### 3.807 $$\int \frac{(d^2-e^2 x^2)^{7/2}}{(d+e x)^5} \, dx$$

Optimal. Leaf size=132 $-\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^4}-\frac{14 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^2}-\frac{35 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac{35}{2} d x \sqrt{d^2-e^2 x^2}-\frac{35 d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e}$

[Out]

(-35*d*x*Sqrt[d^2 - e^2*x^2])/2 - (35*(d^2 - e^2*x^2)^(3/2))/(3*e) - (14*(d^2 - e^2*x^2)^(5/2))/(e*(d + e*x)^2
) - (2*(d^2 - e^2*x^2)^(7/2))/(e*(d + e*x)^4) - (35*d^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e)

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Rubi [A]  time = 0.0508519, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.208, Rules used = {663, 665, 195, 217, 203} $-\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^4}-\frac{14 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^2}-\frac{35 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac{35}{2} d x \sqrt{d^2-e^2 x^2}-\frac{35 d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d^2 - e^2*x^2)^(7/2)/(d + e*x)^5,x]

[Out]

(-35*d*x*Sqrt[d^2 - e^2*x^2])/2 - (35*(d^2 - e^2*x^2)^(3/2))/(3*e) - (14*(d^2 - e^2*x^2)^(5/2))/(e*(d + e*x)^2
) - (2*(d^2 - e^2*x^2)^(7/2))/(e*(d + e*x)^4) - (35*d^3*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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[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 )^{7/2}}{(d+e x)^5} \, dx &=-\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^4}-7 \int \frac{\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^3} \, dx\\ &=-\frac{14 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^2}-\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^4}-35 \int \frac{\left (d^2-e^2 x^2\right )^{3/2}}{d+e x} \, dx\\ &=-\frac{35 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac{14 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^2}-\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^4}-(35 d) \int \sqrt{d^2-e^2 x^2} \, dx\\ &=-\frac{35}{2} d x \sqrt{d^2-e^2 x^2}-\frac{35 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac{14 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^2}-\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^4}-\frac{1}{2} \left (35 d^3\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=-\frac{35}{2} d x \sqrt{d^2-e^2 x^2}-\frac{35 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac{14 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^2}-\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^4}-\frac{1}{2} \left (35 d^3\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{35}{2} d x \sqrt{d^2-e^2 x^2}-\frac{35 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac{14 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^2}-\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^4}-\frac{35 d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e}\\ \end{align*}

Mathematica [A]  time = 0.0993727, size = 85, normalized size = 0.64 $\frac{1}{6} \sqrt{d^2-e^2 x^2} \left (-\frac{96 d^3}{e (d+e x)}-\frac{70 d^2}{e}+15 d x-2 e x^2\right )-\frac{35 d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{2 e}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d^2 - e^2*x^2)^(7/2)/(d + e*x)^5,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*((-70*d^2)/e + 15*d*x - 2*e*x^2 - (96*d^3)/(e*(d + e*x))))/6 - (35*d^3*ArcTan[(e*x)/Sqrt[
d^2 - e^2*x^2]])/(2*e)

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Maple [B]  time = 0.051, size = 364, normalized size = 2.8 \begin{align*} -{\frac{1}{{e}^{6}d} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{9}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-5}}-4\,{\frac{1}{{e}^{5}{d}^{2}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{9/2} \left ({\frac{d}{e}}+x \right ) ^{-4}}-{\frac{20}{3\,{e}^{4}{d}^{3}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{9}{2}}} \left ({\frac{d}{e}}+x \right ) ^{-3}}-8\,{\frac{1}{{e}^{3}{d}^{4}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{9/2} \left ({\frac{d}{e}}+x \right ) ^{-2}}-8\,{\frac{1}{e{d}^{4}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{7/2}}-{\frac{28\,x}{3\,{d}^{3}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{35\,x}{3\,d} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{35\,dx}{2}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}-{\frac{35\,{d}^{3}}{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)^(7/2)/(e*x+d)^5,x)

[Out]

-1/e^6/d/(d/e+x)^5*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(9/2)-4/e^5/d^2/(d/e+x)^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(9/
2)-20/3/e^4/d^3/(d/e+x)^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(9/2)-8/e^3/d^4/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+
x))^(9/2)-8/e/d^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)-28/3/d^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)*x-35/3/d*
(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x-35/2*d*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2)*x-35/2*d^3/(e^2)^(1/2)*arct
an((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)^(7/2)/(e*x+d)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.23631, size = 244, normalized size = 1.85 \begin{align*} -\frac{166 \, d^{3} e x + 166 \, d^{4} - 210 \,{\left (d^{3} e x + d^{4}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (2 \, e^{3} x^{3} - 13 \, d e^{2} x^{2} + 55 \, d^{2} e x + 166 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{6 \,{\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)^(7/2)/(e*x+d)^5,x, algorithm="fricas")

[Out]

-1/6*(166*d^3*e*x + 166*d^4 - 210*(d^3*e*x + d^4)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (2*e^3*x^3 - 13*
d*e^2*x^2 + 55*d^2*e*x + 166*d^3)*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{7}{2}}}{\left (d + e x\right )^{5}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e**2*x**2+d**2)**(7/2)/(e*x+d)**5,x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**(7/2)/(d + e*x)**5, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-e^2*x^2+d^2)^(7/2)/(e*x+d)^5,x, algorithm="giac")

[Out]

sage0*x