3.1330 \(\int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx\)

Optimal. Leaf size=137 \[ \frac {\sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{3 c^2 d^{5/2} \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {684, 691, 689, 221} \[ \frac {\sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{3 c^2 d^{5/2} \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[a + b*x + c*x^2]/(3*c*d*(b*d + 2*c*d*x)^(3/2)) + ((b^2 - 4*a*c)^(1/4)*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])/(3*c^2*d^(5/2)*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 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 689

Int[1/(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[1/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]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx &=-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}}+\frac {\int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx}{6 c d^2}\\ &=-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}}+\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\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}{6 c d^2 \sqrt {a+b x+c x^2}}\\ &=-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}}+\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \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 )}{3 c^2 d^3 \sqrt {a+b x+c x^2}}\\ &=-\frac {\sqrt {a+b x+c x^2}}{3 c d (b d+2 c d x)^{3/2}}+\frac {\sqrt [4]{b^2-4 a c} \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 )}{3 c^2 d^{5/2} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 91, normalized size = 0.66 \[ -\frac {\sqrt {a+x (b+c x)} \, _2F_1\left (-\frac {3}{4},-\frac {1}{2};\frac {1}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{6 c d \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}} (d (b+2 c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-3/4, -1/2, 1/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c*d*(d*(b + 2*c*
x))^(3/2)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{8 \, c^{3} d^{3} x^{3} + 12 \, b c^{2} d^{3} x^{2} + 6 \, b^{2} c d^{3} x + b^{3} d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)/(8*c^3*d^3*x^3 + 12*b*c^2*d^3*x^2 + 6*b^2*c*d^3*x + b^3*d^3
), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.14, size = 319, normalized size = 2.33 \[ \frac {\left (-2 c^{2} x^{2}-2 b c x +2 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, c x \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )-2 a c +\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, b \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )\right ) \sqrt {\left (2 c x +b \right ) d}}{6 \sqrt {c \,x^{2}+b x +a}\, \left (2 c x +b \right )^{2} c^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(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)*EllipticF(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))*(-4*a*c+b^2)^(1/2)*x*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)*EllipticF(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))*(-4*a*c+b^2)^(1/2)*b-2*c^2*x^2-2*b*c*x-2*a*c
)/d^3*((2*c*x+b)*d)^(1/2)/(c*x^2+b*x+a)^(1/2)/(2*c*x+b)^2/c^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c\,x^2+b\,x+a}}{{\left (b\,d+2\,c\,d\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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