3.1359 \(\int \frac {(a+b x+c x^2)^{5/2}}{(b d+2 c d x)^{3/2}} \, dx\)

Optimal. Leaf size=316 \[ -\frac {\left (b^2-4 a c\right )^{11/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{12 c^4 d^{3/2} \sqrt {a+b x+c x^2}}+\frac {\left (b^2-4 a c\right )^{11/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{12 c^4 d^{3/2} \sqrt {a+b x+c x^2}}-\frac {\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}}{12 c^3 d^3}+\frac {5 \left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{3/2}}{18 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt {b d+2 c d x}} \]

[Out]

5/18*(2*c*d*x+b*d)^(3/2)*(c*x^2+b*x+a)^(3/2)/c^2/d^3-(c*x^2+b*x+a)^(5/2)/c/d/(2*c*d*x+b*d)^(1/2)-1/12*(-4*a*c+
b^2)*(2*c*d*x+b*d)^(3/2)*(c*x^2+b*x+a)^(1/2)/c^3/d^3+1/12*(-4*a*c+b^2)^(11/4)*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^(3/2)/(c*x^2+b*x+a)^(1/2)-1/12*(-4*a*c
+b^2)^(11/4)*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^(3/2)/(c*x^2+b*x+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.29, antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {684, 685, 691, 690, 307, 221, 1199, 424} \[ -\frac {\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}}{12 c^3 d^3}-\frac {\left (b^2-4 a c\right )^{11/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{12 c^4 d^{3/2} \sqrt {a+b x+c x^2}}+\frac {\left (b^2-4 a c\right )^{11/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{12 c^4 d^{3/2} \sqrt {a+b x+c x^2}}+\frac {5 \left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^{3/2}}{18 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt {b d+2 c d x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(3/2),x]

[Out]

-((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/(12*c^3*d^3) + (5*(b*d + 2*c*d*x)^(3/2)*(a + b*x
+ c*x^2)^(3/2))/(18*c^2*d^3) - (a + b*x + c*x^2)^(5/2)/(c*d*Sqrt[b*d + 2*c*d*x]) + ((b^2 - 4*a*c)^(11/4)*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^(3/2)*Sqrt[a + b*x + c*x^2]) - ((b^2 - 4*a*c)^(11/4)*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])/(12*c^4*d^(3/2)*Sqrt[a + b*x + c*x
^2])

Rule 221

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 307

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 424

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

Rule 684

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 685

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 + 2*p + 1)), x] - Dist[(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] && 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] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && IntegerQ[2*p]

Rule 690

Int[Sqrt[(d_) + (e_.)*(x_)]/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(4*Sqrt[-(c/(b^2 - 4*a*
c))])/e, 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 691

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 1199

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*} \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{3/2}} \, dx &=-\frac {\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt {b d+2 c d x}}+\frac {5 \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^{3/2} \, dx}{2 c d^2}\\ &=\frac {5 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt {b d+2 c d x}}-\frac {\left (5 \left (b^2-4 a c\right )\right ) \int \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2} \, dx}{12 c^2 d^2}\\ &=-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{12 c^3 d^3}+\frac {5 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt {b d+2 c d x}}+\frac {\left (b^2-4 a c\right )^2 \int \frac {\sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \, dx}{24 c^3 d^2}\\ &=-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{12 c^3 d^3}+\frac {5 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt {b d+2 c d x}}+\frac {\left (\left (b^2-4 a c\right )^2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \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^2 \sqrt {a+b x+c x^2}}\\ &=-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{12 c^3 d^3}+\frac {5 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt {b d+2 c d x}}+\frac {\left (\left (b^2-4 a c\right )^2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname {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^3 \sqrt {a+b x+c x^2}}\\ &=-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{12 c^3 d^3}+\frac {5 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt {b d+2 c d x}}-\frac {\left (\left (b^2-4 a c\right )^{5/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname {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^2 \sqrt {a+b x+c x^2}}+\frac {\left (\left (b^2-4 a c\right )^{5/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname {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^2 \sqrt {a+b x+c x^2}}\\ &=-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{12 c^3 d^3}+\frac {5 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt {b d+2 c d x}}-\frac {\left (b^2-4 a c\right )^{11/4} \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^{3/2} \sqrt {a+b x+c x^2}}+\frac {\left (\left (b^2-4 a c\right )^{5/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname {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^2 \sqrt {a+b x+c x^2}}\\ &=-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{12 c^3 d^3}+\frac {5 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^{3/2}}{18 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{c d \sqrt {b d+2 c d x}}+\frac {\left (b^2-4 a c\right )^{11/4} \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^{3/2} \sqrt {a+b x+c x^2}}-\frac {\left (b^2-4 a c\right )^{11/4} \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^{3/2} \sqrt {a+b x+c x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 101, normalized size = 0.32 \[ -\frac {\left (b^2-4 a c\right )^2 \sqrt {a+x (b+c x)} \, _2F_1\left (-\frac {5}{2},-\frac {1}{4};\frac {3}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{32 c^3 d \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}} \sqrt {d (b+2 c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(3/2),x]

