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

Optimal. Leaf size=190 \[ \frac{1}{128} d^6 (128 d+125 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{192} d^4 (64 d+125 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{240} d^2 (48 d+125 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{3}{7} d \left (d^2-e^2 x^2\right )^{7/2}-\frac{1}{8} e x \left (d^2-e^2 x^2\right )^{7/2}+\frac{125}{128} d^8 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-d^8 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

[Out]

(d^6*(128*d + 125*e*x)*Sqrt[d^2 - e^2*x^2])/128 + (d^4*(64*d + 125*e*x)*(d^2 - e^2*x^2)^(3/2))/192 + (d^2*(48*
d + 125*e*x)*(d^2 - e^2*x^2)^(5/2))/240 - (3*d*(d^2 - e^2*x^2)^(7/2))/7 - (e*x*(d^2 - e^2*x^2)^(7/2))/8 + (125
*d^8*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/128 - d^8*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

________________________________________________________________________________________

Rubi [A]  time = 0.30707, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {1809, 815, 844, 217, 203, 266, 63, 208} \[ \frac{1}{128} d^6 (128 d+125 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{192} d^4 (64 d+125 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{240} d^2 (48 d+125 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{3}{7} d \left (d^2-e^2 x^2\right )^{7/2}-\frac{1}{8} e x \left (d^2-e^2 x^2\right )^{7/2}+\frac{125}{128} d^8 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-d^8 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^6*(128*d + 125*e*x)*Sqrt[d^2 - e^2*x^2])/128 + (d^4*(64*d + 125*e*x)*(d^2 - e^2*x^2)^(3/2))/192 + (d^2*(48*
d + 125*e*x)*(d^2 - e^2*x^2)^(5/2))/240 - (3*d*(d^2 - e^2*x^2)^(7/2))/7 - (e*x*(d^2 - e^2*x^2)^(7/2))/8 + (125
*d^8*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/128 - d^8*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

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} \, dx &=-\frac{1}{8} e x \left (d^2-e^2 x^2\right )^{7/2}-\frac{\int \frac{\left (d^2-e^2 x^2\right )^{5/2} \left (-8 d^3 e^2-25 d^2 e^3 x-24 d e^4 x^2\right )}{x} \, dx}{8 e^2}\\ &=-\frac{3}{7} d \left (d^2-e^2 x^2\right )^{7/2}-\frac{1}{8} e x \left (d^2-e^2 x^2\right )^{7/2}+\frac{\int \frac{\left (56 d^3 e^4+175 d^2 e^5 x\right ) \left (d^2-e^2 x^2\right )^{5/2}}{x} \, dx}{56 e^4}\\ &=\frac{1}{240} d^2 (48 d+125 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{3}{7} d \left (d^2-e^2 x^2\right )^{7/2}-\frac{1}{8} e x \left (d^2-e^2 x^2\right )^{7/2}-\frac{\int \frac{\left (-336 d^5 e^6-875 d^4 e^7 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x} \, dx}{336 e^6}\\ &=\frac{1}{192} d^4 (64 d+125 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{240} d^2 (48 d+125 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{3}{7} d \left (d^2-e^2 x^2\right )^{7/2}-\frac{1}{8} e x \left (d^2-e^2 x^2\right )^{7/2}+\frac{\int \frac{\left (1344 d^7 e^8+2625 d^6 e^9 x\right ) \sqrt{d^2-e^2 x^2}}{x} \, dx}{1344 e^8}\\ &=\frac{1}{128} d^6 (128 d+125 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{192} d^4 (64 d+125 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{240} d^2 (48 d+125 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{3}{7} d \left (d^2-e^2 x^2\right )^{7/2}-\frac{1}{8} e x \left (d^2-e^2 x^2\right )^{7/2}-\frac{\int \frac{-2688 d^9 e^{10}-2625 d^8 e^{11} x}{x \sqrt{d^2-e^2 x^2}} \, dx}{2688 e^{10}}\\ &=\frac{1}{128} d^6 (128 d+125 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{192} d^4 (64 d+125 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{240} d^2 (48 d+125 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{3}{7} d \left (d^2-e^2 x^2\right )^{7/2}-\frac{1}{8} e x \left (d^2-e^2 x^2\right )^{7/2}+d^9 \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx+\frac{1}{128} \left (125 d^8 e\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=\frac{1}{128} d^6 (128 d+125 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{192} d^4 (64 d+125 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{240} d^2 (48 d+125 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{3}{7} d \left (d^2-e^2 x^2\right )^{7/2}-\frac{1}{8} e x \left (d^2-e^2 x^2\right )^{7/2}+\frac{1}{2} d^9 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )+\frac{1}{128} \left (125 d^8 e\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{1}{128} d^6 (128 d+125 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{192} d^4 (64 d+125 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{240} d^2 (48 d+125 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{3}{7} d \left (d^2-e^2 x^2\right )^{7/2}-\frac{1}{8} e x \left (d^2-e^2 x^2\right )^{7/2}+\frac{125}{128} d^8 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-\frac{d^9 \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^2}\\ &=\frac{1}{128} d^6 (128 d+125 e x) \sqrt{d^2-e^2 x^2}+\frac{1}{192} d^4 (64 d+125 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac{1}{240} d^2 (48 d+125 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac{3}{7} d \left (d^2-e^2 x^2\right )^{7/2}-\frac{1}{8} e x \left (d^2-e^2 x^2\right )^{7/2}+\frac{125}{128} d^8 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-d^8 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )\\ \end{align*}

Mathematica [A]  time = 0.380262, size = 168, normalized size = 0.88 \[ \frac{\sqrt{d^2-e^2 x^2} \left (7424 d^5 e^2 x^2-17710 d^4 e^3 x^3-14592 d^3 e^4 x^4+1960 d^2 e^5 x^5+27195 d^6 e x+14848 d^7+5760 d e^6 x^6+1680 e^7 x^7\right )}{13440}+\frac{125 d^7 \sqrt{d^2-e^2 x^2} \sin ^{-1}\left (\frac{e x}{d}\right )}{128 \sqrt{1-\frac{e^2 x^2}{d^2}}}+d^8 \left (-\tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(14848*d^7 + 27195*d^6*e*x + 7424*d^5*e^2*x^2 - 17710*d^4*e^3*x^3 - 14592*d^3*e^4*x^4 + 1
960*d^2*e^5*x^5 + 5760*d*e^6*x^6 + 1680*e^7*x^7))/13440 + (125*d^7*Sqrt[d^2 - e^2*x^2]*ArcSin[(e*x)/d])/(128*S
qrt[1 - (e^2*x^2)/d^2]) - d^8*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

________________________________________________________________________________________

Maple [A]  time = 0.056, size = 231, normalized size = 1.2 \begin{align*} -{\frac{ex}{8} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}+{\frac{25\,{d}^{2}ex}{48} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{125\,e{d}^{4}x}{192} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{125\,e{d}^{6}x}{128}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{125\,e{d}^{8}}{128}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{3\,d}{7} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}+{\frac{{d}^{3}}{5} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{{d}^{5}}{3} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{d}^{7}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}-{{d}^{9}\ln \left ({\frac{1}{x} \left ( 2\,{d}^{2}+2\,\sqrt{{d}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}} \right ) } \right ){\frac{1}{\sqrt{{d}^{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,x)

[Out]

-1/8*e*x*(-e^2*x^2+d^2)^(7/2)+25/48*d^2*e*x*(-e^2*x^2+d^2)^(5/2)+125/192*e*d^4*x*(-e^2*x^2+d^2)^(3/2)+125/128*
e*d^6*x*(-e^2*x^2+d^2)^(1/2)+125/128*e*d^8/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-3/7*d*(-e^2*
x^2+d^2)^(7/2)+1/5*d^3*(-e^2*x^2+d^2)^(5/2)+1/3*d^5*(-e^2*x^2+d^2)^(3/2)+d^7*(-e^2*x^2+d^2)^(1/2)-d^9/(d^2)^(1
/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)

________________________________________________________________________________________

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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.98322, size = 358, normalized size = 1.88 \begin{align*} -\frac{125}{64} \, d^{8} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) + d^{8} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) + \frac{1}{13440} \,{\left (1680 \, e^{7} x^{7} + 5760 \, d e^{6} x^{6} + 1960 \, d^{2} e^{5} x^{5} - 14592 \, d^{3} e^{4} x^{4} - 17710 \, d^{4} e^{3} x^{3} + 7424 \, d^{5} e^{2} x^{2} + 27195 \, d^{6} e x + 14848 \, d^{7}\right )} \sqrt{-e^{2} x^{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,x, algorithm="fricas")

[Out]

-125/64*d^8*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + d^8*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + 1/13440*(1680
*e^7*x^7 + 5760*d*e^6*x^6 + 1960*d^2*e^5*x^5 - 14592*d^3*e^4*x^4 - 17710*d^4*e^3*x^3 + 7424*d^5*e^2*x^2 + 2719
5*d^6*e*x + 14848*d^7)*sqrt(-e^2*x^2 + d^2)

________________________________________________________________________________________

Sympy [C]  time = 55.4003, size = 1273, normalized size = 6.7 \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,x)

[Out]

d**7*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*asin(d/(e*x)) + I*e*x/sqrt
(-d**2/(e**2*x**2) + 1), True)) + 3*d**6*e*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)) + d**5*e**2*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**4*e**3*Piecewise((-I*d**4*acosh(e*x/d)/(8*e**3) + I*d**3*x/(8*e**2*sqr
t(-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)) - 5*d**3*e**4*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)) + d**2*e**5*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)/Abs(d**2) > 1), (d**6*asi
n(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)) + 3*d*e**6*Piecewi
se((-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**7*Pie
cewise((-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/(3
84*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*sqrt(-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)/Abs(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))

________________________________________________________________________________________

Giac [A]  time = 1.14547, size = 193, normalized size = 1.02 \begin{align*} \frac{125}{128} \, d^{8} \arcsin \left (\frac{x e}{d}\right ) \mathrm{sgn}\left (d\right ) - d^{8} \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{1}{13440} \,{\left (14848 \, d^{7} +{\left (27195 \, d^{6} e + 2 \,{\left (3712 \, d^{5} e^{2} -{\left (8855 \, d^{4} e^{3} + 4 \,{\left (1824 \, d^{3} e^{4} - 5 \,{\left (49 \, d^{2} e^{5} + 6 \,{\left (7 \, x e^{7} + 24 \, d e^{6}\right )} x\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,x, algorithm="giac")

[Out]

125/128*d^8*arcsin(x*e/d)*sgn(d) - d^8*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x)) + 1/13440
*(14848*d^7 + (27195*d^6*e + 2*(3712*d^5*e^2 - (8855*d^4*e^3 + 4*(1824*d^3*e^4 - 5*(49*d^2*e^5 + 6*(7*x*e^7 +
24*d*e^6)*x)*x)*x)*x)*x)*x)*sqrt(-x^2*e^2 + d^2)