3.458 \(\int \frac{1}{(d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}})^2} \, dx\)

Optimal. Leaf size=151 \[ -\frac{a f^2}{2 d^2 e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}-\frac{\frac{a f^2}{d^2}+1}{2 e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x\right )}-\frac{a f^2 \log \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}{d^3 e}+\frac{a f^2 \log \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x\right )}{d^3 e} \]

[Out]

-(a*f^2)/(2*d^2*e*(e*x + f*Sqrt[a + (e^2*x^2)/f^2])) - (1 + (a*f^2)/d^2)/(2*e*(d + e*x + f*Sqrt[a + (e^2*x^2)/
f^2])) - (a*f^2*Log[e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/(d^3*e) + (a*f^2*Log[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]
])/(d^3*e)

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Rubi [A]  time = 0.112668, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2117, 893} \[ -\frac{a f^2}{2 d^2 e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}-\frac{\frac{a f^2}{d^2}+1}{2 e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x\right )}-\frac{a f^2 \log \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}{d^3 e}+\frac{a f^2 \log \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x\right )}{d^3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*f^2)/(2*d^2*e*(e*x + f*Sqrt[a + (e^2*x^2)/f^2])) - (1 + (a*f^2)/d^2)/(2*e*(d + e*x + f*Sqrt[a + (e^2*x^2)/
f^2])) - (a*f^2*Log[e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/(d^3*e) + (a*f^2*Log[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]
])/(d^3*e)

Rule 2117

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[((g + h*x^n)^p*(d^2 + a*f^2 - 2*d*x + x^2))/(d - x)^2, x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], 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] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

\begin{align*} \int \frac{1}{\left (d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{d^2+a f^2-2 d x+x^2}{(d-x)^2 x^2} \, dx,x,d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}{2 e}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a f^2}{d^2 (d-x)^2}+\frac{2 a f^2}{d^3 (d-x)}+\frac{d^2+a f^2}{d^2 x^2}+\frac{2 a f^2}{d^3 x}\right ) \, dx,x,d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}{2 e}\\ &=-\frac{a f^2}{2 d^2 e \left (e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}-\frac{1+\frac{a f^2}{d^2}}{2 e \left (d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}-\frac{a f^2 \log \left (e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}{d^3 e}+\frac{a f^2 \log \left (d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}{d^3 e}\\ \end{align*}

Mathematica [A]  time = 0.290328, size = 141, normalized size = 0.93 \[ -\frac{\frac{a f^2}{d^2 \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}+\frac{\frac{a f^2}{d^2}+1}{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}+\frac{2 a f^2 \log \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}{d^3}-\frac{2 a f^2 \log \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x\right )}{d^3}}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*f^2)/(d^2*(e*x + f*Sqrt[a + (e^2*x^2)/f^2])) + (1 + (a*f^2)/d^2)/(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]) +
(2*a*f^2*Log[e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/d^3 - (2*a*f^2*Log[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/d^3)/(
2*e)

