{5}+120 \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^{2} c^{2}-30 \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 \,b^{2} c -160 b \,x^{4} c^{4}-336 a \,c^{4} x^{3}-76 b^{2} c^{3} x^{3}-504 a b \,c^{3} x^{2}+46 c^{2} x^{2} b^{3}-240 a^{2} c^{3} x -132 a \,b^{2} c^{2} x +28 b^{4} c x -120 a^{2} b \,c^{2}+18 a \,b^{3} c +b^{5}\right ) d^{5} \sqrt {d \left (2 c x +b \right )}}{3 \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\)	\(958\)
risch	\(\text {Expression too large to display}\)	\(1470\)

Input:

int((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

(d*(2*c*x+b)*(c*x^2+b*x+a))^(1/2)*(d*(2*c*x+b))^(1/2)/(2*c*x+b)/(c*x^2+b*x 
+a)^(1/2)/d*(-2/3*d^5*(16*a^2*c^2-8*a*b^2*c+b^4)/c^2*(2*c^2*d*x^3+3*b*c*d* 
x^2+2*a*c*d*x+b^2*d*x+a*b*d)^(1/2)/(a/c+b/c*x+x^2)^2+52/3*(2*c^2*d*x+b*c*d 
)*(4*a*c-b^2)*d^5/((a/c+b/c*x+x^2)*(2*c^2*d*x+b*c*d))^(1/2)+64/3*c^2*d^5*( 
2*c^2*d*x^3+3*b*c*d*x^2+2*a*c*d*x+b^2*d*x+a*b*d)^(1/2)+2*(-16*c^2*(8*a*c-3 
*b^2)*d^6+52/3*d^6*c^2*(4*a*c-b^2)-64/3*c^2*d^5*(a*c*d+1/2*b^2*d))*(1/2/c* 
(-b+(-4*a*c+b^2)^(1/2))+1/2*(b+(-4*a*c+b^2)^(1/2))/c)*((x+1/2*(b+(-4*a*c+b 
^2)^(1/2))/c)/(1/2/c*(-b+(-4*a*c+b^2)^(1/2))+1/2*(b+(-4*a*c+b^2)^(1/2))/c) 
)^(1/2)*((x+1/2*b/c)/(-1/2*(b+(-4*a*c+b^2)^(1/2))/c+1/2*b/c))^(1/2)*((x-1/ 
2/c*(-b+(-4*a*c+b^2)^(1/2)))/(-1/2*(b+(-4*a*c+b^2)^(1/2))/c-1/2/c*(-b+(-4* 
a*c+b^2)^(1/2))))^(1/2)/(2*c^2*d*x^3+3*b*c*d*x^2+2*a*c*d*x+b^2*d*x+a*b*d)^ 
(1/2)*EllipticF(((x+1/2*(b+(-4*a*c+b^2)^(1/2))/c)/(1/2/c*(-b+(-4*a*c+b^2)^ 
(1/2))+1/2*(b+(-4*a*c+b^2)^(1/2))/c))^(1/2),((-1/2*(b+(-4*a*c+b^2)^(1/2))/ 
c-1/2/c*(-b+(-4*a*c+b^2)^(1/2)))/(-1/2*(b+(-4*a*c+b^2)^(1/2))/c+1/2*b/c))^ 
(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.52 \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (30 \, \sqrt {2} {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{5} x^{4} + 2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d^{5} x^{3} + {\left (b^{4} - 2 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} d^{5} x^{2} + 2 \, {\left (a b^{3} - 4 \, a^{2} b c\right )} d^{5} x + {\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{5}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) + {\left (32 \, c^{4} d^{5} x^{4} + 64 \, b c^{3} d^{5} x^{3} + 6 \, {\left (b^{2} c^{2} + 28 \, a c^{3}\right )} d^{5} x^{2} - 2 \, {\left (13 \, b^{3} c - 84 \, a b c^{2}\right )} d^{5} x - {\left (b^{4} + 18 \, a b^{2} c - 120 \, a^{2} c^{2}\right )} d^{5}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}\right )}}{3 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \] Input:

