### 3.842 $$\int \frac{1}{(d+e x) (d^2-e^2 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=82 $\frac{8 x}{15 d^5 \sqrt{d^2-e^2 x^2}}+\frac{4 x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{1}{5 d e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}$

[Out]

(4*x)/(15*d^3*(d^2 - e^2*x^2)^(3/2)) - 1/(5*d*e*(d + e*x)*(d^2 - e^2*x^2)^(3/2)) + (8*x)/(15*d^5*Sqrt[d^2 - e^
2*x^2])

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Rubi [A]  time = 0.0206295, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {659, 192, 191} $\frac{8 x}{15 d^5 \sqrt{d^2-e^2 x^2}}+\frac{4 x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{1}{5 d e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

(4*x)/(15*d^3*(d^2 - e^2*x^2)^(3/2)) - 1/(5*d*e*(d + e*x)*(d^2 - e^2*x^2)^(3/2)) + (8*x)/(15*d^5*Sqrt[d^2 - e^
2*x^2])

Rule 659

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*c*d*(m + p + 1)), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2
], 0]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{1}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx &=-\frac{1}{5 d e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac{4 \int \frac{1}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac{4 x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{1}{5 d e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac{8 \int \frac{1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^3}\\ &=\frac{4 x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac{1}{5 d e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac{8 x}{15 d^5 \sqrt{d^2-e^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0649542, size = 82, normalized size = 1. $-\frac{\sqrt{d^2-e^2 x^2} \left (-12 d^2 e^2 x^2-12 d^3 e x+3 d^4+8 d e^3 x^3+8 e^4 x^4\right )}{15 d^5 e (d-e x)^2 (d+e x)^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

-(Sqrt[d^2 - e^2*x^2]*(3*d^4 - 12*d^3*e*x - 12*d^2*e^2*x^2 + 8*d*e^3*x^3 + 8*e^4*x^4))/(15*d^5*e*(d - e*x)^2*(
d + e*x)^3)

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Maple [A]  time = 0.045, size = 70, normalized size = 0.9 \begin{align*} -{\frac{ \left ( -ex+d \right ) \left ( 8\,{e}^{4}{x}^{4}+8\,{e}^{3}{x}^{3}d-12\,{e}^{2}{x}^{2}{d}^{2}-12\,x{d}^{3}e+3\,{d}^{4} \right ) }{15\,{d}^{5}e} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x)

[Out]

-1/15*(-e*x+d)*(8*e^4*x^4+8*d*e^3*x^3-12*d^2*e^2*x^2-12*d^3*e*x+3*d^4)/d^5/e/(-e^2*x^2+d^2)^(5/2)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.17436, size = 342, normalized size = 4.17 \begin{align*} -\frac{3 \, e^{5} x^{5} + 3 \, d e^{4} x^{4} - 6 \, d^{2} e^{3} x^{3} - 6 \, d^{3} e^{2} x^{2} + 3 \, d^{4} e x + 3 \, d^{5} +{\left (8 \, e^{4} x^{4} + 8 \, d e^{3} x^{3} - 12 \, d^{2} e^{2} x^{2} - 12 \, d^{3} e x + 3 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{5} e^{6} x^{5} + d^{6} e^{5} x^{4} - 2 \, d^{7} e^{4} x^{3} - 2 \, d^{8} e^{3} x^{2} + d^{9} e^{2} x + d^{10} e\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

-1/15*(3*e^5*x^5 + 3*d*e^4*x^4 - 6*d^2*e^3*x^3 - 6*d^3*e^2*x^2 + 3*d^4*e*x + 3*d^5 + (8*e^4*x^4 + 8*d*e^3*x^3
- 12*d^2*e^2*x^2 - 12*d^3*e*x + 3*d^4)*sqrt(-e^2*x^2 + d^2))/(d^5*e^6*x^5 + d^6*e^5*x^4 - 2*d^7*e^4*x^3 - 2*d^
8*e^3*x^2 + d^9*e^2*x + d^10*e)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{5}{2}} \left (d + e x\right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral(1/((-(-d + e*x)*(d + e*x))**(5/2)*(d + e*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, 1\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

[undef, undef, undef, undef, undef, 1]