\(\int \frac {a+b x+c x^2}{(d+e x) (f+g x)^{3/2}} \, dx\) [830]

Optimal result

Integrand size = 27, antiderivative size = 122 \[ \int \frac {a+b x+c x^2}{(d+e x) (f+g x)^{3/2}} \, dx=\frac {2 \left (c f^2-b f g+a g^2\right )}{g^2 (e f-d g) \sqrt {f+g x}}+\frac {2 c \sqrt {f+g x}}{e g^2}-\frac {2 \left (c d^2-b d e+a e^2\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{3/2} (e f-d g)^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {911, 1275, 214} \[ \int \frac {a+b x+c x^2}{(d+e x) (f+g x)^{3/2}} \, dx=-\frac {2 \left (a e^2-b d e+c d^2\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{3/2} (e f-d g)^{3/2}}+\frac {2 \left (a g^2-b f g+c f^2\right )}{g^2 \sqrt {f+g x} (e f-d g)}+\frac {2 c \sqrt {f+g x}}{e g^2} \]

[In]

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

[Out]

(2*(c*f^2 - b*f*g + a*g^2))/(g^2*(e*f - d*g)*Sqrt[f + g*x]) + (2*c*Sqrt[f + g*x])/(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^(3/2)*(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 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 1275

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

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int \frac {a+b x+c x^2}{(d+e x) (f+g x)^{3/2}} \, dx=\frac {2 (e g (b f-a g)+c d g (f+g x)-c e f (2 f+g x))}{e g^2 (-e f+d g) \sqrt {f+g x}}-\frac {2 \left (c d^2+e (-b d+a e)\right ) \arctan \left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right )}{e^{3/2} (-e f+d g)^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {\frac {2 c \sqrt {g x +f}}{e}-\frac {2 \left (e^{2} a -b d e +c \,d^{2}\right ) g^{2} \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{e \left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}-\frac {2 \left (a \,g^{2}-b f g +c \,f^{2}\right )}{\left (d g -e f \right ) \sqrt {g x +f}}}{g^{2}}\)	\(122\)
default	\(\frac {\frac {2 c \sqrt {g x +f}}{e}-\frac {2 \left (e^{2} a -b d e +c \,d^{2}\right ) g^{2} \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{e \left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}-\frac {2 \left (a \,g^{2}-b f g +c \,f^{2}\right )}{\left (d g -e f \right ) \sqrt {g x +f}}}{g^{2}}\)	\(122\)
pseudoelliptic	\(\frac {\frac {2 c \sqrt {g x +f}}{e}-\frac {2 \left (e^{2} a -b d e +c \,d^{2}\right ) g^{2} \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{e \left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}-\frac {2 \left (a \,g^{2}-b f g +c \,f^{2}\right )}{\left (d g -e f \right ) \sqrt {g x +f}}}{g^{2}}\)	\(122\)
risch	\(\frac {2 c \sqrt {g x +f}}{e \,g^{2}}+\frac {-\frac {2 g^{2} \left (e^{2} a -b d e +c \,d^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}-\frac {2 e \left (a \,g^{2}-b f g +c \,f^{2}\right )}{\left (d g -e f \right ) \sqrt {g x +f}}}{g^{2} e}\)	\(128\)

[In]

int((c*x^2+b*x+a)/(e*x+d)/(g*x+f)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (108) = 216\).

