3.2.100 \(\int \frac {x^3 (d^2-e^2 x^2)^{5/2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=192 \[ \frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}+\frac {27 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4}+\frac {101 d^4 \sqrt {d^2-e^2 x^2}}{5 e^4}-\frac {19 d^3 x \sqrt {d^2-e^2 x^2}}{2 e^3} \]

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Rubi [A]  time = 0.44, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {852, 1635, 1815, 641, 217, 203} \begin {gather*} \frac {101 d^4 \sqrt {d^2-e^2 x^2}}{5 e^4}-\frac {19 d^3 x \sqrt {d^2-e^2 x^2}}{2 e^3}+\frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}+\frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {27 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^4,x]

[Out]

(d^2*(d - e*x)^4)/(e^4*Sqrt[d^2 - e^2*x^2]) + (101*d^4*Sqrt[d^2 - e^2*x^2])/(5*e^4) - (19*d^3*x*Sqrt[d^2 - e^2
*x^2])/(2*e^3) + (18*d^2*x^2*Sqrt[d^2 - e^2*x^2])/(5*e^2) - (d*x^3*Sqrt[d^2 - e^2*x^2])/e + (x^4*Sqrt[d^2 - e^
2*x^2])/5 + (27*d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e^4)

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 852

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^4} \, dx &=\int \frac {x^3 (d-e x)^4}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx\\ &=\frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {(d-e x)^3 \left (-\frac {4 d^3}{e^3}+\frac {d^2 x}{e^2}-\frac {d x^2}{e}\right )}{\sqrt {d^2-e^2 x^2}} \, dx}{d}\\ &=\frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {\int \frac {\frac {20 d^6}{e}-65 d^5 x+80 d^4 e x^2-54 d^3 e^2 x^3+20 d^2 e^3 x^4}{\sqrt {d^2-e^2 x^2}} \, dx}{5 d e^2}\\ &=\frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {-80 d^6 e+260 d^5 e^2 x-380 d^4 e^3 x^2+216 d^3 e^4 x^3}{\sqrt {d^2-e^2 x^2}} \, dx}{20 d e^4}\\ &=\frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}+\frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {\int \frac {240 d^6 e^3-1212 d^5 e^4 x+1140 d^4 e^5 x^2}{\sqrt {d^2-e^2 x^2}} \, dx}{60 d e^6}\\ &=\frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}-\frac {19 d^3 x \sqrt {d^2-e^2 x^2}}{2 e^3}+\frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {-1620 d^6 e^5+2424 d^5 e^6 x}{\sqrt {d^2-e^2 x^2}} \, dx}{120 d e^8}\\ &=\frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}+\frac {101 d^4 \sqrt {d^2-e^2 x^2}}{5 e^4}-\frac {19 d^3 x \sqrt {d^2-e^2 x^2}}{2 e^3}+\frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {\left (27 d^5\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^3}\\ &=\frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}+\frac {101 d^4 \sqrt {d^2-e^2 x^2}}{5 e^4}-\frac {19 d^3 x \sqrt {d^2-e^2 x^2}}{2 e^3}+\frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {\left (27 d^5\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3}\\ &=\frac {d^2 (d-e x)^4}{e^4 \sqrt {d^2-e^2 x^2}}+\frac {101 d^4 \sqrt {d^2-e^2 x^2}}{5 e^4}-\frac {19 d^3 x \sqrt {d^2-e^2 x^2}}{2 e^3}+\frac {18 d^2 x^2 \sqrt {d^2-e^2 x^2}}{5 e^2}-\frac {d x^3 \sqrt {d^2-e^2 x^2}}{e}+\frac {1}{5} x^4 \sqrt {d^2-e^2 x^2}+\frac {27 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^4}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 109, normalized size = 0.57 \begin {gather*} \frac {135 d^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {\sqrt {d^2-e^2 x^2} \left (212 d^5+77 d^4 e x-29 d^3 e^2 x^2+16 d^2 e^3 x^3-8 d e^4 x^4+2 e^5 x^5\right )}{d+e x}}{10 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^4,x]

[Out]

