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

Optimal. Leaf size=134 \[ -\frac {e^2 (13 d+8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}+e^4 \left (-\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )\right )+\frac {13}{8} e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

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Rubi [A]  time = 0.22, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {1807, 811, 844, 217, 203, 266, 63, 208} \begin {gather*} -\frac {e^2 (13 d+8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}+e^4 \left (-\tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )\right )+\frac {13}{8} e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^2*(13*d + 8*e*x)*Sqrt[d^2 - e^2*x^2])/(8*x^2) - (d*(d^2 - e^2*x^2)^(3/2))/(4*x^4) - (e*(d^2 - e^2*x^2)^(3/
2))/x^3 - e^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]] + (13*e^4*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/8

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 208

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

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 844

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

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rubi steps

\begin {align*} \int \frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{x^5} \, dx &=-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {\int \frac {\sqrt {d^2-e^2 x^2} \left (-12 d^4 e-13 d^3 e^2 x-4 d^2 e^3 x^2\right )}{x^4} \, dx}{4 d^2}\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}+\frac {\int \frac {\left (39 d^5 e^2+12 d^4 e^3 x\right ) \sqrt {d^2-e^2 x^2}}{x^3} \, dx}{12 d^4}\\ &=-\frac {e^2 (13 d+8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}-\frac {\int \frac {78 d^7 e^4+48 d^6 e^5 x}{x \sqrt {d^2-e^2 x^2}} \, dx}{48 d^6}\\ &=-\frac {e^2 (13 d+8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}-\frac {1}{8} \left (13 d e^4\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx-e^5 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {e^2 (13 d+8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}-\frac {1}{16} \left (13 d e^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )-e^5 \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )\\ &=-\frac {e^2 (13 d+8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}-e^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {1}{8} \left (13 d e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )\\ &=-\frac {e^2 (13 d+8 e x) \sqrt {d^2-e^2 x^2}}{8 x^2}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{x^3}-e^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )+\frac {13}{8} e^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\\ \end {align*}

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Mathematica [C]  time = 0.24, size = 196, normalized size = 1.46 \begin {gather*} -\frac {e \sqrt {d^2-e^2 x^2} \left (6 d^2 e^3 x^3 \sin ^{-1}\left (\frac {e x}{d}\right )+2 e^3 x^3 \left (d^2-e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};1-\frac {e^2 x^2}{d^2}\right )-9 d^2 e^3 x^3 \tanh ^{-1}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )+6 d^5 \sqrt {1-\frac {e^2 x^2}{d^2}}+9 d^4 e x \sqrt {1-\frac {e^2 x^2}{d^2}}\right )}{6 d^3 x^3 \sqrt {1-\frac {e^2 x^2}{d^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(e*Sqrt[d^2 - e^2*x^2]*(6*d^5*Sqrt[1 - (e^2*x^2)/d^2] + 9*d^4*e*x*Sqrt[1 - (e^2*x^2)/d^2] + 6*d^2*e^3*x^3
*ArcSin[(e*x)/d] - 9*d^2*e^3*x^3*ArcTanh[Sqrt[1 - (e^2*x^2)/d^2]] + 2*e^3*x^3*(d^2 - e^2*x^2)*Sqrt[1 - (e^2*x^
2)/d^2]*Hypergeometric2F1[3/2, 3, 5/2, 1 - (e^2*x^2)/d^2]))/(d^3*x^3*Sqrt[1 - (e^2*x^2)/d^2])

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IntegrateAlgebraic [A]  time = 0.56, size = 134, normalized size = 1.00 \begin {gather*} -\frac {13}{4} e^4 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )-\sqrt {-e^2} e^3 \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )+\frac {\sqrt {d^2-e^2 x^2} \left (-2 d^3-8 d^2 e x-11 d e^2 x^2\right )}{8 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-2*d^3 - 8*d^2*e*x - 11*d*e^2*x^2))/(8*x^4) - (13*e^4*ArcTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^
2 - e^2*x^2]/d])/4 - e^3*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]]

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fricas [A]  time = 0.42, size = 111, normalized size = 0.83 \begin {gather*} \frac {16 \, e^{4} x^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - 13 \, e^{4} x^{4} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (11 \, d e^{2} x^{2} + 8 \, d^{2} e x + 2 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{8 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(-e^2*x^2+d^2)^(1/2)/x^5,x, algorithm="fricas")

[Out]

1/8*(16*e^4*x^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - 13*e^4*x^4*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (1
1*d*e^2*x^2 + 8*d^2*e*x + 2*d^3)*sqrt(-e^2*x^2 + d^2))/x^4

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giac [B]  time = 0.27, size = 295, normalized size = 2.20 \begin {gather*} -\arcsin \left (\frac {x e}{d}\right ) e^{4} \mathrm {sgn}\relax (d) + \frac {x^{4} {\left (\frac {8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{8}}{x} + \frac {24 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{6}}{x^{2}} + \frac {8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{4}}{x^{3}} + e^{10}\right )} e^{2}}{64 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4}} - \frac {1}{64} \, {\left (\frac {8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{26}}{x} + \frac {24 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{24}}{x^{2}} + \frac {8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{22}}{x^{3}} + \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{20}}{x^{4}}\right )} e^{\left (-24\right )} + \frac {13}{8} \, e^{4} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(-e^2*x^2+d^2)^(1/2)/x^5,x, algorithm="giac")

[Out]

