3.1367 \(\int \frac {1}{(b d+2 c d x)^{9/2} \sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=188 \[ \frac {10 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{21 c d^{9/2} \left (b^2-4 a c\right )^{7/4} \sqrt {a+b x+c x^2}}+\frac {20 \sqrt {a+b x+c x^2}}{21 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}+\frac {4 \sqrt {a+b x+c x^2}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}} \]

[Out]

4/7*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)/d/(2*c*d*x+b*d)^(7/2)+20/21*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^2/d^3/(2*c*d
*x+b*d)^(3/2)+10/21*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)^(7/4)/d^(9/2)/(c*x^2+b*x+a)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 188, 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 = {693, 691, 689, 221} \[ \frac {20 \sqrt {a+b x+c x^2}}{21 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}+\frac {10 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{21 c d^{9/2} \left (b^2-4 a c\right )^{7/4} \sqrt {a+b x+c x^2}}+\frac {4 \sqrt {a+b x+c x^2}}{7 d \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^(9/2)*Sqrt[a + b*x + c*x^2]),x]

[Out]

(4*Sqrt[a + b*x + c*x^2])/(7*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(7/2)) + (20*Sqrt[a + b*x + c*x^2])/(21*(b^2 - 4*
a*c)^2*d^3*(b*d + 2*c*d*x)^(3/2)) + (10*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])/(21*c*(b^2 - 4*a*c)^(7/4)*d^(9/2)*Sqrt[a + b*x + c*x^2])

Rule 221

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 689

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[(4*Sqrt[-(c/(b^2 -
4*a*c))])/e, 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 691

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 693

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*} \int \frac {1}{(b d+2 c d x)^{9/2} \sqrt {a+b x+c x^2}} \, dx &=\frac {4 \sqrt {a+b x+c x^2}}{7 \left (b^2-4 a c\right ) d (b d+2 c d x)^{7/2}}+\frac {5 \int \frac {1}{(b d+2 c d x)^{5/2} \sqrt {a+b x+c x^2}} \, dx}{7 \left (b^2-4 a c\right ) d^2}\\ &=\frac {4 \sqrt {a+b x+c x^2}}{7 \left (b^2-4 a c\right ) d (b d+2 c d x)^{7/2}}+\frac {20 \sqrt {a+b x+c x^2}}{21 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{3/2}}+\frac {5 \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx}{21 \left (b^2-4 a c\right )^2 d^4}\\ &=\frac {4 \sqrt {a+b x+c x^2}}{7 \left (b^2-4 a c\right ) d (b d+2 c d x)^{7/2}}+\frac {20 \sqrt {a+b x+c x^2}}{21 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{3/2}}+\frac {\left (5 \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}{21 \left (b^2-4 a c\right )^2 d^4 \sqrt {a+b x+c x^2}}\\ &=\frac {4 \sqrt {a+b x+c x^2}}{7 \left (b^2-4 a c\right ) d (b d+2 c d x)^{7/2}}+\frac {20 \sqrt {a+b x+c x^2}}{21 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{3/2}}+\frac {\left (10 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname {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 )}{21 c \left (b^2-4 a c\right )^2 d^5 \sqrt {a+b x+c x^2}}\\ &=\frac {4 \sqrt {a+b x+c x^2}}{7 \left (b^2-4 a c\right ) d (b d+2 c d x)^{7/2}}+\frac {20 \sqrt {a+b x+c x^2}}{21 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{3/2}}+\frac {10 \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 )}{21 c \left (b^2-4 a c\right )^{7/4} d^{9/2} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 99, normalized size = 0.53 \[ -\frac {2 \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}} \sqrt {d (b+2 c x)} \, _2F_1\left (-\frac {7}{4},\frac {1}{2};-\frac {3}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{7 c d^5 (b+2 c x)^4 \sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^(9/2)*Sqrt[a + b*x + c*x^2]),x]

[Out]

(-2*Sqrt[d*(b + 2*c*x)]*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hypergeometric2F1[-7/4, 1/2, -3/4, (b + 2*c
*x)^2/(b^2 - 4*a*c)])/(7*c*d^5*(b + 2*c*x)^4*Sqrt[a + x*(b + c*x)])

