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]

-35/3*(-e^2*x^2+d^2)^(3/2)/e-14*(-e^2*x^2+d^2)^(5/2)/e/(e*x+d)^2-2*(-e^2*x^2+d^2)^(7/2)/e/(e*x+d)^4-35/2*d^3*a
rctan(e*x/(-e^2*x^2+d^2)^(1/2))/e-35/2*d*x*(-e^2*x^2+d^2)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 132, normalized size of antiderivative = 1.00, 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 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.10, 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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fricas [A]  time = 0.65, size = 110, normalized size = 0.83 \[ -\frac {166 \, d^{3} e x + 166 \, d^{4} - 210 \, {\left (d^{3} e x + d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (2 \, e^{3} x^{3} - 13 \, d e^{2} x^{2} + 55 \, d^{2} e x + 166 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, {\left (e^{2} x + d e\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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

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maple [B]  time = 0.06, size = 364, normalized size = 2.76 \[ -\frac {35 d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}}\right )}{2 \sqrt {e^{2}}}-\frac {35 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d x}{2}-\frac {35 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} x}{3 d}-\frac {28 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} x}{3 d^{3}}-\frac {8 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{d^{4} e}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {9}{2}}}{\left (x +\frac {d}{e}\right )^{5} d \,e^{6}}-\frac {4 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {9}{2}}}{\left (x +\frac {d}{e}\right )^{4} d^{2} e^{5}}-\frac {20 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {9}{2}}}{3 \left (x +\frac {d}{e}\right )^{3} d^{3} e^{4}}-\frac {8 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {9}{2}}}{\left (x +\frac {d}{e}\right )^{2} d^{4} e^{3}} \]

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/(x+d/e)^5*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(9/2)-4/e^5/d^2/(x+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(9/2)
-20/3/e^4/d^3/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(9/2)-8/e^3/d^4/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^
(9/2)-8/e/d^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)-28/3/d^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)*x-35/3/d*(2*(x+
d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x-35/2*d*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x-35/2*d^3/(e^2)^(1/2)*arctan((e^2)
^(1/2)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)

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maxima [A]  time = 3.06, size = 197, normalized size = 1.49 \[ -\frac {35 \, d^{3} \arcsin \left (\frac {e x}{d}\right )}{2 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}}{3 \, {\left (e^{5} x^{4} + 4 \, d e^{4} x^{3} + 6 \, d^{2} e^{3} x^{2} + 4 \, d^{3} e^{2} x + d^{4} e\right )}} + \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{6 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} + \frac {35 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{6 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} - \frac {35 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{e^{2} x + d e} \]

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]

-35/2*d^3*arcsin(e*x/d)/e + 1/3*(-e^2*x^2 + d^2)^(7/2)/(e^5*x^4 + 4*d*e^4*x^3 + 6*d^2*e^3*x^2 + 4*d^3*e^2*x +
d^4*e) + 7/6*(-e^2*x^2 + d^2)^(5/2)*d/(e^4*x^3 + 3*d*e^3*x^2 + 3*d^2*e^2*x + d^3*e) + 35/6*(-e^2*x^2 + d^2)^(3
/2)*d^2/(e^3*x^2 + 2*d*e^2*x + d^2*e) - 35*sqrt(-e^2*x^2 + d^2)*d^3/(e^2*x + d*e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d^2-e^2\,x^2\right )}^{7/2}}{{\left (d+e\,x\right )}^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}{\left (d + e x\right )^{5}}\, dx \]

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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