3.373 \(\int \frac{\sqrt{d+e x}}{(b x+c x^2)^2} \, dx\)
Optimal. Leaf size=125 \[ -\frac{(b+2 c x) \sqrt{d+e x}}{b^2 \left (b x+c x^2\right )}+\frac{(4 c d-b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3 \sqrt{d}}-\frac{\sqrt{c} (4 c d-3 b e) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 \sqrt{c d-b e}} \]
[Out]
-(((b + 2*c*x)*Sqrt[d + e*x])/(b^2*(b*x + c*x^2))) + ((4*c*d - b*e)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b^3*Sqrt[
d]) - (Sqrt[c]*(4*c*d - 3*b*e)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b^3*Sqrt[c*d - b*e])
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Rubi [A] time = 0.233964, antiderivative size = 125, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used =
{736, 826, 1166, 208} \[ -\frac{(b+2 c x) \sqrt{d+e x}}{b^2 \left (b x+c x^2\right )}+\frac{(4 c d-b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3 \sqrt{d}}-\frac{\sqrt{c} (4 c d-3 b e) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 \sqrt{c d-b e}} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[d + e*x]/(b*x + c*x^2)^2,x]
[Out]
-(((b + 2*c*x)*Sqrt[d + e*x])/(b^2*(b*x + c*x^2))) + ((4*c*d - b*e)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b^3*Sqrt[
d]) - (Sqrt[c]*(4*c*d - 3*b*e)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b^3*Sqrt[c*d - b*e])
Rule 736
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*(b + 2*
c*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m
- 1)*(b*e*m + 2*c*d*(2*p + 3) + 2*c*e*(m + 2*p + 3)*x)*(a + b*x + c*x^2)^(p + 1), 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] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m
, 0] && (LtQ[m, 1] || (ILtQ[m + 2*p + 3, 0] && NeQ[m, 2])) && IntQuadraticQ[a, b, c, d, e, m, p, x]
Rule 826
Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
Rule 1166
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]
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{\sqrt{d+e x}}{\left (b x+c x^2\right )^2} \, dx &=-\frac{(b+2 c x) \sqrt{d+e x}}{b^2 \left (b x+c x^2\right )}+\frac{\int \frac{-2 c d+\frac{b e}{2}-c e x}{\sqrt{d+e x} \left (b x+c x^2\right )} \, dx}{b^2}\\ &=-\frac{(b+2 c x) \sqrt{d+e x}}{b^2 \left (b x+c x^2\right )}+\frac{2 \operatorname{Subst}\left (\int \frac{c d e+e \left (-2 c d+\frac{b e}{2}\right )-c e x^2}{c d^2-b d e+(-2 c d+b e) x^2+c x^4} \, dx,x,\sqrt{d+e x}\right )}{b^2}\\ &=-\frac{(b+2 c x) \sqrt{d+e x}}{b^2 \left (b x+c x^2\right )}+\frac{(c (4 c d-3 b e)) \operatorname{Subst}\left (\int \frac{1}{\frac{b e}{2}+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{b^3}-\frac{(c (4 c d-b e)) \operatorname{Subst}\left (\int \frac{1}{-\frac{b e}{2}+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{b^3}\\ &=-\frac{(b+2 c x) \sqrt{d+e x}}{b^2 \left (b x+c x^2\right )}+\frac{(4 c d-b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3 \sqrt{d}}-\frac{\sqrt{c} (4 c d-3 b e) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 \sqrt{c d-b e}}\\ \end{align*}
