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

Optimal. Leaf size=207 \[ \frac {1}{24} d^2 e^2 (4 d-85 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac {1}{30} e^2 (3 d-85 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {85}{16} d^6 e^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {1}{2} d^6 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )+\frac {1}{16} d^4 e^2 (8 d-85 e x) \sqrt {d^2-e^2 x^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.31, antiderivative size = 207, normalized size of antiderivative = 1.00, 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 = {1807, 815, 844, 217, 203, 266, 63, 208} \begin {gather*} \frac {1}{16} d^4 e^2 (8 d-85 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{24} d^2 e^2 (4 d-85 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac {1}{30} e^2 (3 d-85 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {85}{16} d^6 e^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {1}{2} d^6 e^2 \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*(d^2 - e^2*x^2)^(5/2))/x^3,x]

[Out]

(d^4*e^2*(8*d - 85*e*x)*Sqrt[d^2 - e^2*x^2])/16 + (d^2*e^2*(4*d - 85*e*x)*(d^2 - e^2*x^2)^(3/2))/24 + (e^2*(3*
d - 85*e*x)*(d^2 - e^2*x^2)^(5/2))/30 - (d*(d^2 - e^2*x^2)^(7/2))/(2*x^2) - (3*e*(d^2 - e^2*x^2)^(7/2))/x - (8
5*d^6*e^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/16 - (d^6*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/2

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 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 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 \left (d^2-e^2 x^2\right )^{5/2}}{x^3} \, dx &=-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{5/2} \left (-6 d^4 e-d^3 e^2 x-2 d^2 e^3 x^2\right )}{x^2} \, dx}{2 d^2}\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac {\int \frac {\left (d^5 e^2-34 d^4 e^3 x\right ) \left (d^2-e^2 x^2\right )^{5/2}}{x} \, dx}{2 d^4}\\ &=\frac {1}{30} e^2 (3 d-85 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{x}-\frac {\int \frac {\left (-6 d^7 e^4+170 d^6 e^5 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x} \, dx}{12 d^4 e^2}\\ &=\frac {1}{24} d^2 e^2 (4 d-85 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {1}{30} e^2 (3 d-85 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac {\int \frac {\left (24 d^9 e^6-510 d^8 e^7 x\right ) \sqrt {d^2-e^2 x^2}}{x} \, dx}{48 d^4 e^4}\\ &=\frac {1}{16} d^4 e^2 (8 d-85 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{24} d^2 e^2 (4 d-85 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {1}{30} e^2 (3 d-85 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{x}-\frac {\int \frac {-48 d^{11} e^8+510 d^{10} e^9 x}{x \sqrt {d^2-e^2 x^2}} \, dx}{96 d^4 e^6}\\ &=\frac {1}{16} d^4 e^2 (8 d-85 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{24} d^2 e^2 (4 d-85 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {1}{30} e^2 (3 d-85 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac {1}{2} \left (d^7 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx-\frac {1}{16} \left (85 d^6 e^3\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {1}{16} d^4 e^2 (8 d-85 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{24} d^2 e^2 (4 d-85 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {1}{30} e^2 (3 d-85 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{x}+\frac {1}{4} \left (d^7 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )-\frac {1}{16} \left (85 d^6 e^3\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}{16} d^4 e^2 (8 d-85 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{24} d^2 e^2 (4 d-85 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {1}{30} e^2 (3 d-85 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{x}-\frac {85}{16} d^6 e^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {1}{2} d^7 \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 {1}{16} d^4 e^2 (8 d-85 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{24} d^2 e^2 (4 d-85 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {1}{30} e^2 (3 d-85 e x) \left (d^2-e^2 x^2\right )^{5/2}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{2 x^2}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{x}-\frac {85}{16} d^6 e^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {1}{2} d^6 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.64, size = 259, normalized size = 1.25 \begin {gather*} -\frac {e \left (5040 d^9 \sqrt {1-\frac {e^2 x^2}{d^2}} \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};\frac {e^2 x^2}{d^2}\right )+e x \left (240 \left (d^2-e^2 x^2\right )^4 \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};1-\frac {e^2 x^2}{d^2}\right )-7 d \left (1104 d^7+165 d^6 e x-1632 d^5 e^2 x^2-295 d^4 e^3 x^3+672 d^3 e^4 x^4+170 d^2 e^5 x^5+75 d^7 \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right )-720 d^6 \sqrt {d^2-e^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )-144 d e^6 x^6-40 e^7 x^7\right )\right )\right )}{1680 d x \sqrt {d^2-e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1680*(e*(5040*d^9*Sqrt[1 - (e^2*x^2)/d^2]*Hypergeometric2F1[-5/2, -1/2, 1/2, (e^2*x^2)/d^2] + e*x*(-7*d*(11
04*d^7 + 165*d^6*e*x - 1632*d^5*e^2*x^2 - 295*d^4*e^3*x^3 + 672*d^3*e^4*x^4 + 170*d^2*e^5*x^5 - 144*d*e^6*x^6
- 40*e^7*x^7 + 75*d^7*Sqrt[1 - (e^2*x^2)/d^2]*ArcSin[(e*x)/d] - 720*d^6*Sqrt[d^2 - e^2*x^2]*ArcTanh[Sqrt[d^2 -
e^2*x^2]/d]) + 240*(d^2 - e^2*x^2)^4*Hypergeometric2F1[2, 7/2, 9/2, 1 - (e^2*x^2)/d^2])))/(d*x*Sqrt[d^2 - e^2
*x^2])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.72, size = 189, normalized size = 0.91 \begin {gather*} -\frac {85}{16} d^6 e \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )+d^6 e^2 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )+\frac {\sqrt {d^2-e^2 x^2} \left (-120 d^7-720 d^6 e x+544 d^5 e^2 x^2-645 d^4 e^3 x^3-448 d^3 e^4 x^4+50 d^2 e^5 x^5+144 d e^6 x^6+40 e^7 x^7\right )}{240 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-120*d^7 - 720*d^6*e*x + 544*d^5*e^2*x^2 - 645*d^4*e^3*x^3 - 448*d^3*e^4*x^4 + 50*d^2*e^
5*x^5 + 144*d*e^6*x^6 + 40*e^7*x^7))/(240*x^2) + d^6*e^2*ArcTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d] - (
85*d^6*e*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/16

