\(\int \frac {\sqrt {b d+2 c d x}}{(a+b x+c x^2)^{3/2}} \, dx\) [1386]

Optimal result

Integrand size = 28, antiderivative size = 231 \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) d \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {d} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{\sqrt [4]{b^2-4 a c} \sqrt {a+b x+c x^2}}-\frac {4 \sqrt {d} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{\sqrt [4]{b^2-4 a c} \sqrt {a+b x+c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {701, 705, 704, 313, 227, 1213, 435} \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {4 \sqrt {d} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{\sqrt [4]{b^2-4 a c} \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {d} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{\sqrt [4]{b^2-4 a c} \sqrt {a+b x+c x^2}}-\frac {2 (b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \]

[In]

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

[Out]

(-2*(b*d + 2*c*d*x)^(3/2))/((b^2 - 4*a*c)*d*Sqrt[a + b*x + c*x^2]) + (4*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)^(1/4)*
Sqrt[a + b*x + c*x^2]) - (4*Sqrt[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)*Sqrt[d])], -1])/((b^2 - 4*a*c)^(1/4)*Sqrt[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 313

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 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && 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 704

Int[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[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 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]

Rule 1213

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*} \text {integral}& = -\frac {2 (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) d \sqrt {a+b x+c x^2}}+\frac {(2 c) \int \frac {\sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \, dx}{b^2-4 a c} \\ & = -\frac {2 (b d+2 c d x)^{3/2}}{\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 {\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 (b d+2 c d x)^{3/2}}{\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 {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 ) d \sqrt {a+b x+c x^2}} \\ & = -\frac {2 (b d+2 c d x)^{3/2}}{\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 )}{\sqrt {b^2-4 a c} \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+\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 )}{\sqrt {b^2-4 a c} \sqrt {a+b x+c x^2}} \\ & = -\frac {2 (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) d \sqrt {a+b x+c x^2}}-\frac {4 \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 )}{\sqrt [4]{b^2-4 a c} \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 {\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 )}{\sqrt {b^2-4 a c} \sqrt {a+b x+c x^2}} \\ & = -\frac {2 (b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) d \sqrt {a+b x+c x^2}}+\frac {4 \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 )}{\sqrt [4]{b^2-4 a c} \sqrt {a+b x+c x^2}}-\frac {4 \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 )}{\sqrt [4]{b^2-4 a c} \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 = 10.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.42 \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {8 (d (b+2 c x))^{3/2} \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 \left (b^2-4 a c\right ) d \sqrt {a+x (b+c x)}} \]

[In]

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

[Out]

(-8*(d*(b + 2*c*x))^(3/2)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hypergeometric2F1[3/4, 3/2, 7/4, (b + 2*c
*x)^2/(b^2 - 4*a*c)])/(3*(b^2 - 4*a*c)*d*Sqrt[a + x*(b + c*x)])

Maple [A] (verified)

Time = 2.43 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.44

method	result	size
default	\(-\frac {2 \sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, \left (4 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) a c -\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) b^{2}-4 c^{2} x^{2}-4 b c x -b^{2}\right )}{\left (2 c^{2} x^{3}+3 c b \,x^{2}+2 a c x +b^{2} x +a b \right ) \left (4 a c -b^{2}\right )}\)	\(332\)
elliptic	\(\text {Expression too large to display}\)	\(1053\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.58 \[ \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (2 \, \sqrt {2} \sqrt {c^{2} d} {\left (c x^{2} + b x + a\right )} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) + \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )}\right )}}{a b^{2} - 4 \, a^{2} c + {\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} + {\left (b^{3} - 4 \, a b c\right )} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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