3.9.23 \(\int \frac {a+b x+c x^2}{(d+e x) \sqrt {f+g x}} \, dx\) [823]

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

[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 = 0.11, 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 = {911, 1167, 214} \begin {gather*} -\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} \end {gather*}

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 214

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

Rule 911

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 1167

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 \text {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 \text {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 ) \text {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 = 0.16, size = 104, normalized size = 0.90 \begin {gather*} \frac {2 \sqrt {f+g x} (3 b e g+c (-2 e f-3 d g+e g x))}{3 e^2 g^2}+\frac {2 \left (c d^2+e (-b d+a e)\right ) \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right )}{e^{5/2} \sqrt {-e f+d g}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 115, normalized size = 0.99

method	result	size
derivativedivides	\(\frac {\frac {2 \left (\frac {c \left (g x +f \right )^{\frac {3}{2}} e}{3}+b e g \sqrt {g x +f}-c d g \sqrt {g x +f}-c e f \sqrt {g x +f}\right )}{e^{2}}+\frac {2 g^{2} \left (a \,e^{2}-b d e +c \,d^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{e^{2} \sqrt {\left (d g -e f \right ) e}}}{g^{2}}\)	\(115\)
default	\(\frac {\frac {2 \left (\frac {c \left (g x +f \right )^{\frac {3}{2}} e}{3}+b e g \sqrt {g x +f}-c d g \sqrt {g x +f}-c e f \sqrt {g x +f}\right )}{e^{2}}+\frac {2 g^{2} \left (a \,e^{2}-b d e +c \,d^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{e^{2} \sqrt {\left (d g -e f \right ) e}}}{g^{2}}\)	\(115\)
risch	\(\frac {2 \left (c e g x +3 b e g -3 d g c -2 c e f \right ) \sqrt {g x +f}}{3 g^{2} e^{2}}+\frac {2 \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right ) a}{\sqrt {\left (d g -e f \right ) e}}-\frac {2 \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right ) b d}{e \sqrt {\left (d g -e f \right ) e}}+\frac {2 \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right ) c \,d^{2}}{e^{2} \sqrt {\left (d g -e f \right ) e}}\)	\(159\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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(4*%e^2*f-4*%e*d*g>0)', see `as
sume?` for m

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Fricas [A]
time = 1.42, size = 334, normalized size = 2.88 \begin {gather*} \left [-\frac {3 \, {\left (c d^{2} g^{2} - b d g^{2} e + a g^{2} e^{2}\right )} \sqrt {-d g e + f e^{2}} \log \left (-\frac {d g - {\left (g x + 2 \, f\right )} e + 2 \, \sqrt {-d g e + f e^{2}} \sqrt {g x + f}}{x e + d}\right ) + 2 \, {\left (3 \, c d^{2} g^{2} e + {\left (c f g x - 2 \, c f^{2} + 3 \, b f g\right )} e^{3} - {\left (c d g^{2} x + c d f g + 3 \, b d g^{2}\right )} e^{2}\right )} \sqrt {g x + f}}{3 \, {\left (d g^{3} e^{3} - f g^{2} e^{4}\right )}}, -\frac {2 \, {\left (3 \, {\left (c d^{2} g^{2} - b d g^{2} e + a g^{2} e^{2}\right )} \sqrt {d g e - f e^{2}} \arctan \left (-\frac {\sqrt {d g e - f e^{2}} \sqrt {g x + f}}{d g - f e}\right ) + {\left (3 \, c d^{2} g^{2} e + {\left (c f g x - 2 \, c f^{2} + 3 \, b f g\right )} e^{3} - {\left (c d g^{2} x + c d f g + 3 \, b d g^{2}\right )} e^{2}\right )} \sqrt {g x + f}\right )}}{3 \, {\left (d g^{3} e^{3} - f g^{2} e^{4}\right )}}\right ] \end {gather*}

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

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Sympy [A]
time = 14.42, size = 112, normalized size = 0.97 \begin {gather*} \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}} \end {gather*}

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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Giac [A]
time = 5.44, size = 128, normalized size = 1.10 \begin {gather*} \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}} \end {gather*}

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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Mupad [B]
time = 0.14, size = 117, normalized size = 1.01 \begin {gather*} \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} \end {gather*}

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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