((Sqrt[d^2 - e^2*x^2]*(212*d^5 + 77*d^4*e*x - 29*d^3*e^2*x^2 + 16*d^2*e^3*x^3 - 8*d*e^4*x^4 + 2*e^5*x^5))/(d +
e*x) + 135*d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(10*e^4)

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IntegrateAlgebraic [A]  time = 0.54, size = 132, normalized size = 0.69 \begin {gather*} \frac {27 d^5 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{2 e^5}+\frac {\sqrt {d^2-e^2 x^2} \left (212 d^5+77 d^4 e x-29 d^3 e^2 x^2+16 d^2 e^3 x^3-8 d e^4 x^4+2 e^5 x^5\right )}{10 e^4 (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^4,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(212*d^5 + 77*d^4*e*x - 29*d^3*e^2*x^2 + 16*d^2*e^3*x^3 - 8*d*e^4*x^4 + 2*e^5*x^5))/(10*e
^4*(d + e*x)) + (27*d^5*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(2*e^5)

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fricas [A]  time = 0.40, size = 134, normalized size = 0.70 \begin {gather*} \frac {212 \, d^{5} e x + 212 \, d^{6} - 270 \, {\left (d^{5} e x + d^{6}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (2 \, e^{5} x^{5} - 8 \, d e^{4} x^{4} + 16 \, d^{2} e^{3} x^{3} - 29 \, d^{3} e^{2} x^{2} + 77 \, d^{4} e x + 212 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{10 \, {\left (e^{5} x + d e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/10*(212*d^5*e*x + 212*d^6 - 270*(d^5*e*x + d^6)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (2*e^5*x^5 - 8*d
*e^4*x^4 + 16*d^2*e^3*x^3 - 29*d^3*e^2*x^2 + 77*d^4*e*x + 212*d^5)*sqrt(-e^2*x^2 + d^2))/(e^5*x + d*e^4)

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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(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (-30*d^5*(-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)^2-6*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/ex
p(2))^5*exp(1)^10*exp(2)^3+432*d^5*(-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)^2+396*d^5*(-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)^3+90*d^5*(-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)^4-102*d^5*(-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)^2+62*d^5*(-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)^3+63*d^5*(-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)^4+9*d^5*(-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)^5+900*d^5*(-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)^3-144*d^5*(-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)^4-648*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/e
xp(2))^4*exp(1)^6*exp(2)^5-180*d^5*(-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)^6+288*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^8*exp(2)^4-312*d^5*(-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)^5-432*d^5*(-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)^6-108*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/e
xp(2))^5*exp(2)^8-1404*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^6*exp(2)^5-118
8*d^5*(-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)^6-11*d^5*exp(1)^8*exp(2)^4
-324*d^5*(-1/2*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^4*exp(2)^8-984*d^5*(-1/2*(-2*d*exp(1)-2*s
qrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(1)^4*exp(2)^6+108*d^5*exp(1)^6*exp(2)^5-756*d^5*(-1/2*(-2*d*exp(1)
-2*sqrt(d^2-x^2*exp(2))*exp(1))/x/exp(2))^3*exp(2)^8+32*d^5*exp(1)^4*exp(2)^6-648*d^5*(-1/2*(-2*d*exp(1)-2*sqr
t(d^2-x^2*exp(2))*exp(1))/x/exp(2))^2*exp(2)^8-324*d^5*exp(2)^8-44*d^5*(-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)+324*d^5*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(2)^8/x/exp(2)+
504*d^5*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^4*exp(2)^6/x/exp(2)-183/2*d^5*(-2*d*exp(1)-2*sqrt(d
^2-x^2*exp(2))*exp(1))*exp(1)^6*exp(2)^5/x/exp(2)-279*d^5*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^8
*exp(2)^4/x/exp(2)+30*d^5*(-2*d*exp(1)-2*sqrt(d^2-x^2*exp(2))*exp(1))*exp(1)^10*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))*exp(1))/x+exp(2
))^3/(3*exp(1)^14+9*exp(1)^10*exp(2)^2+3*exp(1)^8*exp(2)^3+9*exp(1)^12*exp(2))+1/2*(-108*d^5*exp(1)^10*exp(2)^
2+90*d^5*exp(1)^8*exp(2)^3+472*d^5*exp(1)^6*exp(2)^4+72*d^5*exp(1)^4*exp(2)^5-656*d^5*exp(2)^7+4*d^5*exp(1)^12
*exp(2))*atan((-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)^16-3*exp(1)^12*exp(2)^2-exp(1)^10*exp(2)^3-3*exp(1)^14*exp(2))+27/2*d^5*sign(d)*asin(x
*exp(2)/d/exp(1))/exp(1)^4+2*((((24*exp(1)^13*1/240/exp(1)^13*x-120*exp(1)^12*d*1/240/exp(1)^13)*x+312*exp(1)^
11*d^2*1/240/exp(1)^13)*x-660*exp(1)^10*d^3*1/240/exp(1)^13)*x+1584*exp(1)^9*d^4*1/240/exp(1)^13)*sqrt(d^2-x^2
*exp(2))