Mathematica [A] time = 0.216863, size = 169, normalized size = 1.35 \[ \frac{\sqrt{d} \left (b (b+2 c x) \sqrt{d+e x} (c d-b e)+\sqrt{c} x (b+c x) (4 c d-3 b e) \sqrt{c d-b e} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )\right )-x (b+c x) \left (b^2 e^2-5 b c d e+4 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3 \sqrt{d} x (b+c x) (b e-c d)} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[d + e*x]/(b*x + c*x^2)^2,x]
[Out]
(-((4*c^2*d^2 - 5*b*c*d*e + b^2*e^2)*x*(b + c*x)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]) + Sqrt[d]*(b*(c*d - b*e)*(b +
2*c*x)*Sqrt[d + e*x] + Sqrt[c]*(4*c*d - 3*b*e)*Sqrt[c*d - b*e]*x*(b + c*x)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sq
rt[c*d - b*e]]))/(b^3*Sqrt[d]*(-(c*d) + b*e)*x*(b + c*x))
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Maple [A] time = 0.221, size = 167, normalized size = 1.3 \begin{align*} -{\frac{ce}{{b}^{2} \left ( cex+be \right ) }\sqrt{ex+d}}-3\,{\frac{ce}{{b}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+4\,{\frac{{c}^{2}d}{{b}^{3}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-{\frac{1}{{b}^{2}x}\sqrt{ex+d}}-{\frac{e}{{b}^{2}}{\it Artanh} \left ({\sqrt{ex+d}{\frac{1}{\sqrt{d}}}} \right ){\frac{1}{\sqrt{d}}}}+4\,{\frac{\sqrt{d}c}{{b}^{3}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*x+d)^(1/2)/(c*x^2+b*x)^2,x)
[Out]
-e/b^2*c*(e*x+d)^(1/2)/(c*e*x+b*e)-3*e/b^2*c/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))+4
/b^3*c^2/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*d-1/b^2*(e*x+d)^(1/2)/x-e/b^2/d^(1/2)
*arctanh((e*x+d)^(1/2)/d^(1/2))+4/b^3*d^(1/2)*arctanh((e*x+d)^(1/2)/d^(1/2))*c
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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((e*x+d)^(1/2)/(c*x^2+b*x)^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 2.45067, size = 1750, normalized size = 14. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^(1/2)/(c*x^2+b*x)^2,x, algorithm="fricas")
[Out]
[-1/2*(((4*c^2*d^2 - 3*b*c*d*e)*x^2 + (4*b*c*d^2 - 3*b^2*d*e)*x)*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b*e
+ 2*(c*d - b*e)*sqrt(e*x + d)*sqrt(c/(c*d - b*e)))/(c*x + b)) + ((4*c^2*d - b*c*e)*x^2 + (4*b*c*d - b^2*e)*x)*
sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*(2*b*c*d*x + b^2*d)*sqrt(e*x + d))/(b^3*c*d*x^2 + b^4
*d*x), -1/2*(2*((4*c^2*d^2 - 3*b*c*d*e)*x^2 + (4*b*c*d^2 - 3*b^2*d*e)*x)*sqrt(-c/(c*d - b*e))*arctan(-(c*d - b
*e)*sqrt(e*x + d)*sqrt(-c/(c*d - b*e))/(c*e*x + c*d)) + ((4*c^2*d - b*c*e)*x^2 + (4*b*c*d - b^2*e)*x)*sqrt(d)*
log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*(2*b*c*d*x + b^2*d)*sqrt(e*x + d))/(b^3*c*d*x^2 + b^4*d*x), -
1/2*(2*((4*c^2*d - b*c*e)*x^2 + (4*b*c*d - b^2*e)*x)*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) + ((4*c^2*d^2 -
3*b*c*d*e)*x^2 + (4*b*c*d^2 - 3*b^2*d*e)*x)*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b*e + 2*(c*d - b*e)*sqrt