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 179, normalized size = 0.86 \begin {gather*} \frac {2550 \, d^{6} e^{2} x^{2} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + 120 \, d^{6} e^{2} x^{2} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + 544 \, d^{6} e^{2} x^{2} + {\left (40 \, e^{7} x^{7} + 144 \, d e^{6} x^{6} + 50 \, d^{2} e^{5} x^{5} - 448 \, d^{3} e^{4} x^{4} - 645 \, d^{4} e^{3} x^{3} + 544 \, d^{5} e^{2} x^{2} - 720 \, d^{6} e x - 120 \, d^{7}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, x^{2}} \end {gather*}

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^3,x, algorithm="fricas")

[Out]

1/240*(2550*d^6*e^2*x^2*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + 120*d^6*e^2*x^2*log(-(d - sqrt(-e^2*x^2 +
d^2))/x) + 544*d^6*e^2*x^2 + (40*e^7*x^7 + 144*d*e^6*x^6 + 50*d^2*e^5*x^5 - 448*d^3*e^4*x^4 - 645*d^4*e^3*x^3
+ 544*d^5*e^2*x^2 - 720*d^6*e*x - 120*d^7)*sqrt(-e^2*x^2 + d^2))/x^2

________________________________________________________________________________________

giac [A]  time = 0.25, size = 262, normalized size = 1.27 \begin {gather*} -\frac {85}{16} \, d^{6} \arcsin \left (\frac {x e}{d}\right ) e^{2} \mathrm {sgn}\relax (d) - \frac {1}{2} \, d^{6} e^{2} \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}{8} \, {\left (\frac {12 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{6} e^{8}}{x} + \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{6} e^{6}}{x^{2}}\right )} e^{\left (-8\right )} + \frac {1}{240} \, {\left (544 \, d^{5} e^{2} - {\left (645 \, d^{4} e^{3} + 2 \, {\left (224 \, d^{3} e^{4} - {\left (25 \, d^{2} e^{5} + 4 \, {\left (5 \, x e^{7} + 18 \, d e^{6}\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} + \frac {{\left (d^{6} e^{6} + \frac {12 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{6} e^{4}}{x}\right )} x^{2}}{8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2}} \end {gather*}

