\(\int \frac {(d+e x)^5}{(d^2-e^2 x^2)^{5/2}} \, dx\) [837]

Optimal result

Integrand size = 24, antiderivative size = 108 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {10 (d+e x)^2}{3 e \sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{e}+\frac {5 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \]

[Out]

2/3*(e*x+d)^4/e/(-e^2*x^2+d^2)^(3/2)+5*d*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e-10/3*(e*x+d)^2/e/(-e^2*x^2+d^2)^(1
/2)-5*(-e^2*x^2+d^2)^(1/2)/e

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 108, 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 = {683, 655, 223, 209} \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {5 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}+\frac {2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {10 (d+e x)^2}{3 e \sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{e} \]

[In]

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

[Out]

(2*(d + e*x)^4)/(3*e*(d^2 - e^2*x^2)^(3/2)) - (10*(d + e*x)^2)/(3*e*Sqrt[d^2 - e^2*x^2]) - (5*Sqrt[d^2 - e^2*x
^2])/e + (5*d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^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 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 683

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*(p + 1))), x] - Dist[e^2*((m + p)/(c*(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] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \frac {2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {5}{3} \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx \\ & = \frac {2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {10 (d+e x)^2}{3 e \sqrt {d^2-e^2 x^2}}+5 \int \frac {d+e x}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = \frac {2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {10 (d+e x)^2}{3 e \sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{e}+(5 d) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = \frac {2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {10 (d+e x)^2}{3 e \sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{e}+(5 d) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right ) \\ & = \frac {2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {10 (d+e x)^2}{3 e \sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{e}+\frac {5 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.87 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\left (-23 d^2+34 d e x-3 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{3 e (-d+e x)^2}-\frac {5 d \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{\sqrt {-e^2}} \]

[In]

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

[Out]

((-23*d^2 + 34*d*e*x - 3*e^2*x^2)*Sqrt[d^2 - e^2*x^2])/(3*e*(-d + e*x)^2) - (5*d*Log[-(Sqrt[-e^2]*x) + Sqrt[d^
2 - e^2*x^2]])/Sqrt[-e^2]

Maple [A] (verified)

Time = 2.34 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.34

method	result	size
risch	\(-\frac {\sqrt {-x^{2} e^{2}+d^{2}}}{e}+\frac {5 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}+\frac {8 d^{2} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 \left (x -\frac {d}{e}\right ) d e}}{3 e^{3} \left (x -\frac {d}{e}\right )^{2}}+\frac {28 d \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 \left (x -\frac {d}{e}\right ) d e}}{3 e^{2} \left (x -\frac {d}{e}\right )}\)	\(145\)
default	\(d^{5} \left (\frac {x}{3 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-x^{2} e^{2}+d^{2}}}\right )+e^{5} \left (-\frac {x^{4}}{e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {4 d^{2} \left (\frac {x^{2}}{e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {2 d^{2}}{3 e^{4} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}\right )}{e^{2}}\right )+5 d \,e^{4} \left (\frac {x^{3}}{3 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-x^{2} e^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}\right )+\frac {5 d^{4}}{3 e \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+10 d^{2} e^{3} \left (\frac {x^{2}}{e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {2 d^{2}}{3 e^{4} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}\right )+10 d^{3} e^{2} \left (\frac {x}{2 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d^{2} \left (\frac {x}{3 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-x^{2} e^{2}+d^{2}}}\right )}{2 e^{2}}\right )\)	\(364\)

[In]

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

[Out]

-(-e^2*x^2+d^2)^(1/2)/e+5*d/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+8/3*d^2/e^3/(x-d/e)^2*(-(x-
d/e)^2*e^2-2*(x-d/e)*d*e)^(1/2)+28/3*d/e^2/(x-d/e)*(-(x-d/e)^2*e^2-2*(x-d/e)*d*e)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.19 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {23 \, d e^{2} x^{2} - 46 \, d^{2} e x + 23 \, d^{3} + 30 \, {\left (d e^{2} x^{2} - 2 \, d^{2} e x + d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (3 \, e^{2} x^{2} - 34 \, d e x + 23 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, {\left (e^{3} x^{2} - 2 \, d e^{2} x + d^{2} e\right )}} \]

[In]

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

[Out]

-1/3*(23*d*e^2*x^2 - 46*d^2*e*x + 23*d^3 + 30*(d*e^2*x^2 - 2*d^2*e*x + d^3)*arctan(-(d - sqrt(-e^2*x^2 + d^2))
/(e*x)) + (3*e^2*x^2 - 34*d*e*x + 23*d^2)*sqrt(-e^2*x^2 + d^2))/(e^3*x^2 - 2*d*e^2*x + d^2*e)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.67 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {5}{3} \, d e^{4} x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )} - \frac {e^{3} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {14 \, d^{2} e x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {11 \, d^{3} x}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {23 \, d^{4}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e} - \frac {13 \, d x}{3 \, \sqrt {-e^{2} x^{2} + d^{2}}} + \frac {5 \, d \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}}} \]

[In]

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

[Out]

5/3*d*e^4*x*(3*x^2/((-e^2*x^2 + d^2)^(3/2)*e^2) - 2*d^2/((-e^2*x^2 + d^2)^(3/2)*e^4)) - e^3*x^4/(-e^2*x^2 + d^
2)^(3/2) + 14*d^2*e*x^2/(-e^2*x^2 + d^2)^(3/2) + 11/3*d^3*x/(-e^2*x^2 + d^2)^(3/2) - 23/3*d^4/((-e^2*x^2 + d^2
)^(3/2)*e) - 13/3*d*x/sqrt(-e^2*x^2 + d^2) + 5*d*arcsin(e^2*x/(d*sqrt(e^2)))/sqrt(e^2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.31 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {5 \, d \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{{\left | e \right |}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e} - \frac {8 \, {\left (5 \, d - \frac {12 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d}{e^{2} x} + \frac {3 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d}{e^{4} x^{2}}\right )}}{3 \, {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{3} {\left | e \right |}} \]

[In]

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

[Out]

5*d*arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) - sqrt(-e^2*x^2 + d^2)/e - 8/3*(5*d - 12*(d*e + sqrt(-e^2*x^2 + d^2)*ab
s(e))*d/(e^2*x) + 3*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*d/(e^4*x^2))/(((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(
e^2*x) - 1)^3*abs(e))

Mupad [F(-1)]

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

[In]

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

[Out]

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