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Maple [B]  time = 0.023, size = 4136, normalized size = 27.4 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*f^5/e/d^2/(a^2*f^4+2*a*d^2*f^2+d^4)*(4*e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+4*e*(a*f^2-d^2)/f^2/d*(x+1/2*(-
a*f^2+d^2)/d/e)+(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2)*a^2-1/4*f^5/d^3/(a^2*f^4+2*a*d^2*f^2+d^4)*ln((1/2*e*(
a*f^2-d^2)/f^2/d+e^2*(x+1/2*(-a*f^2+d^2)/d/e)/f^2)/(e^2/f^2)^(1/2)+(e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+e*(a*f^
2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/4*(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2))/(e^2/f^2)^(1/2)*a^3-f^3/d/
(a^2*f^4+2*a*d^2*f^2+d^4)*(e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/4
*(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2)*x*a+1/4*f^3/e/d^4/((a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2)*ln((1/2*
(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2+e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/2*((a^2*f^4+2*a*d^2*f^2+d^4)/
f^2/d^2)^(1/2)*(4*e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+4*e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+(a^2*f^4+2
*a*d^2*f^2+d^4)/f^2/d^2)^(1/2))/(x+1/2*(-a*f^2+d^2)/d/e))*a^2+1/2*f/e/d^2/((a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^
(1/2)*ln((1/2*(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2+e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/2*((a^2*f^4+2*a
*d^2*f^2+d^4)/f^2/d^2)^(1/2)*(4*e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+4*e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d
/e)+(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2))/(x+1/2*(-a*f^2+d^2)/d/e))*a+d*f^3/e^2/(a^2*f^4+2*a*d^2*f^2+d^4)/
(x-1/2/e/d*a*f^2+1/2*d/e)*(e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/4
*(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(3/2)-3/4/d*f^3/(a^2*f^4+2*a*d^2*f^2+d^4)*ln((1/2*e*(a*f^2-d^2)/f^2/d+e^2*
(x+1/2*(-a*f^2+d^2)/d/e)/f^2)/(e^2/f^2)^(1/2)+(e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+e*(a*f^2-d^2)/f^2/d*(x+1/2*(
-a*f^2+d^2)/d/e)+1/4*(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2))/(e^2/f^2)^(1/2)*a^2-3/4*d*f/(a^2*f^4+2*a*d^2*f^
2+d^4)*ln((1/2*e*(a*f^2-d^2)/f^2/d+e^2*(x+1/2*(-a*f^2+d^2)/d/e)/f^2)/(e^2/f^2)^(1/2)+(e^2*(x+1/2*(-a*f^2+d^2)/
d/e)^2/f^2+e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/4*(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2))/(e^2/f^2
)^(1/2)*a-1/4*d^4/f/e/(a^2*f^4+2*a*d^2*f^2+d^4)/((a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2)*ln((1/2*(a^2*f^4+2*a
*d^2*f^2+d^4)/f^2/d^2+e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/2*((a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/
2)*(4*e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+4*e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+(a^2*f^4+2*a*d^2*f^2+d
^4)/f^2/d^2)^(1/2))/(x+1/2*(-a*f^2+d^2)/d/e))+1/4*f^7/e/d^4/(a^2*f^4+2*a*d^2*f^2+d^4)/((a^2*f^4+2*a*d^2*f^2+d^
4)/f^2/d^2)^(1/2)*ln((1/2*(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2+e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/2*(
(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2)*(4*e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+4*e*(a*f^2-d^2)/f^2/d*(x+1/2*(-
a*f^2+d^2)/d/e)+(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2))/(x+1/2*(-a*f^2+d^2)/d/e))*a^4-1/4*f/e/d^2*(4*e^2*(x+
1/2*(-a*f^2+d^2)/d/e)^2/f^2+4*e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^
(1/2)+1/4/f/d*ln((1/2*e*(a*f^2-d^2)/f^2/d+e^2*(x+1/2*(-a*f^2+d^2)/d/e)/f^2)/(e^2/f^2)^(1/2)+(e^2*(x+1/2*(-a*f^
2+d^2)/d/e)^2/f^2+e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/4*(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2))/(
e^2/f^2)^(1/2)+1/4/f/e/((a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2)*ln((1/2*(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2+e*(
a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/2*((a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2)*(4*e^2*(x+1/2*(-a*f^2+
d^2)/d/e)^2/f^2+4*e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2))/(x+1/
2*(-a*f^2+d^2)/d/e))+1/2/d^2*x+1/2/d^2*f^5/e/(a^2*f^4+2*a*d^2*f^2+d^4)/((a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/
2)*ln((1/2*(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2+e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/2*((a^2*f^4+2*a*d^
2*f^2+d^4)/f^2/d^2)^(1/2)*(4*e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+4*e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)
+(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2))/(x+1/2*(-a*f^2+d^2)/d/e))*a^3-1/2*d^2*f/e/(a^2*f^4+2*a*d^2*f^2+d^4)
/((a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2)*ln((1/2*(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2+e*(a*f^2-d^2)/f^2/d*(x+1/
2*(-a*f^2+d^2)/d/e)+1/2*((a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2)*(4*e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+4*e*(a
*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2))/(x+1/2*(-a*f^2+d^2)/d/e))*a
+f^5/e^2/d/(a^2*f^4+2*a*d^2*f^2+d^4)/(x-1/2/e/d*a*f^2+1/2*d/e)*(e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+e*(a*f^2-d^
2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/4*(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(3/2)*a-1/2*a*f^2/(-a*f^2+2*d*e*x+d^2
)/d/e+1/4*d^2*f/e/(a^2*f^4+2*a*d^2*f^2+d^4)*(4*e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+4*e*(a*f^2-d^2)/f^2/d*(x+1/2
*(-a*f^2+d^2)/d/e)+(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2)-1/4*d^3/f/(a^2*f^4+2*a*d^2*f^2+d^4)*ln((1/2*e*(a*f
^2-d^2)/f^2/d+e^2*(x+1/2*(-a*f^2+d^2)/d/e)/f^2)/(e^2/f^2)^(1/2)+(e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+e*(a*f^2-d
^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/4*(a^2*f^4+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2))/(e^2/f^2)^(1/2)-d*f/(a^2*f^4+
2*a*d^2*f^2+d^4)*(e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/4*(a^2*f^4
+2*a*d^2*f^2+d^4)/f^2/d^2)^(1/2)*x-1/4*f/d^3*ln((1/2*e*(a*f^2-d^2)/f^2/d+e^2*(x+1/2*(-a*f^2+d^2)/d/e)/f^2)/(e^
2/f^2)^(1/2)+(e^2*(x+1/2*(-a*f^2+d^2)/d/e)^2/f^2+e*(a*f^2-d^2)/f^2/d*(x+1/2*(-a*f^2+d^2)/d/e)+1/4*(a^2*f^4+2*a
*d^2*f^2+d^4)/f^2/d^2)^(1/2))/(e^2/f^2)^(1/2)*a-1/4/e/d^3/(-a*f^2+2*d*e*x+d^2)*a^2*f^4+1/2/e/d^3*ln(-a*f^2+2*d
*e*x+d^2)*a*f^2-1/4*d/(-a*f^2+2*d*e*x+d^2)/e

