3.806 \(\int \frac{(d^2-e^2 x^2)^{7/2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=136 \[ \frac{35}{8} d^2 x \sqrt{d^2-e^2 x^2}+\frac{35 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}+\frac{7 (d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac{35 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e} \]

[Out]

(35*d^2*x*Sqrt[d^2 - e^2*x^2])/8 + (35*d*(d^2 - e^2*x^2)^(3/2))/(12*e) + (7*(d - e*x)*(d^2 - e^2*x^2)^(3/2))/(
4*e) + (2*(d^2 - e^2*x^2)^(7/2))/(e*(d + e*x)^3) + (35*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e)

________________________________________________________________________________________

Rubi [A]  time = 0.0566787, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {663, 655, 671, 641, 195, 217, 203} \[ \frac{35}{8} d^2 x \sqrt{d^2-e^2 x^2}+\frac{35 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}+\frac{7 (d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac{35 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e} \]

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^(7/2)/(d + e*x)^4,x]

[Out]

(35*d^2*x*Sqrt[d^2 - e^2*x^2])/8 + (35*d*(d^2 - e^2*x^2)^(3/2))/(12*e) + (7*(d - e*x)*(d^2 - e^2*x^2)^(3/2))/(
4*e) + (2*(d^2 - e^2*x^2)^(7/2))/(e*(d + e*x)^3) + (35*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*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 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)^4} \, dx &=\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}+7 \int \frac{\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx\\ &=\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}+7 \int (d-e x)^2 \sqrt{d^2-e^2 x^2} \, dx\\ &=\frac{7 (d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}+\frac{1}{4} (35 d) \int (d-e x) \sqrt{d^2-e^2 x^2} \, dx\\ &=\frac{35 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac{7 (d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}+\frac{1}{4} \left (35 d^2\right ) \int \sqrt{d^2-e^2 x^2} \, dx\\ &=\frac{35}{8} d^2 x \sqrt{d^2-e^2 x^2}+\frac{35 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac{7 (d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}+\frac{1}{8} \left (35 d^4\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{35}{8} d^2 x \sqrt{d^2-e^2 x^2}+\frac{35 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac{7 (d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}+\frac{1}{8} \left (35 d^4\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}{8} d^2 x \sqrt{d^2-e^2 x^2}+\frac{35 d \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac{7 (d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}+\frac{2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}+\frac{35 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e}\\ \end{align*}

Mathematica [A]  time = 0.0784545, size = 80, normalized size = 0.59 \[ \frac{\sqrt{d^2-e^2 x^2} \left (-81 d^2 e x+160 d^3+32 d e^2 x^2-6 e^3 x^3\right )+105 d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{24 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(d^2 - e^2*x^2)^(7/2)/(d + e*x)^4,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(160*d^3 - 81*d^2*e*x + 32*d*e^2*x^2 - 6*e^3*x^3) + 105*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x
^2]])/(24*e)

________________________________________________________________________________________

Maple [B]  time = 0.05, size = 317, normalized size = 2.3 \begin{align*}{\frac{1}{{e}^{5}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 ) ^{-4}}+{\frac{5}{3\,{e}^{4}{d}^{2}} \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}}+2\,{\frac{1}{{e}^{3}{d}^{3}} \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}}+2\,{\frac{1}{e{d}^{3}} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{7/2}}+{\frac{7\,x}{3\,{d}^{2}} \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}{12} \left ( - \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{35\,{d}^{2}x}{8}\sqrt{- \left ({\frac{d}{e}}+x \right ) ^{2}{e}^{2}+2\,de \left ({\frac{d}{e}}+x \right ) }}+{\frac{35\,{d}^{4}}{8}\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)^4,x)

[Out]

1/e^5/d/(d/e+x)^4*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(9/2)+5/3/e^4/d^2/(d/e+x)^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(9
/2)+2/e^3/d^3/(d/e+x)^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(9/2)+2/e/d^3*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(7/2)+7/3/
d^2*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(5/2)*x+35/12*(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(3/2)*x+35/8*d^2*(-(d/e+x)^2*e
^2+2*d*e*(d/e+x))^(1/2)*x+35/8*d^4/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(d/e+x)^2*e^2+2*d*e*(d/e+x))^(1/2))

________________________________________________________________________________________

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)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.19882, size = 181, normalized size = 1.33 \begin{align*} -\frac{210 \, d^{4} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (6 \, e^{3} x^{3} - 32 \, d e^{2} x^{2} + 81 \, d^{2} e x - 160 \, d^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{24 \, 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)^4,x, algorithm="fricas")

[Out]

-1/24*(210*d^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (6*e^3*x^3 - 32*d*e^2*x^2 + 81*d^2*e*x - 160*d^3)*s
qrt(-e^2*x^2 + d^2))/e

________________________________________________________________________________________

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 )^{4}}\, 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)**4,x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**(7/2)/(d + e*x)**4, x)

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \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)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError