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

Optimal. Leaf size=179 \[ -\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) (b f h-2 c e h+2 c f g)}{2 c^{3/2} h^2}+\frac{\left (f g^2-h (e g-d h)\right ) \tanh ^{-1}\left (\frac{-2 a h+x (2 c g-b h)+b g}{2 \sqrt{a+b x+c x^2} \sqrt{a h^2-b g h+c g^2}}\right )}{h^2 \sqrt{a h^2-b g h+c g^2}}+\frac{f \sqrt{a+b x+c x^2}}{c h} \]

[Out]

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

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Rubi [A]  time = 0.293372, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {1653, 843, 621, 206, 724} \[ -\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) (b f h-2 c e h+2 c f g)}{2 c^{3/2} h^2}+\frac{\left (f g^2-h (e g-d h)\right ) \tanh ^{-1}\left (\frac{-2 a h+x (2 c g-b h)+b g}{2 \sqrt{a+b x+c x^2} \sqrt{a h^2-b g h+c g^2}}\right )}{h^2 \sqrt{a h^2-b g h+c g^2}}+\frac{f \sqrt{a+b x+c x^2}}{c h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.285503, size = 172, normalized size = 0.96 \[ -\frac{\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right ) (b f h-2 c e h+2 c f g)}{c^{3/2}}+\frac{2 \left (h (d h-e g)+f g^2\right ) \tanh ^{-1}\left (\frac{2 a h-b g+b h x-2 c g x}{2 \sqrt{a+x (b+c x)} \sqrt{h (a h-b g)+c g^2}}\right )}{\sqrt{h (a h-b g)+c g^2}}-\frac{2 f h \sqrt{a+x (b+c x)}}{c}}{2 h^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.317, size = 599, normalized size = 3.4 \begin{align*}{\frac{f}{ch}\sqrt{c{x}^{2}+bx+a}}-{\frac{bf}{2\,h}\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}{h}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{fg}{{h}^{2}}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{d}{h}\ln \left ({ \left ( 2\,{\frac{a{h}^{2}-bgh+c{g}^{2}}{{h}^{2}}}+{\frac{bh-2\,cg}{h} \left ( x+{\frac{g}{h}} \right ) }+2\,\sqrt{{\frac{a{h}^{2}-bgh+c{g}^{2}}{{h}^{2}}}}\sqrt{ \left ( x+{\frac{g}{h}} \right ) ^{2}c+{\frac{bh-2\,cg}{h} \left ( x+{\frac{g}{h}} \right ) }+{\frac{a{h}^{2}-bgh+c{g}^{2}}{{h}^{2}}}} \right ) \left ( x+{\frac{g}{h}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{h}^{2}-bgh+c{g}^{2}}{{h}^{2}}}}}}}+{\frac{eg}{{h}^{2}}\ln \left ({ \left ( 2\,{\frac{a{h}^{2}-bgh+c{g}^{2}}{{h}^{2}}}+{\frac{bh-2\,cg}{h} \left ( x+{\frac{g}{h}} \right ) }+2\,\sqrt{{\frac{a{h}^{2}-bgh+c{g}^{2}}{{h}^{2}}}}\sqrt{ \left ( x+{\frac{g}{h}} \right ) ^{2}c+{\frac{bh-2\,cg}{h} \left ( x+{\frac{g}{h}} \right ) }+{\frac{a{h}^{2}-bgh+c{g}^{2}}{{h}^{2}}}} \right ) \left ( x+{\frac{g}{h}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{h}^{2}-bgh+c{g}^{2}}{{h}^{2}}}}}}}-{\frac{f{g}^{2}}{{h}^{3}}\ln \left ({ \left ( 2\,{\frac{a{h}^{2}-bgh+c{g}^{2}}{{h}^{2}}}+{\frac{bh-2\,cg}{h} \left ( x+{\frac{g}{h}} \right ) }+2\,\sqrt{{\frac{a{h}^{2}-bgh+c{g}^{2}}{{h}^{2}}}}\sqrt{ \left ( x+{\frac{g}{h}} \right ) ^{2}c+{\frac{bh-2\,cg}{h} \left ( x+{\frac{g}{h}} \right ) }+{\frac{a{h}^{2}-bgh+c{g}^{2}}{{h}^{2}}}} \right ) \left ( x+{\frac{g}{h}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{h}^{2}-bgh+c{g}^{2}}{{h}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError