3.203 \(\int \frac{(d^2-e^2 x^2)^{5/2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=113 \[ -\frac{2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac{5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac{15 d \sqrt{d^2-e^2 x^2}}{2 e}-\frac{15 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e} \]

[Out]

(-15*d*Sqrt[d^2 - e^2*x^2])/(2*e) - (5*(d^2 - e^2*x^2)^(3/2))/(2*e*(d + e*x)) - (2*(d^2 - e^2*x^2)^(5/2))/(e*(
d + e*x)^3) - (15*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e)

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Rubi [A]  time = 0.0480503, antiderivative size = 113, normalized size of antiderivative = 1., 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 = {663, 665, 217, 203} \[ -\frac{2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac{5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac{15 d \sqrt{d^2-e^2 x^2}}{2 e}-\frac{15 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^(5/2)/(d + e*x)^4,x]

[Out]

(-15*d*Sqrt[d^2 - e^2*x^2])/(2*e) - (5*(d^2 - e^2*x^2)^(3/2))/(2*e*(d + e*x)) - (2*(d^2 - e^2*x^2)^(5/2))/(e*(
d + e*x)^3) - (15*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e)

Rule 663

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 + p + 1)), x] - Dist[(c*p)/(e^2*(m + 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] && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1
, 0] && IntegerQ[2*p]

Rule 665

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]

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 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])

Rubi steps

\begin{align*} \int \frac{\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx &=-\frac{2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-5 \int \frac{\left (d^2-e^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx\\ &=-\frac{5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac{2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac{1}{2} (15 d) \int \frac{\sqrt{d^2-e^2 x^2}}{d+e x} \, dx\\ &=-\frac{15 d \sqrt{d^2-e^2 x^2}}{2 e}-\frac{5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac{2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac{1}{2} \left (15 d^2\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=-\frac{15 d \sqrt{d^2-e^2 x^2}}{2 e}-\frac{5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac{2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac{1}{2} \left (15 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )\\ &=-\frac{15 d \sqrt{d^2-e^2 x^2}}{2 e}-\frac{5 \left (d^2-e^2 x^2\right )^{3/2}}{2 e (d+e x)}-\frac{2 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^3}-\frac{15 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e}\\ \end{align*}

Mathematica [A]  time = 0.0763588, size = 75, normalized size = 0.66 \[ \sqrt{d^2-e^2 x^2} \left (-\frac{8 d^2}{e (d+e x)}-\frac{4 d}{e}+\frac{x}{2}\right )-\frac{15 d^2 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(d^2 - e^2*x^2)^(5/2)/(d + e*x)^4,x]

[Out]

Sqrt[d^2 - e^2*x^2]*((-4*d)/e + x/2 - (8*d^2)/(e*(d + e*x))) - (15*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e
)

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Maple [B]  time = 0.056, size = 284, normalized size = 2.5 \begin{align*} -{\frac{1}{{e}^{5}d} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{7}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-4}}-3\,{\frac{1}{{e}^{4}{d}^{2}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{7/2} \left ({\frac{d}{e}}+x \right ) ^{-3}}-4\,{\frac{1}{{e}^{3}{d}^{3}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{7/2} \left ({\frac{d}{e}}+x \right ) ^{-2}}-4\,{\frac{1}{e{d}^{3}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{5/2}}-5\,{\frac{x}{{d}^{2}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{3/2}}-{\frac{15\,x}{2}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-{\frac{15\,{d}^{2}}{2}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x)

[Out]

-1/e^5/d/(d/e+x)^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)-3/e^4/d^2/(d/e+x)^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/
2)-4/e^3/d^3/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)-4/e/d^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)-5/d^2
*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x-15/2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)*x-15/2*d^2/(e^2)^(1/2)*arcta
n((e^2)^(1/2)*x/(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64021, size = 212, normalized size = 1.88 \begin{align*} -\frac{24 \, d^{2} e x + 24 \, d^{3} - 30 \,{\left (d^{2} e x + d^{3}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) -{\left (e^{2} x^{2} - 7 \, d e x - 24 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{2 \,{\left (e^{2} x + d e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

-1/2*(24*d^2*e*x + 24*d^3 - 30*(d^2*e*x + d^3)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (e^2*x^2 - 7*d*e*x
- 24*d^2)*sqrt(-e^2*x^2 + d^2))/(e^2*x + d*e)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{5}{2}}}{\left (d + e x\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(5/2)/(e*x+d)**4,x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**(5/2)/(d + e*x)**4, x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError