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

Optimal. Leaf size=193 \[ \frac{3}{16} d^5 e (16 d-5 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac{1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac{15}{16} d^7 e \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-3 d^7 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

[Out]

(3*d^5*e*(16*d - 5*e*x)*Sqrt[d^2 - e^2*x^2])/16 + (d^3*e*(8*d - 5*e*x)*(d^2 - e^2*x^2)^(3/2))/8 + (d*e*(6*d -
5*e*x)*(d^2 - e^2*x^2)^(5/2))/10 - (e*(d^2 - e^2*x^2)^(7/2))/7 - (d*(d^2 - e^2*x^2)^(7/2))/x - (15*d^7*e*ArcTa
n[(e*x)/Sqrt[d^2 - e^2*x^2]])/16 - 3*d^7*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

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Rubi [A]  time = 0.305454, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {1807, 1809, 815, 844, 217, 203, 266, 63, 208} \[ \frac{3}{16} d^5 e (16 d-5 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac{1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac{15}{16} d^7 e \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-3 d^7 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d^5*e*(16*d - 5*e*x)*Sqrt[d^2 - e^2*x^2])/16 + (d^3*e*(8*d - 5*e*x)*(d^2 - e^2*x^2)^(3/2))/8 + (d*e*(6*d -
5*e*x)*(d^2 - e^2*x^2)^(5/2))/10 - (e*(d^2 - e^2*x^2)^(7/2))/7 - (d*(d^2 - e^2*x^2)^(7/2))/x - (15*d^7*e*ArcTa
n[(e*x)/Sqrt[d^2 - e^2*x^2]])/16 - 3*d^7*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

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])

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 815

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

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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{(d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}}{x^2} \, dx &=-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{x}-\frac{\int \frac{\left (d^2-e^2 x^2\right )^{5/2} \left (-3 d^4 e+3 d^3 e^2 x-d^2 e^3 x^2\right )}{x} \, dx}{d^2}\\ &=-\frac{1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac{\int \frac{\left (21 d^4 e^3-21 d^3 e^4 x\right ) \left (d^2-e^2 x^2\right )^{5/2}}{x} \, dx}{7 d^2 e^2}\\ &=\frac{1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{x}-\frac{\int \frac{\left (-126 d^6 e^5+105 d^5 e^6 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x} \, dx}{42 d^2 e^4}\\ &=\frac{1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac{\int \frac{\left (504 d^8 e^7-315 d^7 e^8 x\right ) \sqrt{d^2-e^2 x^2}}{x} \, dx}{168 d^2 e^6}\\ &=\frac{3}{16} d^5 e (16 d-5 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{x}-\frac{\int \frac{-1008 d^{10} e^9+315 d^9 e^{10} x}{x \sqrt{d^2-e^2 x^2}} \, dx}{336 d^2 e^8}\\ &=\frac{3}{16} d^5 e (16 d-5 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{x}+\left (3 d^8 e\right ) \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx-\frac{1}{16} \left (15 d^7 e^2\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{3}{16} d^5 e (16 d-5 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac{1}{2} \left (3 d^8 e\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )-\frac{1}{16} \left (15 d^7 e^2\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{3}{16} d^5 e (16 d-5 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{x}-\frac{15}{16} d^7 e \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-\frac{\left (3 d^8\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 )}{e}\\ &=\frac{3}{16} d^5 e (16 d-5 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{8} d^3 e (8 d-5 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{10} d e (6 d-5 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{1}{7} e \left (d^2-e^2 x^2\right )^{7/2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{x}-\frac{15}{16} d^7 e \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-3 d^7 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )\\ \end{align*}

Mathematica [C]  time = 0.576062, size = 221, normalized size = 1.15 \[ -\frac{d^7 \sqrt{d^2-e^2 x^2} \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x \sqrt{1-\frac{e^2 x^2}{d^2}}}+\frac{1}{560} e \sqrt{d^2-e^2 x^2} \left (-992 d^4 e^2 x^2-910 d^3 e^3 x^3+96 d^2 e^4 x^4+1155 d^5 e x+2496 d^6+280 d e^5 x^5+80 e^6 x^6\right )+\frac{15 d^6 e \sqrt{d^2-e^2 x^2} \sin ^{-1}\left (\frac{e x}{d}\right )}{16 \sqrt{1-\frac{e^2 x^2}{d^2}}}-3 d^7 e \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[d^2 - e^2*x^2]*(2496*d^6 + 1155*d^5*e*x - 992*d^4*e^2*x^2 - 910*d^3*e^3*x^3 + 96*d^2*e^4*x^4 + 280*d*e
^5*x^5 + 80*e^6*x^6))/560 + (15*d^6*e*Sqrt[d^2 - e^2*x^2]*ArcSin[(e*x)/d])/(16*Sqrt[1 - (e^2*x^2)/d^2]) - 3*d^
7*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d] - (d^7*Sqrt[d^2 - e^2*x^2]*Hypergeometric2F1[-5/2, -1/2, 1/2, (e^2*x^2)/d^2
])/(x*Sqrt[1 - (e^2*x^2)/d^2])

