\(\int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx\) [1332]

Optimal result

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

[Out]

-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*EllipticF((2*c*d*x+b*d)^(1/2)/(
-4*a*c+b^2)^(1/4)/d^(1/2),I)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)/c^2/(-4*a*c+b^2)^(7/4)/d^(13/2)/(c*x^2+b*x+
a)^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {698, 707, 705, 703, 227} \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx=\frac {5 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{231 c^2 d^{13/2} \left (b^2-4 a c\right )^{7/4} \sqrt {a+b x+c x^2}}+\frac {10 \sqrt {a+b x+c x^2}}{231 c d^5 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}+\frac {2 \sqrt {a+b x+c x^2}}{77 c d^3 \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}-\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}} \]

[In]

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

[Out]

-1/11*Sqrt[a + b*x + c*x^2]/(c*d*(b*d + 2*c*d*x)^(11/2)) + (2*Sqrt[a + b*x + c*x^2])/(77*c*(b^2 - 4*a*c)*d^3*(
b*d + 2*c*d*x)^(7/2)) + (10*Sqrt[a + b*x + c*x^2])/(231*c*(b^2 - 4*a*c)^2*d^5*(b*d + 2*c*d*x)^(3/2)) + (5*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])/(231*c^2*(b^2 - 4*a*c)^(7/4)*d^(13/2)*Sqrt[a + b*x + c*x^2])

Rule 227

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

Rule 698

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 1))), x] - Dist[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] && NeQ[b^2 - 4*a*c, 0] && 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 703

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[(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] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]

Rule 705

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[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] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && EqQ[m^2, 1/4]

Rule 707

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> 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] + Dist[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] && NeQ[b^2 - 4*a*
c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {a+b x+c x^2}}{11 c d (b d+2 c d x)^{11/2}}+\frac {\int \frac {1}{(b d+2 c d x)^{9/2} \sqrt {a+b x+c x^2}} \, dx}{22 c d^2} \\ & = -\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 {5 \int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx}{154 c \left (b^2-4 a c\right ) d^4} \\ & = -\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 \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx}{462 c \left (b^2-4 a c\right )^2 d^6} \\ & = -\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 {\left (5 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {-\frac {a c}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {c^2 x^2}{b^2-4 a c}}} \, dx}{462 c \left (b^2-4 a c\right )^2 d^6 \sqrt {a+b x+c x^2}} \\ & = -\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 {\left (5 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{231 c^2 \left (b^2-4 a c\right )^2 d^7 \sqrt {a+b x+c x^2}} \\ & = -\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}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\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}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 10.05 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.43 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx=-\frac {\sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {1}{2},-\frac {7}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{22 c d^7 (b+2 c x)^6 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \]

[In]

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

[Out]

-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)])

Maple [B] (verified)

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

Time = 2.81 (sec) , antiderivative size = 613, normalized size of antiderivative = 2.65

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

[In]

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

[Out]

(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/c*b)^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/c*b)^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/c*b)^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
/c*b)/(-1/2*(b+(-4*a*c+b^2)^(1/2))/c+1/2/c*b))^(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)*Ellipti
cF(((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/c*b))^(1/2)))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.23 (sec) , antiderivative size = 452, normalized size of antiderivative = 1.96 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx=\frac {5 \, \sqrt {2} {\left (64 \, c^{6} x^{6} + 192 \, b c^{5} x^{5} + 240 \, b^{2} c^{4} x^{4} + 160 \, b^{3} c^{3} x^{3} + 60 \, b^{4} c^{2} x^{2} + 12 \, b^{5} c x + b^{6}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) + 2 \, {\left (160 \, c^{6} x^{4} + 320 \, b c^{5} x^{3} - 5 \, b^{4} c^{2} + 144 \, a b^{2} c^{3} - 336 \, a^{2} c^{4} + 24 \, {\left (11 \, b^{2} c^{4} - 4 \, a c^{5}\right )} x^{2} + 8 \, {\left (13 \, b^{3} c^{3} - 12 \, a b c^{4}\right )} x\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{462 \, {\left (64 \, {\left (b^{4} c^{9} - 8 \, a b^{2} c^{10} + 16 \, a^{2} c^{11}\right )} d^{7} x^{6} + 192 \, {\left (b^{5} c^{8} - 8 \, a b^{3} c^{9} + 16 \, a^{2} b c^{10}\right )} d^{7} x^{5} + 240 \, {\left (b^{6} c^{7} - 8 \, a b^{4} c^{8} + 16 \, a^{2} b^{2} c^{9}\right )} d^{7} x^{4} + 160 \, {\left (b^{7} c^{6} - 8 \, a b^{5} c^{7} + 16 \, a^{2} b^{3} c^{8}\right )} d^{7} x^{3} + 60 \, {\left (b^{8} c^{5} - 8 \, a b^{6} c^{6} + 16 \, a^{2} b^{4} c^{7}\right )} d^{7} x^{2} + 12 \, {\left (b^{9} c^{4} - 8 \, a b^{7} c^{5} + 16 \, a^{2} b^{5} c^{6}\right )} d^{7} x + {\left (b^{10} c^{3} - 8 \, a b^{8} c^{4} + 16 \, a^{2} b^{6} c^{5}\right )} d^{7}\right )}} \]

[In]

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

[Out]

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)*weierstrassPInverse((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 + 60*(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. \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{13/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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