\(\int \frac {(d^2-e^2 x^2)^{5/2}}{(d+e x)^4} \, dx\) [203]

Optimal result

Integrand size = 24, antiderivative size = 113 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {15 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac {15 d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {677, 679, 223, 209} \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {15 d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac {15 d \sqrt {d^2-e^2 x^2}}{2 e} \]

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(d + e*x)^4,x]

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 677

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

Rule 679

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-5 \int \frac {\left (d^2-e^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx \\ & = -\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac {1}{2} (15 d) \int \frac {\sqrt {d^2-e^2 x^2}}{d+e x} \, dx \\ & = -\frac {15 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac {1}{2} \left (15 d^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = -\frac {15 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac {1}{2} \left (15 d^2\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right ) \\ & = -\frac {15 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac {2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac {15 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.79 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-24 d^2-7 d e x+e^2 x^2\right )}{2 e (d+e x)}+\frac {15 d^2 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e} \]

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(d + e*x)^4,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-24*d^2 - 7*d*e*x + e^2*x^2))/(2*e*(d + e*x)) + (15*d^2*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d
^2 - e^2*x^2])])/e

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.94

method	result	size
risch	\(-\frac {\left (-e x +8 d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{2 e}-\frac {15 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}-\frac {8 d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{2} \left (x +\frac {d}{e}\right )}\)	\(106\)
default	\(\frac {-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{d e \left (x +\frac {d}{e}\right )^{4}}-\frac {3 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{d e \left (x +\frac {d}{e}\right )^{3}}+\frac {4 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{3 d e \left (x +\frac {d}{e}\right )^{2}}+\frac {5 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )\right )}{3 d}\right )}{d}\right )}{d}}{e^{4}}\)	\(347\)

[In]

int((-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

-1/2*(-e*x+8*d)/e*(-e^2*x^2+d^2)^(1/2)-15/2*d^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-8*d^2/e
^2/(x+d/e)*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.88 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {24 \, d^{2} e x + 24 \, d^{3} - 30 \, {\left (d^{2} e x + d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (e^{2} x^{2} - 7 \, d e x - 24 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{2 \, {\left (e^{2} x + d e\right )}} \]

[In]

integrate((-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

-1/2*(24*d^2*e*x + 24*d^3 - 30*(d^2*e*x + d^3)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (e^2*x^2 - 7*d*e*x
- 24*d^2)*sqrt(-e^2*x^2 + d^2))/(e^2*x + d*e)

Sympy [F]

\[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \]

[In]

integrate((-e**2*x**2+d**2)**(5/2)/(e*x+d)**4,x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**(5/2)/(d + e*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.19 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {15 \, d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{2 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} - \frac {15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{e^{2} x + d e} \]

[In]

integrate((-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.76 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {15 \, d^{2} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{2 \, {\left | e \right |}} + \frac {1}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (x - \frac {8 \, d}{e}\right )} + \frac {16 \, d^{2}}{{\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )} {\left | e \right |}} \]

[In]

integrate((-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

-15/2*d^2*arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) + 1/2*sqrt(-e^2*x^2 + d^2)*(x - 8*d/e) + 16*d^2/(((d*e + sqrt(-e^
2*x^2 + d^2)*abs(e))/(e^2*x) + 1)*abs(e))

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \]

[In]

int((d^2 - e^2*x^2)^(5/2)/(d + e*x)^4,x)

[Out]

int((d^2 - e^2*x^2)^(5/2)/(d + e*x)^4, x)