3.476 \(\int \frac{1}{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}} \, dx\)
Optimal. Leaf size=215 \[ -\frac{f^2 \left (4 a e^2-b^2 f^2\right )}{2 e \left (2 d e-b f^2\right ) \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}-\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \log \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}{2 e \left (2 d e-b f^2\right )^2}+\frac{2 \left (a e f^2-b d f^2+d^2 e\right ) \log \left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )}{\left (2 d e-b f^2\right )^2} \]
[Out]
-(f^2*(4*a*e^2 - b^2*f^2))/(2*e*(2*d*e - b*f^2)*(b*f^2 + 2*e*(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x))/f^2]))) + (
2*(d^2*e - b*d*f^2 + a*e*f^2)*Log[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]])/(2*d*e - b*f^2)^2 - (f^2*(4*a*e^
2 - b^2*f^2)*Log[b*f^2 + 2*e*(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x))/f^2])])/(2*e*(2*d*e - b*f^2)^2)
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Rubi [A] time = 0.192901, antiderivative size = 215, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used =
{2116, 893} \[ -\frac{f^2 \left (4 a e^2-b^2 f^2\right )}{2 e \left (2 d e-b f^2\right ) \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}-\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \log \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}{2 e \left (2 d e-b f^2\right )^2}+\frac{2 \left (a e f^2-b d f^2+d^2 e\right ) \log \left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )}{\left (2 d e-b f^2\right )^2} \]
Antiderivative was successfully verified.
[In]
Int[(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(-1),x]
[Out]
-(f^2*(4*a*e^2 - b^2*f^2))/(2*e*(2*d*e - b*f^2)*(b*f^2 + 2*e*(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x))/f^2]))) + (
2*(d^2*e - b*d*f^2 + a*e*f^2)*Log[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]])/(2*d*e - b*f^2)^2 - (f^2*(4*a*e^
2 - b^2*f^2)*Log[b*f^2 + 2*e*(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x))/f^2])])/(2*e*(2*d*e - b*f^2)^2)
Rule 2116
Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol]
:> Dist[2, Subst[Int[((g + h*x^n)^p*(d^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e*x^2))/(-2*d*e + b*f^2 +
2*e*x)^2, x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, 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}{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}} \, dx &=2 \operatorname{Subst}\left (\int \frac{d^2 e-(b d-a e) f^2-\left (2 d e-b f^2\right ) x+e x^2}{x \left (-2 d e+b f^2+2 e x\right )^2} \, dx,x,d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )\\ &=2 \operatorname{Subst}\left (\int \left (\frac{d^2 e-b d f^2+a e f^2}{\left (2 d e-b f^2\right )^2 x}+\frac{4 a e^2 f^2-b^2 f^4}{2 \left (2 d e-b f^2\right ) \left (2 d e-b f^2-2 e x\right )^2}+\frac{4 a e^2 f^2-b^2 f^4}{2 \left (2 d e-b f^2\right )^2 \left (2 d e-b f^2-2 e x\right )}\right ) \, dx,x,d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )\\ &=-\frac{f^2 \left (4 a e^2-b^2 f^2\right )}{2 e \left (2 d e-b f^2\right ) \left (b f^2+2 e \left (e x+f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}+\frac{2 \left (d^2 e-b d f^2+a e f^2\right ) \log \left (d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )}{\left (2 d e-b f^2\right )^2}-\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \log \left (b f^2+2 e \left (e x+f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}{2 e \left (2 d e-b f^2\right )^2}\\ \end{align*}
Mathematica [A] time = 0.220899, size = 187, normalized size = 0.87 \[ \frac{-\frac{f^2 \left (b^2 f^2-4 a e^2\right ) \left (b f^2-2 d e\right )}{e \left (2 e \left (f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+e x\right )+b f^2\right )}+\frac{f^2 \left (b^2 f^2-4 a e^2\right ) \log \left (-2 e \left (f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+e x\right )-b f^2\right )}{e}+4 \left (a e f^2-b d f^2+d^2 e\right ) \log \left (f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x\right )}{2 \left (b f^2-2 d e\right )^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(-1),x]
[Out]
(-((f^2*(-2*d*e + b*f^2)*(-4*a*e^2 + b^2*f^2))/(e*(b*f^2 + 2*e*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])))) + 4*
(d^2*e - b*d*f^2 + a*e*f^2)*Log[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]] + (f^2*(-4*a*e^2 + b^2*f^2)*Log[-(b
*f^2) - 2*e*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])])/e)/(2*(-2*d*e + b*f^2)^2)
