3.9.24 \(\int \frac {a+b x+c x^2}{(d+e x)^2 \sqrt {f+g x}} \, dx\) [824]

Optimal. Leaf size=140 \[ \frac {2 c \sqrt {f+g x}}{e^2 g}-\frac {\left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{(e f-d g) (d+e x)}+\frac {(c d (4 e f-3 d g)-e (2 b e f-b d g-a e g)) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{5/2} (e f-d g)^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 144, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {911, 1171, 396, 214} \begin {gather*} -\frac {\sqrt {f+g x} \left (a e^2-b d e+c d^2\right )}{e^2 (d+e x) (e f-d g)}+\frac {\tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) (c d (4 e f-3 d g)-e (-a e g-b d g+2 b e f))}{e^{5/2} (e f-d g)^{3/2}}+\frac {2 c \sqrt {f+g x}}{e^2 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*Sqrt[f + g*x])/(e^2*g) - ((c*d^2 - b*d*e + a*e^2)*Sqrt[f + g*x])/(e^2*(e*f - d*g)*(d + e*x)) + ((c*d*(4*e
*f - 3*d*g) - e*(2*b*e*f - b*d*g - a*e*g))*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(e^(5/2)*(e*f - d
*g)^(3/2))

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 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

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 1171

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1
)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], 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] && LtQ[q, -1]

Rubi steps

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

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Mathematica [A]
time = 0.48, size = 150, normalized size = 1.07 \begin {gather*} \frac {\sqrt {f+g x} \left (e (b d-a e) g+c \left (-3 d^2 g+2 e^2 f x+2 d e (f-g x)\right )\right )}{e^2 g (e f-d g) (d+e x)}-\frac {(c d (-4 e f+3 d g)+e (2 b e f-b d g-a e g)) \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right )}{e^{5/2} (-e f+d g)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[f + g*x]*(e*(b*d - a*e)*g + c*(-3*d^2*g + 2*e^2*f*x + 2*d*e*(f - g*x))))/(e^2*g*(e*f - d*g)*(d + e*x)) -
((c*d*(-4*e*f + 3*d*g) + e*(2*b*e*f - b*d*g - a*e*g))*ArcTan[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[-(e*f) + d*g]])/(e^
(5/2)*(-(e*f) + d*g)^(3/2))

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Maple [A]
time = 0.13, size = 156, normalized size = 1.11

