\(\int \frac {1}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx\) [833]

Optimal result

Integrand size = 24, antiderivative size = 100 \[ \int \frac {1}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2}}{5 d e (d+e x)^3}-\frac {2 \sqrt {d^2-e^2 x^2}}{15 d^2 e (d+e x)^2}-\frac {2 \sqrt {d^2-e^2 x^2}}{15 d^3 e (d+e x)} \]

[Out]

-1/5*(-e^2*x^2+d^2)^(1/2)/d/e/(e*x+d)^3-2/15*(-e^2*x^2+d^2)^(1/2)/d^2/e/(e*x+d)^2-2/15*(-e^2*x^2+d^2)^(1/2)/d^
3/e/(e*x+d)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {673, 665} \[ \int \frac {1}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {2 \sqrt {d^2-e^2 x^2}}{15 d^2 e (d+e x)^2}-\frac {\sqrt {d^2-e^2 x^2}}{5 d e (d+e x)^3}-\frac {2 \sqrt {d^2-e^2 x^2}}{15 d^3 e (d+e x)} \]

[In]

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

[Out]

-1/5*Sqrt[d^2 - e^2*x^2]/(d*e*(d + e*x)^3) - (2*Sqrt[d^2 - e^2*x^2])/(15*d^2*e*(d + e*x)^2) - (2*Sqrt[d^2 - e^
2*x^2])/(15*d^3*e*(d + e*x))

Rule 665

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*(p + 1))), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 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 {\sqrt {d^2-e^2 x^2}}{5 d e (d+e x)^3}+\frac {2 \int \frac {1}{(d+e x)^2 \sqrt {d^2-e^2 x^2}} \, dx}{5 d} \\ & = -\frac {\sqrt {d^2-e^2 x^2}}{5 d e (d+e x)^3}-\frac {2 \sqrt {d^2-e^2 x^2}}{15 d^2 e (d+e x)^2}+\frac {2 \int \frac {1}{(d+e x) \sqrt {d^2-e^2 x^2}} \, dx}{15 d^2} \\ & = -\frac {\sqrt {d^2-e^2 x^2}}{5 d e (d+e x)^3}-\frac {2 \sqrt {d^2-e^2 x^2}}{15 d^2 e (d+e x)^2}-\frac {2 \sqrt {d^2-e^2 x^2}}{15 d^3 e (d+e x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.52 \[ \int \frac {1}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {\left (-7 d^2-6 d e x-2 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{15 d^3 e (d+e x)^3} \]

[In]

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

[Out]

((-7*d^2 - 6*d*e*x - 2*e^2*x^2)*Sqrt[d^2 - e^2*x^2])/(15*d^3*e*(d + e*x)^3)

Maple [A] (verified)

Time = 2.66 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.49

method	result	size
trager	\(-\frac {\left (2 x^{2} e^{2}+6 d e x +7 d^{2}\right ) \sqrt {-x^{2} e^{2}+d^{2}}}{15 d^{3} \left (e x +d \right )^{3} e}\)	\(49\)
gosper	\(-\frac {\left (-e x +d \right ) \left (2 x^{2} e^{2}+6 d e x +7 d^{2}\right )}{15 \left (e x +d \right )^{2} d^{3} e \sqrt {-x^{2} e^{2}+d^{2}}}\)	\(55\)
default	\(\frac {-\frac {\sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (-\frac {\sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d e \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}{3 d^{2} \left (x +\frac {d}{e}\right )}\right )}{5 d}}{e^{3}}\)	\(145\)

[In]

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

[Out]

-1/15*(2*e^2*x^2+6*d*e*x+7*d^2)/d^3/(e*x+d)^3/e*(-e^2*x^2+d^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {7 \, e^{3} x^{3} + 21 \, d e^{2} x^{2} + 21 \, d^{2} e x + 7 \, d^{3} + {\left (2 \, e^{2} x^{2} + 6 \, d e x + 7 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{3} e^{4} x^{3} + 3 \, d^{4} e^{3} x^{2} + 3 \, d^{5} e^{2} x + d^{6} e\right )}} \]

[In]

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

[Out]

-1/15*(7*e^3*x^3 + 21*d*e^2*x^2 + 21*d^2*e*x + 7*d^3 + (2*e^2*x^2 + 6*d*e*x + 7*d^2)*sqrt(-e^2*x^2 + d^2))/(d^
3*e^4*x^3 + 3*d^4*e^3*x^2 + 3*d^5*e^2*x + d^6*e)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d e^{4} x^{3} + 3 \, d^{2} e^{3} x^{2} + 3 \, d^{3} e^{2} x + d^{4} e\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{2} e^{3} x^{2} + 2 \, d^{3} e^{2} x + d^{4} e\right )}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{3} e^{2} x + d^{4} e\right )}} \]

[In]

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

[Out]

-1/5*sqrt(-e^2*x^2 + d^2)/(d*e^4*x^3 + 3*d^2*e^3*x^2 + 3*d^3*e^2*x + d^4*e) - 2/15*sqrt(-e^2*x^2 + d^2)/(d^2*e
^3*x^2 + 2*d^3*e^2*x + d^4*e) - 2/15*sqrt(-e^2*x^2 + d^2)/(d^3*e^2*x + d^4*e)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.65 \[ \int \frac {1}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=\frac {2 \, {\left (\frac {20 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} + \frac {40 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + \frac {30 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{6} x^{3}} + \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{8} x^{4}} + 7\right )}}{15 \, d^{3} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]

[In]

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

[Out]

2/15*(20*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 40*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2/(e^4*x^2) + 30
*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3/(e^6*x^3) + 15*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^4/(e^8*x^4) + 7)/(d^
3*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) + 1)^5*abs(e))

Mupad [B] (verification not implemented)

Time = 9.51 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.48 \[ \int \frac {1}{(d+e x)^3 \sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2\,x^2}\,\left (7\,d^2+6\,d\,e\,x+2\,e^2\,x^2\right )}{15\,d^3\,e\,{\left (d+e\,x\right )}^3} \]

[In]

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

[Out]

-((d^2 - e^2*x^2)^(1/2)*(7*d^2 + 2*e^2*x^2 + 6*d*e*x))/(15*d^3*e*(d + e*x)^3)