3.7.47 \(\int \frac {(d^2-e^2 x^2)^{7/2}}{(d+e x)^4} \, dx\)

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

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Rubi [A]  time = 0.06, 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 = {663, 655, 671, 641, 195, 217, 203} \begin {gather*} \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}+\frac {35 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 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 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 641

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 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 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 671

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 ) \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}{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*}

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Mathematica [A]  time = 0.08, size = 80, normalized size = 0.59 \begin {gather*} \frac {105 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\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} \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) + 105*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x
^2]])/(24*e)

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IntegrateAlgebraic [A]  time = 0.37, size = 103, normalized size = 0.76 \begin {gather*} \frac {35 d^4 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{8 e^2}+\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.41, size = 83, normalized size = 0.61 \begin {gather*} -\frac {210 \, d^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\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} \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(-e^2*x^2 + d^2))/(e*x)) + (6*e^3*x^3 - 32*d*e^2*x^2 + 81*d^2*e*x - 160*d^3)*s
qrt(-e^2*x^2 + d^2))/e

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \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]

Exception raised: NotImplementedError >> Unable to parse Giac output: (12*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^
2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^14*exp(2)^2+6*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp
(2))^5*exp(1)^12*exp(2)^3-8*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^14*exp(2)
^2+36*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^12*exp(2)^3+18*d^4*(-1/2*(-2*d*
exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^10*exp(2)^4+12*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*
exp(2))*exp(1))/x/exp(2))^2*exp(1)^14*exp(2)^2+568*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp
(2))^3*exp(1)^12*exp(2)^3+540*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^10*exp(
2)^4+126*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^8*exp(2)^5-60*d^4*(-1/2*(-2*
d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^12*exp(2)^3+296*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x
^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^10*exp(2)^4+228*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/
exp(2))^4*exp(1)^8*exp(2)^5+42*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(1)^6*exp(
2)^6+1134*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^10*exp(2)^4+330*d^4*(-1/2*(
-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^8*exp(2)^5-351*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2
-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^6*exp(2)^6-117*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x
/exp(2))^5*exp(1)^4*exp(2)^7+690*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^8*ex
p(2)^5+102*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^6*exp(2)^6-249*d^4*(-1/2*(
-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(1)^4*exp(2)^7+2*d^4*exp(1)^10*exp(2)^4-75*d^4*(-1/2
*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^5*exp(2)^9-744*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*ex
p(2))*exp(1))/x/exp(2))^2*exp(1)^6*exp(2)^6-756*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2)
)^3*exp(1)^4*exp(2)^7-2*d^4*exp(1)^8*exp(2)^5-216*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(
2))^4*exp(2)^9-600*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^4*exp(2)^7+139*d^4
*exp(1)^6*exp(2)^6-540*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(2)^9+77*d^4*exp(1
)^4*exp(2)^7-432*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)^9-216*d^4*exp(2)^9+8
*d^4*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(1)^16*exp(2)+465/2*d^4*(-2*d*exp(1)-2*s
qrt(d^2-x^2*exp(2))*exp(1))*exp(2)^9/x/exp(2)+639/2*d^4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^4*e
xp(2)^7/x/exp(2)-210*d^4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^6*exp(2)^6/x/exp(2)-354*d^4*(-2*d*
exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^8*exp(2)^5/x/exp(2)+15*d^4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*ex
p(1))*exp(1)^10*exp(2)^4/x/exp(2)-3*d^4*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^12*exp(2)^3/x/exp(2
))/((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)-(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*e
xp(1))/x+exp(2))^3/(3*exp(1)^13+9*exp(1)^9*exp(2)^2+3*exp(1)^7*exp(2)^3+9*exp(1)^11*exp(2))+1/2*(96*d^4*exp(1)
^10*exp(2)^3-64*d^4*exp(1)^8*exp(2)^4-346*d^4*exp(1)^6*exp(2)^5-86*d^4*exp(1)^4*exp(2)^6+400*d^4*exp(2)^8)*ata
n((-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x+exp(2))/sqrt(-exp(1)^4+exp(2)^2))/sqrt(-exp(1)^4+exp(2)^
2)/(exp(1)^15+3*exp(1)^11*exp(2)^2+exp(1)^9*exp(2)^3+3*exp(1)^13*exp(2))+35/8*d^4*sign(d)*asin(x*exp(2)/d/exp(
1))/exp(1)+2*(((-12*exp(1)^6*1/96/exp(1)^4*x+64*exp(1)^5*d*1/96/exp(1)^4)*x-162*exp(1)^4*d^2*1/96/exp(1)^4)*x+
320*exp(1)^3*d^3*1/96/exp(1)^4)*sqrt(d^2-x^2*exp(2))

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maple [B]  time = 0.06, size = 317, normalized size = 2.33 \begin {gather*} \frac {35 d^{4} \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 )}{8 \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^{2} x}{8}+\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}{12}+\frac {7 \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^{2}}+\frac {2 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{d^{3} 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 )^{4} d \,e^{5}}+\frac {5 \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^{2} e^{4}}+\frac {2 \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^{3} e^{3}} \end {gather*}

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)

[Out]

1/e^5/d/(x+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(9/2)+5/3/e^4/d^2/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(9/2
)+2/e^3/d^3/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(9/2)+2/e/d^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)+7/3/d^2*
(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)*x+35/12*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x+35/8*d^2*(2*(x+d/e)*d*e-(x+d
/e)^2*e^2)^(1/2)*x+35/8*d^4/(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 = 3.02, size = 156, normalized size = 1.15 \begin {gather*} \frac {35 \, d^{4} \arcsin \left (\frac {e x}{d}\right )}{8 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}}{4 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} + \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{12 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} + \frac {35 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{24 \, {\left (e^{2} x + d e\right )}} + \frac {35 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{8 \, 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)^4,x, algorithm="maxima")

[Out]

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

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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 )}^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)

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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 )^{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)

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