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

Optimal. Leaf size=209 \[ -\frac{45}{8} d^2 e^4 (d-e x) \sqrt{d^2-e^2 x^2}+\frac{15 d e^3 (2 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{8 x}-\frac{3 e^2 (3 d+2 e x) \left (d^2-e^2 x^2\right )^{5/2}}{8 x^2}-\frac{e \left (d^2-e^2 x^2\right )^{7/2}}{x^3}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}+\frac{45}{8} d^4 e^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+\frac{45}{8} d^4 e^4 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

[Out]

(-45*d^2*e^4*(d - e*x)*Sqrt[d^2 - e^2*x^2])/8 + (15*d*e^3*(2*d - e*x)*(d^2 - e^2*x^2)^(3/2))/(8*x) - (3*e^2*(3
*d + 2*e*x)*(d^2 - e^2*x^2)^(5/2))/(8*x^2) - (d*(d^2 - e^2*x^2)^(7/2))/(4*x^4) - (e*(d^2 - e^2*x^2)^(7/2))/x^3
+ (45*d^4*e^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/8 + (45*d^4*e^4*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/8

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Rubi [A]  time = 0.319215, antiderivative size = 209, 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, 813, 815, 844, 217, 203, 266, 63, 208} \[ -\frac{45}{8} d^2 e^4 (d-e x) \sqrt{d^2-e^2 x^2}+\frac{15 d e^3 (2 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{8 x}-\frac{3 e^2 (3 d+2 e x) \left (d^2-e^2 x^2\right )^{5/2}}{8 x^2}-\frac{e \left (d^2-e^2 x^2\right )^{7/2}}{x^3}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}+\frac{45}{8} d^4 e^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+\frac{45}{8} d^4 e^4 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-45*d^2*e^4*(d - e*x)*Sqrt[d^2 - e^2*x^2])/8 + (15*d*e^3*(2*d - e*x)*(d^2 - e^2*x^2)^(3/2))/(8*x) - (3*e^2*(3
*d + 2*e*x)*(d^2 - e^2*x^2)^(5/2))/(8*x^2) - (d*(d^2 - e^2*x^2)^(7/2))/(4*x^4) - (e*(d^2 - e^2*x^2)^(7/2))/x^3
+ (45*d^4*e^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/8 + (45*d^4*e^4*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/8

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 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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^5} \, dx &=-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}-\frac{\int \frac{\left (d^2-e^2 x^2\right )^{5/2} \left (-12 d^4 e-9 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 )^{7/2}}{4 x^4}-\frac{e \left (d^2-e^2 x^2\right )^{7/2}}{x^3}+\frac{\int \frac{\left (27 d^5 e^2-36 d^4 e^3 x\right ) \left (d^2-e^2 x^2\right )^{5/2}}{x^3} \, dx}{12 d^4}\\ &=-\frac{3 e^2 (3 d+2 e x) \left (d^2-e^2 x^2\right )^{5/2}}{8 x^2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}-\frac{e \left (d^2-e^2 x^2\right )^{7/2}}{x^3}-\frac{5 \int \frac{\left (144 d^6 e^3+216 d^5 e^4 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x^2} \, dx}{192 d^4}\\ &=\frac{15 d e^3 (2 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{8 x}-\frac{3 e^2 (3 d+2 e x) \left (d^2-e^2 x^2\right )^{5/2}}{8 x^2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}-\frac{e \left (d^2-e^2 x^2\right )^{7/2}}{x^3}+\frac{5 \int \frac{\left (-432 d^7 e^4+864 d^6 e^5 x\right ) \sqrt{d^2-e^2 x^2}}{x} \, dx}{384 d^4}\\ &=-\frac{45}{8} d^2 e^4 (d-e x) \sqrt{d^2-e^2 x^2}+\frac{15 d e^3 (2 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{8 x}-\frac{3 e^2 (3 d+2 e x) \left (d^2-e^2 x^2\right )^{5/2}}{8 x^2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}-\frac{e \left (d^2-e^2 x^2\right )^{7/2}}{x^3}-\frac{5 \int \frac{864 d^9 e^6-864 d^8 e^7 x}{x \sqrt{d^2-e^2 x^2}} \, dx}{768 d^4 e^2}\\ &=-\frac{45}{8} d^2 e^4 (d-e x) \sqrt{d^2-e^2 x^2}+\frac{15 d e^3 (2 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{8 x}-\frac{3 e^2 (3 d+2 e x) \left (d^2-e^2 x^2\right )^{5/2}}{8 x^2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}-\frac{e \left (d^2-e^2 x^2\right )^{7/2}}{x^3}-\frac{1}{8} \left (45 d^5 e^4\right ) \int \frac{1}{x \sqrt{d^2-e^2 x^2}} \, dx+\frac{1}{8} \left (45 d^4 e^5\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=-\frac{45}{8} d^2 e^4 (d-e x) \sqrt{d^2-e^2 x^2}+\frac{15 d e^3 (2 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{8 x}-\frac{3 e^2 (3 d+2 e x) \left (d^2-e^2 x^2\right )^{5/2}}{8 x^2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}-\frac{e \left (d^2-e^2 x^2\right )^{7/2}}{x^3}-\frac{1}{16} \left (45 d^5 e^4\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{d^2-e^2 x}} \, dx,x,x^2\right )+\frac{1}{8} \left (45 d^4 e^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 )\\ &=-\frac{45}{8} d^2 e^4 (d-e x) \sqrt{d^2-e^2 x^2}+\frac{15 d e^3 (2 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{8 x}-\frac{3 e^2 (3 d+2 e x) \left (d^2-e^2 x^2\right )^{5/2}}{8 x^2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}-\frac{e \left (d^2-e^2 x^2\right )^{7/2}}{x^3}+\frac{45}{8} d^4 e^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+\frac{1}{8} \left (45 d^5 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{45}{8} d^2 e^4 (d-e x) \sqrt{d^2-e^2 x^2}+\frac{15 d e^3 (2 d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{8 x}-\frac{3 e^2 (3 d+2 e x) \left (d^2-e^2 x^2\right )^{5/2}}{8 x^2}-\frac{d \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}-\frac{e \left (d^2-e^2 x^2\right )^{7/2}}{x^3}+\frac{45}{8} d^4 e^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )+\frac{45}{8} d^4 e^4 \tanh ^{-1}\left (\frac{\sqrt{d^2-e^2 x^2}}{d}\right )\\ \end{align*}