Time = 0.43 (sec) , antiderivative size = 540, normalized size of antiderivative = 4.43 \[ \int \frac {a+b x+c x^2}{(d+e x) (f+g x)^{3/2}} \, dx=\left [-\frac {{\left ({\left (c d^{2} - b d e + a e^{2}\right )} g^{3} x + {\left (c d^{2} - b d e + a e^{2}\right )} f g^{2}\right )} \sqrt {e^{2} f - d e g} \log \left (\frac {e g x + 2 \, e f - d g + 2 \, \sqrt {e^{2} f - d e g} \sqrt {g x + f}}{e x + d}\right ) - 2 \, {\left (2 \, c e^{3} f^{3} - a d e^{2} g^{3} - {\left (3 \, c d e^{2} + b e^{3}\right )} f^{2} g + {\left (c d^{2} e + b d e^{2} + a e^{3}\right )} f g^{2} + {\left (c e^{3} f^{2} g - 2 \, c d e^{2} f g^{2} + c d^{2} e g^{3}\right )} x\right )} \sqrt {g x + f}}{e^{4} f^{3} g^{2} - 2 \, d e^{3} f^{2} g^{3} + d^{2} e^{2} f g^{4} + {\left (e^{4} f^{2} g^{3} - 2 \, d e^{3} f g^{4} + d^{2} e^{2} g^{5}\right )} x}, \frac {2 \, {\left ({\left ({\left (c d^{2} - b d e + a e^{2}\right )} g^{3} x + {\left (c d^{2} - b d e + a e^{2}\right )} f g^{2}\right )} \sqrt {-e^{2} f + d e g} \arctan \left (\frac {\sqrt {-e^{2} f + d e g} \sqrt {g x + f}}{e g x + e f}\right ) + {\left (2 \, c e^{3} f^{3} - a d e^{2} g^{3} - {\left (3 \, c d e^{2} + b e^{3}\right )} f^{2} g + {\left (c d^{2} e + b d e^{2} + a e^{3}\right )} f g^{2} + {\left (c e^{3} f^{2} g - 2 \, c d e^{2} f g^{2} + c d^{2} e g^{3}\right )} x\right )} \sqrt {g x + f}\right )}}{e^{4} f^{3} g^{2} - 2 \, d e^{3} f^{2} g^{3} + d^{2} e^{2} f g^{4} + {\left (e^{4} f^{2} g^{3} - 2 \, d e^{3} f g^{4} + d^{2} e^{2} g^{5}\right )} x}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 6.45 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.41 \[ \int \frac {a+b x+c x^2}{(d+e x) (f+g x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {c \sqrt {f + g x}}{e g} - \frac {a g^{2} - b f g + c f^{2}}{g \sqrt {f + g x} \left (d g - e f\right )} - \frac {g \left (a e^{2} - b d e + c d^{2}\right ) \operatorname {atan}{\left (\frac {\sqrt {f + g x}}{\sqrt {\frac {d g - e f}{e}}} \right )}}{e^{2} \sqrt {\frac {d g - e f}{e}} \left (d g - e f\right )}\right )}{g} & \text {for}\: g \neq 0 \\\frac {\frac {c x^{2}}{2 e} + \frac {x \left (b e - c d\right )}{e^{2}} + \frac {\left (a e^{2} - b d e + c d^{2}\right ) \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{2}}}{f^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(c*sqrt(f + g*x)/(e*g) - (a*g**2 - b*f*g + c*f**2)/(g*sqrt(f + g*x)*(d*g - e*f)) - g*(a*e**2 - b*
d*e + c*d**2)*atan(sqrt(f + g*x)/sqrt((d*g - e*f)/e))/(e**2*sqrt((d*g - e*f)/e)*(d*g - e*f)))/g, Ne(g, 0)), ((
c*x**2/(2*e) + x*(b*e - c*d)/e**2 + (a*e**2 - b*d*e + c*d**2)*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True
))/e**2)/f**(3/2), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b x+c x^2}{(d+e x) (f+g x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((c*x^2+b*x+a)/(e*x+d)/(g*x+f)^(3/2),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(e*(d*g-e*f)>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int \frac {a+b x+c x^2}{(d+e x) (f+g x)^{3/2}} \, dx=\frac {2 \, {\left (c d^{2} - b d e + a e^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {-e^{2} f + d e g}}\right )}{{\left (e^{2} f - d e g\right )} \sqrt {-e^{2} f + d e g}} + \frac {2 \, {\left (c f^{2} - b f g + a g^{2}\right )}}{{\left (e f g^{2} - d g^{3}\right )} \sqrt {g x + f}} + \frac {2 \, \sqrt {g x + f} c}{e g^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.83 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.33 \[ \int \frac {a+b x+c x^2}{(d+e x) (f+g x)^{3/2}} \, dx=\frac {2\,c\,\sqrt {f+g\,x}}{e\,g^2}+\frac {2\,\mathrm {atan}\left (\frac {2\,\sqrt {f+g\,x}\,\left (e^2\,f-d\,e\,g\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{\sqrt {e}\,{\left (d\,g-e\,f\right )}^{3/2}\,\left (2\,c\,d^2-2\,b\,d\,e+2\,a\,e^2\right )}\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{e^{3/2}\,{\left (d\,g-e\,f\right )}^{3/2}}-\frac {2\,\left (c\,e\,f^2-b\,e\,f\,g+a\,e\,g^2\right )}{e\,g^2\,\sqrt {f+g\,x}\,\left (d\,g-e\,f\right )} \]

[In]

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

[Out]

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