### 3.856 $$\int \frac{1}{(d+e x)^2 (d^2-e^2 x^2)^{7/2}} \, dx$$

Optimal. Leaf size=139 $\frac{16 x}{45 d^8 \sqrt{d^2-e^2 x^2}}+\frac{8 x}{45 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 x}{15 d^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{9 d^2 e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{9 d e (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}$

[Out]

(2*x)/(15*d^4*(d^2 - e^2*x^2)^(5/2)) - 1/(9*d*e*(d + e*x)^2*(d^2 - e^2*x^2)^(5/2)) - 1/(9*d^2*e*(d + e*x)*(d^2
- e^2*x^2)^(5/2)) + (8*x)/(45*d^6*(d^2 - e^2*x^2)^(3/2)) + (16*x)/(45*d^8*Sqrt[d^2 - e^2*x^2])

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Rubi [A]  time = 0.0466924, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {659, 192, 191} $\frac{16 x}{45 d^8 \sqrt{d^2-e^2 x^2}}+\frac{8 x}{45 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 x}{15 d^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{9 d^2 e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{9 d e (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*x)/(15*d^4*(d^2 - e^2*x^2)^(5/2)) - 1/(9*d*e*(d + e*x)^2*(d^2 - e^2*x^2)^(5/2)) - 1/(9*d^2*e*(d + e*x)*(d^2
- e^2*x^2)^(5/2)) + (8*x)/(45*d^6*(d^2 - e^2*x^2)^(3/2)) + (16*x)/(45*d^8*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)^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=-\frac{1}{9 d e (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{7 \int \frac{1}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx}{9 d}\\ &=-\frac{1}{9 d e (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{9 d^2 e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac{2 \int \frac{1}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx}{3 d^2}\\ &=\frac{2 x}{15 d^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{9 d e (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{9 d^2 e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac{8 \int \frac{1}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{15 d^4}\\ &=\frac{2 x}{15 d^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{9 d e (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{9 d^2 e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac{8 x}{45 d^6 \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}{45 d^6}\\ &=\frac{2 x}{15 d^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{9 d e (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{1}{9 d^2 e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac{8 x}{45 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{16 x}{45 d^8 \sqrt{d^2-e^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0757517, size = 115, normalized size = 0.83 $\frac{\sqrt{d^2-e^2 x^2} \left (60 d^5 e^2 x^2-10 d^4 e^3 x^3-80 d^3 e^4 x^4-24 d^2 e^5 x^5+25 d^6 e x-10 d^7+32 d e^6 x^6+16 e^7 x^7\right )}{45 d^8 e (d-e x)^3 (d+e x)^5}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-10*d^7 + 25*d^6*e*x + 60*d^5*e^2*x^2 - 10*d^4*e^3*x^3 - 80*d^3*e^4*x^4 - 24*d^2*e^5*x^5
+ 32*d*e^6*x^6 + 16*e^7*x^7))/(45*d^8*e*(d - e*x)^3*(d + e*x)^5)

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Maple [A]  time = 0.048, size = 110, normalized size = 0.8 \begin{align*} -{\frac{ \left ( -ex+d \right ) \left ( -16\,{e}^{7}{x}^{7}-32\,{e}^{6}{x}^{6}d+24\,{e}^{5}{x}^{5}{d}^{2}+80\,{e}^{4}{x}^{4}{d}^{3}+10\,{e}^{3}{x}^{3}{d}^{4}-60\,{e}^{2}{x}^{2}{d}^{5}-25\,x{d}^{6}e+10\,{d}^{7} \right ) }{ \left ( 45\,ex+45\,d \right ){d}^{8}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)^2/(-e^2*x^2+d^2)^(7/2),x)

[Out]

-1/45*(-e*x+d)*(-16*e^7*x^7-32*d*e^6*x^6+24*d^2*e^5*x^5+80*d^3*e^4*x^4+10*d^4*e^3*x^3-60*d^5*e^2*x^2-25*d^6*e*
x+10*d^7)/(e*x+d)/d^8/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)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.84771, size = 524, normalized size = 3.77 \begin{align*} -\frac{10 \, e^{8} x^{8} + 20 \, d e^{7} x^{7} - 20 \, d^{2} e^{6} x^{6} - 60 \, d^{3} e^{5} x^{5} + 60 \, d^{5} e^{3} x^{3} + 20 \, d^{6} e^{2} x^{2} - 20 \, d^{7} e x - 10 \, d^{8} +{\left (16 \, e^{7} x^{7} + 32 \, d e^{6} x^{6} - 24 \, d^{2} e^{5} x^{5} - 80 \, d^{3} e^{4} x^{4} - 10 \, d^{4} e^{3} x^{3} + 60 \, d^{5} e^{2} x^{2} + 25 \, d^{6} e x - 10 \, d^{7}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{45 \,{\left (d^{8} e^{9} x^{8} + 2 \, d^{9} e^{8} x^{7} - 2 \, d^{10} e^{7} x^{6} - 6 \, d^{11} e^{6} x^{5} + 6 \, d^{13} e^{4} x^{3} + 2 \, d^{14} e^{3} x^{2} - 2 \, d^{15} e^{2} x - d^{16} e\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

-1/45*(10*e^8*x^8 + 20*d*e^7*x^7 - 20*d^2*e^6*x^6 - 60*d^3*e^5*x^5 + 60*d^5*e^3*x^3 + 20*d^6*e^2*x^2 - 20*d^7*
e*x - 10*d^8 + (16*e^7*x^7 + 32*d*e^6*x^6 - 24*d^2*e^5*x^5 - 80*d^3*e^4*x^4 - 10*d^4*e^3*x^3 + 60*d^5*e^2*x^2
+ 25*d^6*e*x - 10*d^7)*sqrt(-e^2*x^2 + d^2))/(d^8*e^9*x^8 + 2*d^9*e^8*x^7 - 2*d^10*e^7*x^6 - 6*d^11*e^6*x^5 +
6*d^13*e^4*x^3 + 2*d^14*e^3*x^2 - 2*d^15*e^2*x - d^16*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 )^{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**2/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(1/((-(-d + e*x)*(d + e*x))**(7/2)*(d + e*x)**2), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError