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

Optimal. Leaf size=172 \[ -\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac {3 d^8 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{128 e^4}-\frac {3 d^6 x \sqrt {d^2-e^2 x^2}}{128 e^3}-\frac {d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3} \]

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Rubi [A]  time = 0.12, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {850, 833, 780, 195, 217, 203} \begin {gather*} -\frac {3 d^6 x \sqrt {d^2-e^2 x^2}}{128 e^3}-\frac {d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac {d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {3 d^8 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{128 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d^6*x*Sqrt[d^2 - e^2*x^2])/(128*e^3) - (d^4*x*(d^2 - e^2*x^2)^(3/2))/(64*e^3) - (d*x^2*(d^2 - e^2*x^2)^(5/
2))/(7*e^2) + (x^3*(d^2 - e^2*x^2)^(5/2))/(8*e) - (d^2*(32*d - 35*e*x)*(d^2 - e^2*x^2)^(5/2))/(560*e^4) - (3*d
^8*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(128*e^4)

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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 833

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

Rule 850

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

Rubi steps

\begin {align*} \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx &=\int x^3 (d-e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx\\ &=\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {\int x^2 \left (3 d^2 e-8 d e^2 x\right ) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{8 e^2}\\ &=-\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}+\frac {\int x \left (16 d^3 e^2-21 d^2 e^3 x\right ) \left (d^2-e^2 x^2\right )^{3/2} \, dx}{56 e^4}\\ &=-\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac {d^4 \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{16 e^3}\\ &=-\frac {d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac {\left (3 d^6\right ) \int \sqrt {d^2-e^2 x^2} \, dx}{64 e^3}\\ &=-\frac {3 d^6 x \sqrt {d^2-e^2 x^2}}{128 e^3}-\frac {d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac {\left (3 d^8\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{128 e^3}\\ &=-\frac {3 d^6 x \sqrt {d^2-e^2 x^2}}{128 e^3}-\frac {d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac {\left (3 d^8\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{128 e^3}\\ &=-\frac {3 d^6 x \sqrt {d^2-e^2 x^2}}{128 e^3}-\frac {d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac {3 d^8 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{128 e^4}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 124, normalized size = 0.72 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-256 d^7+105 d^6 e x-128 d^5 e^2 x^2+70 d^4 e^3 x^3+1024 d^3 e^4 x^4-840 d^2 e^5 x^5-640 d e^6 x^6+560 e^7 x^7\right )-105 d^8 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{4480 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-256*d^7 + 105*d^6*e*x - 128*d^5*e^2*x^2 + 70*d^4*e^3*x^3 + 1024*d^3*e^4*x^4 - 840*d^2*e
^5*x^5 - 640*d*e^6*x^6 + 560*e^7*x^7) - 105*d^8*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(4480*e^4)

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IntegrateAlgebraic [A]  time = 0.39, size = 147, normalized size = 0.85 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-256 d^7+105 d^6 e x-128 d^5 e^2 x^2+70 d^4 e^3 x^3+1024 d^3 e^4 x^4-840 d^2 e^5 x^5-640 d e^6 x^6+560 e^7 x^7\right )}{4480 e^4}-\frac {3 d^8 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{128 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-256*d^7 + 105*d^6*e*x - 128*d^5*e^2*x^2 + 70*d^4*e^3*x^3 + 1024*d^3*e^4*x^4 - 840*d^2*e
^5*x^5 - 640*d*e^6*x^6 + 560*e^7*x^7))/(4480*e^4) - (3*d^8*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2
]])/(128*e^5)

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fricas [A]  time = 0.40, size = 127, normalized size = 0.74 \begin {gather*} \frac {210 \, d^{8} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (560 \, e^{7} x^{7} - 640 \, d e^{6} x^{6} - 840 \, d^{2} e^{5} x^{5} + 1024 \, d^{3} e^{4} x^{4} + 70 \, d^{4} e^{3} x^{3} - 128 \, d^{5} e^{2} x^{2} + 105 \, d^{6} e x - 256 \, d^{7}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{4480 \, 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),x, algorithm="fricas")

[Out]

1/4480*(210*d^8*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (560*e^7*x^7 - 640*d*e^6*x^6 - 840*d^2*e^5*x^5 + 1
024*d^3*e^4*x^4 + 70*d^4*e^3*x^3 - 128*d^5*e^2*x^2 + 105*d^6*e*x - 256*d^7)*sqrt(-e^2*x^2 + d^2))/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),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/2*(-12*d^8*exp(1)^4*exp(2)^2+8*d^8*exp
(2)^4+4*d^8*exp(1)^6*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+e
xp(2)^2))/sqrt(-exp(1)^4+exp(2)^2)/exp(1)^9/exp(1)-3/128*d^8*sign(d)*asin(x*exp(2)/d/exp(1))/exp(1)^4+2*((((((
(40320*exp(1)^15*1/645120/exp(1)^12*x-46080*exp(1)^14*d*1/645120/exp(1)^12)*x-60480*exp(1)^13*d^2*1/645120/exp
(1)^12)*x+73728*exp(1)^12*d^3*1/645120/exp(1)^12)*x+5040*exp(1)^11*d^4*1/645120/exp(1)^12)*x-9216*exp(1)^10*d^
5*1/645120/exp(1)^12)*x+7560*exp(1)^9*d^6*1/645120/exp(1)^12)*x-18432*exp(1)^8*d^7*1/645120/exp(1)^12)*sqrt(d^
2-x^2*exp(2))

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maple [B]  time = 0.01, size = 305, normalized size = 1.77 \begin {gather*} -\frac {3 d^{8} \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}}\, e^{3}}+\frac {45 d^{8} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{128 \sqrt {e^{2}}\, e^{3}}+\frac {45 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{6} x}{128 e^{3}}-\frac {3 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{6} x}{8 e^{3}}+\frac {15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{4} x}{64 e^{3}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d^{4} x}{4 e^{3}}+\frac {3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{2} x}{16 e^{3}}-\frac {\left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {5}{2}} d^{3}}{5 e^{4}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} x}{8 e^{3}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} d}{7 e^{4}} \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),x)

[Out]

-1/8/e^3*x*(-e^2*x^2+d^2)^(7/2)+3/16*(-e^2*x^2+d^2)^(5/2)*d^2/e^3*x+15/64*(-e^2*x^2+d^2)^(3/2)*d^4/e^3*x+45/12
8*(-e^2*x^2+d^2)^(1/2)*d^6/e^3*x+45/128/(e^2)^(1/2)*d^8/e^3*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)+1/7*d/e
^4*(-e^2*x^2+d^2)^(7/2)-1/5*d^3/e^4*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(5/2)-1/4*d^4/e^3*(2*(x+d/e)*d*e-(x+d/e)^2*e
^2)^(3/2)*x-3/8*d^6/e^3*(2*(x+d/e)*d*e-(x+d/e)^2*e^2)^(1/2)*x-3/8*d^8/e^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 [C]  time = 1.03, size = 221, normalized size = 1.28 \begin {gather*} \frac {3 i \, d^{8} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e^{4}} + \frac {45 \, d^{8} \arcsin \left (\frac {e x}{d}\right )}{128 \, e^{4}} - \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{6} x}{8 \, e^{3}} + \frac {45 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{6} x}{128 \, e^{3}} - \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{7}}{4 \, e^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4} x}{64 \, e^{3}} + \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2} x}{16 \, e^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}}{5 \, e^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} x}{8 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} d}{7 \, 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),x, algorithm="maxima")

[Out]

3/8*I*d^8*arcsin(e*x/d + 2)/e^4 + 45/128*d^8*arcsin(e*x/d)/e^4 - 3/8*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^6*x/e^3
+ 45/128*sqrt(-e^2*x^2 + d^2)*d^6*x/e^3 - 3/4*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^7/e^4 - 1/64*(-e^2*x^2 + d^2)
^(3/2)*d^4*x/e^3 + 3/16*(-e^2*x^2 + d^2)^(5/2)*d^2*x/e^3 - 1/5*(-e^2*x^2 + d^2)^(5/2)*d^3/e^4 - 1/8*(-e^2*x^2
+ d^2)^(7/2)*x/e^3 + 1/7*(-e^2*x^2 + d^2)^(7/2)*d/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}}{d+e\,x} \,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),x)

[Out]

