\(\int \frac {(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{11/2}} \, dx\) [257]

Optimal result

Integrand size = 28, antiderivative size = 310 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{11/2}} \, dx=-\frac {\sqrt {a+b x+c x^2}}{12 c^3 d^5 \sqrt {b d+2 c d x}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3 (b d+2 c d x)^{5/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{9 c d (b d+2 c d x)^{9/2}}+\frac {\left (b^2-4 a c\right )^{3/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{12 c^4 d^{11/2} \sqrt {a+b x+c x^2}}-\frac {\left (b^2-4 a c\right )^{3/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{12 c^4 d^{11/2} \sqrt {a+b x+c x^2}} \] Output:

-1/12*(c*x^2+b*x+a)^(1/2)/c^3/d^5/(2*c*d*x+b*d)^(1/2)-1/18*(c*x^2+b*x+a)^( 
3/2)/c^2/d^3/(2*c*d*x+b*d)^(5/2)-1/9*(c*x^2+b*x+a)^(5/2)/c/d/(2*c*d*x+b*d) 
^(9/2)+1/12*(-4*a*c+b^2)^(3/4)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)*Ellip 
ticE((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)/c^4/d^(11/2)/(c*x^2 
+b*x+a)^(1/2)-1/12*(-4*a*c+b^2)^(3/4)*(-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^4/d^(11/2) 
/(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 = 109, normalized size of antiderivative = 0.35 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{11/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2 \sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {9}{4},-\frac {5}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{288 c^3 d^6 (b+2 c x)^5 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

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

Output:

-1/288*((b^2 - 4*a*c)^2*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeo 
metric2F1[-5/2, -9/4, -5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c^3*d^6*(b + 2* 
c*x)^5*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.92, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1108, 1108, 1108, 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.

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

\(\Big \downarrow \) 1108

\(\displaystyle \frac {5 \int \frac {\left (c x^2+b x+a\right )^{3/2}}{(b d+2 c x d)^{7/2}}dx}{18 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{9 c d (b d+2 c d x)^{9/2}}\)

\(\Big \downarrow \) 1108

\(\displaystyle \frac {5 \left (\frac {3 \int \frac {\sqrt {c x^2+b x+a}}{(b d+2 c x d)^{3/2}}dx}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{5 c d (b d+2 c d x)^{5/2}}\right )}{18 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{9 c d (b d+2 c d x)^{9/2}}\)

\(\Big \downarrow \) 1108

\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\int \frac {\sqrt {b d+2 c x d}}{\sqrt {c x^2+b x+a}}dx}{2 c d^2}-\frac {\sqrt {a+b x+c x^2}}{c d \sqrt {b d+2 c d x}}\right )}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{5 c d (b d+2 c d x)^{5/2}}\right )}{18 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{9 c d (b d+2 c d x)^{9/2}}\)

\(\Big \downarrow \) 1115

\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {\sqrt {b d+2 c x d}}{\sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{2 c d^2 \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{c d \sqrt {b d+2 c d x}}\right )}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{5 c d (b d+2 c d x)^{5/2}}\right )}{18 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{9 c d (b d+2 c d x)^{9/2}}\)

\(\Big \downarrow \) 1114

\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {b d+2 c x d}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{c^2 d^3 \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{c d \sqrt {b d+2 c d x}}\right )}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{5 c d (b d+2 c d x)^{5/2}}\right )}{18 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{9 c d (b d+2 c d x)^{9/2}}\)

\(\Big \downarrow \) 836

\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d \sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{d \sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d \sqrt {b^2-4 a c} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}\right )}{c^2 d^3 \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{c d \sqrt {b d+2 c d x}}\right )}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{5 c d (b d+2 c d x)^{5/2}}\right )}{18 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{9 c d (b d+2 c d x)^{9/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (\sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d \sqrt {b^2-4 a c} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}\right )}{c^2 d^3 \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{c d \sqrt {b d+2 c d x}}\right )}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{5 c d (b d+2 c d x)^{5/2}}\right )}{18 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{9 c d (b d+2 c d x)^{9/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (\sqrt {b^2-4 a c} \int \frac {d+\frac {b d+2 c x d}{\sqrt {b^2-4 a c}}}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{c^2 d^3 \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{c d \sqrt {b d+2 c d x}}\right )}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{5 c d (b d+2 c d x)^{5/2}}\right )}{18 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{9 c d (b d+2 c d x)^{9/2}}\)

\(\Big \downarrow \) 1389

\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d \sqrt {b^2-4 a c} \int \frac {\sqrt {\frac {b d+2 c x d}{\sqrt {b^2-4 a c} d}+1}}{\sqrt {1-\frac {b d+2 c x d}{\sqrt {b^2-4 a c} d}}}d\sqrt {b d+2 c x d}-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{c^2 d^3 \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{c d \sqrt {b d+2 c d x}}\right )}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{5 c d (b d+2 c d x)^{5/2}}\right )}{18 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{9 c d (b d+2 c d x)^{9/2}}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (d^{3/2} \left (b^2-4 a c\right )^{3/4} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )-d^{3/2} \left (b^2-4 a c\right )^{3/4} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )\right )}{c^2 d^3 \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{c d \sqrt {b d+2 c d x}}\right )}{10 c d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{5 c d (b d+2 c d x)^{5/2}}\right )}{18 c d^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{9 c d (b d+2 c d x)^{9/2}}\)

Input:

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

Output:

-1/9*(a + b*x + c*x^2)^(5/2)/(c*d*(b*d + 2*c*d*x)^(9/2)) + (5*(-1/5*(a + b 
*x + c*x^2)^(3/2)/(c*d*(b*d + 2*c*d*x)^(5/2)) + (3*(-(Sqrt[a + b*x + c*x^2 
]/(c*d*Sqrt[b*d + 2*c*d*x])) + (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)*EllipticF[Arc 
Sin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1]))/(c^2*d^3*Sqr 
t[a + b*x + c*x^2])))/(10*c*d^2)))/(18*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 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. \(1143\) vs. \(2(260)=520\).

