3.1382 \(\int \frac {1}{\sqrt {b d+2 c d x} (a+b x+c x^2)^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.11, 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 = {687, 691, 689, 221} \[ -\frac {2 \sqrt {b d+2 c d x}}{d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {4 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{\sqrt {d} \left (b^2-4 a c\right )^{3/4} \sqrt {a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)^(3/2)),x]

[Out]

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

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Mathematica [C]  time = 0.07, size = 100, normalized size = 0.73 \[ -\frac {2 \sqrt {d (b+2 c x)} \left (2 \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )+1\right )}{d \left (b^2-4 a c\right ) \sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)^(3/2)),x]

[Out]

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

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{2 \, c^{3} d x^{5} + 5 \, b c^{2} d x^{4} + 4 \, {\left (b^{2} c + a c^{2}\right )} d x^{3} + a^{2} b d + {\left (b^{3} + 6 \, a b c\right )} d x^{2} + 2 \, {\left (a b^{2} + a^{2} c\right )} d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 \, c d x + b d} {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(2*c*d*x + b*d)*(c*x^2 + b*x + a)^(3/2)), x)

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maple [A]  time = 0.09, size = 206, normalized size = 1.50 \[ \frac {2 \left (2 c x +b +\sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )\right ) \sqrt {c \,x^{2}+b x +a}\, \sqrt {\left (2 c x +b \right ) d}}{\left (4 a c -b^{2}\right ) \left (2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right ) d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 \, c d x + b d} {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(2*c*d*x + b*d)*(c*x^2 + b*x + a)^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {b\,d+2\,c\,d\,x}\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*d + 2*c*d*x)^(1/2)*(a + b*x + c*x^2)^(3/2)),x)

[Out]

int(1/((b*d + 2*c*d*x)^(1/2)*(a + b*x + c*x^2)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {d \left (b + 2 c x\right )} \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**(1/2)/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral(1/(sqrt(d*(b + 2*c*x))*(a + b*x + c*x**2)**(3/2)), x)

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