\(\int \frac {(d^2-e^2 x^2)^{5/2}}{d+e x} \, dx\) [107]

Optimal result

Integrand size = 24, antiderivative size = 100 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {3}{8} d^3 x \sqrt {d^2-e^2 x^2}+\frac {1}{4} d x \left (d^2-e^2 x^2\right )^{3/2}+\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e}+\frac {3 d^5 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 100, 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 = {679, 201, 223, 209} \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {3 d^5 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e}+\frac {1}{4} d x \left (d^2-e^2 x^2\right )^{3/2}+\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e}+\frac {3}{8} d^3 x \sqrt {d^2-e^2 x^2} \]

[In]

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

[Out]

(3*d^3*x*Sqrt[d^2 - e^2*x^2])/8 + (d*x*(d^2 - e^2*x^2)^(3/2))/4 + (d^2 - e^2*x^2)^(5/2)/(5*e) + (3*d^5*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 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 {\left (d^2-e^2 x^2\right )^{5/2}}{5 e}+d \int \left (d^2-e^2 x^2\right )^{3/2} \, dx \\ & = \frac {1}{4} d x \left (d^2-e^2 x^2\right )^{3/2}+\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e}+\frac {1}{4} \left (3 d^3\right ) \int \sqrt {d^2-e^2 x^2} \, dx \\ & = \frac {3}{8} d^3 x \sqrt {d^2-e^2 x^2}+\frac {1}{4} d x \left (d^2-e^2 x^2\right )^{3/2}+\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e}+\frac {1}{8} \left (3 d^5\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = \frac {3}{8} d^3 x \sqrt {d^2-e^2 x^2}+\frac {1}{4} d x \left (d^2-e^2 x^2\right )^{3/2}+\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e}+\frac {1}{8} \left (3 d^5\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 {3}{8} d^3 x \sqrt {d^2-e^2 x^2}+\frac {1}{4} d x \left (d^2-e^2 x^2\right )^{3/2}+\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 e}+\frac {3 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.11 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (8 d^4+25 d^3 e x-16 d^2 e^2 x^2-10 d e^3 x^3+8 e^4 x^4\right )}{40 e}-\frac {3 d^5 \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{8 \sqrt {-e^2}} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(8*d^4 + 25*d^3*e*x - 16*d^2*e^2*x^2 - 10*d*e^3*x^3 + 8*e^4*x^4))/(40*e) - (3*d^5*Log[-(S
qrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(8*Sqrt[-e^2])

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.94

method	result	size
risch	\(\frac {\left (8 e^{4} x^{4}-10 d \,e^{3} x^{3}-16 d^{2} e^{2} x^{2}+25 d^{3} e x +8 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{40 e}+\frac {3 d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}\)	\(94\)
default	\(\frac {\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{e}\)	\(192\)

[In]

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

[Out]

1/40*(8*e^4*x^4-10*d*e^3*x^3-16*d^2*e^2*x^2+25*d^3*e*x+8*d^4)/e*(-e^2*x^2+d^2)^(1/2)+3/8*d^5/(e^2)^(1/2)*arcta
n((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 = 95, normalized size of antiderivative = 0.95 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=-\frac {30 \, d^{5} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (8 \, e^{4} x^{4} - 10 \, d e^{3} x^{3} - 16 \, d^{2} e^{2} x^{2} + 25 \, d^{3} e x + 8 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{40 \, e} \]

[In]

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

[Out]

-1/40*(30*d^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (8*e^4*x^4 - 10*d*e^3*x^3 - 16*d^2*e^2*x^2 + 25*d^3*
e*x + 8*d^4)*sqrt(-e^2*x^2 + d^2))/e

Sympy [A] (verification not implemented)

Time = 1.30 (sec) , antiderivative size = 313, normalized size of antiderivative = 3.13 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=d^{3} \left (\begin {cases} \frac {d^{2} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {d^{2} - e^{2} x^{2}}}{2} & \text {for}\: e^{2} \neq 0 \\x \sqrt {d^{2}} & \text {otherwise} \end {cases}\right ) - d^{2} e \left (\begin {cases} \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{2}}{3 e^{2}} + \frac {x^{2}}{3}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{2} \sqrt {d^{2}}}{2} & \text {otherwise} \end {cases}\right ) - d e^{2} \left (\begin {cases} \frac {d^{4} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8 e^{2}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{2} x}{8 e^{2}} + \frac {x^{3}}{4}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{3} \sqrt {d^{2}}}{3} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {2 d^{4}}{15 e^{4}} - \frac {d^{2} x^{2}}{15 e^{2}} + \frac {x^{4}}{5}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{4} \sqrt {d^{2}}}{4} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

d**3*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)) - d*
*2*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)) - d
*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)) + e**3*Piecewise((sqrt(d**2 - e**2*x**2)*(-2*d**4/(15*e**4) - d**2*x**2/(15*e
**2) + x**4/5), Ne(e**2, 0)), (x**4*sqrt(d**2)/4, True))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.09 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=-\frac {3 i \, d^{5} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e} + \frac {3}{8} \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3} x + \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4}}{4 \, e} + \frac {1}{4} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d x + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{5 \, e} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {3 \, d^{5} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{8 \, {\left | e \right |}} + \frac {1}{40} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (\frac {8 \, d^{4}}{e} + {\left (25 \, d^{3} - 2 \, {\left (8 \, d^{2} e - {\left (4 \, e^{3} x - 5 \, d e^{2}\right )} x\right )} x\right )} x\right )} \]

[In]

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

[Out]

3/8*d^5*arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) + 1/40*sqrt(-e^2*x^2 + d^2)*(8*d^4/e + (25*d^3 - 2*(8*d^2*e - (4*e^
3*x - 5*d*e^2)*x)*x)*x)

Mupad [F(-1)]

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

[In]

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

[Out]

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