(e*x + d)*sqrt(c/(c*d - b*e)))/(c*x + b)) + 2*(2*b*c*d*x + b^2*d)*sqrt(e*x + d))/(b^3*c*d*x^2 + b^4*d*x), -(((
4*c^2*d^2 - 3*b*c*d*e)*x^2 + (4*b*c*d^2 - 3*b^2*d*e)*x)*sqrt(-c/(c*d - b*e))*arctan(-(c*d - b*e)*sqrt(e*x + d)
*sqrt(-c/(c*d - b*e))/(c*e*x + c*d)) + ((4*c^2*d - b*c*e)*x^2 + (4*b*c*d - b^2*e)*x)*sqrt(-d)*arctan(sqrt(e*x
+ d)*sqrt(-d)/d) + (2*b*c*d*x + b^2*d)*sqrt(e*x + d))/(b^3*c*d*x^2 + b^4*d*x)]
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Sympy [B] time = 49.2941, size = 790, normalized size = 6.32 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)**(1/2)/(c*x**2+b*x)**2,x)
[Out]
2*c**2*d*e*sqrt(d + e*x)/(2*b**4*e**2 - 2*b**3*c*d*e + 2*b**3*c*e**2*x - 2*b**2*c**2*d*e*x) - 2*c*e**2*sqrt(d
+ e*x)/(2*b**3*e**2 - 2*b**2*c*d*e + 2*b**2*c*e**2*x - 2*b*c**2*d*e*x) + c*e**2*sqrt(-1/(c*(b*e - c*d)**3))*lo
g(-b**2*e**2*sqrt(-1/(c*(b*e - c*d)**3)) + 2*b*c*d*e*sqrt(-1/(c*(b*e - c*d)**3)) - c**2*d**2*sqrt(-1/(c*(b*e -
c*d)**3)) + sqrt(d + e*x))/(2*b) - c*e**2*sqrt(-1/(c*(b*e - c*d)**3))*log(b**2*e**2*sqrt(-1/(c*(b*e - c*d)**3
)) - 2*b*c*d*e*sqrt(-1/(c*(b*e - c*d)**3)) + c**2*d**2*sqrt(-1/(c*(b*e - c*d)**3)) + sqrt(d + e*x))/(2*b) - c*
*2*d*e*sqrt(-1/(c*(b*e - c*d)**3))*log(-b**2*e**2*sqrt(-1/(c*(b*e - c*d)**3)) + 2*b*c*d*e*sqrt(-1/(c*(b*e - c*
d)**3)) - c**2*d**2*sqrt(-1/(c*(b*e - c*d)**3)) + sqrt(d + e*x))/(2*b**2) + c**2*d*e*sqrt(-1/(c*(b*e - c*d)**3
))*log(b**2*e**2*sqrt(-1/(c*(b*e - c*d)**3)) - 2*b*c*d*e*sqrt(-1/(c*(b*e - c*d)**3)) + c**2*d**2*sqrt(-1/(c*(b
*e - c*d)**3)) + sqrt(d + e*x))/(2*b**2) - d*e*sqrt(d**(-3))*log(-d**2*sqrt(d**(-3)) + sqrt(d + e*x))/(2*b**2)
+ d*e*sqrt(d**(-3))*log(d**2*sqrt(d**(-3)) + sqrt(d + e*x))/(2*b**2) - 2*e*atan(sqrt(d + e*x)/sqrt(b*e/c - d)
)/(b**2*sqrt(b*e/c - d)) + 2*e*atan(sqrt(d + e*x)/sqrt(-d))/(b**2*sqrt(-d)) - sqrt(d + e*x)/(b**2*x) + 4*c*d*a
tan(sqrt(d + e*x)/sqrt(b*e/c - d))/(b**3*sqrt(b*e/c - d)) - 4*c*d*atan(sqrt(d + e*x)/sqrt(-d))/(b**3*sqrt(-d))
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Giac [A] time = 1.32419, size = 244, normalized size = 1.95 \begin{align*} \frac{{\left (4 \, c^{2} d - 3 \, b c e\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{\sqrt{-c^{2} d + b c e} b^{3}} - \frac{{\left (4 \, c d - b e\right )} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b^{3} \sqrt{-d}} - \frac{2 \,{\left (x e + d\right )}^{\frac{3}{2}} c e - 2 \, \sqrt{x e + d} c d e + \sqrt{x e + d} b e^{2}}{{\left ({\left (x e + d\right )}^{2} c - 2 \,{\left (x e + d\right )} c d + c d^{2} +{\left (x e + d\right )} b e - b d e\right )} b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^(1/2)/(c*x^2+b*x)^2,x, algorithm="giac")
[Out]
(4*c^2*d - 3*b*c*e)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c*e)*b^3) - (4*c*d - b*e)*ar
ctan(sqrt(x*e + d)/sqrt(-d))/(b^3*sqrt(-d)) - (2*(x*e + d)^(3/2)*c*e - 2*sqrt(x*e + d)*c*d*e + sqrt(x*e + d)*b
*e^2)/(((x*e + d)^2*c - 2*(x*e + d)*c*d + c*d^2 + (x*e + d)*b*e - b*d*e)*b^2)