\(\int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx\) [228]

Optimal result

Integrand size = 28, antiderivative size = 231 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx=-\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}}+\frac {2 \sqrt {a+b x+c x^2}}{77 c \left (b^2-4 a c\right ) d^3 (b d+2 c d x)^{7/2}}+\frac {10 \sqrt {a+b x+c x^2}}{231 c \left (b^2-4 a c\right )^2 d^5 (b d+2 c d x)^{3/2}}+\frac {5 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{231 c^2 \left (b^2-4 a c\right )^{7/4} d^{13/2} \sqrt {a+b x+c x^2}} \] Output:

-1/11*(c*x^2+b*x+a)^(1/2)/c/d/(2*c*d*x+b*d)^(11/2)+2/77*(c*x^2+b*x+a)^(1/2 
)/c/(-4*a*c+b^2)/d^3/(2*c*d*x+b*d)^(7/2)+10/231*(c*x^2+b*x+a)^(1/2)/c/(-4* 
a*c+b^2)^2/d^5/(2*c*d*x+b*d)^(3/2)+5/231*(-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^2/(-4*a 
*c+b^2)^(7/4)/d^(13/2)/(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 = 10.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.43 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx=-\frac {\sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {1}{2},-\frac {7}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{22 c d^7 (b+2 c x)^6 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

Integrate[Sqrt[a + b*x + c*x^2]/(b*d + 2*c*d*x)^(13/2),x]
 

Output:

-1/22*(Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-11/4, 
-1/2, -7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c*d^7*(b + 2*c*x)^6*Sqrt[(c*(a 
+ x*(b + c*x)))/(-b^2 + 4*a*c)])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1108, 1117, 1117, 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.

\(\displaystyle \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx\)

\(\Big \downarrow \) 1108

\(\displaystyle \frac {\int \frac {1}{(b d+2 c x d)^{9/2} \sqrt {c x^2+b x+a}}dx}{22 c d^2}-\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}}\)

\(\Big \downarrow \) 1117

\(\displaystyle \frac {\frac {5 \int \frac {1}{(b d+2 c x d)^{5/2} \sqrt {c x^2+b x+a}}dx}{7 d^2 \left (b^2-4 a c\right )}+\frac {4 \sqrt {a+b x+c x^2}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}}{22 c d^2}-\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}}\)

\(\Big \downarrow \) 1117

\(\displaystyle \frac {\frac {5 \left (\frac {\int \frac {1}{\sqrt {b d+2 c x d} \sqrt {c x^2+b x+a}}dx}{3 d^2 \left (b^2-4 a c\right )}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{7 d^2 \left (b^2-4 a c\right )}+\frac {4 \sqrt {a+b x+c x^2}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}}{22 c d^2}-\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}}\)

\(\Big \downarrow \) 1115

\(\displaystyle \frac {\frac {5 \left (\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{3 d^2 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{7 d^2 \left (b^2-4 a c\right )}+\frac {4 \sqrt {a+b x+c x^2}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}}{22 c d^2}-\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}}\)

\(\Big \downarrow \) 1113

\(\displaystyle \frac {\frac {5 \left (\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{3 c d^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{7 d^2 \left (b^2-4 a c\right )}+\frac {4 \sqrt {a+b x+c x^2}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}}{22 c d^2}-\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {\frac {5 \left (\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c d^{5/2} \left (b^2-4 a c\right )^{3/4} \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}\right )}{7 d^2 \left (b^2-4 a c\right )}+\frac {4 \sqrt {a+b x+c x^2}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}}{22 c d^2}-\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}}\)

Input:

Int[Sqrt[a + b*x + c*x^2]/(b*d + 2*c*d*x)^(13/2),x]
 

Output:

-1/11*Sqrt[a + b*x + c*x^2]/(c*d*(b*d + 2*c*d*x)^(11/2)) + ((4*Sqrt[a + b* 
x + c*x^2])/(7*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(7/2)) + (5*((4*Sqrt[a + b* 
x + c*x^2])/(3*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)) + (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*(b^2 - 4*a*c)^(3/4)*d^(5/2)*Sqrt[a + b* 
x + c*x^2])))/(7*(b^2 - 4*a*c)*d^2))/(22*c*d^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 1108

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 1))), x] - Si 
mp[b*(p/(d*e*(m + 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] && NeQ[m + 2*p + 
3, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !(IntegerQ[m/2] && LtQ[m + 2*p + 3, 0] 
) && 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 1117

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(612\) vs. \(2(197)=394\).

