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

Optimal. Leaf size=165 \[ -\frac{\sqrt{f+g x} \left (a e^2-b d e+c d^2\right )}{e (d+e x) (e f-d g)^2}+\frac{\tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) (c d (4 e f-d g)-e (-3 a e g+b d g+2 b e f))}{e^{3/2} (e f-d g)^{5/2}}-\frac{2 \left (a g^2-b f g+c f^2\right )}{g \sqrt{f+g x} (e f-d g)^2} \]

[Out]

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

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Rubi [A]  time = 0.366759, antiderivative size = 165, normalized size of antiderivative = 1., 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 = {897, 1259, 453, 208} \[ -\frac{\sqrt{f+g x} \left (a e^2-b d e+c d^2\right )}{e (d+e x) (e f-d g)^2}+\frac{\tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) (c d (4 e f-d g)-e (-3 a e g+b d g+2 b e f))}{e^{3/2} (e f-d g)^{5/2}}-\frac{2 \left (a g^2-b f g+c f^2\right )}{g \sqrt{f+g x} (e f-d g)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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]

Rubi steps

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

Mathematica [A]  time = 0.465803, size = 176, normalized size = 1.07 \[ -\frac{e g (2 a d g+a e (f+3 g x)-b (3 d f+d g x+2 e f x))+c \left (d^2 g (f+g x)+2 d e f^2+2 e^2 f^2 x\right )}{e g (d+e x) \sqrt{f+g x} (e f-d g)^2}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right ) (e (-3 a e g+b d g+2 b e f)+c d (d g-4 e f))}{e^{3/2} (e f-d g)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.244, size = 418, normalized size = 2.5 \begin{align*} -2\,{\frac{ag}{ \left ( dg-ef \right ) ^{2}\sqrt{gx+f}}}+2\,{\frac{bf}{ \left ( dg-ef \right ) ^{2}\sqrt{gx+f}}}-2\,{\frac{c{f}^{2}}{g \left ( dg-ef \right ) ^{2}\sqrt{gx+f}}}-{\frac{aeg}{ \left ( dg-ef \right ) ^{2} \left ( egx+dg \right ) }\sqrt{gx+f}}+{\frac{bdg}{ \left ( dg-ef \right ) ^{2} \left ( egx+dg \right ) }\sqrt{gx+f}}-{\frac{c{d}^{2}g}{ \left ( dg-ef \right ) ^{2}e \left ( egx+dg \right ) }\sqrt{gx+f}}-3\,{\frac{aeg}{ \left ( dg-ef \right ) ^{2}\sqrt{ \left ( dg-ef \right ) e}}\arctan \left ({\frac{e\sqrt{gx+f}}{\sqrt{ \left ( dg-ef \right ) e}}} \right ) }+{\frac{bdg}{ \left ( dg-ef \right ) ^{2}}\arctan \left ({e\sqrt{gx+f}{\frac{1}{\sqrt{ \left ( dg-ef \right ) e}}}} \right ){\frac{1}{\sqrt{ \left ( dg-ef \right ) e}}}}+2\,{\frac{bef}{ \left ( dg-ef \right ) ^{2}\sqrt{ \left ( dg-ef \right ) e}}\arctan \left ({\frac{e\sqrt{gx+f}}{\sqrt{ \left ( dg-ef \right ) e}}} \right ) }+{\frac{c{d}^{2}g}{ \left ( dg-ef \right ) ^{2}e}\arctan \left ({e\sqrt{gx+f}{\frac{1}{\sqrt{ \left ( dg-ef \right ) e}}}} \right ){\frac{1}{\sqrt{ \left ( dg-ef \right ) e}}}}-4\,{\frac{cdf}{ \left ( dg-ef \right ) ^{2}\sqrt{ \left ( dg-ef \right ) e}}\arctan \left ({\frac{e\sqrt{gx+f}}{\sqrt{ \left ( dg-ef \right ) e}}} \right ) } \end{align*}

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)^(3/2),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.78997, size = 2184, normalized size = 13.24 \begin{align*} \text{result too large to display} \end{align*}

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)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [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((c*x**2+b*x+a)/(e*x+d)**2/(g*x+f)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.16058, size = 381, normalized size = 2.31 \begin{align*} \frac{{\left (c d^{2} g - 4 \, c d f e + b d g e + 2 \, b f e^{2} - 3 \, a g e^{2}\right )} \arctan \left (\frac{\sqrt{g x + f} e}{\sqrt{d g e - f e^{2}}}\right )}{{\left (d^{2} g^{2} e - 2 \, d f g e^{2} + f^{2} e^{3}\right )} \sqrt{d g e - f e^{2}}} - \frac{{\left (g x + f\right )} c d^{2} g^{2} + 2 \, c d f^{2} g e -{\left (g x + f\right )} b d g^{2} e - 2 \, b d f g^{2} e + 2 \, a d g^{3} e + 2 \,{\left (g x + f\right )} c f^{2} e^{2} - 2 \, c f^{3} e^{2} - 2 \,{\left (g x + f\right )} b f g e^{2} + 2 \, b f^{2} g e^{2} + 3 \,{\left (g x + f\right )} a g^{2} e^{2} - 2 \, a f g^{2} e^{2}}{{\left (d^{2} g^{3} e - 2 \, d f g^{2} e^{2} + f^{2} g e^{3}\right )}{\left (\sqrt{g x + f} d g +{\left (g x + f\right )}^{\frac{3}{2}} e - \sqrt{g x + f} f e\right )}} \end{align*}

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)^(3/2),x, algorithm="giac")

[Out]

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