3.282 \(\int \frac {d+e x+f x^2+g x^3}{\sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=177 \[ \frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-8 c^2 (a f+b e)+6 b c (2 a g+b f)-5 b^3 g+16 c^3 d\right )}{16 c^{7/2}}+\frac {\sqrt {a+b x+c x^2} \left (-16 a c g+15 b^2 g-18 b c f+24 c^2 e\right )}{24 c^3}+\frac {x \sqrt {a+b x+c x^2} (6 c f-5 b g)}{12 c^2}+\frac {g x^2 \sqrt {a+b x+c x^2}}{3 c} \]

[Out]

1/16*(16*c^3*d-8*c^2*(a*f+b*e)-5*b^3*g+6*b*c*(2*a*g+b*f))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c
^(7/2)+1/24*(-16*a*c*g+15*b^2*g-18*b*c*f+24*c^2*e)*(c*x^2+b*x+a)^(1/2)/c^3+1/12*(-5*b*g+6*c*f)*x*(c*x^2+b*x+a)
^(1/2)/c^2+1/3*g*x^2*(c*x^2+b*x+a)^(1/2)/c

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Rubi [A]  time = 0.24, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1661, 640, 621, 206} \[ \frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-8 c^2 (a f+b e)+6 b c (2 a g+b f)-5 b^3 g+16 c^3 d\right )}{16 c^{7/2}}+\frac {\sqrt {a+b x+c x^2} \left (-16 a c g+15 b^2 g-18 b c f+24 c^2 e\right )}{24 c^3}+\frac {x \sqrt {a+b x+c x^2} (6 c f-5 b g)}{12 c^2}+\frac {g x^2 \sqrt {a+b x+c x^2}}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((24*c^2*e - 18*b*c*f + 15*b^2*g - 16*a*c*g)*Sqrt[a + b*x + c*x^2])/(24*c^3) + ((6*c*f - 5*b*g)*x*Sqrt[a + b*x
+ c*x^2])/(12*c^2) + (g*x^2*Sqrt[a + b*x + c*x^2])/(3*c) + ((16*c^3*d - 8*c^2*(b*e + a*f) - 5*b^3*g + 6*b*c*(
b*f + 2*a*g))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(7/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 1661

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[(e*x^(q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*(q + 2*p + 1)), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {d+e x+f x^2+g x^3}{\sqrt {a+b x+c x^2}} \, dx &=\frac {g x^2 \sqrt {a+b x+c x^2}}{3 c}+\frac {\int \frac {3 c d+(3 c e-2 a g) x+\frac {1}{2} (6 c f-5 b g) x^2}{\sqrt {a+b x+c x^2}} \, dx}{3 c}\\ &=\frac {(6 c f-5 b g) x \sqrt {a+b x+c x^2}}{12 c^2}+\frac {g x^2 \sqrt {a+b x+c x^2}}{3 c}+\frac {\int \frac {\frac {1}{2} \left (12 c^2 d-6 a c f+5 a b g\right )+\frac {1}{4} \left (24 c^2 e-18 b c f+15 b^2 g-16 a c g\right ) x}{\sqrt {a+b x+c x^2}} \, dx}{6 c^2}\\ &=\frac {\left (24 c^2 e-18 b c f+15 b^2 g-16 a c g\right ) \sqrt {a+b x+c x^2}}{24 c^3}+\frac {(6 c f-5 b g) x \sqrt {a+b x+c x^2}}{12 c^2}+\frac {g x^2 \sqrt {a+b x+c x^2}}{3 c}+\frac {\left (16 c^3 d-8 c^2 (b e+a f)-5 b^3 g+6 b c (b f+2 a g)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{16 c^3}\\ &=\frac {\left (24 c^2 e-18 b c f+15 b^2 g-16 a c g\right ) \sqrt {a+b x+c x^2}}{24 c^3}+\frac {(6 c f-5 b g) x \sqrt {a+b x+c x^2}}{12 c^2}+\frac {g x^2 \sqrt {a+b x+c x^2}}{3 c}+\frac {\left (16 c^3 d-8 c^2 (b e+a f)-5 b^3 g+6 b c (b f+2 a g)\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{8 c^3}\\ &=\frac {\left (24 c^2 e-18 b c f+15 b^2 g-16 a c g\right ) \sqrt {a+b x+c x^2}}{24 c^3}+\frac {(6 c f-5 b g) x \sqrt {a+b x+c x^2}}{12 c^2}+\frac {g x^2 \sqrt {a+b x+c x^2}}{3 c}+\frac {\left (16 c^3 d-8 c^2 (b e+a f)-5 b^3 g+6 b c (b f+2 a g)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 141, normalized size = 0.80 \[ \frac {3 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right ) \left (-8 c^2 (a f+b e)+6 b c (2 a g+b f)-5 b^3 g+16 c^3 d\right )+2 \sqrt {c} \sqrt {a+x (b+c x)} \left (-2 c (8 a g+9 b f+5 b g x)+15 b^2 g+4 c^2 (6 e+x (3 f+2 g x))\right )}{48 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(15*b^2*g - 2*c*(9*b*f + 8*a*g + 5*b*g*x) + 4*c^2*(6*e + x*(3*f + 2*g*x))) +
3*(16*c^3*d - 8*c^2*(b*e + a*f) - 5*b^3*g + 6*b*c*(b*f + 2*a*g))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b
+ c*x)])])/(48*c^(7/2))

