3.852 \(\int \frac{(d+e x)^3}{(d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=103 \[ \frac{2 \sqrt{d^2-e^2 x^2}}{15 d^3 e (d-e x)}+\frac{2 \sqrt{d^2-e^2 x^2}}{15 d^2 e (d-e x)^2}+\frac{\sqrt{d^2-e^2 x^2}}{5 d e (d-e x)^3} \]

[Out]

Sqrt[d^2 - e^2*x^2]/(5*d*e*(d - e*x)^3) + (2*Sqrt[d^2 - e^2*x^2])/(15*d^2*e*(d - e*x)^2) + (2*Sqrt[d^2 - e^2*x
^2])/(15*d^3*e*(d - e*x))

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Rubi [A]  time = 0.0436404, antiderivative size = 103, 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 = {655, 659, 651} \[ \frac{2 \sqrt{d^2-e^2 x^2}}{15 d^3 e (d-e x)}+\frac{2 \sqrt{d^2-e^2 x^2}}{15 d^2 e (d-e x)^2}+\frac{\sqrt{d^2-e^2 x^2}}{5 d e (d-e x)^3} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(d^2 - e^2*x^2)^(7/2),x]

[Out]

Sqrt[d^2 - e^2*x^2]/(5*d*e*(d - e*x)^3) + (2*Sqrt[d^2 - e^2*x^2])/(15*d^2*e*(d - e*x)^2) + (2*Sqrt[d^2 - e^2*x
^2])/(15*d^3*e*(d - e*x))

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 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 651

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*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rubi steps

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

Mathematica [A]  time = 0.0625678, size = 58, normalized size = 0.56 \[ \frac{(d+e x) \left (7 d^2-6 d e x+2 e^2 x^2\right )}{15 d^3 e (d-e x)^2 \sqrt{d^2-e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3/(d^2 - e^2*x^2)^(7/2),x]

[Out]

((d + e*x)*(7*d^2 - 6*d*e*x + 2*e^2*x^2))/(15*d^3*e*(d - e*x)^2*Sqrt[d^2 - e^2*x^2])

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Maple [A]  time = 0.043, size = 55, normalized size = 0.5 \begin{align*}{\frac{ \left ( ex+d \right ) ^{4} \left ( -ex+d \right ) \left ( 2\,{e}^{2}{x}^{2}-6\,dex+7\,{d}^{2} \right ) }{15\,{d}^{3}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((e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x)

[Out]

1/15*(e*x+d)^4*(-e*x+d)*(2*e^2*x^2-6*d*e*x+7*d^2)/d^3/e/(-e^2*x^2+d^2)^(7/2)

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Maxima [A]  time = 1.30242, size = 136, normalized size = 1.32 \begin{align*} \frac{e x^{2}}{3 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} + \frac{4 \, d x}{5 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}}} + \frac{7 \, d^{2}}{15 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{5}{2}} e} + \frac{x}{15 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} d} + \frac{2 \, x}{15 \, \sqrt{-e^{2} x^{2} + d^{2}} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/3*e*x^2/(-e^2*x^2 + d^2)^(5/2) + 4/5*d*x/(-e^2*x^2 + d^2)^(5/2) + 7/15*d^2/((-e^2*x^2 + d^2)^(5/2)*e) + 1/15
*x/((-e^2*x^2 + d^2)^(3/2)*d) + 2/15*x/(sqrt(-e^2*x^2 + d^2)*d^3)

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Fricas [A]  time = 2.13638, size = 215, normalized size = 2.09 \begin{align*} \frac{7 \, e^{3} x^{3} - 21 \, d e^{2} x^{2} + 21 \, d^{2} e x - 7 \, d^{3} -{\left (2 \, e^{2} x^{2} - 6 \, d e x + 7 \, d^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{3} e^{4} x^{3} - 3 \, d^{4} e^{3} x^{2} + 3 \, d^{5} e^{2} x - d^{6} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(7*e^3*x^3 - 21*d*e^2*x^2 + 21*d^2*e*x - 7*d^3 - (2*e^2*x^2 - 6*d*e*x + 7*d^2)*sqrt(-e^2*x^2 + d^2))/(d^3
*e^4*x^3 - 3*d^4*e^3*x^2 + 3*d^5*e^2*x - d^6*e)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral((d + e*x)**3/(-(-d + e*x)*(d + e*x))**(7/2), x)

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Giac [A]  time = 1.27872, size = 95, normalized size = 0.92 \begin{align*} -\frac{\sqrt{-x^{2} e^{2} + d^{2}}{\left (7 \, d^{2} e^{\left (-1\right )} +{\left ({\left (x{\left (\frac{2 \, x^{2} e^{4}}{d^{3}} - \frac{5 \, e^{2}}{d}\right )} + 5 \, e\right )} x + 15 \, d\right )} x\right )}}{15 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-1/15*sqrt(-x^2*e^2 + d^2)*(7*d^2*e^(-1) + ((x*(2*x^2*e^4/d^3 - 5*e^2/d) + 5*e)*x + 15*d)*x)/(x^2*e^2 - d^2)^3