Time = 2.86 (sec) , antiderivative size = 613, normalized size of antiderivative = 2.65

method	result	size
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{704 d^{7} c^{7} \left (x +\frac {b}{2 c}\right )^{6}}-\frac {\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{616 c^{5} \left (4 a c -b^{2}\right ) d^{7} \left (x +\frac {b}{2 c}\right )^{4}}+\frac {5 \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}{462 c^{3} \left (4 a c -b^{2}\right )^{2} d^{7} \left (x +\frac {b}{2 c}\right )^{2}}+\frac {5 \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{231 c \left (4 a c -b^{2}\right )^{2} d^{6} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(613\)
default	\(\text {Expression too large to display}\)	\(1016\)

Input:

int((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^(13/2),x,method=_RETURNVERBOSE)
 

Output:

(d*(2*c*x+b)*(c*x^2+b*x+a))^(1/2)/(d*(2*c*x+b))^(1/2)/(c*x^2+b*x+a)^(1/2)* 
(-1/704/d^7/c^7*(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)/(x 
+1/2*b/c)^6-1/616/c^5/(4*a*c-b^2)/d^7*(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)/(x+1/2*b/c)^4+5/462/c^3/(4*a*c-b^2)^2/d^7*(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)/(x+1/2*b/c)^2+5/231/c/(4*a*c-b 
^2)^2/d^6*(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 452 vs. \(2 (196) = 392\).

Time = 0.11 (sec) , antiderivative size = 452, normalized size of antiderivative = 1.96 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx=\frac {5 \, \sqrt {2} {\left (64 \, c^{6} x^{6} + 192 \, b c^{5} x^{5} + 240 \, b^{2} c^{4} x^{4} + 160 \, b^{3} c^{3} x^{3} + 60 \, b^{4} c^{2} x^{2} + 12 \, b^{5} c x + b^{6}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) + 2 \, {\left (160 \, c^{6} x^{4} + 320 \, b c^{5} x^{3} - 5 \, b^{4} c^{2} + 144 \, a b^{2} c^{3} - 336 \, a^{2} c^{4} + 24 \, {\left (11 \, b^{2} c^{4} - 4 \, a c^{5}\right )} x^{2} + 8 \, {\left (13 \, b^{3} c^{3} - 12 \, a b c^{4}\right )} x\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{462 \, {\left (64 \, {\left (b^{4} c^{9} - 8 \, a b^{2} c^{10} + 16 \, a^{2} c^{11}\right )} d^{7} x^{6} + 192 \, {\left (b^{5} c^{8} - 8 \, a b^{3} c^{9} + 16 \, a^{2} b c^{10}\right )} d^{7} x^{5} + 240 \, {\left (b^{6} c^{7} - 8 \, a b^{4} c^{8} + 16 \, a^{2} b^{2} c^{9}\right )} d^{7} x^{4} + 160 \, {\left (b^{7} c^{6} - 8 \, a b^{5} c^{7} + 16 \, a^{2} b^{3} c^{8}\right )} d^{7} x^{3} + 60 \, {\left (b^{8} c^{5} - 8 \, a b^{6} c^{6} + 16 \, a^{2} b^{4} c^{7}\right )} d^{7} x^{2} + 12 \, {\left (b^{9} c^{4} - 8 \, a b^{7} c^{5} + 16 \, a^{2} b^{5} c^{6}\right )} d^{7} x + {\left (b^{10} c^{3} - 8 \, a b^{8} c^{4} + 16 \, a^{2} b^{6} c^{5}\right )} d^{7}\right )}} \] Input:

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^(13/2),x, algorithm="fricas")
 

Output:

1/462*(5*sqrt(2)*(64*c^6*x^6 + 192*b*c^5*x^5 + 240*b^2*c^4*x^4 + 160*b^3*c 
^3*x^3 + 60*b^4*c^2*x^2 + 12*b^5*c*x + b^6)*sqrt(c^2*d)*weierstrassPInvers 
e((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c) + 2*(160*c^6*x^4 + 320*b*c^5*x^ 
3 - 5*b^4*c^2 + 144*a*b^2*c^3 - 336*a^2*c^4 + 24*(11*b^2*c^4 - 4*a*c^5)*x^ 
2 + 8*(13*b^3*c^3 - 12*a*b*c^4)*x)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + 
a))/(64*(b^4*c^9 - 8*a*b^2*c^10 + 16*a^2*c^11)*d^7*x^6 + 192*(b^5*c^8 - 8* 
a*b^3*c^9 + 16*a^2*b*c^10)*d^7*x^5 + 240*(b^6*c^7 - 8*a*b^4*c^8 + 16*a^2*b 
^2*c^9)*d^7*x^4 + 160*(b^7*c^6 - 8*a*b^5*c^7 + 16*a^2*b^3*c^8)*d^7*x^3 + 6 
0*(b^8*c^5 - 8*a*b^6*c^6 + 16*a^2*b^4*c^7)*d^7*x^2 + 12*(b^9*c^4 - 8*a*b^7 
*c^5 + 16*a^2*b^5*c^6)*d^7*x + (b^10*c^3 - 8*a*b^8*c^4 + 16*a^2*b^6*c^5)*d 
^7)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx=\text {Timed out} \] Input:

integrate((c*x**2+b*x+a)**(1/2)/(2*c*d*x+b*d)**(13/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx=\int { \frac {\sqrt {c x^{2} + b x + a}}{{\left (2 \, c d x + b d\right )}^{\frac {13}{2}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^(13/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(c*x^2 + b*x + a)/(2*c*d*x + b*d)^(13/2), x)
 

Giac [F]

\[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx=\int { \frac {\sqrt {c x^{2} + b x + a}}{{\left (2 \, c d x + b d\right )}^{\frac {13}{2}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^(13/2),x, algorithm="giac")
 

Output:

integrate(sqrt(c*x^2 + b*x + a)/(2*c*d*x + b*d)^(13/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx=\int \frac {\sqrt {c\,x^2+b\,x+a}}{{\left (b\,d+2\,c\,d\,x\right )}^{13/2}} \,d x \] Input:

int((a + b*x + c*x^2)^(1/2)/(b*d + 2*c*d*x)^(13/2),x)
 

Output:

int((a + b*x + c*x^2)^(1/2)/(b*d + 2*c*d*x)^(13/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx=\text {too large to display} \] Input:

int((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^(13/2),x)
 

Output:

(sqrt(d)*( - 2*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a + 88*int((sqrt(b + 
 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(22*a**2*b**7*c + 308*a**2*b**6*c**2* 
x + 1848*a**2*b**5*c**3*x**2 + 6160*a**2*b**4*c**4*x**3 + 12320*a**2*b**3* 
c**5*x**4 + 14784*a**2*b**2*c**6*x**5 + 9856*a**2*b*c**7*x**6 + 2816*a**2* 
c**8*x**7 - a*b**9 + 8*a*b**8*c*x + 246*a*b**7*c**2*x**2 + 1876*a*b**6*c** 
3*x**3 + 7448*a*b**5*c**4*x**4 + 17808*a*b**4*c**5*x**5 + 26656*a*b**3*c** 
6*x**6 + 24512*a*b**2*c**7*x**7 + 12672*a*b*c**8*x**8 + 2816*a*c**9*x**9 - 
 b**10*x - 15*b**9*c*x**2 - 98*b**8*c**2*x**3 - 364*b**7*c**3*x**4 - 840*b 
**6*c**4*x**5 - 1232*b**5*c**5*x**6 - 1120*b**4*c**6*x**7 - 576*b**3*c**7* 
x**8 - 128*b**2*c**8*x**9),x)*a**2*b**6*c**3 + 1056*int((sqrt(b + 2*c*x)*s 
qrt(a + b*x + c*x**2)*x**2)/(22*a**2*b**7*c + 308*a**2*b**6*c**2*x + 1848* 
a**2*b**5*c**3*x**2 + 6160*a**2*b**4*c**4*x**3 + 12320*a**2*b**3*c**5*x**4 
 + 14784*a**2*b**2*c**6*x**5 + 9856*a**2*b*c**7*x**6 + 2816*a**2*c**8*x**7 
 - a*b**9 + 8*a*b**8*c*x + 246*a*b**7*c**2*x**2 + 1876*a*b**6*c**3*x**3 + 
7448*a*b**5*c**4*x**4 + 17808*a*b**4*c**5*x**5 + 26656*a*b**3*c**6*x**6 + 
24512*a*b**2*c**7*x**7 + 12672*a*b*c**8*x**8 + 2816*a*c**9*x**9 - b**10*x 
- 15*b**9*c*x**2 - 98*b**8*c**2*x**3 - 364*b**7*c**3*x**4 - 840*b**6*c**4* 
x**5 - 1232*b**5*c**5*x**6 - 1120*b**4*c**6*x**7 - 576*b**3*c**7*x**8 - 12 
8*b**2*c**8*x**9),x)*a**2*b**5*c**4*x + 5280*int((sqrt(b + 2*c*x)*sqrt(a + 
 b*x + c*x**2)*x**2)/(22*a**2*b**7*c + 308*a**2*b**6*c**2*x + 1848*a**2...