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

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sympy [A]  time = 23.01, size = 775, normalized size = 4.51 \begin {gather*} d^{3} \left (\begin {cases} - \frac {2 d^{4} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{4}} - \frac {d^{2} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{2}} + \frac {x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {x^{4} \sqrt {d^{2}}}{4} & \text {otherwise} \end {cases}\right ) - d^{2} e \left (\begin {cases} - \frac {i d^{6} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{16 e^{5}} + \frac {i d^{5} x}{16 e^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i d^{3} x^{3}}{48 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {5 i d x^{5}}{24 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{7}}{6 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{6} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{16 e^{5}} - \frac {d^{5} x}{16 e^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {d^{3} x^{3}}{48 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {5 d x^{5}}{24 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{7}}{6 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) - d e^{2} \left (\begin {cases} - \frac {8 d^{6} \sqrt {d^{2} - e^{2} x^{2}}}{105 e^{6}} - \frac {4 d^{4} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{105 e^{4}} - \frac {d^{2} x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{35 e^{2}} + \frac {x^{6} \sqrt {d^{2} - e^{2} x^{2}}}{7} & \text {for}\: e \neq 0 \\\frac {x^{6} \sqrt {d^{2}}}{6} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} - \frac {5 i d^{8} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{128 e^{7}} + \frac {5 i d^{7} x}{128 e^{6} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {5 i d^{5} x^{3}}{384 e^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i d^{3} x^{5}}{192 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {7 i d x^{7}}{48 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{9}}{8 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {5 d^{8} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{128 e^{7}} - \frac {5 d^{7} x}{128 e^{6} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {5 d^{5} x^{3}}{384 e^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {d^{3} x^{5}}{192 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {7 d x^{7}}{48 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{9}}{8 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\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),x)

[Out]

d**3*Piecewise((-2*d**4*sqrt(d**2 - e**2*x**2)/(15*e**4) - d**2*x**2*sqrt(d**2 - e**2*x**2)/(15*e**2) + x**4*s
qrt(d**2 - e**2*x**2)/5, Ne(e, 0)), (x**4*sqrt(d**2)/4, True)) - d**2*e*Piecewise((-I*d**6*acosh(e*x/d)/(16*e*
*5) + I*d**5*x/(16*e**4*sqrt(-1 + e**2*x**2/d**2)) - I*d**3*x**3/(48*e**2*sqrt(-1 + e**2*x**2/d**2)) - 5*I*d*x
**5/(24*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**7/(6*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d
**6*asin(e*x/d)/(16*e**5) - d**5*x/(16*e**4*sqrt(1 - e**2*x**2/d**2)) + d**3*x**3/(48*e**2*sqrt(1 - e**2*x**2/
d**2)) + 5*d*x**5/(24*sqrt(1 - e**2*x**2/d**2)) - e**2*x**7/(6*d*sqrt(1 - e**2*x**2/d**2)), True)) - d*e**2*Pi
ecewise((-8*d**6*sqrt(d**2 - e**2*x**2)/(105*e**6) - 4*d**4*x**2*sqrt(d**2 - e**2*x**2)/(105*e**4) - d**2*x**4
*sqrt(d**2 - e**2*x**2)/(35*e**2) + x**6*sqrt(d**2 - e**2*x**2)/7, Ne(e, 0)), (x**6*sqrt(d**2)/6, True)) + e**
3*Piecewise((-5*I*d**8*acosh(e*x/d)/(128*e**7) + 5*I*d**7*x/(128*e**6*sqrt(-1 + e**2*x**2/d**2)) - 5*I*d**5*x*
*3/(384*e**4*sqrt(-1 + e**2*x**2/d**2)) - I*d**3*x**5/(192*e**2*sqrt(-1 + e**2*x**2/d**2)) - 7*I*d*x**7/(48*sq
rt(-1 + e**2*x**2/d**2)) + I*e**2*x**9/(8*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (5*d**8*asin
(e*x/d)/(128*e**7) - 5*d**7*x/(128*e**6*sqrt(1 - e**2*x**2/d**2)) + 5*d**5*x**3/(384*e**4*sqrt(1 - e**2*x**2/d
**2)) + d**3*x**5/(192*e**2*sqrt(1 - e**2*x**2/d**2)) + 7*d*x**7/(48*sqrt(1 - e**2*x**2/d**2)) - e**2*x**9/(8*
d*sqrt(1 - e**2*x**2/d**2)), True))

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