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]

((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))

________________________________________________________________________________________

Rubi [A]  time = 0.235167, antiderivative size = 177, normalized size of antiderivative = 1., 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.271913, size = 141, normalized size = 0.8 \[ \frac{3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right ) \left (-8 c^2 (a f+b e)+6 b c (2 a g+b f)-5 b^3 g+16 c^3 d\right )+2 \sqrt{c} \sqrt{a+x (b+c x)} \left (-2 c (8 a g+9 b f+5 b g x)+15 b^2 g+4 c^2 (6 e+x (3 f+2 g x))\right )}{48 c^{7/2}} \]

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))

________________________________________________________________________________________

Maple [B]  time = 0.086, size = 333, normalized size = 1.9 \begin{align*}{\frac{g{x}^{2}}{3\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{5\,bgx}{12\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,{b}^{2}g}{8\,{c}^{3}}\sqrt{c{x}^{2}+bx+a}}-{\frac{5\,{b}^{3}g}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{7}{2}}}}+{\frac{3\,bga}{4}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{2\,ag}{3\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{fx}{2\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,bf}{4\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,{b}^{2}f}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{af}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{e}{c}\sqrt{c{x}^{2}+bx+a}}-{\frac{be}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{d\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}} \end{align*}

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

________________________________________________________________________________________

Fricas [A]  time = 1.53496, size = 803, normalized size = 4.54 \begin{align*} \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 ] \end{align*}

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]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d + e x + f x^{2} + g x^{3}}{\sqrt{a + b x + c x^{2}}}\, dx \end{align*}

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)

________________________________________________________________________________________

Giac [A]  time = 1.18691, size = 201, normalized size = 1.14 \begin{align*} \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}}} \end{align*}

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)