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (e x + \sqrt{\frac{e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + sqrt(e^2*x^2/f^2 + a)*f + d)^(-2), x)

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Fricas [B]  time = 1.12494, size = 579, normalized size = 3.83 \begin{align*} \frac{a^{2} f^{4} - 2 \, d^{2} e^{2} x^{2} + a d^{2} f^{2} - 2 \, d^{3} e x +{\left (a^{2} f^{4} - 2 \, a d e f^{2} x - a d^{2} f^{2}\right )} \log \left (-a e f^{2} x + 2 \, d e^{2} x^{2} + a d f^{2} +{\left (a f^{3} - 2 \, d e f x\right )} \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) +{\left (a^{2} f^{4} - 2 \, a d e f^{2} x - a d^{2} f^{2}\right )} \log \left (-a f^{2} + 2 \, d e x + d^{2}\right ) -{\left (a^{2} f^{4} - 2 \, a d e f^{2} x - a d^{2} f^{2}\right )} \log \left (-e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} - d\right ) - 2 \,{\left (a d f^{3} - d^{2} e f x\right )} \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}}}{2 \,{\left (a d^{3} e f^{2} - 2 \, d^{4} e^{2} x - d^{5} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2*f^4 - 2*d^2*e^2*x^2 + a*d^2*f^2 - 2*d^3*e*x + (a^2*f^4 - 2*a*d*e*f^2*x - a*d^2*f^2)*log(-a*e*f^2*x +
2*d*e^2*x^2 + a*d*f^2 + (a*f^3 - 2*d*e*f*x)*sqrt((e^2*x^2 + a*f^2)/f^2)) + (a^2*f^4 - 2*a*d*e*f^2*x - a*d^2*f^
2)*log(-a*f^2 + 2*d*e*x + d^2) - (a^2*f^4 - 2*a*d*e*f^2*x - a*d^2*f^2)*log(-e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2
) - d) - 2*(a*d*f^3 - d^2*e*f*x)*sqrt((e^2*x^2 + a*f^2)/f^2))/(a*d^3*e*f^2 - 2*d^4*e^2*x - d^5*e)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d + e x + f \sqrt{a + \frac{e^{2} x^{2}}{f^{2}}}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x + f*sqrt(a + e**2*x**2/f**2))**(-2), x)

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Giac [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(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^2,x, algorithm="giac")

[Out]

Timed out