3.9.6 \(\int \frac {(d^2-e^2 x^2)^{7/2}}{(d+e x)^4} \, dx\) [806]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 136, normalized size of antiderivative = 1.00, 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 = {677, 669, 685, 655, 201, 223, 209} \begin {gather*} \frac {35 d^4 \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e}+\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} \end {gather*}

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 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 655

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 669

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 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 685

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
]

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 ) \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}{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 = 0.27, size = 100, normalized size = 0.74 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (160 d^3-81 d^2 e x+32 d e^2 x^2-6 e^3 x^3\right )}{24 e}-\frac {35 d^4 \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{8 \sqrt {-e^2}} \end {gather*}

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))/(24*e) - (35*d^4*Log[-(Sqrt[-e^2]*x) +
Sqrt[d^2 - e^2*x^2]])/(8*Sqrt[-e^2])

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(402\) vs. \(2(118)=236\).
time = 0.48, size = 403, normalized size = 2.96

method	result	size
risch	\(\frac {\left (-6 e^{3} x^{3}+32 d \,e^{2} x^{2}-81 d^{2} e x +160 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{24 e}+\frac {35 d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}\)	\(83\)
default	\(\frac {\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {9}{2}}}{d e \left (x +\frac {d}{e}\right )^{4}}+\frac {5 e \left (\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {9}{2}}}{3 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {9}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{2}}+\frac {7 e \left (\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{7}+d e \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{12 e^{2}}+\frac {5 d^{2} \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{6}\right )\right )}{5 d}\right )}{d}\right )}{d}}{e^{4}}\)	\(403\)

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,method=_RETURNVERBOSE)

[Out]

1/e^4*(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)*arctan((e^2)^(1/2)*x/(-e^2*(x+d/e)^2+2*d*e*(x+d/e))^(1/2)))))))))

________________________________________________________________________________________

Maxima [A]
time = 0.54, size = 148, normalized size = 1.09 \begin {gather*} \frac {35}{8} \, d^{4} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} + \frac {35}{8} \, \sqrt {-x^{2} e^{2} + d^{2}} d^{3} e^{\left (-1\right )} + \frac {{\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {7}{2}}}{4 \, {\left (x^{3} e^{4} + 3 \, d x^{2} e^{3} + 3 \, d^{2} x e^{2} + d^{3} e\right )}} + \frac {7 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {5}{2}} d}{12 \, {\left (x^{2} e^{3} + 2 \, d x e^{2} + d^{2} e\right )}} + \frac {35 \, {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{24 \, {\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)^4,x, algorithm="maxima")

[Out]

35/8*d^4*arcsin(x*e/d)*e^(-1) + 35/8*sqrt(-x^2*e^2 + d^2)*d^3*e^(-1) + 1/4*(-x^2*e^2 + d^2)^(7/2)/(x^3*e^4 + 3
*d*x^2*e^3 + 3*d^2*x*e^2 + d^3*e) + 7/12*(-x^2*e^2 + d^2)^(5/2)*d/(x^2*e^3 + 2*d*x*e^2 + d^2*e) + 35/24*(-x^2*
e^2 + d^2)^(3/2)*d^2/(x*e^2 + d*e)

________________________________________________________________________________________

Fricas [A]
time = 1.86, size = 78, normalized size = 0.57 \begin {gather*} -\frac {1}{24} \, {\left (210 \, d^{4} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (6 \, x^{3} e^{3} - 32 \, d x^{2} e^{2} + 81 \, d^{2} x e - 160 \, d^{3}\right )} \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\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)^4,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.91, size = 64, normalized size = 0.47 \begin {gather*} \frac {35}{8} \, d^{4} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\left (d\right ) + \frac {1}{24} \, {\left (160 \, d^{3} e^{\left (-1\right )} - {\left (81 \, d^{2} + 2 \, {\left (3 \, x e^{2} - 16 \, d e\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \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)^4,x, algorithm="giac")

[Out]

35/8*d^4*arcsin(x*e/d)*e^(-1)*sgn(d) + 1/24*(160*d^3*e^(-1) - (81*d^2 + 2*(3*x*e^2 - 16*d*e)*x)*x)*sqrt(-x^2*e
^2 + d^2)

________________________________________________________________________________________

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 )}^4} \,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)^4,x)

[Out]

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

________________________________________________________________________________________