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Maple [A]  time = 0.061, size = 243, normalized size = 1.3 \begin{align*} -{\frac{e}{7} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-{\frac{d{e}^{2}x}{2} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{d}^{3}{e}^{2}x}{8} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{15\,{d}^{5}{e}^{2}x}{16}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{15\,{d}^{7}{e}^{2}}{16}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+{\frac{3\,{d}^{2}e}{5} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{d}^{4}e \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}+3\,{d}^{6}e\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}-3\,{\frac{{d}^{8}e}{\sqrt{{d}^{2}}}\ln \left ({\frac{2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}{x}} \right ) }-{\frac{d}{x} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*e*(-e^2*x^2+d^2)^(7/2)-1/2*d*e^2*x*(-e^2*x^2+d^2)^(5/2)-5/8*d^3*e^2*x*(-e^2*x^2+d^2)^(3/2)-15/16*d^5*e^2*
x*(-e^2*x^2+d^2)^(1/2)-15/16*d^7*e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+3/5*d^2*e*(-e^2*x^
2+d^2)^(5/2)+d^4*e*(-e^2*x^2+d^2)^(3/2)+3*d^6*e*(-e^2*x^2+d^2)^(1/2)-3*d^8*e/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/
2)*(-e^2*x^2+d^2)^(1/2))/x)-d*(-e^2*x^2+d^2)^(7/2)/x

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.9589, size = 377, normalized size = 1.95 \begin{align*} \frac{1050 \, d^{7} e x \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) + 1680 \, d^{7} e x \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) + 2496 \, d^{7} e x +{\left (80 \, e^{7} x^{7} + 280 \, d e^{6} x^{6} + 96 \, d^{2} e^{5} x^{5} - 770 \, d^{3} e^{4} x^{4} - 992 \, d^{4} e^{3} x^{3} + 525 \, d^{5} e^{2} x^{2} + 2496 \, d^{6} e x - 560 \, d^{7}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{560 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/560*(1050*d^7*e*x*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + 1680*d^7*e*x*log(-(d - sqrt(-e^2*x^2 + d^2))/x
) + 2496*d^7*e*x + (80*e^7*x^7 + 280*d*e^6*x^6 + 96*d^2*e^5*x^5 - 770*d^3*e^4*x^4 - 992*d^4*e^3*x^3 + 525*d^5*
e^2*x^2 + 2496*d^6*e*x - 560*d^7)*sqrt(-e^2*x^2 + d^2))/x

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Sympy [C]  time = 22.3394, size = 1068, normalized size = 5.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**7*Piecewise((I*d/(x*sqrt(-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)/Abs(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)) + 3*d**6*e*Piecewise((d**2/(e*x*sqrt(d**2/(e**2*x**2) - 1)) - d*acosh(d/(e*x)) - e*x/sqrt(d**
2/(e**2*x**2) - 1), Abs(d**2)/(Abs(e**2)*Abs(x**2)) > 1), (-I*d**2/(e*x*sqrt(-d**2/(e**2*x**2) + 1)) + I*d*asi
n(d/(e*x)) + I*e*x/sqrt(-d**2/(e**2*x**2) + 1), True)) + d**5*e**2*Piecewise((-I*d**2*acosh(e*x/d)/(2*e) - I*d
*x/(2*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**3/(2*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1),
(d**2*asin(e*x/d)/(2*e) + d*x*sqrt(1 - e**2*x**2/d**2)/2, True)) - 5*d**4*e**3*Piecewise((x**2*sqrt(d**2)/2,
Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) - 5*d**3*e**4*Piecewise((-I*d**4*acosh(e*x/d)/(8*e*
*3) + I*d**3*x/(8*e**2*sqrt(-1 + e**2*x**2/d**2)) - 3*I*d*x**3/(8*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**5/(4*
d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (d**4*asin(e*x/d)/(8*e**3) - d**3*x/(8*e**2*sqrt(
1 - e**2*x**2/d**2)) + 3*d*x**3/(8*sqrt(1 - e**2*x**2/d**2)) - e**2*x**5/(4*d*sqrt(1 - e**2*x**2/d**2)), True)
) + d**2*e**5*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*sqrt(d**2 - e**2*x**2)/5, Ne(e, 0)), (x**4*sqrt(d**2)/4, True)) + 3*d*e**6*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)/Ab
s(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*sqr
t(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)), Tr
ue)) + e**7*Piecewise((-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))

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Giac [A]  time = 1.17445, size = 269, normalized size = 1.39 \begin{align*} -\frac{15}{16} \, d^{7} \arcsin \left (\frac{x e}{d}\right ) e \mathrm{sgn}\left (d\right ) - 3 \, d^{7} e \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 ) + \frac{d^{7} x e^{3}}{2 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}} - \frac{{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} d^{7} e^{\left (-1\right )}}{2 \, x} + \frac{1}{560} \,{\left (2496 \, d^{6} e +{\left (525 \, d^{5} e^{2} - 2 \,{\left (496 \, d^{4} e^{3} +{\left (385 \, d^{3} e^{4} - 4 \,{\left (12 \, d^{2} e^{5} + 5 \,{\left (2 \, x e^{7} + 7 \, d e^{6}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt{-x^{2} e^{2} + d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/16*d^7*arcsin(x*e/d)*e*sgn(d) - 3*d^7*e*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x)) + 1/
2*d^7*x*e^3/(d*e + sqrt(-x^2*e^2 + d^2)*e) - 1/2*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^7*e^(-1)/x + 1/560*(2496*d^6
*e + (525*d^5*e^2 - 2*(496*d^4*e^3 + (385*d^3*e^4 - 4*(12*d^2*e^5 + 5*(2*x*e^7 + 7*d*e^6)*x)*x)*x)*x)*x)*sqrt(
-x^2*e^2 + d^2)