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

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 = 0.04, 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 = {669, 641, 217, 203} \[ \frac {2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {10 (d+e x)^2}{3 e \sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{e}+\frac {5 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \]

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 203

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

Rule 217

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 641

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] :> 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) \operatorname {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 = 0.16, size = 109, normalized size = 1.01 \[ \frac {(d+e x) \left (\left (23 d^2-34 d e x+3 e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}}-15 (d-e x)^2 \sin ^{-1}\left (\frac {e x}{d}\right )\right )}{3 e (e x-d) \sqrt {d^2-e^2 x^2} \sqrt {1-\frac {e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)*((23*d^2 - 34*d*e*x + 3*e^2*x^2)*Sqrt[1 - (e^2*x^2)/d^2] - 15*(d - e*x)^2*ArcSin[(e*x)/d]))/(3*e*(-
d + e*x)*Sqrt[d^2 - e^2*x^2]*Sqrt[1 - (e^2*x^2)/d^2])

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fricas [A]  time = 1.08, size = 128, normalized size = 1.19 \[ -\frac {23 \, d e^{2} x^{2} - 46 \, d^{2} e x + 23 \, d^{3} + 30 \, {\left (d e^{2} x^{2} - 2 \, d^{2} e x + d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (3 \, e^{2} x^{2} - 34 \, d e x + 23 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, {\left (e^{3} x^{2} - 2 \, d e^{2} x + d^{2} e\right )}} \]

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*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)

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giac [A]  time = 0.33, size = 86, normalized size = 0.80 \[ 5 \, d \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\relax (d) - \frac {{\left (23 \, d^{4} e^{\left (-1\right )} + {\left (12 \, d^{3} - {\left (42 \, d^{2} e - {\left (3 \, x e^{3} - 28 \, d e^{2}\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{3 \, {\left (x^{2} e^{2} - d^{2}\right )}^{2}} \]

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) - 1/3*(23*d^4*e^(-1) + (12*d^3 - (42*d^2*e - (3*x*e^3 - 28*d*e^2)*x)*x)*x)*sqr
t(-x^2*e^2 + d^2)/(x^2*e^2 - d^2)^2

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maple [A]  time = 0.07, size = 160, normalized size = 1.48 \[ -\frac {e^{3} x^{4}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {5 d \,e^{2} x^{3}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {14 d^{2} e \,x^{2}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {11 d^{3} x}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {23 d^{4}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e}-\frac {23 d x}{3 \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {5 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}} \]

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)

[Out]

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

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maxima [A]  time = 2.96, size = 171, normalized size = 1.58 \[ \frac {5}{3} \, d e^{4} x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )} - \frac {e^{3} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {14 \, d^{2} e x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {11 \, d^{3} x}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {23 \, d^{4}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e} - \frac {13 \, d x}{3 \, \sqrt {-e^{2} x^{2} + d^{2}}} + \frac {5 \, d \arcsin \left (\frac {e x}{d}\right )}{e} \]

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*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*x/d)/e

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

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

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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