3.1398 \(\int \frac {\sqrt {b d+2 c d x}}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=278 \[ \frac {4 c (b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 (b d+2 c d x)^{3/2}}{3 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {8 c \sqrt {d} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{\left (b^2-4 a c\right )^{5/4} \sqrt {a+b x+c x^2}}-\frac {8 c \sqrt {d} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{\left (b^2-4 a c\right )^{5/4} \sqrt {a+b x+c x^2}} \]

[Out]

-2/3*(2*c*d*x+b*d)^(3/2)/(-4*a*c+b^2)/d/(c*x^2+b*x+a)^(3/2)+4*c*(2*c*d*x+b*d)^(3/2)/(-4*a*c+b^2)^2/d/(c*x^2+b*
x+a)^(1/2)-8*c*EllipticE((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)*d^(1/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b
^2))^(1/2)/(-4*a*c+b^2)^(5/4)/(c*x^2+b*x+a)^(1/2)+8*c*EllipticF((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2)
,I)*d^(1/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)/(-4*a*c+b^2)^(5/4)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {687, 691, 690, 307, 221, 1199, 424} \[ \frac {4 c (b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 (b d+2 c d x)^{3/2}}{3 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {8 c \sqrt {d} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{\left (b^2-4 a c\right )^{5/4} \sqrt {a+b x+c x^2}}-\frac {8 c \sqrt {d} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{\left (b^2-4 a c\right )^{5/4} \sqrt {a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*d + 2*c*d*x)^(3/2))/(3*(b^2 - 4*a*c)*d*(a + b*x + c*x^2)^(3/2)) + (4*c*(b*d + 2*c*d*x)^(3/2))/((b^2 - 4
*a*c)^2*d*Sqrt[a + b*x + c*x^2]) - (8*c*Sqrt[d]*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])/((b^2 - 4*a*c)^(5/4)*Sqrt[a + b*x + c*x^2]) + (8*c*Sq
rt[d]*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)*S
qrt[d])], -1])/((b^2 - 4*a*c)^(5/4)*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 687

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*c*e*(m + 2*p + 3))/(e*(p + 1)*(b^2 - 4*a
*c)), 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] && LtQ[p, -1] &&  !GtQ[m, 1] && 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 {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 (b d+2 c d x)^{3/2}}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}-\frac {(2 c) \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^{3/2}} \, dx}{b^2-4 a c}\\ &=-\frac {2 (b d+2 c d x)^{3/2}}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}+\frac {4 c (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right )^2 d \sqrt {a+b x+c x^2}}-\frac {\left (4 c^2\right ) \int \frac {\sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right )^2}\\ &=-\frac {2 (b d+2 c d x)^{3/2}}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}+\frac {4 c (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right )^2 d \sqrt {a+b x+c x^2}}-\frac {\left (4 c^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}{\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}\\ &=-\frac {2 (b d+2 c d x)^{3/2}}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}+\frac {4 c (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right )^2 d \sqrt {a+b x+c x^2}}-\frac {\left (8 c \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 )}{\left (b^2-4 a c\right )^2 d \sqrt {a+b x+c x^2}}\\ &=-\frac {2 (b d+2 c d x)^{3/2}}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}+\frac {4 c (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right )^2 d \sqrt {a+b x+c x^2}}+\frac {\left (8 c \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 )}{\left (b^2-4 a c\right )^{3/2} \sqrt {a+b x+c x^2}}-\frac {\left (8 c \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 )}{\left (b^2-4 a c\right )^{3/2} \sqrt {a+b x+c x^2}}\\ &=-\frac {2 (b d+2 c d x)^{3/2}}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}+\frac {4 c (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right )^2 d \sqrt {a+b x+c x^2}}+\frac {8 c \sqrt {d} \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 )}{\left (b^2-4 a c\right )^{5/4} \sqrt {a+b x+c x^2}}-\frac {\left (8 c \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 )}{\left (b^2-4 a c\right )^{3/2} \sqrt {a+b x+c x^2}}\\ &=-\frac {2 (b d+2 c d x)^{3/2}}{3 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^{3/2}}+\frac {4 c (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right )^2 d \sqrt {a+b x+c x^2}}-\frac {8 c \sqrt {d} \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 )}{\left (b^2-4 a c\right )^{5/4} \sqrt {a+b x+c x^2}}+\frac {8 c \sqrt {d} \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 )}{\left (b^2-4 a c\right )^{5/4} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 99, normalized size = 0.36 \[ \frac {32 c \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}} (d (b+2 c x))^{3/2} \, _2F_1\left (\frac {3}{4},\frac {5}{2};\frac {7}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 d \left (b^2-4 a c\right )^2 \sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*c*(d*(b + 2*c*x))^(3/2)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hypergeometric2F1[3/4, 5/2, 7/4, (b + 2
*c*x)^2/(b^2 - 4*a*c)])/(3*(b^2 - 4*a*c)^2*d*Sqrt[a + x*(b + c*x)])

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \, {\left (b^{2} c + a c^{2}\right )} x^{4} + 3 \, a^{2} b x + {\left (b^{3} + 6 \, a b c\right )} x^{3} + a^{3} + 3 \, {\left (a b^{2} + a^{2} c\right )} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)/(c^3*x^6 + 3*b*c^2*x^5 + 3*(b^2*c + a*c^2)*x^4 + 3*a^2*b*x
+ (b^3 + 6*a*b*c)*x^3 + a^3 + 3*(a*b^2 + a^2*c)*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.08, size = 866, normalized size = 3.12 \[ -\frac {2 \left (-24 c^{4} x^{4}+24 \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 \,c^{3} x^{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 )-6 \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^{2} c^{2} x^{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 )-48 b \,c^{3} x^{3}+24 \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 \,c^{2} x \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 \,c^{3} x^{2}-6 \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^{3} c x \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 )-26 b^{2} c^{2} x^{2}+24 \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} 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 )-6 \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^{2} 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 \,c^{2} x -2 b^{3} c x -10 a \,b^{2} c +b^{4}\right ) \sqrt {\left (2 c x +b \right ) d}}{3 \left (4 a c -b^{2}\right )^{2} \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(5/2),x)

[Out]

-2/3*(24*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))*x^2*a*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)-6*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))*x^2*b^2*c^2*((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)+24*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))*x*a*b*c^2*((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)-6*Elliptic
E(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))*x*b^3*c*((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)+24*((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*c^2-6*((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^2*c-24*c^4*x^4-48*b*c^3*x^3-40*a*c^3*x^
2-26*b^2*c^2*x^2-40*a*b*c^2*x-2*b^3*c*x-10*a*b^2*c+b^4)*((2*c*x+b)*d)^(1/2)/(4*a*c-b^2)^2/(2*c*x+b)/(c*x^2+b*x
+a)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^(1/2)/(a + b*x + c*x^2)^(5/2),x)

[Out]

int((b*d + 2*c*d*x)^(1/2)/(a + b*x + c*x^2)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**(1/2)/(c*x**2+b*x+a)**(5/2),x)

[Out]

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

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