\(\int \frac {(b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}} \, dx\) [1364]

Optimal result

Integrand size = 28, antiderivative size = 132 \[ \int \frac {(b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}} \, dx=\frac {4}{3} d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {2 \left (b^2-4 a c\right )^{5/4} d^{3/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c \sqrt {a+b x+c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 132, 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 = {706, 705, 703, 227} \[ \int \frac {(b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}} \, dx=\frac {2 d^{3/2} \left (b^2-4 a c\right )^{5/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x} \]

[In]

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

[Out]

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

Rule 706

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*d*(d + e*x)^(m - 1
)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] + Dist[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1))), In
t[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[
2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

Rubi steps \begin{align*} \text {integral}& = \frac {4}{3} d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {1}{3} \left (\left (b^2-4 a c\right ) d^2\right ) \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx \\ & = \frac {4}{3} d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {\left (\left (b^2-4 a c\right ) d^2 \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}{3 \sqrt {a+b x+c x^2}} \\ & = \frac {4}{3} d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {\left (2 \left (b^2-4 a c\right ) d \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 )}{3 c \sqrt {a+b x+c x^2}} \\ & = \frac {4}{3} d \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}+\frac {2 \left (b^2-4 a c\right )^{5/4} d^{3/2} \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 \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 = 10.08 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.84 \[ \int \frac {(b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}} \, dx=\frac {2 d \sqrt {d (b+2 c x)} \left (2 c (a+x (b+c x))+\left (b^2-4 a c\right ) \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{3 c \sqrt {a+x (b+c x)}} \]

[In]

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

[Out]

(2*d*Sqrt[d*(b + 2*c*x)]*(2*c*(a + x*(b + c*x)) + (b^2 - 4*a*c)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hyp
ergeometric2F1[1/4, 1/2, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(3*c*Sqrt[a + x*(b + c*x)])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(361\) vs. \(2(110)=220\).

Time = 3.00 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.74

method	result	size
default	\(-\frac {\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, d \left (4 \sqrt {-4 a c +b^{2}}\, \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, F\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) a c -\sqrt {-4 a c +b^{2}}\, \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, F\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) b^{2}-8 c^{3} x^{3}-12 b \,c^{2} x^{2}-8 a \,c^{2} x -4 b^{2} c x -4 a b c \right )}{3 c \left (2 c^{2} x^{3}+3 c b \,x^{2}+2 a c x +b^{2} x +a b \right )}\)	\(362\)
risch	\(\frac {4 \left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}\, d^{2}}{3 \sqrt {d \left (2 c x +b \right )}}+\frac {2 \left (\frac {b^{2}}{3}-\frac {4 a c}{3}\right ) \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, F\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right ) d^{2} \sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}}{\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}\, \sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(454\)
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \sqrt {d \left (2 c x +b \right )}\, \left (\frac {4 d \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}{3}+\frac {2 \left (b^{2} d^{2}-\frac {4 d \left (a c d +\frac {1}{2} b^{2} d \right )}{3}\right ) \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, F\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}\right )}{\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}\, d}\)	\(481\)

[In]

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

[Out]

-1/3*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*d*(4*(-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))*a*c-(-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))*b^2-8*c^3*x^3-12*b*c^2*x^2-8*a*c^2*x-4*b^2*c*x-4*a*b*c)/c/(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 = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.61 \[ \int \frac {(b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}} \, dx=\frac {4 \, \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a} c^{2} d + \sqrt {2} \sqrt {c^{2} d} {\left (b^{2} - 4 \, a c\right )} d {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )}{3 \, c^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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