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

Optimal. Leaf size=117 \[ -\frac {a f^2 \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}{2 d^2 e}+\frac {\left (\frac {a f^2}{d^2}+1\right ) \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )}{2 e}-\frac {a f^2}{2 d e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )} \]

[Out]

-1/2*a*f^2*ln(e*x+f*(a+e^2*x^2/f^2)^(1/2))/d^2/e+1/2*(1+a*f^2/d^2)*ln(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))/e-1/2*a*f
^2/d/e/(e*x+f*(a+e^2*x^2/f^2)^(1/2))

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Rubi [A]  time = 0.09, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2117, 893} \[ -\frac {a f^2 \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}{2 d^2 e}+\frac {\left (\frac {a f^2}{d^2}+1\right ) \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )}{2 e}-\frac {a f^2}{2 d e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))

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]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 109, normalized size = 0.93 \[ \frac {-\frac {a f^2 \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}{d^2}+\left (\frac {a f^2}{d^2}+1\right ) \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )+\frac {a f^2}{d \left (f \left (-\sqrt {a+\frac {e^2 x^2}{f^2}}\right )-e x\right )}}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 187, normalized size = 1.60 \[ \frac {2 \, d e x - 2 \, d f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + {\left (a f^{2} + d^{2}\right )} \log \left (a f^{2} - d e x + d f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) + {\left (a f^{2} + d^{2}\right )} \log \left (-a f^{2} + 2 \, d e x + d^{2}\right ) - {\left (a f^{2} + d^{2}\right )} \log \left (-e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} - d\right ) + {\left (a f^{2} - d^{2}\right )} \log \left (-e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right )}{4 \, d^{2} e} \]

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)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \left [\mathit {undef}, +\infty , 1\right ] \]

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)),x, algorithm="giac")

[Out]

[undef, +Infinity, 1]

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maple [B]  time = 0.04, size = 1325, normalized size = 11.32 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a} f + d}\,{d x} \]

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)),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{d+e\,x+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}}} \,d x \]

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)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{d + e x + f \sqrt {a + \frac {e^{2} x^{2}}{f^{2}}}}\, dx \]

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)),x)

[Out]

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

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