[Out]

-1/32*((b^2 - 4*a*c)^2*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-5/2, -1/4, 3/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/
(c^3*d*Sqrt[d*(b + 2*c*x)]*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

________________________________________________________________________________________

fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{4 \, c^{2} d^{2} x^{2} + 4 \, b c d^{2} x + b^{2} d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(3/2),x, algorithm="fricas")

[Out]

integral((c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)/(
4*c^2*d^2*x^2 + 4*b*c*d^2*x + b^2*d^2), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(3/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(3/2), x)

________________________________________________________________________________________

maple [B]  time = 0.10, size = 700, normalized size = 2.22 \[ \frac {\sqrt {c \,x^{2}+b x +a}\, \sqrt {\left (2 c x +b \right ) d}\, \left (8 c^{6} x^{6}+24 b \,c^{5} x^{5}+40 a \,c^{5} x^{4}+20 b^{2} c^{4} x^{4}+80 a b \,c^{4} x^{3}-40 a^{2} c^{4} x^{2}+80 a \,b^{2} c^{3} x^{2}-10 b^{4} c^{2} x^{2}+192 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, a^{3} c^{3} \EllipticE \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )-144 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, a^{2} b^{2} c^{2} \EllipticE \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )-40 a^{2} b \,c^{3} x +36 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, a \,b^{4} c \EllipticE \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )+40 a \,b^{3} c^{2} x -3 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, b^{6} \EllipticE \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )-6 b^{5} c x -72 a^{3} c^{3}+44 a^{2} b^{2} c^{2}-6 a \,b^{4} c \right )}{72 \left (2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right ) c^{4} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(3/2),x)

[Out]

1/72*(c*x^2+b*x+a)^(1/2)*((2*c*x+b)*d)^(1/2)*(8*c^6*x^6+24*b*c^5*x^5+192*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c
+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1
/2)*EllipticE(1/2*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*a^3*c^3-144*((2*c*x
+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x-b+(-4*a*c+b^2)
^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticE(1/2*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2
),2^(1/2))*a^2*b^2*c^2+36*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/
2))^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticE(1/2*((2*c*x+b+(-4*a*c+b^2)^(1/2))
/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*a*b^4*c-3*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*
(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticE(1/2*(
(2*c*x+b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2),2^(1/2))*b^6+40*a*c^5*x^4+20*b^2*c^4*x^4+80*a*b
*c^4*x^3-40*a^2*c^4*x^2+80*a*b^2*c^3*x^2-10*b^4*c^2*x^2-40*a^2*b*c^3*x+40*a*b^3*c^2*x-6*b^5*c*x-72*a^3*c^3+44*
a^2*b^2*c^2-6*a*b^4*c)/d^2/c^4/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{2} + b x + a\right )}^{\frac {5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(2*c*d*x+b*d)^(3/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(5/2)/(2*c*d*x + b*d)^(3/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(3/2),x)

[Out]

int((a + b*x + c*x^2)^(5/2)/(b*d + 2*c*d*x)^(3/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x + c x^{2}\right )^{\frac {5}{2}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(5/2)/(2*c*d*x+b*d)**(3/2),x)

[Out]

Integral((a + b*x + c*x**2)**(5/2)/(d*(b + 2*c*x))**(3/2), x)

________________________________________________________________________________________