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fricas [A]  time = 0.76, size = 341, normalized size = 1.93 \[ \left [\frac {3 \, {\left (16 \, c^{3} d - 8 \, b c^{2} e + 2 \, {\left (3 \, b^{2} c - 4 \, a c^{2}\right )} f - {\left (5 \, b^{3} - 12 \, a b c\right )} g\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (8 \, c^{3} g x^{2} + 24 \, c^{3} e - 18 \, b c^{2} f + {\left (15 \, b^{2} c - 16 \, a c^{2}\right )} g + 2 \, {\left (6 \, c^{3} f - 5 \, b c^{2} g\right )} x\right )} \sqrt {c x^{2} + b x + a}}{96 \, c^{4}}, -\frac {3 \, {\left (16 \, c^{3} d - 8 \, b c^{2} e + 2 \, {\left (3 \, b^{2} c - 4 \, a c^{2}\right )} f - {\left (5 \, b^{3} - 12 \, a b c\right )} g\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (8 \, c^{3} g x^{2} + 24 \, c^{3} e - 18 \, b c^{2} f + {\left (15 \, b^{2} c - 16 \, a c^{2}\right )} g + 2 \, {\left (6 \, c^{3} f - 5 \, b c^{2} g\right )} x\right )} \sqrt {c x^{2} + b x + a}}{48 \, c^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(3*(16*c^3*d - 8*b*c^2*e + 2*(3*b^2*c - 4*a*c^2)*f - (5*b^3 - 12*a*b*c)*g)*sqrt(c)*log(-8*c^2*x^2 - 8*b*
c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(8*c^3*g*x^2 + 24*c^3*e - 18*b*c^2*f + (1
5*b^2*c - 16*a*c^2)*g + 2*(6*c^3*f - 5*b*c^2*g)*x)*sqrt(c*x^2 + b*x + a))/c^4, -1/48*(3*(16*c^3*d - 8*b*c^2*e
+ 2*(3*b^2*c - 4*a*c^2)*f - (5*b^3 - 12*a*b*c)*g)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-
c)/(c^2*x^2 + b*c*x + a*c)) - 2*(8*c^3*g*x^2 + 24*c^3*e - 18*b*c^2*f + (15*b^2*c - 16*a*c^2)*g + 2*(6*c^3*f -
5*b*c^2*g)*x)*sqrt(c*x^2 + b*x + a))/c^4]

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giac [A]  time = 0.27, size = 149, normalized size = 0.84 \[ \frac {1}{24} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (\frac {4 \, g x}{c} + \frac {6 \, c^{2} f - 5 \, b c g}{c^{3}}\right )} x - \frac {18 \, b c f - 15 \, b^{2} g + 16 \, a c g - 24 \, c^{2} e}{c^{3}}\right )} - \frac {{\left (16 \, c^{3} d + 6 \, b^{2} c f - 8 \, a c^{2} f - 5 \, b^{3} g + 12 \, a b c g - 8 \, b c^{2} e\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{16 \, c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/24*sqrt(c*x^2 + b*x + a)*(2*(4*g*x/c + (6*c^2*f - 5*b*c*g)/c^3)*x - (18*b*c*f - 15*b^2*g + 16*a*c*g - 24*c^2
*e)/c^3) - 1/16*(16*c^3*d + 6*b^2*c*f - 8*a*c^2*f - 5*b^3*g + 12*a*b*c*g - 8*b*c^2*e)*log(abs(-2*(sqrt(c)*x -
sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)

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maple [B]  time = 0.01, size = 333, normalized size = 1.88 \[ \frac {\sqrt {c \,x^{2}+b x +a}\, g \,x^{2}}{3 c}+\frac {3 a b g \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {5}{2}}}-\frac {a f \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}-\frac {5 b^{3} g \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {7}{2}}}+\frac {3 b^{2} f \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {5}{2}}}-\frac {b e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}+\frac {d \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}-\frac {5 \sqrt {c \,x^{2}+b x +a}\, b g x}{12 c^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, f x}{2 c}-\frac {2 \sqrt {c \,x^{2}+b x +a}\, a g}{3 c^{2}}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, b^{2} g}{8 c^{3}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, b f}{4 c^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, e}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*g*x^2*(c*x^2+b*x+a)^(1/2)/c-5/12*g/c^2*b*x*(c*x^2+b*x+a)^(1/2)+5/8*g/c^3*b^2*(c*x^2+b*x+a)^(1/2)-5/16*g/c^
(7/2)*b^3*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+3/4*g/c^(5/2)*b*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(
1/2))-2/3*g*a/c^2*(c*x^2+b*x+a)^(1/2)+1/2*(c*x^2+b*x+a)^(1/2)/c*f*x-3/4*(c*x^2+b*x+a)^(1/2)*b/c^2*f+3/8*b^2/c^
(5/2)*f*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/2*a/c^(3/2)*f*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2)
)+(c*x^2+b*x+a)^(1/2)/c*e-1/2*b/c^(3/2)*e*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/c^(1/2)*d*ln((c*x+1/2*
b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {g\,x^3+f\,x^2+e\,x+d}{\sqrt {c\,x^2+b\,x+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d + e x + f x^{2} + g x^{3}}{\sqrt {a + b x + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x**3+f*x**2+e*x+d)/(c*x**2+b*x+a)**(1/2),x)

[Out]

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

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