method	result	size
derivativedivides	\(\frac {\frac {2 c \sqrt {g x +f}}{e^{2}}+\frac {2 g \left (\frac {g \left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {g x +f}}{2 \left (d g -e f \right ) \left (e \left (g x +f \right )+d g -e f \right )}+\frac {\left (a \,e^{2} g +b d e g -2 b \,e^{2} f -3 c \,d^{2} g +4 c d e f \right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{2 \left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}\right )}{e^{2}}}{g}\)	\(156\)
default	\(\frac {\frac {2 c \sqrt {g x +f}}{e^{2}}+\frac {2 g \left (\frac {g \left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {g x +f}}{2 \left (d g -e f \right ) \left (e \left (g x +f \right )+d g -e f \right )}+\frac {\left (a \,e^{2} g +b d e g -2 b \,e^{2} f -3 c \,d^{2} g +4 c d e f \right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{2 \left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}\right )}{e^{2}}}{g}\)	\(156\)
risch	\(\frac {2 c \sqrt {g x +f}}{e^{2} g}+\frac {g \sqrt {g x +f}\, a}{\left (d g -e f \right ) \left (e g x +d g \right )}-\frac {g \sqrt {g x +f}\, b d}{e \left (d g -e f \right ) \left (e g x +d g \right )}+\frac {g \sqrt {g x +f}\, c \,d^{2}}{e^{2} \left (d g -e f \right ) \left (e g x +d g \right )}+\frac {\arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right ) a g}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}+\frac {\arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right ) b d g}{e \left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}-\frac {2 \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right ) b f}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}-\frac {3 \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right ) c \,d^{2} g}{e^{2} \left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}+\frac {4 \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right ) c d f}{e \left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}\)	\(371\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/g*(c/e^2*(g*x+f)^(1/2)+g/e^2*(1/2*g*(a*e^2-b*d*e+c*d^2)/(d*g-e*f)*(g*x+f)^(1/2)/(e*(g*x+f)+d*g-e*f)+1/2*(a*e
^2*g+b*d*e*g-2*b*e^2*f-3*c*d^2*g+4*c*d*e*f)/(d*g-e*f)/((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)^2/(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 [B] Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (129) = 258\).
time = 1.11, size = 617, normalized size = 4.41 \begin {gather*} \left [-\frac {{\left (3 \, c d^{3} g^{2} + {\left (2 \, b f g - a g^{2}\right )} x e^{3} + {\left (2 \, b d f g - a d g^{2} - {\left (4 \, c d f g + b d g^{2}\right )} x\right )} e^{2} + {\left (3 \, c d^{2} g^{2} x - 4 \, c d^{2} f g - b d^{2} g^{2}\right )} e\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^{3} g^{2} e + {\left (2 \, c f^{2} x - a f g\right )} e^{4} - {\left (4 \, c d f g x - 2 \, c d f^{2} - b d f g - a d g^{2}\right )} e^{3} + {\left (2 \, c d^{2} g^{2} x - 5 \, c d^{2} f g - b d^{2} g^{2}\right )} e^{2}\right )} \sqrt {g x + f}}{2 \, {\left (d^{3} g^{3} e^{3} + f^{2} g x e^{6} - {\left (2 \, d f g^{2} x - d f^{2} g\right )} e^{5} + {\left (d^{2} g^{3} x - 2 \, d^{2} f g^{2}\right )} e^{4}\right )}}, \frac {{\left (3 \, c d^{3} g^{2} + {\left (2 \, b f g - a g^{2}\right )} x e^{3} + {\left (2 \, b d f g - a d g^{2} - {\left (4 \, c d f g + b d g^{2}\right )} x\right )} e^{2} + {\left (3 \, c d^{2} g^{2} x - 4 \, c d^{2} f g - b d^{2} g^{2}\right )} e\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^{3} g^{2} e + {\left (2 \, c f^{2} x - a f g\right )} e^{4} - {\left (4 \, c d f g x - 2 \, c d f^{2} - b d f g - a d g^{2}\right )} e^{3} + {\left (2 \, c d^{2} g^{2} x - 5 \, c d^{2} f g - b d^{2} g^{2}\right )} e^{2}\right )} \sqrt {g x + f}}{d^{3} g^{3} e^{3} + f^{2} g x e^{6} - {\left (2 \, d f g^{2} x - d f^{2} g\right )} e^{5} + {\left (d^{2} g^{3} x - 2 \, d^{2} f g^{2}\right )} e^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((3*c*d^3*g^2 + (2*b*f*g - a*g^2)*x*e^3 + (2*b*d*f*g - a*d*g^2 - (4*c*d*f*g + b*d*g^2)*x)*e^2 + (3*c*d^2
*g^2*x - 4*c*d^2*f*g - b*d^2*g^2)*e)*sqrt(-d*g*e + f*e^2)*log(-(d*g - (g*x + 2*f)*e - 2*sqrt(-d*g*e + f*e^2)*s
qrt(g*x + f))/(x*e + d)) - 2*(3*c*d^3*g^2*e + (2*c*f^2*x - a*f*g)*e^4 - (4*c*d*f*g*x - 2*c*d*f^2 - b*d*f*g - a
*d*g^2)*e^3 + (2*c*d^2*g^2*x - 5*c*d^2*f*g - b*d^2*g^2)*e^2)*sqrt(g*x + f))/(d^3*g^3*e^3 + f^2*g*x*e^6 - (2*d*
f*g^2*x - d*f^2*g)*e^5 + (d^2*g^3*x - 2*d^2*f*g^2)*e^4), ((3*c*d^3*g^2 + (2*b*f*g - a*g^2)*x*e^3 + (2*b*d*f*g
- a*d*g^2 - (4*c*d*f*g + b*d*g^2)*x)*e^2 + (3*c*d^2*g^2*x - 4*c*d^2*f*g - b*d^2*g^2)*e)*sqrt(d*g*e - f*e^2)*ar
ctan(-sqrt(d*g*e - f*e^2)*sqrt(g*x + f)/(d*g - f*e)) + (3*c*d^3*g^2*e + (2*c*f^2*x - a*f*g)*e^4 - (4*c*d*f*g*x
- 2*c*d*f^2 - b*d*f*g - a*d*g^2)*e^3 + (2*c*d^2*g^2*x - 5*c*d^2*f*g - b*d^2*g^2)*e^2)*sqrt(g*x + f))/(d^3*g^3
*e^3 + f^2*g*x*e^6 - (2*d*f*g^2*x - d*f^2*g)*e^5 + (d^2*g^3*x - 2*d^2*f*g^2)*e^4)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b x + c x^{2}}{\left (d + e x\right )^{2} \sqrt {f + g x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 5.11, size = 175, normalized size = 1.25 \begin {gather*} \frac {2 \, \sqrt {g x + f} c e^{\left (-2\right )}}{g} - \frac {{\left (3 \, c d^{2} g - 4 \, c d f e - b d g e + 2 \, b f e^{2} - a g e^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {d g e - f e^{2}}}\right )}{{\left (d g e^{2} - f e^{3}\right )} \sqrt {d g e - f e^{2}}} + \frac {\sqrt {g x + f} c d^{2} g - \sqrt {g x + f} b d g e + \sqrt {g x + f} a g e^{2}}{{\left (d g e^{2} - f e^{3}\right )} {\left (d g + {\left (g x + f\right )} e - f e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.23, size = 146, normalized size = 1.04 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {e}\,\sqrt {f+g\,x}}{\sqrt {d\,g-e\,f}}\right )\,\left (a\,e^2\,g-2\,b\,e^2\,f-3\,c\,d^2\,g+b\,d\,e\,g+4\,c\,d\,e\,f\right )}{e^{5/2}\,{\left (d\,g-e\,f\right )}^{3/2}}+\frac {\sqrt {f+g\,x}\,\left (c\,g\,d^2-b\,g\,d\,e+a\,g\,e^2\right )}{\left (d\,g-e\,f\right )\,\left (e^3\,\left (f+g\,x\right )-e^3\,f+d\,e^2\,g\right )}+\frac {2\,c\,\sqrt {f+g\,x}}{e^2\,g} \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)^2),x)

[Out]

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

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