### 3.236 $$\int \frac{d+e x+f x^2}{(a+b x+c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=111 $\frac{2 \left (c \left (2 a e-b \left (\frac{a f}{c}+d\right )\right )-x \left (-2 a c f+b^2 f-b c e+2 c^2 d\right )\right )}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{f \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{3/2}}$

[Out]

(2*(c*(2*a*e - b*(d + (a*f)/c)) - (2*c^2*d - b*c*e + b^2*f - 2*a*c*f)*x))/(c*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^
2]) + (f*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/c^(3/2)

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Rubi [A]  time = 0.0658969, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.16, Rules used = {1660, 12, 621, 206} $\frac{2 \left (c \left (2 a e-b \left (\frac{a f}{c}+d\right )\right )-x \left (-2 a c f+b^2 f-b c e+2 c^2 d\right )\right )}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{f \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(c*(2*a*e - b*(d + (a*f)/c)) - (2*c^2*d - b*c*e + b^2*f - 2*a*c*f)*x))/(c*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^
2]) + (f*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/c^(3/2)

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.319114, size = 113, normalized size = 1.02 $\frac{\frac{2 \sqrt{c} \left (a b f-2 a c (e+f x)+b^2 f x+b c (d-e x)+2 c^2 d x\right )}{\sqrt{a+x (b+c x)}}-f \left (b^2-4 a c\right ) \log \left (2 \sqrt{c} \sqrt{a+x (b+c x)}+b+2 c x\right )}{c^{3/2} \left (4 a c-b^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((2*Sqrt[c]*(a*b*f + 2*c^2*d*x + b^2*f*x + b*c*(d - e*x) - 2*a*c*(e + f*x)))/Sqrt[a + x*(b + c*x)] - (b^2 - 4*
a*c)*f*Log[b + 2*c*x + 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(c^(3/2)*(-b^2 + 4*a*c))

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Maple [B]  time = 0.056, size = 249, normalized size = 2.2 \begin{align*} -{\frac{fx}{c}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{\frac{bf}{2\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{\frac{{b}^{2}fx}{c \left ( 4\,ac-{b}^{2} \right ) }{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{\frac{f{b}^{3}}{2\,{c}^{2} \left ( 4\,ac-{b}^{2} \right ) }{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+{f\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}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}-2\,{\frac{bxe}{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}-{\frac{{b}^{2}e}{c \left ( 4\,ac-{b}^{2} \right ) }{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}+2\,{\frac{d \left ( 2\,cx+b \right ) }{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.56513, size = 941, normalized size = 8.48 \begin{align*} \left [\frac{{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} f x^{2} +{\left (b^{3} - 4 \, a b c\right )} f x +{\left (a b^{2} - 4 \, a^{2} c\right )} f\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 (b c^{2} d - 2 \, a c^{2} e + a b c f +{\left (2 \, c^{3} d - b c^{2} e +{\left (b^{2} c - 2 \, a c^{2}\right )} f\right )} x\right )} \sqrt{c x^{2} + b x + a}}{2 \,{\left (a b^{2} c^{2} - 4 \, a^{2} c^{3} +{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} +{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x\right )}}, -\frac{{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} f x^{2} +{\left (b^{3} - 4 \, a b c\right )} f x +{\left (a b^{2} - 4 \, a^{2} c\right )} f\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 (b c^{2} d - 2 \, a c^{2} e + a b c f +{\left (2 \, c^{3} d - b c^{2} e +{\left (b^{2} c - 2 \, a c^{2}\right )} f\right )} x\right )} \sqrt{c x^{2} + b x + a}}{a b^{2} c^{2} - 4 \, a^{2} c^{3} +{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} +{\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(((b^2*c - 4*a*c^2)*f*x^2 + (b^3 - 4*a*b*c)*f*x + (a*b^2 - 4*a^2*c)*f)*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*(b*c^2*d - 2*a*c^2*e + a*b*c*f + (2*c^3*d - b*
c^2*e + (b^2*c - 2*a*c^2)*f)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^2 + (b^3
*c^2 - 4*a*b*c^3)*x), -(((b^2*c - 4*a*c^2)*f*x^2 + (b^3 - 4*a*b*c)*f*x + (a*b^2 - 4*a^2*c)*f)*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*(b*c^2*d - 2*a*c^2*e + a*b*c*f + (
2*c^3*d - b*c^2*e + (b^2*c - 2*a*c^2)*f)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4
)*x^2 + (b^3*c^2 - 4*a*b*c^3)*x)]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21096, size = 165, normalized size = 1.49 \begin{align*} -\frac{2 \,{\left (\frac{{\left (2 \, c^{2} d + b^{2} f - 2 \, a c f - b c e\right )} x}{b^{2} c - 4 \, a c^{2}} + \frac{b c d + a b f - 2 \, a c e}{b^{2} c - 4 \, a c^{2}}\right )}}{\sqrt{c x^{2} + b x + a}} - \frac{f \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{c^{\frac{3}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((2*c^2*d + b^2*f - 2*a*c*f - b*c*e)*x/(b^2*c - 4*a*c^2) + (b*c*d + a*b*f - 2*a*c*e)/(b^2*c - 4*a*c^2))/sqr
t(c*x^2 + b*x + a) - f*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(3/2)