\(\int \frac {(d^2-e^2 x^2)^{5/2}}{(d+e x)^2} \, dx\) [162]

Optimal result

Integrand size = 24, antiderivative size = 108 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {5}{8} d^2 x \sqrt {d^2-e^2 x^2}+\frac {5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac {5 d^4 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e} \]

[Out]

5/12*d*(-e^2*x^2+d^2)^(3/2)/e+1/4*(-e*x+d)*(-e^2*x^2+d^2)^(3/2)/e+5/8*d^4*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e+5
/8*d^2*x*(-e^2*x^2+d^2)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {669, 685, 655, 201, 223, 209} \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {5 d^4 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e}+\frac {5}{8} d^2 x \sqrt {d^2-e^2 x^2}+\frac {5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e} \]

[In]

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

[Out]

(5*d^2*x*Sqrt[d^2 - e^2*x^2])/8 + (5*d*(d^2 - e^2*x^2)^(3/2))/(12*e) + ((d - e*x)*(d^2 - e^2*x^2)^(3/2))/(4*e)
+ (5*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e)

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 655

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

Rule 669

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

Rule 685

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

Rubi steps \begin{align*} \text {integral}& = \int (d-e x)^2 \sqrt {d^2-e^2 x^2} \, dx \\ & = \frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac {1}{4} (5 d) \int (d-e x) \sqrt {d^2-e^2 x^2} \, dx \\ & = \frac {5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac {1}{4} \left (5 d^2\right ) \int \sqrt {d^2-e^2 x^2} \, dx \\ & = \frac {5}{8} d^2 x \sqrt {d^2-e^2 x^2}+\frac {5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac {1}{8} \left (5 d^4\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = \frac {5}{8} d^2 x \sqrt {d^2-e^2 x^2}+\frac {5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac {1}{8} \left (5 d^4\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 {5}{8} d^2 x \sqrt {d^2-e^2 x^2}+\frac {5 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac {5 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.85 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (16 d^3+9 d^2 e x-16 d e^2 x^2+6 e^3 x^3\right )-30 d^4 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{24 e} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(16*d^3 + 9*d^2*e*x - 16*d*e^2*x^2 + 6*e^3*x^3) - 30*d^4*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d
^2 - e^2*x^2])])/(24*e)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.77

method	result	size
risch	\(\frac {\left (6 e^{3} x^{3}-16 d \,e^{2} x^{2}+9 d^{2} e x +16 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{24 e}+\frac {5 d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}\)	\(83\)
default	\(\frac {\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{3 d e \left (x +\frac {d}{e}\right )^{2}}+\frac {5 e \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )\right )}{3 d}}{e^{2}}\)	\(244\)

[In]

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

[Out]

1/24*(6*e^3*x^3-16*d*e^2*x^2+9*d^2*e*x+16*d^3)/e*(-e^2*x^2+d^2)^(1/2)+5/8*d^4/(e^2)^(1/2)*arctan((e^2)^(1/2)*x
/(-e^2*x^2+d^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.78 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=-\frac {30 \, d^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (6 \, e^{3} x^{3} - 16 \, d e^{2} x^{2} + 9 \, d^{2} e x + 16 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, e} \]

[In]

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

[Out]

-1/24*(30*d^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (6*e^3*x^3 - 16*d*e^2*x^2 + 9*d^2*e*x + 16*d^3)*sqrt
(-e^2*x^2 + d^2))/e

Sympy [A] (verification not implemented)

Time = 1.66 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.34 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=d^{2} \left (\begin {cases} \frac {d^{2} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {d^{2} - e^{2} x^{2}}}{2} & \text {for}\: e^{2} \neq 0 \\x \sqrt {d^{2}} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{2}}{3 e^{2}} + \frac {x^{2}}{3}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{2} \sqrt {d^{2}}}{2} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} \frac {d^{4} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8 e^{2}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{2} x}{8 e^{2}} + \frac {x^{3}}{4}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{3} \sqrt {d^{2}}}{3} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

d**2*Piecewise((d**2*Piecewise((log(-2*e**2*x + 2*sqrt(-e**2)*sqrt(d**2 - e**2*x**2))/sqrt(-e**2), Ne(d**2, 0)
), (x*log(x)/sqrt(-e**2*x**2), True))/2 + x*sqrt(d**2 - e**2*x**2)/2, Ne(e**2, 0)), (x*sqrt(d**2), True)) - 2*
d*e*Piecewise((sqrt(d**2 - e**2*x**2)*(-d**2/(3*e**2) + x**2/3), Ne(e**2, 0)), (x**2*sqrt(d**2)/2, True)) + e*
*2*Piecewise((d**4*Piecewise((log(-2*e**2*x + 2*sqrt(-e**2)*sqrt(d**2 - e**2*x**2))/sqrt(-e**2), Ne(d**2, 0)),
(x*log(x)/sqrt(-e**2*x**2), True))/(8*e**2) + sqrt(d**2 - e**2*x**2)*(-d**2*x/(8*e**2) + x**3/4), Ne(e**2, 0)
), (x**3*sqrt(d**2)/3, True))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.10 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=-\frac {5 i \, d^{4} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e} + \frac {5}{8} \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{2} x + \frac {5 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3}}{4 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{4 \, {\left (e^{2} x + d e\right )}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{12 \, e} \]

[In]

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

[Out]

-5/8*I*d^4*arcsin(e*x/d + 2)/e + 5/8*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^2*x + 5/4*sqrt(e^2*x^2 + 4*d*e*x + 3*d^
2)*d^3/e + 1/4*(-e^2*x^2 + d^2)^(5/2)/(e^2*x + d*e) + 5/12*(-e^2*x^2 + d^2)^(3/2)*d/e

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.70 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx=-\frac {{\left (240 \, d^{5} e^{5} \arctan \left (\sqrt {\frac {2 \, d}{e x + d} - 1}\right ) \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + \frac {{\left (15 \, d^{5} e^{5} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 73 \, d^{5} e^{5} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 55 \, d^{5} e^{5} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 15 \, d^{5} e^{5} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )\right )} {\left (e x + d\right )}^{4}}{d^{4}}\right )} {\left | e \right |}}{192 \, d e^{7}} \]

[In]

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

[Out]

-1/192*(240*d^5*e^5*arctan(sqrt(2*d/(e*x + d) - 1))*sgn(1/(e*x + d))*sgn(e) + (15*d^5*e^5*(2*d/(e*x + d) - 1)^
(7/2)*sgn(1/(e*x + d))*sgn(e) - 73*d^5*e^5*(2*d/(e*x + d) - 1)^(5/2)*sgn(1/(e*x + d))*sgn(e) - 55*d^5*e^5*(2*d
/(e*x + d) - 1)^(3/2)*sgn(1/(e*x + d))*sgn(e) - 15*d^5*e^5*sqrt(2*d/(e*x + d) - 1)*sgn(1/(e*x + d))*sgn(e))*(e
*x + d)^4/d^4)*abs(e)/(d*e^7)

Mupad [F(-1)]

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

[In]

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

[Out]

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