3.9.7 \(\int \frac {(d^2-e^2 x^2)^{7/2}}{(d+e x)^5} \, dx\) [807]

Optimal. Leaf size=132 \[ -\frac {35}{2} d x \sqrt {d^2-e^2 x^2}-\frac {35 \left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac {14 \left (d^2-e^2 x^2\right )^{5/2}}{e (d+e x)^2}-\frac {2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^4}-\frac {35 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \]

[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 = 0.04, 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 = {677, 679, 201, 223, 209} \begin {gather*} -\frac {35 d^3 \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}-\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} \end {gather*}

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 201

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

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 677

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 679

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 ) \text {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 = 0.35, size = 107, normalized size = 0.81 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-166 d^3-55 d^2 e x+13 d e^2 x^2-2 e^3 x^3\right )}{6 e (d+e x)}+\frac {35 d^3 \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{2 \sqrt {-e^2}} \end {gather*}

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]*(-166*d^3 - 55*d^2*e*x + 13*d*e^2*x^2 - 2*e^3*x^3))/(6*e*(d + e*x)) + (35*d^3*Log[-(Sqrt[
-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(2*Sqrt[-e^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(454\) vs. \(2(116)=232\).
time = 0.50, size = 455, normalized size = 3.45

method	result	size
risch	\(-\frac {\left (2 e^{2} x^{2}-15 d x e +70 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{6 e}-\frac {35 d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}-\frac {16 d^{3} \sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}{e^{2} \left (x +\frac {d}{e}\right )}\)	\(117\)
default	\(\frac {-\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {9}{2}}}{d e \left (x +\frac {d}{e}\right )^{5}}-\frac {4 e \left (\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {9}{2}}}{d e \left (x +\frac {d}{e}\right )^{4}}+\frac {5 e \left (\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {9}{2}}}{3 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {9}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{2}}+\frac {7 e \left (\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{7}+d e \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{12 e^{2}}+\frac {5 d^{2} \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{6}\right )\right )}{5 d}\right )}{d}\right )}{d}\right )}{d}}{e^{5}}\)	\(455\)

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,method=_RETURNVERBOSE)

[Out]

1/e^5*(-1/d/e/(x+d/e)^5*(-e^2*(x+d/e)^2+2*d*e*(x+d/e))^(9/2)-4*e/d*(1/d/e/(x+d/e)^4*(-e^2*(x+d/e)^2+2*d*e*(x+d
/e))^(9/2)+5*e/d*(1/3/d/e/(x+d/e)^3*(-e^2*(x+d/e)^2+2*d*e*(x+d/e))^(9/2)+2*e/d*(1/5/d/e/(x+d/e)^2*(-e^2*(x+d/e
)^2+2*d*e*(x+d/e))^(9/2)+7/5*e/d*(1/7*(-e^2*(x+d/e)^2+2*d*e*(x+d/e))^(7/2)+d*e*(-1/12*(-2*e^2*(x+d/e)+2*d*e)/e
^2*(-e^2*(x+d/e)^2+2*d*e*(x+d/e))^(5/2)+5/6*d^2*(-1/8*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-e^2*(x+d/e)^2+2*d*e*(x+d/e)
)^(3/2)+3/4*d^2*(-1/4*(-2*e^2*(x+d/e)+2*d*e)/e^2*(-e^2*(x+d/e)^2+2*d*e*(x+d/e))^(1/2)+1/2*d^2/(e^2)^(1/2)*arct
an((e^2)^(1/2)*x/(-e^2*(x+d/e)^2+2*d*e*(x+d/e))^(1/2))))))))))

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Maxima [A]
time = 0.57, size = 187, normalized size = 1.42 \begin {gather*} -\frac {35}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} + \frac {{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {7}{2}}}{3 \, {\left (x^{4} e^{5} + 4 \, d x^{3} e^{4} + 6 \, d^{2} x^{2} e^{3} + 4 \, d^{3} x e^{2} + d^{4} e\right )}} + \frac {7 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} d}{6 \, {\left (x^{3} e^{4} + 3 \, d x^{2} e^{3} + 3 \, d^{2} x e^{2} + d^{3} e\right )}} + \frac {35 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{6 \, {\left (x^{2} e^{3} + 2 \, d x e^{2} + d^{2} e\right )}} - \frac {35 \, \sqrt {-x^{2} e^{2} + d^{2}} d^{3}}{x e^{2} + d e} \end {gather*}

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

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Fricas [A]
time = 2.17, size = 108, normalized size = 0.82 \begin {gather*} -\frac {166 \, d^{3} x e + 166 \, d^{4} - 210 \, {\left (d^{3} x e + d^{4}\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (2 \, x^{3} e^{3} - 13 \, d x^{2} e^{2} + 55 \, d^{2} x e + 166 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{6 \, {\left (x e^{2} + d e\right )}} \end {gather*}

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*x*e + 166*d^4 - 210*(d^3*x*e + d^4)*arctan(-(d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x) + (2*x^3*e^3 -
13*d*x^2*e^2 + 55*d^2*x*e + 166*d^3)*sqrt(-x^2*e^2 + d^2))/(x*e^2 + d*e)

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

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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Giac [A]
time = 1.15, size = 178, normalized size = 1.35 \begin {gather*} \frac {{\left (840 \, d^{4} \arctan \left (\sqrt {\frac {2 \, d}{x e + d} - 1}\right ) e^{4} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) - 384 \, d^{4} \sqrt {\frac {2 \, d}{x e + d} - 1} e^{4} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) - \frac {{\left (87 \, d^{4} {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {5}{2}} e^{4} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) + 136 \, d^{4} {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {3}{2}} e^{4} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) + 57 \, d^{4} \sqrt {\frac {2 \, d}{x e + d} - 1} e^{4} \mathrm {sgn}\left (\frac {1}{x e + d}\right )\right )} {\left (x e + d\right )}^{3}}{d^{3}}\right )} e^{\left (-5\right )}}{24 \, d} \end {gather*}

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]

1/24*(840*d^4*arctan(sqrt(2*d/(x*e + d) - 1))*e^4*sgn(1/(x*e + d)) - 384*d^4*sqrt(2*d/(x*e + d) - 1)*e^4*sgn(1
/(x*e + d)) - (87*d^4*(2*d/(x*e + d) - 1)^(5/2)*e^4*sgn(1/(x*e + d)) + 136*d^4*(2*d/(x*e + d) - 1)^(3/2)*e^4*s
gn(1/(x*e + d)) + 57*d^4*sqrt(2*d/(x*e + d) - 1)*e^4*sgn(1/(x*e + d)))*(x*e + d)^3/d^3)*e^(-5)/d

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

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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