3.14.41 \(\int \frac {(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^{9/2}} \, dx\) [1341]

Optimal. Leaf size=174 \[ -\frac {\sqrt {a+b x+c x^2}}{14 c^2 d^3 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}+\frac {\sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{14 c^3 d^{9/2} \sqrt {a+b x+c x^2}} \]

[Out]

-1/7*(c*x^2+b*x+a)^(3/2)/c/d/(2*c*d*x+b*d)^(7/2)-1/14*(c*x^2+b*x+a)^(1/2)/c^2/d^3/(2*c*d*x+b*d)^(3/2)+1/14*(-4
*a*c+b^2)^(1/4)*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)/c^3/d^(9/2)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {698, 705, 703, 227} \begin {gather*} \frac {\sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\text {ArcSin}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{14 c^3 d^{9/2} \sqrt {a+b x+c x^2}}-\frac {\sqrt {a+b x+c x^2}}{14 c^2 d^3 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(9/2),x]

[Out]

-1/14*Sqrt[a + b*x + c*x^2]/(c^2*d^3*(b*d + 2*c*d*x)^(3/2)) - (a + b*x + c*x^2)^(3/2)/(7*c*d*(b*d + 2*c*d*x)^(
7/2)) + ((b^2 - 4*a*c)^(1/4)*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])/(14*c^3*d^(9/2)*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 698

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 1))), x] - Dist[b*(p/(d*e*(m + 1))), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1
), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] &&
GtQ[p, 0] && LtQ[m, -1] &&  !(IntegerQ[m/2] && LtQ[m + 2*p + 3, 0]) && 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*} \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{9/2}} \, dx &=-\frac {\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}+\frac {3 \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{5/2}} \, dx}{14 c d^2}\\ &=-\frac {\sqrt {a+b x+c x^2}}{14 c^2 d^3 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}+\frac {\int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx}{28 c^2 d^4}\\ &=-\frac {\sqrt {a+b x+c x^2}}{14 c^2 d^3 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}+\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}{28 c^2 d^4 \sqrt {a+b x+c x^2}}\\ &=-\frac {\sqrt {a+b x+c x^2}}{14 c^2 d^3 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}+\frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \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 )}{14 c^3 d^5 \sqrt {a+b x+c x^2}}\\ &=-\frac {\sqrt {a+b x+c x^2}}{14 c^2 d^3 (b d+2 c d x)^{3/2}}-\frac {\left (a+b x+c x^2\right )^{3/2}}{7 c d (b d+2 c d x)^{7/2}}+\frac {\sqrt [4]{b^2-4 a c} \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 )}{14 c^3 d^{9/2} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.06, size = 107, normalized size = 0.61 \begin {gather*} \frac {\left (b^2-4 a c\right ) \sqrt {d (b+2 c x)} \sqrt {a+x (b+c x)} \, _2F_1\left (-\frac {7}{4},-\frac {3}{2};-\frac {3}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{56 c^2 d^5 (b+2 c x)^4 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(9/2),x]

[Out]

((b^2 - 4*a*c)*Sqrt[d*(b + 2*c*x)]*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-7/4, -3/2, -3/4, (b + 2*c*x)^2/(b^
2 - 4*a*c)])/(56*c^2*d^5*(b + 2*c*x)^4*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(677\) vs. \(2(146)=292\).
time = 0.78, size = 678, normalized size = 3.90

method	result	size
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {\left (4 a c -b^{2}\right ) \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a c d x +b^{2} d x +a b d}}{448 c^{6} d^{5} \left (x +\frac {b}{2 c}\right )^{4}}-\frac {3 \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a c d x +b^{2} d x +a b d}}{112 d^{5} c^{4} \left (x +\frac {b}{2 c}\right )^{2}}+\frac {\left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \EllipticF \left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{14 d^{4} c^{2} \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a c d x +b^{2} d x +a b d}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}\)	\(533\)
default	\(\frac {\sqrt {c \,x^{2}+b x +a}\, \sqrt {d \left (2 c x +b \right )}\, \left (8 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \EllipticF \left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, c^{3} x^{3}+12 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \EllipticF \left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, b \,c^{2} x^{2}+6 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \EllipticF \left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, b^{2} c x +\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \EllipticF \left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) \sqrt {-4 a c +b^{2}}\, b^{3}-12 c^{4} x^{4}-24 b \,c^{3} x^{3}-16 x^{2} c^{3} a -14 b^{2} c^{2} x^{2}-16 x a b \,c^{2}-2 b^{3} c x -4 a^{2} c^{2}-2 a c \,b^{2}\right )}{28 d^{5} \left (2 c^{2} x^{3}+3 c \,x^{2} b +2 a c x +b^{2} x +a b \right ) \left (2 c x +b \right )^{3} c^{3}}\)	\(678\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(9/2),x,method=_RETURNVERBOSE)

[Out]

1/28*(c*x^2+b*x+a)^(1/2)*(d*(2*c*x+b))^(1/2)*(8*((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^3*x^3+12*((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*c^2*x^2+6*((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^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^3-12*c^4*x^4-24*b*c^3*x^3-16*x^2*c^3*a-14*b^2*c^2*x^2-16*x*a*b*c^2-2*b^3*c*x-4*a^2*c^2-2*a*c*b^2)/d^5
/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)/(2*c*x+b)^3/c^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(9/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(3/2)/(2*c*d*x + b*d)^(9/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.43, size = 194, normalized size = 1.11 \begin {gather*} \frac {\sqrt {2} {\left (16 \, c^{4} x^{4} + 32 \, b c^{3} x^{3} + 24 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + b^{4}\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 ) - 2 \, {\left (6 \, c^{4} x^{2} + 6 \, b c^{3} x + b^{2} c^{2} + 2 \, a c^{3}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{28 \, {\left (16 \, c^{8} d^{5} x^{4} + 32 \, b c^{7} d^{5} x^{3} + 24 \, b^{2} c^{6} d^{5} x^{2} + 8 \, b^{3} c^{5} d^{5} x + b^{4} c^{4} d^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(9/2),x, algorithm="fricas")

[Out]

1/28*(sqrt(2)*(16*c^4*x^4 + 32*b*c^3*x^3 + 24*b^2*c^2*x^2 + 8*b^3*c*x + b^4)*sqrt(c^2*d)*weierstrassPInverse((
b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c) - 2*(6*c^4*x^2 + 6*b*c^3*x + b^2*c^2 + 2*a*c^3)*sqrt(2*c*d*x + b*d)*sq
rt(c*x^2 + b*x + a))/(16*c^8*d^5*x^4 + 32*b*c^7*d^5*x^3 + 24*b^2*c^6*d^5*x^2 + 8*b^3*c^5*d^5*x + b^4*c^4*d^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}{\left (d \left (b + 2 c x\right )\right )^{\frac {9}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(3/2)/(2*c*d*x+b*d)**(9/2),x)

[Out]

Integral((a + b*x + c*x**2)**(3/2)/(d*(b + 2*c*x))**(9/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^(9/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(3/2)/(2*c*d*x + b*d)^(9/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\left (b\,d+2\,c\,d\,x\right )}^{9/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(9/2),x)

[Out]

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^(9/2), x)

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