Mathematica [C]  time = 0.10926, size = 195, normalized size = 0.93 \[ \frac{e \sqrt{d^2-e^2 x^2} \left (-\frac{7 d^9 \, _2F_1\left (-\frac{5}{2},-\frac{3}{2};-\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x^3 \sqrt{1-\frac{e^2 x^2}{d^2}}}-\frac{7 d^7 e^2 \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x \sqrt{1-\frac{e^2 x^2}{d^2}}}+3 \left (e^3 x^2-d^2 e\right )^3 \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};1-\frac{e^2 x^2}{d^2}\right )+\left (e^3 x^2-d^2 e\right )^3 \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};1-\frac{e^2 x^2}{d^2}\right )\right )}{7 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[d^2 - e^2*x^2]*((-7*d^9*Hypergeometric2F1[-5/2, -3/2, -1/2, (e^2*x^2)/d^2])/(x^3*Sqrt[1 - (e^2*x^2)/d^
2]) - (7*d^7*e^2*Hypergeometric2F1[-5/2, -1/2, 1/2, (e^2*x^2)/d^2])/(x*Sqrt[1 - (e^2*x^2)/d^2]) + 3*(-(d^2*e)
+ e^3*x^2)^3*Hypergeometric2F1[2, 7/2, 9/2, 1 - (e^2*x^2)/d^2] + (-(d^2*e) + e^3*x^2)^3*Hypergeometric2F1[3, 7
/2, 9/2, 1 - (e^2*x^2)/d^2]))/(7*d^3)

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Maple [A]  time = 0.074, size = 302, normalized size = 1.4 \begin{align*} -{\frac{e}{{x}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}+3\,{\frac{{e}^{3} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{7/2}}{{d}^{2}x}}+3\,{\frac{{e}^{5}x \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{5/2}}{{d}^{2}}}+{\frac{15\,{e}^{5}x}{4} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{45\,{d}^{2}{e}^{5}x}{8}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{45\,{d}^{4}{e}^{5}}{8}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{d}{4\,{x}^{4}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-{\frac{9\,{e}^{2}}{8\,d{x}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{7}{2}}}}-{\frac{9\,{e}^{4}}{8\,d} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{15\,d{e}^{4}}{8} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{45\,{d}^{3}{e}^{4}}{8}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{45\,{d}^{5}{e}^{4}}{8}\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^5,x)

[Out]