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^3,x, algorithm="giac")

[Out]

-85/16*d^6*arcsin(x*e/d)*e^2*sgn(d) - 1/2*d^6*e^2*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x)
) - 1/8*(12*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^6*e^8/x + (d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^6*e^6/x^2)*e^(-8) +
1/240*(544*d^5*e^2 - (645*d^4*e^3 + 2*(224*d^3*e^4 - (25*d^2*e^5 + 4*(5*x*e^7 + 18*d*e^6)*x)*x)*x)*x)*sqrt(-x^
2*e^2 + d^2) + 1/8*(d^6*e^6 + 12*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^6*e^4/x)*x^2/(d*e + sqrt(-x^2*e^2 + d^2)*e)^
2

________________________________________________________________________________________

maple [A]  time = 0.02, size = 252, normalized size = 1.22 \begin {gather*} -\frac {d^{7} e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}}-\frac {85 d^{6} e^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 \sqrt {e^{2}}}-\frac {85 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{4} e^{3} x}{16}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, d^{5} e^{2}}{2}-\frac {85 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{2} e^{3} x}{24}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{3} e^{2}}{6}-\frac {17 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3} x}{6}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d \,e^{2}}{10}-\frac {3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e}{x}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} d}{2 x^{2}} \end {gather*}

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^3,x)

[Out]

-17/6*e^3*x*(-e^2*x^2+d^2)^(5/2)-85/24*e^3*d^2*x*(-e^2*x^2+d^2)^(3/2)-85/16*e^3*d^4*x*(-e^2*x^2+d^2)^(1/2)-85/
16*e^3*d^6/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-1/2*d*(-e^2*x^2+d^2)^(7/2)/x^2+1/10*d*e^2*(-
e^2*x^2+d^2)^(5/2)+1/6*d^3*e^2*(-e^2*x^2+d^2)^(3/2)+1/2*d^5*e^2*(-e^2*x^2+d^2)^(1/2)-1/2*d^7*e^2/(d^2)^(1/2)*l
n((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-3*e*(-e^2*x^2+d^2)^(7/2)/x

________________________________________________________________________________________

maxima [A]  time = 1.00, size = 229, normalized size = 1.11 \begin {gather*} -\frac {85}{16} \, d^{6} e^{2} \arcsin \left (\frac {e x}{d}\right ) - \frac {1}{2} \, d^{6} e^{2} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) - \frac {85}{16} \, \sqrt {-e^{2} x^{2} + d^{2}} d^{4} e^{3} x + \frac {1}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5} e^{2} - \frac {85}{24} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} e^{3} x + \frac {1}{6} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3} e^{2} + \frac {1}{6} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3} x + \frac {1}{10} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d e^{2} - \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2} e}{x} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} d}{2 \, x^{2}} \end {gather*}

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^3,x, algorithm="maxima")

[Out]

-85/16*d^6*e^2*arcsin(e*x/d) - 1/2*d^6*e^2*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x)) - 85/16*sqrt(-e
^2*x^2 + d^2)*d^4*e^3*x + 1/2*sqrt(-e^2*x^2 + d^2)*d^5*e^2 - 85/24*(-e^2*x^2 + d^2)^(3/2)*d^2*e^3*x + 1/6*(-e^
2*x^2 + d^2)^(3/2)*d^3*e^2 + 1/6*(-e^2*x^2 + d^2)^(5/2)*e^3*x + 1/10*(-e^2*x^2 + d^2)^(5/2)*d*e^2 - 3*(-e^2*x^
2 + d^2)^(5/2)*d^2*e/x - 1/2*(-e^2*x^2 + d^2)^(7/2)*d/x^2

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 22.22, size = 1059, normalized size = 5.12

result too large to display

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**3,x)

[Out]

d**7*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)) + 3*d**6*e*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/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)) + d**5*e**2*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/(e**2*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)) - 5*d**4*e**3*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/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**3*e**4*Piecewise((x**2*sqrt(d**2)/2,
Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) + d**2*e**5*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/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)) + 3*d
*e**6*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)) + e**7*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))

________________________________________________________________________________________