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maple [A]  time = 0.02, size = 285, normalized size = 1.48 \begin {gather*} \frac {27 d^{5} \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}}\, e^{3}}+\frac {27 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{3} x}{2 e^{3}}+\frac {9 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d x}{e^{3}}+\frac {36 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}}}{5 e^{4}}+\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}} d^{2}}{\left (x +\frac {d}{e}\right )^{4} e^{8}}+\frac {6 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}} d}{\left (x +\frac {d}{e}\right )^{3} e^{7}}+\frac {7 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {7}{2}}}{\left (x +\frac {d}{e}\right )^{2} e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x)

[Out]

d^2/e^8/(x+d/e)^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)+6*d/e^7/(x+d/e)^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)+7/
e^6/(x+d/e)^2*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(7/2)+36/5/e^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)+9*d/e^3*(2*(x+d
/e)*d*e-(x+d/e)^2*e^2)^(3/2)*x+27/2*d^3/e^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x+27/2*d^5/e^3/(e^2)^(1/2)*arc
tan((e^2)^(1/2)/(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x)

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maxima [C]  time = 1.03, size = 407, normalized size = 2.12 \begin {gather*} -\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}}{2 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} + \frac {15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5}}{e^{5} x + d e^{4}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}}{e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}}{2 \, {\left (e^{5} x + d e^{4}\right )}} - \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{4 \, {\left (e^{5} x + d e^{4}\right )}} + \frac {3 i \, d^{5} \arcsin \left (\frac {e x}{d} + 2\right )}{2 \, e^{4}} + \frac {15 \, d^{5} \arcsin \left (\frac {e x}{d}\right )}{e^{4}} - \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{3} x}{2 \, e^{3}} - \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{4}}{e^{4}} + \frac {15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4}}{2 \, e^{4}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d x}{4 \, e^{3}} - \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{4 \, e^{4}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{5 \, e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

-1/2*(-e^2*x^2 + d^2)^(5/2)*d^3/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4) - 5/2*(-e^2*x^2 + d^2)^(3/2)*d
^4/(e^6*x^2 + 2*d*e^5*x + d^2*e^4) + 15*sqrt(-e^2*x^2 + d^2)*d^5/(e^5*x + d*e^4) + (-e^2*x^2 + d^2)^(5/2)*d^2/
(e^6*x^2 + 2*d*e^5*x + d^2*e^4) + 5/2*(-e^2*x^2 + d^2)^(3/2)*d^3/(e^5*x + d*e^4) - 3/4*(-e^2*x^2 + d^2)^(5/2)*
d/(e^5*x + d*e^4) + 3/2*I*d^5*arcsin(e*x/d + 2)/e^4 + 15*d^5*arcsin(e*x/d)/e^4 - 3/2*sqrt(e^2*x^2 + 4*d*e*x +
3*d^2)*d^3*x/e^3 - 3*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^4/e^4 + 15/2*sqrt(-e^2*x^2 + d^2)*d^4/e^4 + 1/4*(-e^2*x
^2 + d^2)^(3/2)*d*x/e^3 - 5/4*(-e^2*x^2 + d^2)^(3/2)*d^2/e^4 + 1/5*(-e^2*x^2 + d^2)^(5/2)/e^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^4,x)

[Out]

int((x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(-e**2*x**2+d**2)**(5/2)/(e*x+d)**4,x)

[Out]

Integral(x**3*(-(-d + e*x)*(d + e*x))**(5/2)/(d + e*x)**4, x)

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