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

Optimal. Leaf size=116 \[ -\frac {2 \left (a e^2-b d e+c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{5/2} \sqrt {e f-d g}}+\frac {2 \sqrt {f+g x} (b e g-c (d g+e f))}{e^2 g^2}+\frac {2 c (f+g x)^{3/2}}{3 e g^2} \]

[Out]

2/3*c*(g*x+f)^(3/2)/e/g^2-2*(a*e^2-b*d*e+c*d^2)*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))/e^(5/2)/(-d*g+
e*f)^(1/2)+2*(b*e*g-c*(d*g+e*f))*(g*x+f)^(1/2)/e^2/g^2

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Rubi [A]  time = 0.17, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {897, 1153, 208} \[ -\frac {2 \left (a e^2-b d e+c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{5/2} \sqrt {e f-d g}}+\frac {2 \sqrt {f+g x} (b e g-c (d g+e f))}{e^2 g^2}+\frac {2 c (f+g x)^{3/2}}{3 e g^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*e*g - c*(e*f + d*g))*Sqrt[f + g*x])/(e^2*g^2) + (2*c*(f + g*x)^(3/2))/(3*e*g^2) - (2*(c*d^2 - b*d*e + a*
e^2)*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(e^(5/2)*Sqrt[e*f - d*g])

Rule 208

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

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 118, normalized size = 1.02 \[ \frac {2 \left (-\frac {g^2 \left (c d^2-e (b d-a e)\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{5/2} \sqrt {e f-d g}}+\frac {\sqrt {f+g x} (b e g-c (d g+e f))}{e^2}+\frac {c (f+g x)^{3/2}}{3 e}\right )}{g^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(((b*e*g - c*(e*f + d*g))*Sqrt[f + g*x])/e^2 + (c*(f + g*x)^(3/2))/(3*e) - ((c*d^2 - e*(b*d - a*e))*g^2*Arc
Tanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(e^(5/2)*Sqrt[e*f - d*g])))/g^2

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fricas [A]  time = 0.87, size = 341, normalized size = 2.94 \[ \left [\frac {3 \, {\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {e^{2} f - d e g} g^{2} \log \left (\frac {e g x + 2 \, e f - d g - 2 \, \sqrt {e^{2} f - d e g} \sqrt {g x + f}}{e x + d}\right ) - 2 \, {\left (2 \, c e^{3} f^{2} + {\left (c d e^{2} - 3 \, b e^{3}\right )} f g - 3 \, {\left (c d^{2} e - b d e^{2}\right )} g^{2} - {\left (c e^{3} f g - c d e^{2} g^{2}\right )} x\right )} \sqrt {g x + f}}{3 \, {\left (e^{4} f g^{2} - d e^{3} g^{3}\right )}}, \frac {2 \, {\left (3 \, {\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-e^{2} f + d e g} g^{2} \arctan \left (\frac {\sqrt {-e^{2} f + d e g} \sqrt {g x + f}}{e g x + e f}\right ) - {\left (2 \, c e^{3} f^{2} + {\left (c d e^{2} - 3 \, b e^{3}\right )} f g - 3 \, {\left (c d^{2} e - b d e^{2}\right )} g^{2} - {\left (c e^{3} f g - c d e^{2} g^{2}\right )} x\right )} \sqrt {g x + f}\right )}}{3 \, {\left (e^{4} f g^{2} - d e^{3} g^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 128, normalized size = 1.10 \[ \frac {2 \, {\left (c d^{2} - b d e + a e^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {d g e - f e^{2}}}\right ) e^{\left (-2\right )}}{\sqrt {d g e - f e^{2}}} - \frac {2 \, {\left (3 \, \sqrt {g x + f} c d g^{5} e - {\left (g x + f\right )}^{\frac {3}{2}} c g^{4} e^{2} + 3 \, \sqrt {g x + f} c f g^{4} e^{2} - 3 \, \sqrt {g x + f} b g^{5} e^{2}\right )} e^{\left (-3\right )}}{3 \, g^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(c*d^2 - b*d*e + a*e^2)*arctan(sqrt(g*x + f)*e/sqrt(d*g*e - f*e^2))*e^(-2)/sqrt(d*g*e - f*e^2) - 2/3*(3*sqrt
(g*x + f)*c*d*g^5*e - (g*x + f)^(3/2)*c*g^4*e^2 + 3*sqrt(g*x + f)*c*f*g^4*e^2 - 3*sqrt(g*x + f)*b*g^5*e^2)*e^(
-3)/g^6

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maple [A]  time = 0.01, size = 189, normalized size = 1.63 \[ \frac {2 a \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\sqrt {\left (d g -e f \right ) e}}-\frac {2 b d \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\sqrt {\left (d g -e f \right ) e}\, e}+\frac {2 c \,d^{2} \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\sqrt {\left (d g -e f \right ) e}\, e^{2}}+\frac {2 \sqrt {g x +f}\, b}{e g}-\frac {2 \sqrt {g x +f}\, c d}{e^{2} g}-\frac {2 \sqrt {g x +f}\, c f}{e \,g^{2}}+\frac {2 \left (g x +f \right )^{\frac {3}{2}} c}{3 e \,g^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((c*x^2+b*x+a)/(e*x+d)/(g*x+f)^(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(d*g-e*f>0)', see `assume?` for
more details)Is d*g-e*f positive or negative?

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mupad [B]  time = 0.14, size = 117, normalized size = 1.01 \[ \sqrt {f+g\,x}\,\left (\frac {2\,b\,g-4\,c\,f}{e\,g^2}-\frac {2\,c\,\left (d\,g^3-e\,f\,g^2\right )}{e^2\,g^4}\right )+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {e}\,\sqrt {f+g\,x}}{\sqrt {d\,g-e\,f}}\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{e^{5/2}\,\sqrt {d\,g-e\,f}}+\frac {2\,c\,{\left (f+g\,x\right )}^{3/2}}{3\,e\,g^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 37.63, size = 112, normalized size = 0.97 \[ \frac {2 c \left (f + g x\right )^{\frac {3}{2}}}{3 e g^{2}} - \frac {2 \left (a e^{2} - b d e + c d^{2}\right ) \operatorname {atan}{\left (\frac {1}{\sqrt {\frac {e}{d g - e f}} \sqrt {f + g x}} \right )}}{e^{2} \sqrt {\frac {e}{d g - e f}} \left (d g - e f\right )} + \frac {2 \sqrt {f + g x} \left (b e g - c d g - c e f\right )}{e^{2} g^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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