\(\int \frac {1}{\sqrt {b d+2 c d x} (a+b x+c x^2)^{3/2}} \, dx\) [1382]

Optimal result

Integrand size = 28, antiderivative size = 137 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \sqrt {b d+2 c d x}}{\left (b^2-4 a c\right ) d \sqrt {a+b x+c x^2}}-\frac {4 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{\left (b^2-4 a c\right )^{3/4} \sqrt {d} \sqrt {a+b x+c x^2}} \]

[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)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {701, 705, 703, 227} \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {4 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{\sqrt {d} \left (b^2-4 a c\right )^{3/4} \sqrt {a+b x+c x^2}}-\frac {2 \sqrt {b d+2 c d x}}{d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \]

[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 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 701

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 703

Int[1/(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[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 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]

Rubi steps \begin{align*} \text {integral}& = -\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 ) \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 )}{\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*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.06 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \sqrt {d (b+2 c x)} \left (1+2 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{\left (b^2-4 a c\right ) d \sqrt {a+x (b+c x)}} \]

[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)])

Maple [A] (verified)

Time = 4.65 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.50

method	result	size
default	\(\frac {2 \left (\sqrt {-4 a c +b^{2}}\, \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, F\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )+2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}\, \sqrt {d \left (2 c x +b \right )}}{\left (4 a c -b^{2}\right ) d \left (2 c^{2} x^{3}+3 c b \,x^{2}+2 a c x +b^{2} x +a b \right )}\)	\(206\)
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (\frac {4 c^{2} d x +2 b c d}{\left (4 a c -b^{2}\right ) d c \sqrt {\left (\frac {a}{c}+\frac {b x}{c}+x^{2}\right ) \left (2 c^{2} d x +b c d \right )}}+\frac {4 c \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, F\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{\left (4 a c -b^{2}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(484\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (\sqrt {2} \sqrt {c^{2} d} {\left (c x^{2} + b x + a\right )} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right ) + \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a} c\right )}}{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d x^{2} + {\left (b^{3} c - 4 \, a b c^{2}\right )} d x + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} d} \]

[In]

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

[Out]

-2*(sqrt(2)*sqrt(c^2*d)*(c*x^2 + b*x + a)*weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c) + sqrt(
2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)*c)/((b^2*c^2 - 4*a*c^3)*d*x^2 + (b^3*c - 4*a*b*c^2)*d*x + (a*b^2*c - 4*a^
2*c^2)*d)

Sympy [F]

\[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{\sqrt {d \left (b + 2 c x\right )} \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{\sqrt {2 \, c d x + b d} {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{\sqrt {2 \, c d x + b d} {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{\sqrt {b\,d+2\,c\,d\,x}\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

[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)