Time = 8.48 (sec) , antiderivative size = 1144, normalized size of antiderivative = 3.69

method	result	size
elliptic	\(\text {Expression too large to display}\)	\(1144\)
default	\(\text {Expression too large to display}\)	\(1489\)

Input:

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(11/2),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)* 
(-1/4608*(16*a^2*c^2-8*a*b^2*c+b^4)/c^8/d^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)^5-1/288/c^6*(4*a*c-b^2)/d^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)^3-5/96*(2*c^2* 
d*x^2+2*b*c*d*x+2*a*c*d)/c^4/d^6/((x+1/2*b/c)*(2*c^2*d*x^2+2*b*c*d*x+2*a*c 
*d))^(1/2)+1/12*b/d^5/c^3*(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))+1/6/c^2/d^5*(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)*((-1/2*(b 
+(-4*a*c+b^2)^(1/2))/c+1/2*b/c)*EllipticE(((x+1/2*(b+(-4*a*c+b^2)^(1/2)...
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 284, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{11/2}} \, dx=-\frac {3 \, \sqrt {2} {\left (32 \, c^{5} x^{5} + 80 \, b c^{4} x^{4} + 80 \, b^{2} c^{3} x^{3} + 40 \, b^{3} c^{2} x^{2} + 10 \, b^{4} c x + b^{5}\right )} \sqrt {c^{2} d} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) + {\left (60 \, c^{5} x^{4} + 120 \, b c^{4} x^{3} + 3 \, b^{4} c + 2 \, a b^{2} c^{2} + 4 \, a^{2} c^{3} + 2 \, {\left (43 \, b^{2} c^{3} + 8 \, a c^{4}\right )} x^{2} + 2 \, {\left (13 \, b^{3} c^{2} + 8 \, a b c^{3}\right )} x\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{36 \, {\left (32 \, c^{9} d^{6} x^{5} + 80 \, b c^{8} d^{6} x^{4} + 80 \, b^{2} c^{7} d^{6} x^{3} + 40 \, b^{3} c^{6} d^{6} x^{2} + 10 \, b^{4} c^{5} d^{6} x + b^{5} c^{4} d^{6}\right )}} \] Input:

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

Output:

-1/36*(3*sqrt(2)*(32*c^5*x^5 + 80*b*c^4*x^4 + 80*b^2*c^3*x^3 + 40*b^3*c^2* 
x^2 + 10*b^4*c*x + b^5)*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)) + (60*c^5*x^ 
4 + 120*b*c^4*x^3 + 3*b^4*c + 2*a*b^2*c^2 + 4*a^2*c^3 + 2*(43*b^2*c^3 + 8* 
a*c^4)*x^2 + 2*(13*b^3*c^2 + 8*a*b*c^3)*x)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 
+ b*x + a))/(32*c^9*d^6*x^5 + 80*b*c^8*d^6*x^4 + 80*b^2*c^7*d^6*x^3 + 40*b 
^3*c^6*d^6*x^2 + 10*b^4*c^5*d^6*x + b^5*c^4*d^6)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(d)*( - 24*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**3*c**2 - 32*sqrt 
(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a**2*b**2*c - 288*sqrt(b + 2*c*x)*sqrt( 
a + b*x + c*x**2)*a**2*b*c**2*x - 288*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x** 
2)*a**2*c**3*x**2 + 20*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**4 + 196 
*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*b**3*c*x + 412*sqrt(b + 2*c*x)*s 
qrt(a + b*x + c*x**2)*a*b**2*c**2*x**2 + 432*sqrt(b + 2*c*x)*sqrt(a + b*x 
+ c*x**2)*a*b*c**3*x**3 + 216*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*c** 
4*x**4 - 10*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**5*x - 22*sqrt(b + 2* 
c*x)*sqrt(a + b*x + c*x**2)*b**4*c*x**2 - 24*sqrt(b + 2*c*x)*sqrt(a + b*x 
+ c*x**2)*b**3*c**2*x**3 - 12*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b**2* 
c**3*x**4 - 17280*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(18*a* 
*2*b**6*c + 216*a**2*b**5*c**2*x + 1080*a**2*b**4*c**3*x**2 + 2880*a**2*b* 
*3*c**4*x**3 + 4320*a**2*b**2*c**5*x**4 + 3456*a**2*b*c**6*x**5 + 1152*a** 
2*c**7*x**6 - a*b**8 + 6*a*b**7*c*x + 174*a*b**6*c**2*x**2 + 1136*a*b**5*c 
**3*x**3 + 3720*a*b**4*c**4*x**4 + 7008*a*b**3*c**5*x**5 + 7712*a*b**2*c** 
6*x**6 + 4608*a*b*c**7*x**7 + 1152*a*c**8*x**8 - b**9*x - 13*b**8*c*x**2 - 
 72*b**7*c**2*x**3 - 220*b**6*c**3*x**4 - 400*b**5*c**4*x**5 - 432*b**4*c* 
*5*x**6 - 256*b**3*c**6*x**7 - 64*b**2*c**7*x**8),x)*a**4*b**5*c**5 - 1728 
00*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(18*a**2*b**6*c + 216 
*a**2*b**5*c**2*x + 1080*a**2*b**4*c**3*x**2 + 2880*a**2*b**3*c**4*x**3...