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")
 

Output:

2/3*(30*sqrt(2)*((b^2*c^2 - 4*a*c^3)*d^5*x^4 + 2*(b^3*c - 4*a*b*c^2)*d^5*x 
^3 + (b^4 - 2*a*b^2*c - 8*a^2*c^2)*d^5*x^2 + 2*(a*b^3 - 4*a^2*b*c)*d^5*x + 
 (a^2*b^2 - 4*a^3*c)*d^5)*sqrt(c^2*d)*weierstrassPInverse((b^2 - 4*a*c)/c^ 
2, 0, 1/2*(2*c*x + b)/c) + (32*c^4*d^5*x^4 + 64*b*c^3*d^5*x^3 + 6*(b^2*c^2 
 + 28*a*c^3)*d^5*x^2 - 2*(13*b^3*c - 84*a*b*c^2)*d^5*x - (b^4 + 18*a*b^2*c 
 - 120*a^2*c^2)*d^5)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/(c^2*x^4 + 
 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((2*c*d*x+b*d)**(11/2)/(c*x**2+b*x+a)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {11}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")
 

Output:

integrate((2*c*d*x + b*d)^(11/2)/(c*x^2 + b*x + a)^(5/2), x)
 

Giac [F]

\[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {11}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")
 

Output:

integrate((2*c*d*x + b*d)^(11/2)/(c*x^2 + b*x + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^{11/2}}{{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \] Input:

int((b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^(5/2),x)
 

Output:

int((b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {too large to display} \] Input:

int((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^(5/2),x)
 

Output:

(2*sqrt(d)*d**5*( - 576*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**2*b**2*c 
**2 + 576*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**2*b*c**3*x + 576*sqrt( 
b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**2*c**4*x**2 + 112*sqrt(b + 2*c*x)*sqr 
t(a + b*x + c*x**2)*a*b**4*c - 976*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)* 
a*b**3*c**2*x - 912*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**2*c**3*x** 
2 + 128*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b*c**4*x**3 + 64*sqrt(b + 
 2*c*x)*sqrt(a + b*x + c*x**2)*a*c**5*x**4 + 3*sqrt(b + 2*c*x)*sqrt(a + b* 
x + c*x**2)*b**6 + 168*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**5*c*x + 7 
2*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**4*c**2*x**2 - 192*sqrt(b + 2*c 
*x)*sqrt(a + b*x + c*x**2)*b**3*c**3*x**3 - 96*sqrt(b + 2*c*x)*sqrt(a + b* 
x + c*x**2)*b**2*c**4*x**4 - 5760*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x* 
*2)*x**2)/(2*a**4*b*c + 4*a**4*c**2*x - 3*a**3*b**3 + 18*a**3*b*c**2*x**2 
+ 12*a**3*c**3*x**3 - 9*a**2*b**4*x - 21*a**2*b**3*c*x**2 + 6*a**2*b**2*c* 
*2*x**3 + 30*a**2*b*c**3*x**4 + 12*a**2*c**4*x**5 - 9*a*b**5*x**2 - 34*a*b 
**4*c*x**3 - 35*a*b**3*c**2*x**4 + 14*a*b*c**4*x**6 + 4*a*c**5*x**7 - 3*b* 
*6*x**3 - 15*b**5*c*x**4 - 27*b**4*c**2*x**5 - 21*b**3*c**3*x**6 - 6*b**2* 
c**4*x**7),x)*a**6*c**6 + 12960*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2 
)*x**2)/(2*a**4*b*c + 4*a**4*c**2*x - 3*a**3*b**3 + 18*a**3*b*c**2*x**2 + 
12*a**3*c**3*x**3 - 9*a**2*b**4*x - 21*a**2*b**3*c*x**2 + 6*a**2*b**2*c**2 
*x**3 + 30*a**2*b*c**3*x**4 + 12*a**2*c**4*x**5 - 9*a*b**5*x**2 - 34*a*...