### 3.804 $$\int \frac{(d^2-e^2 x^2)^{7/2}}{(d+e x)^2} \, dx$$

Optimal. Leaf size=132 $\frac{7}{16} d^4 x \sqrt{d^2-e^2 x^2}+\frac{7}{24} d^2 x \left (d^2-e^2 x^2\right )^{3/2}+\frac{7 d \left (d^2-e^2 x^2\right )^{5/2}}{30 e}+\frac{(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac{7 d^6 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{16 e}$

[Out]

(7*d^4*x*Sqrt[d^2 - e^2*x^2])/16 + (7*d^2*x*(d^2 - e^2*x^2)^(3/2))/24 + (7*d*(d^2 - e^2*x^2)^(5/2))/(30*e) + (
(d - e*x)*(d^2 - e^2*x^2)^(5/2))/(6*e) + (7*d^6*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(16*e)

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Rubi [A]  time = 0.0508933, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {655, 671, 641, 195, 217, 203} $\frac{7}{16} d^4 x \sqrt{d^2-e^2 x^2}+\frac{7}{24} d^2 x \left (d^2-e^2 x^2\right )^{3/2}+\frac{7 d \left (d^2-e^2 x^2\right )^{5/2}}{30 e}+\frac{(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac{7 d^6 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{16 e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(7*d^4*x*Sqrt[d^2 - e^2*x^2])/16 + (7*d^2*x*(d^2 - e^2*x^2)^(3/2))/24 + (7*d*(d^2 - e^2*x^2)^(5/2))/(30*e) + (
(d - e*x)*(d^2 - e^2*x^2)^(5/2))/(6*e) + (7*d^6*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(16*e)

Rule 655

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

Rule 671

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

Rule 641

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

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)^2} \, dx &=\int (d-e x)^2 \left (d^2-e^2 x^2\right )^{3/2} \, dx\\ &=\frac{(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac{1}{6} (7 d) \int (d-e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx\\ &=\frac{7 d \left (d^2-e^2 x^2\right )^{5/2}}{30 e}+\frac{(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac{1}{6} \left (7 d^2\right ) \int \left (d^2-e^2 x^2\right )^{3/2} \, dx\\ &=\frac{7}{24} d^2 x \left (d^2-e^2 x^2\right )^{3/2}+\frac{7 d \left (d^2-e^2 x^2\right )^{5/2}}{30 e}+\frac{(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac{1}{8} \left (7 d^4\right ) \int \sqrt{d^2-e^2 x^2} \, dx\\ &=\frac{7}{16} d^4 x \sqrt{d^2-e^2 x^2}+\frac{7}{24} d^2 x \left (d^2-e^2 x^2\right )^{3/2}+\frac{7 d \left (d^2-e^2 x^2\right )^{5/2}}{30 e}+\frac{(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac{1}{16} \left (7 d^6\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{7}{16} d^4 x \sqrt{d^2-e^2 x^2}+\frac{7}{24} d^2 x \left (d^2-e^2 x^2\right )^{3/2}+\frac{7 d \left (d^2-e^2 x^2\right )^{5/2}}{30 e}+\frac{(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac{1}{16} \left (7 d^6\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{7}{16} d^4 x \sqrt{d^2-e^2 x^2}+\frac{7}{24} d^2 x \left (d^2-e^2 x^2\right )^{3/2}+\frac{7 d \left (d^2-e^2 x^2\right )^{5/2}}{30 e}+\frac{(d-e x) \left (d^2-e^2 x^2\right )^{5/2}}{6 e}+\frac{7 d^6 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{16 e}\\ \end{align*}

Mathematica [A]  time = 0.0778846, size = 102, normalized size = 0.77 $\frac{\sqrt{d^2-e^2 x^2} \left (-192 d^3 e^2 x^2+10 d^2 e^3 x^3+135 d^4 e x+96 d^5+96 d e^4 x^4-40 e^5 x^5\right )+105 d^6 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{240 e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(96*d^5 + 135*d^4*e*x - 192*d^3*e^2*x^2 + 10*d^2*e^3*x^3 + 96*d*e^4*x^4 - 40*e^5*x^5) + 1
05*d^6*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(240*e)

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Maple [B]  time = 0.05, size = 228, normalized size = 1.7 \begin{align*}{\frac{1}{5\,{e}^{3}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 ) ^{-2}}+{\frac{1}{5\,de} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{7}{2}}}}+{\frac{7\,x}{30} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{7\,{d}^{2}x}{24} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{7\,{d}^{4}x}{16}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}+{\frac{7\,{d}^{6}}{16}\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)^2,x)

[Out]

1/5/e^3/d/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(9/2)+1/5/e/d*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)+7/30*(-(
d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)*x+7/24*d^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x+7/16*d^4*(-(d/e+x)^2*e^2+2
*d*e*(d/e+x))^(1/2)*x+7/16*d^6/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2))

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Maxima [C]  time = 1.58455, size = 188, normalized size = 1.42 \begin{align*} -\frac{7 i \, d^{6} \arcsin \left (\frac{e x}{d} + 2\right )}{16 \, e} + \frac{7}{16} \, \sqrt{e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4} x + \frac{7 \, \sqrt{e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{5}}{8 \, e} + \frac{7}{24} \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} d^{2} x + \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{7}{2}}}{6 \,{\left (e^{2} x + d e\right )}} + \frac{7 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} d}{30 \, e} \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)^2,x, algorithm="maxima")

