\(\int \frac {(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{11/2}} \, dx\) [1361]

Optimal result

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

[Out]

-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*(c*
x^2+b*x+a)^(1/2)/c^3/d^5/(2*c*d*x+b*d)^(1/2)+1/12*(-4*a*c+b^2)^(3/4)*EllipticE((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^4/d^(11/2)/(c*x^2+b*x+a)^(1/2)-1/12*(-4*a*c+b^2)^(3
/4)*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^4/d^(1
1/2)/(c*x^2+b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {698, 705, 704, 313, 227, 1213, 435} \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{11/2}} \, dx=-\frac {\left (b^2-4 a c\right )^{3/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{12 c^4 d^{11/2} \sqrt {a+b x+c x^2}}+\frac {\left (b^2-4 a c\right )^{3/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{12 c^4 d^{11/2} \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{12 c^3 d^5 \sqrt {b d+2 c d x}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3 (b d+2 c d x)^{5/2}}-\frac {\left (a+b x+c x^2\right )^{5/2}}{9 c d (b d+2 c d x)^{9/2}} \]

[In]

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

[Out]

-1/12*Sqrt[a + b*x + c*x^2]/(c^3*d^5*Sqrt[b*d + 2*c*d*x]) - (a + b*x + c*x^2)^(3/2)/(18*c^2*d^3*(b*d + 2*c*d*x
)^(5/2)) - (a + b*x + c*x^2)^(5/2)/(9*c*d*(b*d + 2*c*d*x)^(9/2)) + ((b^2 - 4*a*c)^(3/4)*Sqrt[-((c*(a + b*x + c
*x^2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(12*c^4*d^(11
/2)*Sqrt[a + b*x + c*x^2]) - ((b^2 - 4*a*c)^(3/4)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticF[ArcSi
n[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(12*c^4*d^(11/2)*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 313

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

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[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 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 704

Int[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[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] && 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 1213

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a+b x+c x^2\right )^{5/2}}{9 c d (b d+2 c d x)^{9/2}}+\frac {5 \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{7/2}} \, dx}{18 c d^2} \\ & = -\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 {\int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{3/2}} \, dx}{12 c^2 d^4} \\ & = -\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 {\int \frac {\sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \, dx}{24 c^3 d^6} \\ & = -\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 {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {\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}{24 c^3 d^6 \sqrt {a+b x+c x^2}} \\ & = -\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 {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{12 c^4 d^7 \sqrt {a+b x+c x^2}} \\ & = -\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 (\sqrt {b^2-4 a c} \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 )}{12 c^4 d^6 \sqrt {a+b x+c x^2}}+\frac {\left (\sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {Subst}\left (\int \frac {1+\frac {x^2}{\sqrt {b^2-4 a c} d}}{\sqrt {1-\frac {x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{12 c^4 d^6 \sqrt {a+b x+c x^2}} \\ & = -\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}} 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 )}{12 c^4 d^{11/2} \sqrt {a+b x+c x^2}}+\frac {\left (\sqrt {b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {x^2}{\sqrt {b^2-4 a c} d}}}{\sqrt {1-\frac {x^2}{\sqrt {b^2-4 a c} d}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{12 c^4 d^6 \sqrt {a+b x+c x^2}} \\ & = -\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 .\sin ^{-1}\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}} 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 )}{12 c^4 d^{11/2} \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 = 10.06 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.35 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{11/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2 \sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {9}{4},-\frac {5}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{288 c^3 d^6 (b+2 c x)^5 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \]

[In]

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

[Out]

-1/288*((b^2 - 4*a*c)^2*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-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)])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1143\) vs. \(2(260)=520\).

Time = 4.91 (sec) , antiderivative size = 1144, normalized size of antiderivative = 3.69

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

[In]

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

Fricas [C] (verification not implemented)

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

Time = 0.17 (sec) , antiderivative size = 284, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{11/2}} \, dx=-\frac {3 \, \sqrt {2} {\left (32 \, c^{5} x^{5} + 80 \, b c^{4} x^{4} + 80 \, b^{2} c^{3} x^{3} + 40 \, b^{3} c^{2} x^{2} + 10 \, b^{4} c x + b^{5}\right )} \sqrt {c^{2} d} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) + {\left (60 \, c^{5} x^{4} + 120 \, b c^{4} x^{3} + 3 \, b^{4} c + 2 \, a b^{2} c^{2} + 4 \, a^{2} c^{3} + 2 \, {\left (43 \, b^{2} c^{3} + 8 \, a c^{4}\right )} x^{2} + 2 \, {\left (13 \, b^{3} c^{2} + 8 \, a b c^{3}\right )} x\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{36 \, {\left (32 \, c^{9} d^{6} x^{5} + 80 \, b c^{8} d^{6} x^{4} + 80 \, b^{2} c^{7} d^{6} x^{3} + 40 \, b^{3} c^{6} d^{6} x^{2} + 10 \, b^{4} c^{5} d^{6} x + b^{5} c^{4} d^{6}\right )}} \]

[In]

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

[Out]

-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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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