3.855 \(\int \frac{1}{(d+e x) (d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=106 \[ \frac{16 x}{35 d^7 \sqrt{d^2-e^2 x^2}}+\frac{8 x}{35 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{6 x}{35 d^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \]

[Out]

(6*x)/(35*d^3*(d^2 - e^2*x^2)^(5/2)) - 1/(7*d*e*(d + e*x)*(d^2 - e^2*x^2)^(5/2)) + (8*x)/(35*d^5*(d^2 - e^2*x^
2)^(3/2)) + (16*x)/(35*d^7*Sqrt[d^2 - e^2*x^2])

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Rubi [A]  time = 0.0277526, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {659, 192, 191} \[ \frac{16 x}{35 d^7 \sqrt{d^2-e^2 x^2}}+\frac{8 x}{35 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{6 x}{35 d^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(6*x)/(35*d^3*(d^2 - e^2*x^2)^(5/2)) - 1/(7*d*e*(d + e*x)*(d^2 - e^2*x^2)^(5/2)) + (8*x)/(35*d^5*(d^2 - e^2*x^
2)^(3/2)) + (16*x)/(35*d^7*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 )^{7/2}} \, dx &=-\frac{1}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac{6 \int \frac{1}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx}{7 d}\\ &=\frac{6 x}{35 d^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac{24 \int \frac{1}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{35 d^3}\\ &=\frac{6 x}{35 d^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac{8 x}{35 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{16 \int \frac{1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{35 d^5}\\ &=\frac{6 x}{35 d^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{7 d e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac{8 x}{35 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{16 x}{35 d^7 \sqrt{d^2-e^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0801594, size = 104, normalized size = 0.98 \[ \frac{\sqrt{d^2-e^2 x^2} \left (30 d^4 e^2 x^2-40 d^3 e^3 x^3-40 d^2 e^4 x^4+30 d^5 e x-5 d^6+16 d e^5 x^5+16 e^6 x^6\right )}{35 d^7 e (d-e x)^3 (d+e x)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-5*d^6 + 30*d^5*e*x + 30*d^4*e^2*x^2 - 40*d^3*e^3*x^3 - 40*d^2*e^4*x^4 + 16*d*e^5*x^5 +
16*e^6*x^6))/(35*d^7*e*(d - e*x)^3*(d + e*x)^4)

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Maple [A]  time = 0.045, size = 92, normalized size = 0.9 \begin{align*} -{\frac{ \left ( -ex+d \right ) \left ( -16\,{e}^{6}{x}^{6}-16\,{e}^{5}{x}^{5}d+40\,{e}^{4}{x}^{4}{d}^{2}+40\,{e}^{3}{x}^{3}{d}^{3}-30\,{e}^{2}{x}^{2}{d}^{4}-30\,x{d}^{5}e+5\,{d}^{6} \right ) }{35\,{d}^{7}e} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{-{\frac{7}{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)^(7/2),x)

[Out]

-1/35*(-e*x+d)*(-16*e^6*x^6-16*d*e^5*x^5+40*d^2*e^4*x^4+40*d^3*e^3*x^3-30*d^4*e^2*x^2-30*d^5*e*x+5*d^6)/d^7/e/
(-e^2*x^2+d^2)^(7/2)

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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)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.05304, size = 487, normalized size = 4.59 \begin{align*} -\frac{5 \, e^{7} x^{7} + 5 \, d e^{6} x^{6} - 15 \, d^{2} e^{5} x^{5} - 15 \, d^{3} e^{4} x^{4} + 15 \, d^{4} e^{3} x^{3} + 15 \, d^{5} e^{2} x^{2} - 5 \, d^{6} e x - 5 \, d^{7} +{\left (16 \, e^{6} x^{6} + 16 \, d e^{5} x^{5} - 40 \, d^{2} e^{4} x^{4} - 40 \, d^{3} e^{3} x^{3} + 30 \, d^{4} e^{2} x^{2} + 30 \, d^{5} e x - 5 \, d^{6}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{35 \,{\left (d^{7} e^{8} x^{7} + d^{8} e^{7} x^{6} - 3 \, d^{9} e^{6} x^{5} - 3 \, d^{10} e^{5} x^{4} + 3 \, d^{11} e^{4} x^{3} + 3 \, d^{12} e^{3} x^{2} - d^{13} e^{2} x - d^{14} 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)^(7/2),x, algorithm="fricas")

[Out]

-1/35*(5*e^7*x^7 + 5*d*e^6*x^6 - 15*d^2*e^5*x^5 - 15*d^3*e^4*x^4 + 15*d^4*e^3*x^3 + 15*d^5*e^2*x^2 - 5*d^6*e*x
- 5*d^7 + (16*e^6*x^6 + 16*d*e^5*x^5 - 40*d^2*e^4*x^4 - 40*d^3*e^3*x^3 + 30*d^4*e^2*x^2 + 30*d^5*e*x - 5*d^6)
*sqrt(-e^2*x^2 + d^2))/(d^7*e^8*x^7 + d^8*e^7*x^6 - 3*d^9*e^6*x^5 - 3*d^10*e^5*x^4 + 3*d^11*e^4*x^3 + 3*d^12*e
^3*x^2 - d^13*e^2*x - d^14*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{7}{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)**(7/2),x)

[Out]

Integral(1/((-(-d + e*x)*(d + e*x))**(7/2)*(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}, \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)^(7/2),x, algorithm="giac")

[Out]

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