-e*(-e^2*x^2+d^2)^(7/2)/x^3+3/d^2*e^3/x*(-e^2*x^2+d^2)^(7/2)+3/d^2*e^5*x*(-e^2*x^2+d^2)^(5/2)+15/4*e^5*x*(-e^2
*x^2+d^2)^(3/2)+45/8*d^2*e^5*x*(-e^2*x^2+d^2)^(1/2)+45/8*d^4*e^5/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^
2)^(1/2))-1/4*d*(-e^2*x^2+d^2)^(7/2)/x^4-9/8/d*e^2/x^2*(-e^2*x^2+d^2)^(7/2)-9/8/d*e^4*(-e^2*x^2+d^2)^(5/2)-15/
8*d*e^4*(-e^2*x^2+d^2)^(3/2)-45/8*d^3*e^4*(-e^2*x^2+d^2)^(1/2)+45/8*d^5*e^4/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2
)*(-e^2*x^2+d^2)^(1/2))/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^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.8945, size = 369, normalized size = 1.77 \begin{align*} -\frac{90 \, d^{4} e^{4} x^{4} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) + 45 \, d^{4} e^{4} x^{4} \log \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{x}\right ) + 48 \, d^{4} e^{4} x^{4} -{\left (2 \, e^{7} x^{7} + 8 \, d e^{6} x^{6} + 3 \, d^{2} e^{5} x^{5} - 48 \, d^{3} e^{4} x^{4} + 48 \, d^{4} e^{3} x^{3} - 3 \, d^{5} e^{2} x^{2} - 8 \, d^{6} e x - 2 \, d^{7}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{8 \, x^{4}} \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^5,x, algorithm="fricas")

[Out]

-1/8*(90*d^4*e^4*x^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + 45*d^4*e^4*x^4*log(-(d - sqrt(-e^2*x^2 + d^2)
)/x) + 48*d^4*e^4*x^4 - (2*e^7*x^7 + 8*d*e^6*x^6 + 3*d^2*e^5*x^5 - 48*d^3*e^4*x^4 + 48*d^4*e^3*x^3 - 3*d^5*e^2
*x^2 - 8*d^6*e*x - 2*d^7)*sqrt(-e^2*x^2 + d^2))/x^4

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Sympy [C]  time = 20.5459, size = 1047, normalized size = 5.01 \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**5,x)

[Out]

d**7*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)/(Abs(e**2)*Abs(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*sq
rt(-d**2/(e**2*x**2) + 1)) - I*e**4*asin(d/(e*x))/(8*d**3), True)) + 3*d**6*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)/(Abs(e**2)*Abs(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)) + d**5*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)/(Abs(e**2)*Abs(x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(2*x) - I*e**2*asin(d/(e*x))/(2*d), Tr
ue)) - 5*d**4*e**3*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*sqr
t(1 - e**2*x**2/d**2)), True)) - 5*d**3*e**4*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)) + d**2*e**5*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)) + 3*d*e**6*Piecewise((x**2*
sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) + e**7*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))

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Giac [B]  time = 1.25771, size = 505, normalized size = 2.42 \begin{align*} \frac{45}{8} \, d^{4} \arcsin \left (\frac{x e}{d}\right ) e^{4} \mathrm{sgn}\left (d\right ) + \frac{45}{8} \, d^{4} 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 ) + \frac{{\left (d^{4} e^{10} + \frac{8 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} d^{4} e^{8}}{x} + \frac{8 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{4} e^{6}}{x^{2}} - \frac{184 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{4} e^{4}}{x^{3}}\right )} x^{4} e^{2}}{64 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{4}} + \frac{1}{64} \,{\left (\frac{184 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )} d^{4} e^{26}}{x} - \frac{8 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{4} e^{24}}{x^{2}} - \frac{8 \,{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{4} e^{22}}{x^{3}} - \frac{{\left (d e + \sqrt{-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{4} e^{20}}{x^{4}}\right )} e^{\left (-24\right )} - \frac{1}{8} \,{\left (48 \, d^{3} e^{4} -{\left (3 \, d^{2} e^{5} + 2 \,{\left (x e^{7} + 4 \, d e^{6}\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^5,x, algorithm="giac")

[Out]

45/8*d^4*arcsin(x*e/d)*e^4*sgn(d) + 45/8*d^4*e^4*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))
+ 1/64*(d^4*e^10 + 8*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^4*e^8/x + 8*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^4*e^6/x^
2 - 184*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d^4*e^4/x^3)*x^4*e^2/(d*e + sqrt(-x^2*e^2 + d^2)*e)^4 + 1/64*(184*(d*
e + sqrt(-x^2*e^2 + d^2)*e)*d^4*e^26/x - 8*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^4*e^24/x^2 - 8*(d*e + sqrt(-x^2*
e^2 + d^2)*e)^3*d^4*e^22/x^3 - (d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d^4*e^20/x^4)*e^(-24) - 1/8*(48*d^3*e^4 - (3*d
^2*e^5 + 2*(x*e^7 + 4*d*e^6)*x)*x)*sqrt(-x^2*e^2 + d^2)