-arcsin(x*e/d)*e^4*sgn(d) + 1/64*x^4*(8*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^8/x + 24*(d*e + sqrt(-x^2*e^2 + d^2)*
e)^2*e^6/x^2 + 8*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*e^4/x^3 + e^10)*e^2/(d*e + sqrt(-x^2*e^2 + d^2)*e)^4 - 1/64*
(8*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^26/x + 24*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e^24/x^2 + 8*(d*e + sqrt(-x^2*e
^2 + d^2)*e)^3*e^22/x^3 + (d*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^20/x^4)*e^(-24) + 13/8*e^4*log(1/2*abs(-2*d*e - 2
*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))

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maple [A]  time = 0.02, size = 212, normalized size = 1.58 \begin {gather*} \frac {13 d \,e^{4} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{8 \sqrt {d^{2}}}-\frac {e^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e^{5} x}{d^{2}}-\frac {13 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{4}}{8 d}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{3}}{d^{2} x}-\frac {13 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{2}}{8 d \,x^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e}{x^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d}{4 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(-e^2*x^2+d^2)^(1/2)/x^5,x)

[Out]

-1/4*d*(-e^2*x^2+d^2)^(3/2)/x^4-13/8/d*e^2/x^2*(-e^2*x^2+d^2)^(3/2)-13/8*(-e^2*x^2+d^2)^(1/2)/d*e^4+13/8/(d^2)
^(1/2)*d*e^4*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-e^3/d^2/x*(-e^2*x^2+d^2)^(3/2)-(-e^2*x^2+d^2)^(1
/2)/d^2*e^5*x-1/(e^2)^(1/2)*e^5*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-e*(-e^2*x^2+d^2)^(3/2)/x^3

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maxima [A]  time = 0.98, size = 159, normalized size = 1.19 \begin {gather*} -e^{4} \arcsin \left (\frac {e x}{d}\right ) + \frac {13}{8} \, e^{4} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) - \frac {13 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{4}}{8 \, d} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} e^{3}}{x} - \frac {13 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}}{8 \, d x^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e}{x^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(-e^2*x^2+d^2)^(1/2)/x^5,x, algorithm="maxima")

[Out]

-e^4*arcsin(e*x/d) + 13/8*e^4*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x)) - 13/8*sqrt(-e^2*x^2 + d^2)*
e^4/d - sqrt(-e^2*x^2 + d^2)*e^3/x - 13/8*(-e^2*x^2 + d^2)^(3/2)*e^2/(d*x^2) - (-e^2*x^2 + d^2)^(3/2)*e/x^3 -
1/4*(-e^2*x^2 + d^2)^(3/2)*d/x^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d^2 - e^2*x^2)^(1/2)*(d + e*x)^3)/x^5,x)

[Out]

int(((d^2 - e^2*x^2)^(1/2)*(d + e*x)^3)/x^5, x)

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sympy [C]  time = 10.03, size = 544, normalized size = 4.06 \begin {gather*} d^{3} \left (\begin {cases} - \frac {d^{2}}{4 e x^{5} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {3 e}{8 x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - \frac {e^{3}}{8 d^{2} x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{4} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{4 e x^{5} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {3 i e}{8 x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e^{3}}{8 d^{2} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{4} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {otherwise} \end {cases}\right ) + 3 d^{2} e \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 x^{2}} + \frac {e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 x^{2}} + \frac {i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2}} & \text {otherwise} \end {cases}\right ) + 3 d e^{2} \left (\begin {cases} - \frac {d^{2}}{2 e x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e}{2 x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{2 x} - \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} \frac {i d}{x \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + i e \operatorname {acosh}{\left (\frac {e x}{d} \right )} - \frac {i e^{2} x}{d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\- \frac {d}{x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - e \operatorname {asin}{\left (\frac {e x}{d} \right )} + \frac {e^{2} x}{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((e*x+d)**3*(-e**2*x**2+d**2)**(1/2)/x**5,x)

[Out]

d**3*Piecewise((-d**2/(4*e*x**5*sqrt(d**2/(e**2*x**2) - 1)) + 3*e/(8*x**3*sqrt(d**2/(e**2*x**2) - 1)) - e**3/(
8*d**2*x*sqrt(d**2/(e**2*x**2) - 1)) + e**4*acosh(d/(e*x))/(8*d**3), Abs(d**2/(e**2*x**2)) > 1), (I*d**2/(4*e*
x**5*sqrt(-d**2/(e**2*x**2) + 1)) - 3*I*e/(8*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**3/(8*d**2*x*sqrt(-d**2/(
e**2*x**2) + 1)) - I*e**4*asin(d/(e*x))/(8*d**3), True)) + 3*d**2*e*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(
3*x**2) + e**3*sqrt(d**2/(e**2*x**2) - 1)/(3*d**2), Abs(d**2/(e**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) +
1)/(3*x**2) + I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(3*d**2), True)) + 3*d*e**2*Piecewise((-d**2/(2*e*x**3*sqrt(
d**2/(e**2*x**2) - 1)) + e/(2*x*sqrt(d**2/(e**2*x**2) - 1)) + e**2*acosh(d/(e*x))/(2*d), Abs(d**2/(e**2*x**2))
> 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(2*x) - I*e**2*asin(d/(e*x))/(2*d), True)) + e**3*Piecewise((I*d/(x*s
qrt(-1 + e**2*x**2/d**2)) + I*e*acosh(e*x/d) - I*e**2*x/(d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1
), (-d/(x*sqrt(1 - e**2*x**2/d**2)) - e*asin(e*x/d) + e**2*x/(d*sqrt(1 - e**2*x**2/d**2)), True))

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