\(\int \frac {(b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}} \, dx\) [259]

Optimal result

Integrand size = 28, antiderivative size = 174 \[ \int \frac {(b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}} \, dx=\frac {20}{21} \left (b^2-4 a c\right ) d^3 \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {4}{7} d (b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}+\frac {10 \left (b^2-4 a c\right )^{9/4} d^{7/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{21 c \sqrt {a+b x+c x^2}} \] Output:

20/21*(-4*a*c+b^2)*d^3*(2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^(1/2)+4/7*d*(2*c* 
d*x+b*d)^(5/2)*(c*x^2+b*x+a)^(1/2)+10/21*(-4*a*c+b^2)^(9/4)*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/(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 = 10.21 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.95 \[ \int \frac {(b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}} \, dx=\frac {2 d^3 \sqrt {d (b+2 c x)} \left (8 c \left (-5 a^2 c+2 a \left (b^2-b c x-c^2 x^2\right )+x \left (2 b^3+5 b^2 c x+6 b c^2 x^2+3 c^3 x^3\right )\right )+5 \left (b^2-4 a c\right )^2 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{21 c \sqrt {a+x (b+c x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1116, 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.

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

\(\Big \downarrow \) 1116

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

\(\Big \downarrow \) 1116

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

\(\Big \downarrow \) 1115

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

\(\Big \downarrow \) 1113

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

\(\Big \downarrow \) 762

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

Input:

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

Output:

(4*d*(b*d + 2*c*d*x)^(5/2)*Sqrt[a + b*x + c*x^2])/7 + (5*(b^2 - 4*a*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[S 
qrt[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])))/7
 

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 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. \(487\) vs. \(2(146)=292\).

Time = 2.58 (sec) , antiderivative size = 488, normalized size of antiderivative = 2.80

method	result	size
risch	\(-\frac {16 \left (-3 c^{2} x^{2}-3 c b x +5 a c -2 b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, \left (2 c x +b \right ) d^{4}}{21 \sqrt {d \left (2 c x +b \right )}}+\frac {2 \left (\frac {5}{21} b^{4}-\frac {40}{21} c a \,b^{2}+\frac {80}{21} a^{2} c^{2}\right ) \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right ) d^{4} \sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}}{\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}\, \sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(488\)
default	\(\frac {\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, d^{3} \left (96 c^{5} x^{5}+80 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) a^{2} c^{2}-40 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) a \,b^{2} c +5 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) b^{4}+240 b \,x^{4} c^{4}-64 a \,c^{4} x^{3}+256 b^{2} c^{3} x^{3}-96 a b \,c^{3} x^{2}+144 c^{2} x^{2} b^{3}-160 a^{2} c^{3} x +32 a \,b^{2} c^{2} x +32 b^{4} c x -80 a^{2} b \,c^{2}+32 a \,b^{3} c \right )}{21 c \left (2 x^{3} c^{2}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right )}\)	\(567\)
elliptic	\(\text {Expression too large to display}\)	\(1257\)

Input:

int((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-16/21*(-3*c^2*x^2-3*b*c*x+5*a*c-2*b^2)*(c*x^2+b*x+a)^(1/2)*(2*c*x+b)*d^4/ 
(d*(2*c*x+b))^(1/2)+2*(5/21*b^4-40/21*c*a*b^2+80/21*a^2*c^2)*(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)) 
*d^4*(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)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.75 \[ \int \frac {(b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}} \, dx=\frac {5 \, \sqrt {2} {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c^{2} d} d^{3} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) + 16 \, {\left (3 \, c^{4} d^{3} x^{2} + 3 \, b c^{3} d^{3} x + {\left (2 \, b^{2} c^{2} - 5 \, a c^{3}\right )} d^{3}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{21 \, c^{2}} \] Input:

integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

1/21*(5*sqrt(2)*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(c^2*d)*d^3*weierstrass 
PInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c) + 16*(3*c^4*d^3*x^2 + 3* 
b*c^3*d^3*x + (2*b^2*c^2 - 5*a*c^3)*d^3)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + 
b*x + a))/c^2
 

Sympy [F]

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

integrate((2*c*d*x+b*d)**(7/2)/(c*x**2+b*x+a)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(1/2),x)
 

Output:

int((b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^(1/2), x)
 

Reduce [F]

\[ \int \frac {(b d+2 c d x)^{7/2}}{\sqrt {a+b x+c x^2}} \, dx=\frac {2 \sqrt {d}\, d^{3} \left (-16 \sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, a \,b^{2} c +16 \sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, a b \,c^{2} x +16 \sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, a \,c^{3} x^{2}+7 \sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, b^{4}+8 \sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, b^{3} c x +8 \sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{2} x^{2}-160 \left (\int \frac {\sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, x^{2}}{4 a \,c^{3} x^{3}+2 b^{2} c^{2} x^{3}+6 a b \,c^{2} x^{2}+3 b^{3} c \,x^{2}+4 a^{2} c^{2} x +4 a \,b^{2} c x +b^{4} x +2 a^{2} b c +a \,b^{3}}d x \right ) a^{3} c^{5}+30 \left (\int \frac {\sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, x^{2}}{4 a \,c^{3} x^{3}+2 b^{2} c^{2} x^{3}+6 a b \,c^{2} x^{2}+3 b^{3} c \,x^{2}+4 a^{2} c^{2} x +4 a \,b^{2} c x +b^{4} x +2 a^{2} b c +a \,b^{3}}d x \right ) a \,b^{4} c^{3}-5 \left (\int \frac {\sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, x^{2}}{4 a \,c^{3} x^{3}+2 b^{2} c^{2} x^{3}+6 a b \,c^{2} x^{2}+3 b^{3} c \,x^{2}+4 a^{2} c^{2} x +4 a \,b^{2} c x +b^{4} x +2 a^{2} b c +a \,b^{3}}d x \right ) b^{6} c^{2}-160 \left (\int \frac {\sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, x}{4 a \,c^{3} x^{3}+2 b^{2} c^{2} x^{3}+6 a b \,c^{2} x^{2}+3 b^{3} c \,x^{2}+4 a^{2} c^{2} x +4 a \,b^{2} c x +b^{4} x +2 a^{2} b c +a \,b^{3}}d x \right ) a^{3} b \,c^{4}+30 \left (\int \frac {\sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, x}{4 a \,c^{3} x^{3}+2 b^{2} c^{2} x^{3}+6 a b \,c^{2} x^{2}+3 b^{3} c \,x^{2}+4 a^{2} c^{2} x +4 a \,b^{2} c x +b^{4} x +2 a^{2} b c +a \,b^{3}}d x \right ) a \,b^{5} c^{2}-5 \left (\int \frac {\sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, x}{4 a \,c^{3} x^{3}+2 b^{2} c^{2} x^{3}+6 a b \,c^{2} x^{2}+3 b^{3} c \,x^{2}+4 a^{2} c^{2} x +4 a \,b^{2} c x +b^{4} x +2 a^{2} b c +a \,b^{3}}d x \right ) b^{7} c \right )}{14 a c +7 b^{2}} \] Input:

int((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^(1/2),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 + 7*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**2*b*c + 4*a**2*c**2*x + a*b**3 + 4*a*b**2* 
c*x + 6*a*b*c**2*x**2 + 4*a*c**3*x**3 + b**4*x + 3*b**3*c*x**2 + 2*b**2*c* 
*2*x**3),x)*a**3*c**5 + 30*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x** 
2)/(2*a**2*b*c + 4*a**2*c**2*x + a*b**3 + 4*a*b**2*c*x + 6*a*b*c**2*x**2 + 
 4*a*c**3*x**3 + b**4*x + 3*b**3*c*x**2 + 2*b**2*c**2*x**3),x)*a*b**4*c**3 
 - 5*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(2*a**2*b*c + 4*a** 
2*c**2*x + a*b**3 + 4*a*b**2*c*x + 6*a*b*c**2*x**2 + 4*a*c**3*x**3 + b**4* 
x + 3*b**3*c*x**2 + 2*b**2*c**2*x**3),x)*b**6*c**2 - 160*int((sqrt(b + 2*c 
*x)*sqrt(a + b*x + c*x**2)*x)/(2*a**2*b*c + 4*a**2*c**2*x + a*b**3 + 4*a*b 
**2*c*x + 6*a*b*c**2*x**2 + 4*a*c**3*x**3 + b**4*x + 3*b**3*c*x**2 + 2*b** 
2*c**2*x**3),x)*a**3*b*c**4 + 30*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x** 
2)*x)/(2*a**2*b*c + 4*a**2*c**2*x + a*b**3 + 4*a*b**2*c*x + 6*a*b*c**2*x** 
2 + 4*a*c**3*x**3 + b**4*x + 3*b**3*c*x**2 + 2*b**2*c**2*x**3),x)*a*b**5*c 
**2 - 5*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x)/(2*a**2*b*c + 4*a** 
2*c**2*x + a*b**3 + 4*a*b**2*c*x + 6*a*b*c**2*x**2 + 4*a*c**3*x**3 + b*...