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

Optimal. Leaf size=108 \[ \frac {2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {10 (d+e x)^2}{3 e \sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{e}+\frac {5 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \]

[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

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Rubi [A]
time = 0.03, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {683, 655, 223, 209} \begin {gather*} \frac {5 d \text {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} \end {gather*}

Antiderivative was successfully verified.

[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*} \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 {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*}

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Mathematica [A]
time = 0.37, size = 94, normalized size = 0.87 \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[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]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(363\) vs. \(2(96)=192\).
time = 0.50, size = 364, normalized size = 3.37

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 160, normalized size = 1.48 \begin {gather*} \frac {5}{3} \, {\left (\frac {3 \, x^{2} e^{\left (-2\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, d^{2} e^{\left (-4\right )}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}}\right )} d x e^{4} - \frac {x^{4} e^{3}}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {14 \, d^{2} x^{2} e}{{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {23 \, d^{4} e^{\left (-1\right )}}{3 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} + 5 \, d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} + \frac {11 \, d^{3} x}{3 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {13 \, d x}{3 \, \sqrt {-x^{2} e^{2} + d^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/3*(3*x^2*e^(-2)/(-x^2*e^2 + d^2)^(3/2) - 2*d^2*e^(-4)/(-x^2*e^2 + d^2)^(3/2))*d*x*e^4 - x^4*e^3/(-x^2*e^2 +
d^2)^(3/2) + 14*d^2*x^2*e/(-x^2*e^2 + d^2)^(3/2) - 23/3*d^4*e^(-1)/(-x^2*e^2 + d^2)^(3/2) + 5*d*arcsin(x*e/d)*
e^(-1) + 11/3*d^3*x/(-x^2*e^2 + d^2)^(3/2) - 13/3*d*x/sqrt(-x^2*e^2 + d^2)

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Fricas [A]
time = 2.71, size = 124, normalized size = 1.15 \begin {gather*} -\frac {23 \, d x^{2} e^{2} - 46 \, d^{2} x e + 23 \, d^{3} + 30 \, {\left (d x^{2} e^{2} - 2 \, d^{2} x e + d^{3}\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (3 \, x^{2} e^{2} - 34 \, d x e + 23 \, d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{3 \, {\left (x^{2} e^{3} - 2 \, d x e^{2} + d^{2} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{5}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]
time = 1.72, size = 132, normalized size = 1.22 \begin {gather*} 5 \, d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\left (d\right ) - \sqrt {-x^{2} e^{2} + d^{2}} e^{\left (-1\right )} + \frac {8 \, {\left (\frac {12 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d e^{\left (-2\right )}}{x} - \frac {3 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d e^{\left (-4\right )}}{x^{2}} - 5 \, d\right )} e^{\left (-1\right )}}{3 \, {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} - 1\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^5}{{\left (d^2-e^2\,x^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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