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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 103, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.06, 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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fricas [A]  time = 0.83, size = 106, normalized size = 1.03 \[ \frac {7 \, e^{3} x^{3} - 21 \, d e^{2} x^{2} + 21 \, d^{2} e x - 7 \, d^{3} - {\left (2 \, e^{2} x^{2} - 6 \, d e x + 7 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{3} e^{4} x^{3} - 3 \, d^{4} e^{3} x^{2} + 3 \, d^{5} e^{2} x - d^{6} e\right )}} \]

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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giac [A]  time = 0.31, size = 70, normalized size = 0.68 \[ -\frac {\sqrt {-x^{2} e^{2} + d^{2}} {\left (7 \, d^{2} e^{\left (-1\right )} + {\left ({\left (x {\left (\frac {2 \, x^{2} e^{4}}{d^{3}} - \frac {5 \, e^{2}}{d}\right )} + 5 \, e\right )} x + 15 \, d\right )} x\right )}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} \]

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

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maple [A]  time = 0.05, size = 55, normalized size = 0.53 \[ \frac {\left (e x +d \right )^{4} \left (-e x +d \right ) \left (2 e^{2} x^{2}-6 d e x +7 d^{2}\right )}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} d^{3} e} \]

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 = 1.39, size = 101, normalized size = 0.98 \[ \frac {e x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {4 \, d x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {7 \, d^{2}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d} + \frac {2 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}} \]

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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mupad [B]  time = 0.52, size = 49, normalized size = 0.48 \[ \frac {\sqrt {d^2-e^2\,x^2}\,\left (7\,d^2-6\,d\,e\,x+2\,e^2\,x^2\right )}{15\,d^3\,e\,{\left (d-e\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]

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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