\(\int \frac {(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{7/2}} \, dx\) [256]

Optimal result

Integrand size = 28, antiderivative size = 310 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{7/2}} \, dx=\frac {3 (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{20 c^3 d^5}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt {b d+2 c d x}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}-\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt {a+b x+c x^2}}+\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{20 c^4 d^{7/2} \sqrt {a+b x+c x^2}} \] Output:

3/20*(2*c*d*x+b*d)^(3/2)*(c*x^2+b*x+a)^(1/2)/c^3/d^5-1/2*(c*x^2+b*x+a)^(3/ 
2)/c^2/d^3/(2*c*d*x+b*d)^(1/2)-1/5*(c*x^2+b*x+a)^(5/2)/c/d/(2*c*d*x+b*d)^( 
5/2)-3/20*(-4*a*c+b^2)^(7/4)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)*Ellipti 
cE((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)/c^4/d^(7/2)/(c*x^2+b* 
x+a)^(1/2)+3/20*(-4*a*c+b^2)^(7/4)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)*E 
llipticF((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)/c^4/d^(7/2)/(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 = 10.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{7/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2 \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {5}{4},-\frac {1}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{160 c^3 d (d (b+2 c x))^{5/2} \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(7/2),x]
 

Output:

-1/160*((b^2 - 4*a*c)^2*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-5/2, -5/4 
, -1/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c^3*d*(d*(b + 2*c*x))^(5/2)*Sqrt[(c 
*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 297, normalized size of antiderivative = 0.96, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1108, 1108, 1109, 1115, 1114, 836, 27, 762, 1389, 327}

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.

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

\(\Big \downarrow \) 1108

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

\(\Big \downarrow \) 1108

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

\(\Big \downarrow \) 1109

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

\(\Big \downarrow \) 1115

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

\(\Big \downarrow \) 1114

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

\(\Big \downarrow \) 836

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 762

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

\(\Big \downarrow \) 1389

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

\(\Big \downarrow \) 327

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

Input:

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(7/2),x]
 

Output:

-1/5*(a + b*x + c*x^2)^(5/2)/(c*d*(b*d + 2*c*d*x)^(5/2)) + (-((a + b*x + c 
*x^2)^(3/2)/(c*d*Sqrt[b*d + 2*c*d*x])) + (3*(((b*d + 2*c*d*x)^(3/2)*Sqrt[a 
 + b*x + c*x^2])/(5*c*d) - ((b^2 - 4*a*c)*Sqrt[-((c*(a + b*x + c*x^2))/(b^ 
2 - 4*a*c))]*((b^2 - 4*a*c)^(3/4)*d^(3/2)*EllipticE[ArcSin[Sqrt[b*d + 2*c* 
d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1] - (b^2 - 4*a*c)^(3/4)*d^(3/2)*Ell 
ipticF[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1]))/(5 
*c^2*d*Sqrt[a + b*x + c*x^2])))/(2*c*d^2))/(2*c*d^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

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 836

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, 
Simp[-q^(-1)   Int[1/Sqrt[a + b*x^4], x], x] + Simp[1/q   Int[(1 + q*x^2)/S 
qrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]
 

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 1109

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 + 2*p + 1))), x 
] - Simp[d*p*((b^2 - 4*a*c)/(b*e*(m + 2*p + 1)))   Int[(d + e*x)^m*(a + b*x 
 + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[2*c*d - b* 
e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1 
)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p])) && RationalQ[m] && IntegerQ[2*p]
 

rule 1114

Int[Sqrt[(d_) + (e_.)*(x_)]/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symb 
ol] :> Simp[(4/e)*Sqrt[-c/(b^2 - 4*a*c)]   Subst[Int[x^2/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 1389

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Simp[d/Sq 
rt[a]   Int[Sqrt[1 + e*(x^2/d)]/Sqrt[1 - e*(x^2/d)], x], x] /; FreeQ[{a, c, 
 d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && NegQ[c/a] && GtQ[a, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1294\) vs. \(2(260)=520\).

Time = 9.44 (sec) , antiderivative size = 1295, normalized size of antiderivative = 4.18

method	result	size
elliptic	\(\text {Expression too large to display}\)	\(1295\)
default	\(\text {Expression too large to display}\)	\(1362\)
risch	\(\text {Expression too large to display}\)	\(3978\)

Input:

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(7/2),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)* 
(-1/640*(16*a^2*c^2-8*a*b^2*c+b^4)/d^4/c^6*(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)^3-3/40*(2*c^2*d*x^2+2*b*c*d*x+2*a*c*d 
)*(4*a*c-b^2)/c^4/d^4/((x+1/2*b/c)*(2*c^2*d*x^2+2*b*c*d*x+2*a*c*d))^(1/2)+ 
1/40/d^4/c^2*x*(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)+1/8 
0/c^3/d^4*b*(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*(1/3 
2*b*(6*a*c-b^2)/d^3/c^3+3/80*b/c^3*(4*a*c-b^2)/d^3-1/40/d^3/c^2*a*b-1/80/c 
^3/d^4*b*(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))+2*(3/8*a/d^3/c+3/40*(4*a*c-b^2)/ 
c^2/d^3-1/40/d^4/c^2*(3*a*c*d+3/2*b^2*d)-3/80/c^2/d^3*b^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*(...
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{7/2}} \, dx=\frac {3 \, \sqrt {2} {\left (b^{5} - 4 \, a b^{3} c + 8 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} + 12 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} + 6 \, {\left (b^{4} c - 4 \, a b^{2} c^{2}\right )} x\right )} \sqrt {c^{2} d} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) + {\left (4 \, c^{5} x^{4} + 8 \, b c^{4} x^{3} + 3 \, b^{4} c - 10 \, a b^{2} c^{2} - 4 \, a^{2} c^{3} + 6 \, {\left (3 \, b^{2} c^{3} - 8 \, a c^{4}\right )} x^{2} + 2 \, {\left (7 \, b^{3} c^{2} - 24 \, a b c^{3}\right )} x\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{20 \, {\left (8 \, c^{7} d^{4} x^{3} + 12 \, b c^{6} d^{4} x^{2} + 6 \, b^{2} c^{5} d^{4} x + b^{3} c^{4} d^{4}\right )}} \] Input:

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(7/2),x, algorithm="fricas")
 

Output:

1/20*(3*sqrt(2)*(b^5 - 4*a*b^3*c + 8*(b^2*c^3 - 4*a*c^4)*x^3 + 12*(b^3*c^2 
 - 4*a*b*c^3)*x^2 + 6*(b^4*c - 4*a*b^2*c^2)*x)*sqrt(c^2*d)*weierstrassZeta 
((b^2 - 4*a*c)/c^2, 0, weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c* 
x + b)/c)) + (4*c^5*x^4 + 8*b*c^4*x^3 + 3*b^4*c - 10*a*b^2*c^2 - 4*a^2*c^3 
 + 6*(3*b^2*c^3 - 8*a*c^4)*x^2 + 2*(7*b^3*c^2 - 24*a*b*c^3)*x)*sqrt(2*c*d* 
x + b*d)*sqrt(c*x^2 + b*x + a))/(8*c^7*d^4*x^3 + 12*b*c^6*d^4*x^2 + 6*b^2* 
c^5*d^4*x + b^3*c^4*d^4)
 

Sympy [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{7/2}} \, dx=\int \frac {\left (a + b x + c x^{2}\right )^{\frac {5}{2}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {7}{2}}}\, dx \] Input:

integrate((c*x**2+b*x+a)**(5/2)/(2*c*d*x+b*d)**(7/2),x)
 

Output:

Integral((a + b*x + c*x**2)**(5/2)/(d*(b + 2*c*x))**(7/2), x)
 

Maxima [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{7/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(7/2),x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(7/2), x)
 

Giac [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{7/2}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(7/2),x, algorithm="giac")
 

Output:

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{7/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{7/2}} \,d x \] Input:

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(7/2),x)
 

Output:

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(7/2), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{7/2}} \, dx=\text {too large to display} \] Input:

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(7/2),x)
 

Output:

(sqrt(d)*( - 80*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**4*c**3 + 232*sqr 
t(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**3*b**2*c**2 + 320*sqrt(b + 2*c*x)*s 
qrt(a + b*x + c*x**2)*a**3*b*c**3*x + 960*sqrt(b + 2*c*x)*sqrt(a + b*x + c 
*x**2)*a**3*c**4*x**2 - 136*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**2*b* 
*4*c - 296*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**2*b**3*c**2*x - 696*s 
qrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**2*b**2*c**3*x**2 + 160*sqrt(b + 2 
*c*x)*sqrt(a + b*x + c*x**2)*a**2*b*c**4*x**3 + 80*sqrt(b + 2*c*x)*sqrt(a 
+ b*x + c*x**2)*a**2*c**5*x**4 + 20*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2) 
*a*b**6 + 48*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**5*c*x + 120*sqrt( 
b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**4*c**2*x**2 - 96*sqrt(b + 2*c*x)*sq 
rt(a + b*x + c*x**2)*a*b**3*c**3*x**3 - 48*sqrt(b + 2*c*x)*sqrt(a + b*x + 
c*x**2)*a*b**2*c**4*x**4 - 6*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**6*c 
*x**2 + 8*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**5*c**2*x**3 + 4*sqrt(b 
 + 2*c*x)*sqrt(a + b*x + c*x**2)*b**4*c**3*x**4 - 12800*int((sqrt(b + 2*c* 
x)*sqrt(a + b*x + c*x**2)*x**3)/(20*a**3*b**4*c**2 + 160*a**3*b**3*c**3*x 
+ 480*a**3*b**2*c**4*x**2 + 640*a**3*b*c**5*x**3 + 320*a**3*c**6*x**4 - 12 
*a**2*b**6*c - 76*a**2*b**5*c**2*x - 108*a**2*b**4*c**3*x**2 + 256*a**2*b* 
*3*c**4*x**3 + 928*a**2*b**2*c**5*x**4 + 960*a**2*b*c**6*x**5 + 320*a**2*c 
**7*x**6 + a*b**8 - 4*a*b**7*c*x - 84*a*b**6*c**2*x**2 - 352*a*b**5*c**3*x 
**3 - 656*a*b**4*c**4*x**4 - 576*a*b**3*c**5*x**5 - 192*a*b**2*c**6*x**...