\(\int \frac {1}{(d+e x) (d^2-e^2 x^2)^{3/2}} \, dx\) [132]

Optimal result

Integrand size = 24, antiderivative size = 58 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {2 x}{3 d^3 \sqrt {d^2-e^2 x^2}}-\frac {1}{3 d e (d+e x) \sqrt {d^2-e^2 x^2}} \]

[Out]

2/3*x/d^3/(-e^2*x^2+d^2)^(1/2)-1/3/d/e/(e*x+d)/(-e^2*x^2+d^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {673, 197} \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {2 x}{3 d^3 \sqrt {d^2-e^2 x^2}}-\frac {1}{3 d e (d+e x) \sqrt {d^2-e^2 x^2}} \]

[In]

Int[1/((d + e*x)*(d^2 - e^2*x^2)^(3/2)),x]

[Out]

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

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 673

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-e)*(d + e*x)^m*((a + c*x^2)^(p +
1)/(2*c*d*(m + p + 1))), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^
p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p +
2], 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3 d e (d+e x) \sqrt {d^2-e^2 x^2}}+\frac {2 \int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{3 d} \\ & = \frac {2 x}{3 d^3 \sqrt {d^2-e^2 x^2}}-\frac {1}{3 d e (d+e x) \sqrt {d^2-e^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-d^2+2 d e x+2 e^2 x^2\right )}{3 d^3 e (d-e x) (d+e x)^2} \]

[In]

Integrate[1/((d + e*x)*(d^2 - e^2*x^2)^(3/2)),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-d^2 + 2*d*e*x + 2*e^2*x^2))/(3*d^3*e*(d - e*x)*(d + e*x)^2)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.79

method	result	size
gosper	\(-\frac {\left (-e x +d \right ) \left (-2 e^{2} x^{2}-2 d e x +d^{2}\right )}{3 d^{3} e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}\)	\(46\)
trager	\(-\frac {\left (-2 e^{2} x^{2}-2 d e x +d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{3 d^{3} \left (e x +d \right )^{2} e \left (-e x +d \right )}\)	\(55\)
default	\(\frac {-\frac {1}{3 d e \left (x +\frac {d}{e}\right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e \,d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}}{e}\)	\(104\)

[In]

int(1/(e*x+d)/(-e^2*x^2+d^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*(-e*x+d)*(-2*e^2*x^2-2*d*e*x+d^2)/d^3/e/(-e^2*x^2+d^2)^(3/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (50) = 100\).

Time = 0.34 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.76 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {e^{3} x^{3} + d e^{2} x^{2} - d^{2} e x - d^{3} + {\left (2 \, e^{2} x^{2} + 2 \, d e x - d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, {\left (d^{3} e^{4} x^{3} + d^{4} e^{3} x^{2} - d^{5} e^{2} x - d^{6} e\right )}} \]

[In]

integrate(1/(e*x+d)/(-e^2*x^2+d^2)^(3/2),x, algorithm="fricas")

[Out]

-1/3*(e^3*x^3 + d*e^2*x^2 - d^2*e*x - d^3 + (2*e^2*x^2 + 2*d*e*x - d^2)*sqrt(-e^2*x^2 + d^2))/(d^3*e^4*x^3 + d
^4*e^3*x^2 - d^5*e^2*x - d^6*e)

Sympy [F]

\[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int \frac {1}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}} \left (d + e x\right )}\, dx \]

[In]

integrate(1/(e*x+d)/(-e**2*x**2+d**2)**(3/2),x)

[Out]

Integral(1/((-(-d + e*x)*(d + e*x))**(3/2)*(d + e*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {1}{3 \, {\left (\sqrt {-e^{2} x^{2} + d^{2}} d e^{2} x + \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e\right )}} + \frac {2 \, x}{3 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}} \]

[In]

integrate(1/(e*x+d)/(-e^2*x^2+d^2)^(3/2),x, algorithm="maxima")

[Out]

-1/3/(sqrt(-e^2*x^2 + d^2)*d*e^2*x + sqrt(-e^2*x^2 + d^2)*d^2*e) + 2/3*x/(sqrt(-e^2*x^2 + d^2)*d^3)

Giac [F]

\[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}} \,d x } \]

[In]

integrate(1/(e*x+d)/(-e^2*x^2+d^2)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((-e^2*x^2 + d^2)^(3/2)*(e*x + d)), x)

Mupad [B] (verification not implemented)

Time = 11.69 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (-d^2+2\,d\,e\,x+2\,e^2\,x^2\right )}{3\,d^3\,e\,{\left (d+e\,x\right )}^2\,\left (d-e\,x\right )} \]

[In]

int(1/((d^2 - e^2*x^2)^(3/2)*(d + e*x)),x)

[Out]

((d^2 - e^2*x^2)^(1/2)*(2*e^2*x^2 - d^2 + 2*d*e*x))/(3*d^3*e*(d + e*x)^2*(d - e*x))