3.1364 \(\int \frac {(b d+2 c d x)^{3/2}}{\sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=132 \[ \frac {2 d^{3/2} \left (b^2-4 a c\right )^{5/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{3 c \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x} \]

[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)

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Rubi [A]  time = 0.12, antiderivative size = 132, 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 = {692, 691, 689, 221} \[ \frac {2 d^{3/2} \left (b^2-4 a c\right )^{5/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{3 c \sqrt {a+b x+c x^2}}+\frac {4}{3} d \sqrt {a+b x+c x^2} \sqrt {b d+2 c d x} \]

Antiderivative was successfully verified.

[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 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 692

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*} \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 {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 ) \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 )}{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*}

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Mathematica [C]  time = 0.11, size = 111, normalized size = 0.84 \[ \frac {2 d \sqrt {d (b+2 c x)} \left (\left (b^2-4 a c\right ) \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )+2 c (a+x (b+c x))\right )}{3 c \sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[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)])

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{\sqrt {c x^{2} + b x + a}}, x\right ) \]

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)^(1/2),x, algorithm="fricas")

[Out]

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

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

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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.08, size = 362, normalized size = 2.74 \[ -\frac {\sqrt {\left (2 c x +b \right ) d}\, \sqrt {c \,x^{2}+b x +a}\, \left (-8 c^{3} x^{3}-12 b \,c^{2} x^{2}-8 a \,c^{2} x -4 b^{2} c x -4 a b c +4 \sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, a c \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )-\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, b^{2} \EllipticF \left (\frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )\right ) d}{3 \left (2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right ) c} \]

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)^(1/2),x)

[Out]

-1/3*((2*c*x+b)*d)^(1/2)*(c*x^2+b*x+a)^(1/2)*d*(4*((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)*a*c-((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^2-8*c^3*x^3-12*x^2*b*c^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)

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

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)^(1/2),x, algorithm="maxima")

[Out]

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

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

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)^(1/2),x)

[Out]

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

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

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)**(1/2),x)

[Out]

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

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