\(\int \frac {a+c x^2}{(d+e x) \sqrt {f+g x}} \, dx\) [593]

Optimal result

Integrand size = 24, antiderivative size = 104 \[ \int \frac {a+c x^2}{(d+e x) \sqrt {f+g x}} \, dx=-\frac {2 c (e f+d g) \sqrt {f+g x}}{e^2 g^2}+\frac {2 c (f+g x)^{3/2}}{3 e g^2}-\frac {2 \left (c d^2+a e^2\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{5/2} \sqrt {e f-d g}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {912, 1167, 214} \[ \int \frac {a+c x^2}{(d+e x) \sqrt {f+g x}} \, dx=-\frac {2 \left (a e^2+c d^2\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{5/2} \sqrt {e f-d g}}-\frac {2 c \sqrt {f+g x} (d g+e f)}{e^2 g^2}+\frac {2 c (f+g x)^{3/2}}{3 e g^2} \]

[In]

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

[Out]

(-2*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 + a*e^2)*ArcTanh[(Sqr
t[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 912

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 + a*e^2)/e^2 - 2*c*
d*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + 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*} \text {integral}& = \frac {2 \text {Subst}\left (\int \frac {\frac {c f^2+a g^2}{g^2}-\frac {2 c f 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 {c (e f+d g)}{e^2 g}+\frac {c x^2}{e g}+\frac {c d^2+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 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 (a+\frac {c d^2}{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 )}{g} \\ & = -\frac {2 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+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*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \frac {a+c x^2}{(d+e x) \sqrt {f+g x}} \, dx=\frac {2 c \sqrt {f+g x} (-2 e f-3 d g+e g x)}{3 e^2 g^2}+\frac {2 \left (c d^2+a e^2\right ) \arctan \left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right )}{e^{5/2} \sqrt {-e f+d g}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.79

method	result	size
risch	\(-\frac {2 c \left (-e g x +3 d g +2 e f \right ) \sqrt {g x +f}}{3 g^{2} e^{2}}+\frac {2 \left (e^{2} a +c \,d^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{e^{2} \sqrt {\left (d g -e f \right ) e}}\)	\(82\)
pseudoelliptic	\(\frac {-\frac {2 c \sqrt {g x +f}\, \left (-e g x +3 d g +2 e f \right )}{3}+\frac {2 g^{2} \left (e^{2} a +c \,d^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{\sqrt {\left (d g -e f \right ) e}}}{g^{2} e^{2}}\)	\(83\)
derivativedivides	\(\frac {-\frac {2 c \left (-\frac {e \left (g x +f \right )^{\frac {3}{2}}}{3}+d g \sqrt {g x +f}+e f \sqrt {g x +f}\right )}{e^{2}}+\frac {2 g^{2} \left (e^{2} a +c \,d^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{e^{2} \sqrt {\left (d g -e f \right ) e}}}{g^{2}}\)	\(96\)
default	\(\frac {-\frac {2 c \left (-\frac {e \left (g x +f \right )^{\frac {3}{2}}}{3}+d g \sqrt {g x +f}+e f \sqrt {g x +f}\right )}{e^{2}}+\frac {2 g^{2} \left (e^{2} a +c \,d^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{e^{2} \sqrt {\left (d g -e f \right ) e}}}{g^{2}}\)	\(96\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.86 \[ \int \frac {a+c x^2}{(d+e x) \sqrt {f+g x}} \, dx=\left [\frac {3 \, {\left (c d^{2} + a e^{2}\right )} \sqrt {e^{2} f - d e g} g^{2} \log \left (\frac {e g x + 2 \, e f - d g - 2 \, \sqrt {e^{2} f - d e g} \sqrt {g x + f}}{e x + d}\right ) - 2 \, {\left (2 \, c e^{3} f^{2} + c d e^{2} f g - 3 \, c d^{2} e g^{2} - {\left (c e^{3} f g - c d e^{2} g^{2}\right )} x\right )} \sqrt {g x + f}}{3 \, {\left (e^{4} f g^{2} - d e^{3} g^{3}\right )}}, \frac {2 \, {\left (3 \, {\left (c d^{2} + a e^{2}\right )} \sqrt {-e^{2} f + d e g} g^{2} \arctan \left (\frac {\sqrt {-e^{2} f + d e g} \sqrt {g x + f}}{e g x + e f}\right ) - {\left (2 \, c e^{3} f^{2} + c d e^{2} f g - 3 \, c d^{2} e g^{2} - {\left (c e^{3} f g - c d e^{2} g^{2}\right )} x\right )} \sqrt {g x + f}\right )}}{3 \, {\left (e^{4} f g^{2} - d e^{3} g^{3}\right )}}\right ] \]

[In]

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

[Out]

[1/3*(3*(c*d^2 + 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*f*g - 3*c*d^2*e*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 + a*e^2)*sqrt(-e^2*f + d*e*g)*g^2*arctan(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*f*g - 3*c*d^2*e*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)]

Sympy [A] (verification not implemented)

Time = 1.79 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.38 \[ \int \frac {a+c x^2}{(d+e x) \sqrt {f+g x}} \, dx=\begin {cases} \frac {2 \left (\frac {c \left (f + g x\right )^{\frac {3}{2}}}{3 e g} + \frac {\sqrt {f + g x} \left (- c d g - c e f\right )}{e^{2} g} + \frac {g \left (a e^{2} + c d^{2}\right ) \operatorname {atan}{\left (\frac {\sqrt {f + g x}}{\sqrt {\frac {d g - e f}{e}}} \right )}}{e^{3} \sqrt {\frac {d g - e f}{e}}}\right )}{g} & \text {for}\: g \neq 0 \\\frac {- \frac {c d x}{e^{2}} + \frac {c x^{2}}{2 e} + \frac {\left (a e^{2} + c d^{2}\right ) \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{2}}}{\sqrt {f}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.05 \[ \int \frac {a+c x^2}{(d+e x) \sqrt {f+g x}} \, dx=\frac {2 \, {\left (c d^{2} + a e^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {-e^{2} f + d e g}}\right )}{\sqrt {-e^{2} f + d e g} e^{2}} + \frac {2 \, {\left ({\left (g x + f\right )}^{\frac {3}{2}} c e^{2} g^{4} - 3 \, \sqrt {g x + f} c e^{2} f g^{4} - 3 \, \sqrt {g x + f} c d e g^{5}\right )}}{3 \, e^{3} g^{6}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.03 \[ \int \frac {a+c x^2}{(d+e x) \sqrt {f+g x}} \, dx=\frac {2\,\mathrm {atan}\left (\frac {\sqrt {e}\,\sqrt {f+g\,x}}{\sqrt {d\,g-e\,f}}\right )\,\left (c\,d^2+a\,e^2\right )}{e^{5/2}\,\sqrt {d\,g-e\,f}}-\sqrt {f+g\,x}\,\left (\frac {2\,c\,\left (d\,g^3-e\,f\,g^2\right )}{e^2\,g^4}+\frac {4\,c\,f}{e\,g^2}\right )+\frac {2\,c\,{\left (f+g\,x\right )}^{3/2}}{3\,e\,g^2} \]

[In]

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

[Out]

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