3.1381 \(\int \frac {(b d+2 c d x)^{3/2}}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=125 \[ \frac {4 d^{3/2} \sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{\sqrt {a+b x+c x^2}}-\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {686, 691, 689, 221} \[ \frac {4 d^{3/2} \sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{\sqrt {a+b x+c x^2}}-\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*Sqrt[b*d + 2*c*d*x])/Sqrt[a + b*x + c*x^2] + (4*(b^2 - 4*a*c)^(1/4)*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])/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 686

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

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]

Rubi steps

\begin {align*} \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}}+\left (2 c d^2\right ) \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}}+\frac {\left (2 c 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}{\sqrt {a+b x+c x^2}}\\ &=-\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}}+\frac {\left (4 d \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 )}{\sqrt {a+b x+c x^2}}\\ &=-\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}}+\frac {4 \sqrt [4]{b^2-4 a c} 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 )}{\sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 88, normalized size = 0.70 \[ \frac {2 d \sqrt {d (b+2 c x)} \left (2 \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )-1\right )}{\sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (2 \, c d x + b d\right )}^{\frac {3}{2}} \sqrt {c x^{2} + b x + a}}{c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 194, normalized size = 1.55 \[ \frac {2 \sqrt {\left (2 c x +b \right ) d}\, \sqrt {c \,x^{2}+b x +a}\, \left (-2 c x -b +\sqrt {-4 a c +b^{2}}\, \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )\right ) d}{2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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