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Maple [B] time = 0.056, size = 4918, normalized size = 22.9 \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+b*x+e^2*x^2/f^2)^(1/2)),x)
[Out]
f/(b*f^2-2*d*e)*(e^2*(x+(a*f^2-d^2)/(b*f^2-2*d*e))^2/f^2-(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f
^2-2*d*e)*(x+(a*f^2-d^2)/(b*f^2-2*d*e))+(a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d
^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2)+1/2*f^3/(b*f^2-2*d*e)^2*ln((-1/2*(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^
2)/f^2/(b*f^2-2*d*e)+e^2*(x+(a*f^2-d^2)/(b*f^2-2*d*e))/f^2)/(e^2/f^2)^(1/2)+(e^2*(x+(a*f^2-d^2)/(b*f^2-2*d*e))
^2/f^2-(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)*(x+(a*f^2-d^2)/(b*f^2-2*d*e))+(a^2*e^2*f
^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2))/(e^2/f^2)^(1/2
)*b^2-f/(b*f^2-2*d*e)^2*ln((-1/2*(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)+e^2*(x+(a*f^2-
d^2)/(b*f^2-2*d*e))/f^2)/(e^2/f^2)^(1/2)+(e^2*(x+(a*f^2-d^2)/(b*f^2-2*d*e))^2/f^2-(-b^2*f^4+2*a*e^2*f^2+2*b*d*
e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)*(x+(a*f^2-d^2)/(b*f^2-2*d*e))+(a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^
2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2))/(e^2/f^2)^(1/2)*a*e^2-f/(b*f^2-2*d*e)^2*ln((-1/2*
(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)+e^2*(x+(a*f^2-d^2)/(b*f^2-2*d*e))/f^2)/(e^2/f^2
)^(1/2)+(e^2*(x+(a*f^2-d^2)/(b*f^2-2*d*e))^2/f^2-(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e
)*(x+(a*f^2-d^2)/(b*f^2-2*d*e))+(a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/
f^2/(b*f^2-2*d*e)^2)^(1/2))/(e^2/f^2)^(1/2)*b*d*e+1/f/(b*f^2-2*d*e)^2*ln((-1/2*(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f
^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)+e^2*(x+(a*f^2-d^2)/(b*f^2-2*d*e))/f^2)/(e^2/f^2)^(1/2)+(e^2*(x+(a*f^2-d^2)/(b*
f^2-2*d*e))^2/f^2-(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)*(x+(a*f^2-d^2)/(b*f^2-2*d*e))
+(a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2))/(e^
2/f^2)^(1/2)*e^2*d^2-f^3/(b*f^2-2*d*e)^3/((a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2
+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2)*ln((2*(a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^
2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2-(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)*(x+(a*f^2-d^2)/(
b*f^2-2*d*e))+2*((a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*
e)^2)^(1/2)*(e^2*(x+(a*f^2-d^2)/(b*f^2-2*d*e))^2/f^2-(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2
*d*e)*(x+(a*f^2-d^2)/(b*f^2-2*d*e))+(a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e
^2)/f^2/(b*f^2-2*d*e)^2)^(1/2))/(x+(a*f^2-d^2)/(b*f^2-2*d*e)))*a^2*e^2+2*f^3/(b*f^2-2*d*e)^3/((a^2*e^2*f^4-2*a
*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2)*ln((2*(a^2*e^2*f^4-2*
a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2-(-b^2*f^4+2*a*e^2*f^2+2*b*d
*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)*(x+(a*f^2-d^2)/(b*f^2-2*d*e))+2*((a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*
a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2)*(e^2*(x+(a*f^2-d^2)/(b*f^2-2*d*e))^2/f^2-(-b^2
*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)*(x+(a*f^2-d^2)/(b*f^2-2*d*e))+(a^2*e^2*f^4-2*a*b*d*e
*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2))/(x+(a*f^2-d^2)/(b*f^2-2*d*
e)))*a*b*d*e-f^3/(b*f^2-2*d*e)^3/((a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2
)/f^2/(b*f^2-2*d*e)^2)^(1/2)*ln((2*(a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^
2)/f^2/(b*f^2-2*d*e)^2-(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)*(x+(a*f^2-d^2)/(b*f^2-2*
d*e))+2*((a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1
/2)*(e^2*(x+(a*f^2-d^2)/(b*f^2-2*d*e))^2/f^2-(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)*(x
+(a*f^2-d^2)/(b*f^2-2*d*e))+(a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/
(b*f^2-2*d*e)^2)^(1/2))/(x+(a*f^2-d^2)/(b*f^2-2*d*e)))*b^2*d^2-2*f/(b*f^2-2*d*e)^3/((a^2*e^2*f^4-2*a*b*d*e*f^4
+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2)*ln((2*(a^2*e^2*f^4-2*a*b*d*e*f^
4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2-(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d
^2*e^2)/f^2/(b*f^2-2*d*e)*(x+(a*f^2-d^2)/(b*f^2-2*d*e))+2*((a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*