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{32 \, c^{6} d^{5} x^{7} + 112 \, b c^{5} d^{5} x^{6} + a b^{5} d^{5} + 32 \, {\left (5 \, b^{2} c^{4} + a c^{5}\right )} d^{5} x^{5} + 40 \, {\left (3 \, b^{3} c^{3} + 2 \, a b c^{4}\right )} d^{5} x^{4} + 10 \, {\left (5 \, b^{4} c^{2} + 8 \, a b^{2} c^{3}\right )} d^{5} x^{3} + {\left (11 \, b^{5} c + 40 \, a b^{3} c^{2}\right )} d^{5} x^{2} + {\left (b^{6} + 10 \, a b^{4} c\right )} d^{5} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)/(32*c^6*d^5*x^7 + 112*b*c^5*d^5*x^6 + a*b^5*d^5 + 32*(5*b^2
*c^4 + a*c^5)*d^5*x^5 + 40*(3*b^3*c^3 + 2*a*b*c^4)*d^5*x^4 + 10*(5*b^4*c^2 + 8*a*b^2*c^3)*d^5*x^3 + (11*b^5*c
+ 40*a*b^3*c^2)*d^5*x^2 + (b^6 + 10*a*b^4*c)*d^5*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (2 \, c d x + b d\right )}^{\frac {9}{2}} \sqrt {c x^{2} + b x + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

integrate(1/((2*c*d*x + b*d)^(9/2)*sqrt(c*x^2 + b*x + a)), x)

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maple [B]  time = 0.08, size = 691, normalized size = 3.68 \[ \frac {\sqrt {\left (2 c x +b \right ) d}\, \sqrt {c \,x^{2}+b x +a}\, \left (80 c^{4} x^{4}+160 b \,c^{3} x^{3}+40 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, c^{3} x^{3} \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )+32 a \,c^{3} x^{2}+112 b^{2} c^{2} x^{2}+60 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, b \,c^{2} x^{2} \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )+32 a b \,c^{2} x +32 b^{3} c x +30 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, b^{2} c x \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )-48 a^{2} c^{2}+32 a \,b^{2} c +5 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, b^{3} \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )\right )}{21 \left (2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right ) \left (4 a c -b^{2}\right )^{2} \left (2 c x +b \right )^{3} c \,d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^(1/2),x)

[Out]

1/21*((2*c*x+b)*d)^(1/2)*(c*x^2+b*x+a)^(1/2)*(40*((2*c*x+b+(-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)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((2*c*
x+b+(-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)*x^3*c^3+60*((2*c*x+b+(-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)*((-2*c*x-b+(-4*a*c+b^2)^(1/2)
)/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((2*c*x+b+(-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)*x^2*b*c^2+30*((2*c*x+b+(-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)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticF(1/2*((2*c*x+b+(-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)*x*b^2*c+5*((2*c*x+b+(-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)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2
)^(1/2))^(1/2)*EllipticF(1/2*((2*c*x+b+(-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+80*c^4*x^4+160*b*c^3*x^3+32*a*c^3*x^2+112*b^2*c^2*x^2+32*a*b*c^2*x+32*b^3*c*x-48*a^2*c^2+32*a*b
^2*c)/d^5/(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/(2*c*x+b)^3/c

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (2 \, c d x + b d\right )}^{\frac {9}{2}} \sqrt {c x^{2} + b x + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((2*c*d*x + b*d)^(9/2)*sqrt(c*x^2 + b*x + a)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (b\,d+2\,c\,d\,x\right )}^{9/2}\,\sqrt {c\,x^2+b\,x+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*d + 2*c*d*x)^(9/2)*(a + b*x + c*x^2)^(1/2)),x)

[Out]

int(1/((b*d + 2*c*d*x)^(9/2)*(a + b*x + c*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d \left (b + 2 c x\right )\right )^{\frac {9}{2}} \sqrt {a + b x + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**(9/2)/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral(1/((d*(b + 2*c*x))**(9/2)*sqrt(a + b*x + c*x**2)), x)

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