\(\int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx\) [1366]

Optimal result

Integrand size = 28, antiderivative size = 144 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx=\frac {4 \sqrt {a+b x+c x^2}}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}+\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c \left (b^2-4 a c\right )^{3/4} d^{5/2} \sqrt {a+b x+c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 144, 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 = {707, 705, 703, 227} \[ \int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx=\frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{3 c d^{5/2} \left (b^2-4 a c\right )^{3/4} \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{3 d \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}} \]

[In]

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

[Out]

(4*Sqrt[a + b*x + c*x^2])/(3*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(3/2)) + (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*(b^2 - 4*a*c)^(3/4)*d^(
5/2)*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 707

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[-2*b*d*(d + e*x)^(m
+ 1)*((a + b*x + c*x^2)^(p + 1)/(d^2*(m + 1)*(b^2 - 4*a*c))), x] + Dist[b^2*((m + 2*p + 3)/(d^2*(m + 1)*(b^2 -
4*a*c))), Int[(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] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

Rubi steps \begin{align*} \text {integral}& = \frac {4 \sqrt {a+b x+c x^2}}{3 \left (b^2-4 a c\right ) 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}{3 \left (b^2-4 a c\right ) d^2} \\ & = \frac {4 \sqrt {a+b x+c x^2}}{3 \left (b^2-4 a c\right ) 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}{3 \left (b^2-4 a c\right ) d^2 \sqrt {a+b x+c x^2}} \\ & = \frac {4 \sqrt {a+b x+c x^2}}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}+\frac {\left (2 \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 \left (b^2-4 a c\right ) d^3 \sqrt {a+b x+c x^2}} \\ & = \frac {4 \sqrt {a+b x+c x^2}}{3 \left (b^2-4 a c\right ) d (b d+2 c d x)^{3/2}}+\frac {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 \left (b^2-4 a c\right )^{3/4} d^{5/2} \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 = 91, normalized size of antiderivative = 0.63 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx=-\frac {2 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 c d (d (b+2 c x))^{3/2} \sqrt {a+x (b+c x)}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(361\) vs. \(2(122)=244\).

Time = 3.09 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.51

method	result	size
default	\(-\frac {\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, \left (2 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, F\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, c x +\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, F\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, b +4 c^{2} x^{2}+4 b c x +4 a c \right )}{3 d^{3} \left (2 c^{2} x^{3}+3 c b \,x^{2}+2 a c x +b^{2} x +a b \right ) \left (4 a c -b^{2}\right ) \left (2 c x +b \right ) c}\)	\(362\)
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}{3 \left (4 a c -b^{2}\right ) d^{3} c^{2} \left (x +\frac {b}{2 c}\right )^{2}}-\frac {2 \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, F\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{3 d^{2} \left (4 a c -b^{2}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(490\)

[In]

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

Fricas [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx=\frac {4 \, \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a} c^{2} + \sqrt {2} {\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )}{3 \, {\left (4 \, {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{3} x^{2} + 4 \, {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} d^{3} x + {\left (b^{4} c^{2} - 4 \, a b^{2} c^{3}\right )} d^{3}\right )}} \]

[In]

integrate(1/(2*c*d*x+b*d)^(5/2)/(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 + sqrt(2)*(4*c^2*x^2 + 4*b*c*x + b^2)*sqrt(c^2*d)*weierst
rassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c))/(4*(b^2*c^4 - 4*a*c^5)*d^3*x^2 + 4*(b^3*c^3 - 4*a*b*c^4
)*d^3*x + (b^4*c^2 - 4*a*b^2*c^3)*d^3)

Sympy [F]

\[ \int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx=\int \frac {1}{\left (d \left (b + 2 c x\right )\right )^{\frac {5}{2}} \sqrt {a + b x + c x^{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx=\int { \frac {1}{{\left (2 \, c d x + b d\right )}^{\frac {5}{2}} \sqrt {c x^{2} + b x + a}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx=\int { \frac {1}{{\left (2 \, c d x + b d\right )}^{\frac {5}{2}} \sqrt {c x^{2} + b x + a}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx=\int \frac {1}{{\left (b\,d+2\,c\,d\,x\right )}^{5/2}\,\sqrt {c\,x^2+b\,x+a}} \,d x \]

[In]

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

[Out]

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