f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2)*(e^2*(x+(a*f^2-d^2)/(b*f^2-2*d*e))^2/f^2-(-b^2*f^4+2*a*e
^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)*(x+(a*f^2-d^2)/(b*f^2-2*d*e))+(a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d
^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2))/(x+(a*f^2-d^2)/(b*f^2-2*d*e)))*a*d^2
*e^2+2*f/(b*f^2-2*d*e)^3/((a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b
*f^2-2*d*e)^2)^(1/2)*ln((2*(a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(
b*f^2-2*d*e)^2-(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)*(x+(a*f^2-d^2)/(b*f^2-2*d*e))+2*
((a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2)*(e^2
*(x+(a*f^2-d^2)/(b*f^2-2*d*e))^2/f^2-(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)*(x+(a*f^2-
d^2)/(b*f^2-2*d*e))+(a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2
*d*e)^2)^(1/2))/(x+(a*f^2-d^2)/(b*f^2-2*d*e)))*b*d^3*e-1/f/(b*f^2-2*d*e)^3/((a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2
*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2)*ln((2*(a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^
2*f^4+2*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2-(-b^2*f^4+2*a*e^2*f^2+2*b*d*e*f^2-2*d^2*e^2)/
f^2/(b*f^2-2*d*e)*(x+(a*f^2-d^2)/(b*f^2-2*d*e))+2*((a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2*a*d^2*e^2*f^2-2*b*
d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2)*(e^2*(x+(a*f^2-d^2)/(b*f^2-2*d*e))^2/f^2-(-b^2*f^4+2*a*e^2*f^2+2
*b*d*e*f^2-2*d^2*e^2)/f^2/(b*f^2-2*d*e)*(x+(a*f^2-d^2)/(b*f^2-2*d*e))+(a^2*e^2*f^4-2*a*b*d*e*f^4+b^2*d^2*f^4+2
*a*d^2*e^2*f^2-2*b*d^3*e*f^2+d^4*e^2)/f^2/(b*f^2-2*d*e)^2)^(1/2))/(x+(a*f^2-d^2)/(b*f^2-2*d*e)))*d^4*e^2-d*ln(
(b*f^2-2*d*e)*x+a*f^2-d^2)/(b*f^2-2*d*e)-e/(b*f^2-2*d*e)*x+e/(b*f^2-2*d*e)^2*ln(b*f^2*x+a*f^2-2*d*e*x-d^2)*a*f
^2-e/(b*f^2-2*d*e)^2*ln(b*f^2*x+a*f^2-2*d*e*x-d^2)*d^2
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{e x + \sqrt{b x + \frac{e^{2} x^{2}}{f^{2}} + a} f + d}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2)),x, algorithm="maxima")
[Out]
integrate(1/(e*x + sqrt(b*x + e^2*x^2/f^2 + a)*f + d), x)
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Fricas [A] time = 20.7166, size = 756, normalized size = 3.52 \begin{align*} -\frac{2 \,{\left (b e^{2} f^{2} - 2 \, d e^{3}\right )} x - 2 \,{\left (d^{2} e^{2} -{\left (b d e - a e^{2}\right )} f^{2}\right )} \log \left ({\left (b d - 2 \, a e\right )} f^{2} -{\left (b e f^{2} - 2 \, d e^{2}\right )} x +{\left (b f^{3} - 2 \, d e f\right )} \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) - 2 \,{\left (d^{2} e^{2} -{\left (b d e - a e^{2}\right )} f^{2}\right )} \log \left (a f^{2} - d^{2} +{\left (b f^{2} - 2 \, d e\right )} x\right ) +{\left (b^{2} f^{4} + 2 \, d^{2} e^{2} - 2 \,{\left (b d e + a e^{2}\right )} f^{2}\right )} \log \left (-b f^{2} - 2 \, e^{2} x + 2 \, e f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) + 2 \,{\left (d^{2} e^{2} -{\left (b d e - a e^{2}\right )} f^{2}\right )} \log \left (-e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} - d\right ) - 2 \,{\left (b e f^{3} - 2 \, d e^{2} f\right )} \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}}{2 \,{\left (b^{2} e f^{4} - 4 \, b d e^{2} f^{2} + 4 \, d^{2} e^{3}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2)),x, algorithm="fricas")
[Out]
-1/2*(2*(b*e^2*f^2 - 2*d*e^3)*x - 2*(d^2*e^2 - (b*d*e - a*e^2)*f^2)*log((b*d - 2*a*e)*f^2 - (b*e*f^2 - 2*d*e^2
)*x + (b*f^3 - 2*d*e*f)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2)) - 2*(d^2*e^2 - (b*d*e - a*e^2)*f^2)*log(a*f^2 -
d^2 + (b*f^2 - 2*d*e)*x) + (b^2*f^4 + 2*d^2*e^2 - 2*(b*d*e + a*e^2)*f^2)*log(-b*f^2 - 2*e^2*x + 2*e*f*sqrt((b
*f^2*x + e^2*x^2 + a*f^2)/f^2)) + 2*(d^2*e^2 - (b*d*e - a*e^2)*f^2)*log(-e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f
^2)/f^2) - d) - 2*(b*e*f^3 - 2*d*e^2*f)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2))/(b^2*e*f^4 - 4*b*d*e^2*f^2 + 4*
d^2*e^3)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{d + e x + f \sqrt{a + b x + \frac{e^{2} x^{2}}{f^{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d+e*x+f*(a+b*x+e**2*x**2/f**2)**(1/2)),x)
[Out]
Integral(1/(d + e*x + f*sqrt(a + b*x + e**2*x**2/f**2)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, +\infty , 1\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2)),x, algorithm="giac")
[Out]
[undef, +Infinity, 1]