### 3.284 $$\int \frac{d+e x+f x^2+g x^3}{x^2 \sqrt{a+b x+c x^2}} \, dx$$

Optimal. Leaf size=139 $\frac{(b d-2 a e) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{2 a^{3/2}}+\frac{(2 c f-b g) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 c^{3/2}}-\frac{d \sqrt{a+b x+c x^2}}{a x}+\frac{g \sqrt{a+b x+c x^2}}{c}$

[Out]

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

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Rubi [A]  time = 0.235271, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {1650, 1653, 843, 621, 206, 724} $\frac{(b d-2 a e) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{2 a^{3/2}}+\frac{(2 c f-b g) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 c^{3/2}}-\frac{d \sqrt{a+b x+c x^2}}{a x}+\frac{g \sqrt{a+b x+c x^2}}{c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 1650

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomia
lQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + b*x + c*
x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^
(m + 1)*(a + b*x + c*x^2)^p*ExpandToSum[(m + 1)*(c*d^2 - b*d*e + a*e^2)*Q + c*d*R*(m + 1) - b*e*R*(m + p + 2)
- c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1]

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 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 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 724

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

Rubi steps

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

Mathematica [A]  time = 0.403244, size = 127, normalized size = 0.91 $\frac{(b d-2 a e) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{2 a^{3/2}}+\frac{(2 c f-b g) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{2 c^{3/2}}+\frac{\sqrt{a+x (b+c x)} (a g x-c d)}{a c x}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.054, size = 173, normalized size = 1.2 \begin{align*}{\frac{g}{c}\sqrt{c{x}^{2}+bx+a}}-{\frac{bg}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{f\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-{e\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}-{\frac{d}{ax}\sqrt{c{x}^{2}+bx+a}}+{\frac{bd}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 12.8985, size = 1658, normalized size = 11.93 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((2*a^2*c*f - a^2*b*g)*sqrt(c)*x*log(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sq
rt(c) - 4*a*c) + (b*c^2*d - 2*a*c^2*e)*sqrt(a)*x*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt(c*x^2 + b*x + a)*(
b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) - 4*(a^2*c*g*x - a*c^2*d)*sqrt(c*x^2 + b*x + a))/(a^2*c^2*x), -1/4*(2*(2*a^2*
c*f - a^2*b*g)*sqrt(-c)*x*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)) + (b*
c^2*d - 2*a*c^2*e)*sqrt(a)*x*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) +
8*a^2)/x^2) - 4*(a^2*c*g*x - a*c^2*d)*sqrt(c*x^2 + b*x + a))/(a^2*c^2*x), -1/4*(2*(b*c^2*d - 2*a*c^2*e)*sqrt(
-a)*x*arctan(1/2*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) + (2*a^2*c*f - a^2*b*g)*s
qrt(c)*x*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*(a^2*c*g*x
- a*c^2*d)*sqrt(c*x^2 + b*x + a))/(a^2*c^2*x), -1/2*((b*c^2*d - 2*a*c^2*e)*sqrt(-a)*x*arctan(1/2*sqrt(c*x^2 +
b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) + (2*a^2*c*f - a^2*b*g)*sqrt(-c)*x*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*(a^2*c*g*x - a*c^2*d)*sqrt(c*x^2 + b*x + a))/(a
^2*c^2*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d + e x + f x^{2} + g x^{3}}{x^{2} \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)/x**2/(c*x**2+b*x+a)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.25539, size = 231, normalized size = 1.66 \begin{align*} \frac{\sqrt{c x^{2} + b x + a} g}{c} - \frac{{\left (b d - 2 \, a e\right )} \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} - \frac{{\left (2 \, c f - b g\right )} \log \left ({\left | 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} + b \right |}\right )}{2 \, c^{\frac{3}{2}}} + \frac{{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} b d + 2 \, a \sqrt{c} d}{{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} - a\right )} a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c*x^2 + b*x + a)*g/c - (b*d - 2*a*e)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt(-a)*a) -
1/2*(2*c*f - b*g)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(3/2) + ((sqrt(c)*x - sqrt(c*
x^2 + b*x + a))*b*d + 2*a*sqrt(c)*d)/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)*a)