[Out]

-7/16*I*d^6*arcsin(e*x/d + 2)/e + 7/16*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^4*x + 7/8*sqrt(e^2*x^2 + 4*d*e*x + 3*
d^2)*d^5/e + 7/24*(-e^2*x^2 + d^2)^(3/2)*d^2*x + 1/6*(-e^2*x^2 + d^2)^(7/2)/(e^2*x + d*e) + 7/30*(-e^2*x^2 + d
^2)^(5/2)*d/e

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Fricas [A]  time = 2.17988, size = 231, normalized size = 1.75 \begin{align*} -\frac{210 \, d^{6} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (40 \, e^{5} x^{5} - 96 \, d e^{4} x^{4} - 10 \, d^{2} e^{3} x^{3} + 192 \, d^{3} e^{2} x^{2} - 135 \, d^{4} e x - 96 \, d^{5}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{240 \, e} \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)^2,x, algorithm="fricas")

[Out]

-1/240*(210*d^6*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (40*e^5*x^5 - 96*d*e^4*x^4 - 10*d^2*e^3*x^3 + 192*
d^3*e^2*x^2 - 135*d^4*e*x - 96*d^5)*sqrt(-e^2*x^2 + d^2))/e

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Sympy [C]  time = 15.2106, size = 498, normalized size = 3.77 \begin{align*} d^{4} \left (\begin{cases} - \frac{i d^{2} \operatorname{acosh}{\left (\frac{e x}{d} \right )}}{2 e} - \frac{i d x}{2 \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} + \frac{i e^{2} x^{3}}{2 d \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} & \text{for}\: \frac{\left |{e^{2} x^{2}}\right |}{\left |{d^{2}}\right |} > 1 \\\frac{d^{2} \operatorname{asin}{\left (\frac{e x}{d} \right )}}{2 e} + \frac{d x \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}}{2} & \text{otherwise} \end{cases}\right ) - 2 d^{3} e \left (\begin{cases} \frac{x^{2} \sqrt{d^{2}}}{2} & \text{for}\: e^{2} = 0 \\- \frac{\left (d^{2} - e^{2} x^{2}\right )^{\frac{3}{2}}}{3 e^{2}} & \text{otherwise} \end{cases}\right ) + 2 d e^{3} \left (\begin{cases} - \frac{2 d^{4} \sqrt{d^{2} - e^{2} x^{2}}}{15 e^{4}} - \frac{d^{2} x^{2} \sqrt{d^{2} - e^{2} x^{2}}}{15 e^{2}} + \frac{x^{4} \sqrt{d^{2} - e^{2} x^{2}}}{5} & \text{for}\: e \neq 0 \\\frac{x^{4} \sqrt{d^{2}}}{4} & \text{otherwise} \end{cases}\right ) - e^{4} \left (\begin{cases} - \frac{i d^{6} \operatorname{acosh}{\left (\frac{e x}{d} \right )}}{16 e^{5}} + \frac{i d^{5} x}{16 e^{4} \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} - \frac{i d^{3} x^{3}}{48 e^{2} \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} - \frac{5 i d x^{5}}{24 \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} + \frac{i e^{2} x^{7}}{6 d \sqrt{-1 + \frac{e^{2} x^{2}}{d^{2}}}} & \text{for}\: \frac{\left |{e^{2} x^{2}}\right |}{\left |{d^{2}}\right |} > 1 \\\frac{d^{6} \operatorname{asin}{\left (\frac{e x}{d} \right )}}{16 e^{5}} - \frac{d^{5} x}{16 e^{4} \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} + \frac{d^{3} x^{3}}{48 e^{2} \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} + \frac{5 d x^{5}}{24 \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} - \frac{e^{2} x^{7}}{6 d \sqrt{1 - \frac{e^{2} x^{2}}{d^{2}}}} & \text{otherwise} \end{cases}\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)**2,x)

[Out]

d**4*Piecewise((-I*d**2*acosh(e*x/d)/(2*e) - I*d*x/(2*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**3/(2*d*sqrt(-1 +
e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (d**2*asin(e*x/d)/(2*e) + d*x*sqrt(1 - e**2*x**2/d**2)/2, Tru
e)) - 2*d**3*e*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) + 2*d*
e**3*Piecewise((-2*d**4*sqrt(d**2 - e**2*x**2)/(15*e**4) - d**2*x**2*sqrt(d**2 - e**2*x**2)/(15*e**2) + x**4*s
qrt(d**2 - e**2*x**2)/5, Ne(e, 0)), (x**4*sqrt(d**2)/4, True)) - e**4*Piecewise((-I*d**6*acosh(e*x/d)/(16*e**5
) + I*d**5*x/(16*e**4*sqrt(-1 + e**2*x**2/d**2)) - I*d**3*x**3/(48*e**2*sqrt(-1 + e**2*x**2/d**2)) - 5*I*d*x**
5/(24*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**7/(6*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1),
(d**6*asin(e*x/d)/(16*e**5) - d**5*x/(16*e**4*sqrt(1 - e**2*x**2/d**2)) + d**3*x**3/(48*e**2*sqrt(1 - e**2*x*
*2/d**2)) + 5*d*x**5/(24*sqrt(1 - e**2*x**2/d**2)) - e**2*x**7/(6*d*sqrt(1 - e**2*x**2/d**2)), True))

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)^2,x, algorithm="giac")

[Out]

Timed out