3.76 \(\int \frac {(d+e x)^3 (d^2-e^2 x^2)^{5/2}}{x^6} \, dx\)

Optimal. Leaf size=216 \[ -\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{5 x^5}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}-\frac {e^2 (52 d+25 e x) \left (d^2-e^2 x^2\right )^{5/2}}{60 x^3}+\frac {d^2 e^4 (52 d+25 e x) \sqrt {d^2-e^2 x^2}}{8 x}+\frac {d e^3 (25 d-52 e x) \left (d^2-e^2 x^2\right )^{3/2}}{24 x^2}+\frac {13}{2} d^3 e^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {25}{8} d^3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

[Out]

1/24*d*e^3*(-52*e*x+25*d)*(-e^2*x^2+d^2)^(3/2)/x^2-1/60*e^2*(25*e*x+52*d)*(-e^2*x^2+d^2)^(5/2)/x^3-1/5*d*(-e^2
*x^2+d^2)^(7/2)/x^5-3/4*e*(-e^2*x^2+d^2)^(7/2)/x^4+13/2*d^3*e^5*arctan(e*x/(-e^2*x^2+d^2)^(1/2))-25/8*d^3*e^5*
arctanh((-e^2*x^2+d^2)^(1/2)/d)+1/8*d^2*e^4*(25*e*x+52*d)*(-e^2*x^2+d^2)^(1/2)/x

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Rubi [A]  time = 0.31, antiderivative size = 216, 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, 813, 844, 217, 203, 266, 63, 208} \[ \frac {d^2 e^4 (52 d+25 e x) \sqrt {d^2-e^2 x^2}}{8 x}+\frac {d e^3 (25 d-52 e x) \left (d^2-e^2 x^2\right )^{3/2}}{24 x^2}-\frac {e^2 (52 d+25 e x) \left (d^2-e^2 x^2\right )^{5/2}}{60 x^3}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{5 x^5}+\frac {13}{2} d^3 e^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {25}{8} d^3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^3*(d^2 - e^2*x^2)^(5/2))/x^6,x]

[Out]

(d^2*e^4*(52*d + 25*e*x)*Sqrt[d^2 - e^2*x^2])/(8*x) + (d*e^3*(25*d - 52*e*x)*(d^2 - e^2*x^2)^(3/2))/(24*x^2) -
(e^2*(52*d + 25*e*x)*(d^2 - e^2*x^2)^(5/2))/(60*x^3) - (d*(d^2 - e^2*x^2)^(7/2))/(5*x^5) - (3*e*(d^2 - e^2*x^
2)^(7/2))/(4*x^4) + (13*d^3*e^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/2 - (25*d^3*e^5*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 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 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^6} \, dx &=-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{5 x^5}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{5/2} \left (-15 d^4 e-13 d^3 e^2 x-5 d^2 e^3 x^2\right )}{x^5} \, dx}{5 d^2}\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{5 x^5}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}+\frac {\int \frac {\left (52 d^5 e^2-25 d^4 e^3 x\right ) \left (d^2-e^2 x^2\right )^{5/2}}{x^4} \, dx}{20 d^4}\\ &=-\frac {e^2 (52 d+25 e x) \left (d^2-e^2 x^2\right )^{5/2}}{60 x^3}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{5 x^5}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}-\frac {\int \frac {\left (150 d^6 e^3+312 d^5 e^4 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x^3} \, dx}{72 d^4}\\ &=\frac {d e^3 (25 d-52 e x) \left (d^2-e^2 x^2\right )^{3/2}}{24 x^2}-\frac {e^2 (52 d+25 e x) \left (d^2-e^2 x^2\right )^{5/2}}{60 x^3}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{5 x^5}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}+\frac {\int \frac {\left (-1248 d^7 e^4+600 d^6 e^5 x\right ) \sqrt {d^2-e^2 x^2}}{x^2} \, dx}{192 d^4}\\ &=\frac {d^2 e^4 (52 d+25 e x) \sqrt {d^2-e^2 x^2}}{8 x}+\frac {d e^3 (25 d-52 e x) \left (d^2-e^2 x^2\right )^{3/2}}{24 x^2}-\frac {e^2 (52 d+25 e x) \left (d^2-e^2 x^2\right )^{5/2}}{60 x^3}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{5 x^5}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}-\frac {\int \frac {-1200 d^8 e^5-2496 d^7 e^6 x}{x \sqrt {d^2-e^2 x^2}} \, dx}{384 d^4}\\ &=\frac {d^2 e^4 (52 d+25 e x) \sqrt {d^2-e^2 x^2}}{8 x}+\frac {d e^3 (25 d-52 e x) \left (d^2-e^2 x^2\right )^{3/2}}{24 x^2}-\frac {e^2 (52 d+25 e x) \left (d^2-e^2 x^2\right )^{5/2}}{60 x^3}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{5 x^5}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}+\frac {1}{8} \left (25 d^4 e^5\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx+\frac {1}{2} \left (13 d^3 e^6\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {d^2 e^4 (52 d+25 e x) \sqrt {d^2-e^2 x^2}}{8 x}+\frac {d e^3 (25 d-52 e x) \left (d^2-e^2 x^2\right )^{3/2}}{24 x^2}-\frac {e^2 (52 d+25 e x) \left (d^2-e^2 x^2\right )^{5/2}}{60 x^3}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{5 x^5}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}+\frac {1}{16} \left (25 d^4 e^5\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )+\frac {1}{2} \left (13 d^3 e^6\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 {d^2 e^4 (52 d+25 e x) \sqrt {d^2-e^2 x^2}}{8 x}+\frac {d e^3 (25 d-52 e x) \left (d^2-e^2 x^2\right )^{3/2}}{24 x^2}-\frac {e^2 (52 d+25 e x) \left (d^2-e^2 x^2\right )^{5/2}}{60 x^3}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{5 x^5}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}+\frac {13}{2} d^3 e^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {1}{8} \left (25 d^4 e^3\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 {d^2 e^4 (52 d+25 e x) \sqrt {d^2-e^2 x^2}}{8 x}+\frac {d e^3 (25 d-52 e x) \left (d^2-e^2 x^2\right )^{3/2}}{24 x^2}-\frac {e^2 (52 d+25 e x) \left (d^2-e^2 x^2\right )^{5/2}}{60 x^3}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{5 x^5}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{4 x^4}+\frac {13}{2} d^3 e^5 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {25}{8} d^3 e^5 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 199, normalized size = 0.92 \[ \frac {\sqrt {d^2-e^2 x^2} \left (5 e^5 \left (e^2 x^2-d^2\right )^3 \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};1-\frac {e^2 x^2}{d^2}\right )+15 e^5 \left (e^2 x^2-d^2\right )^3 \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};1-\frac {e^2 x^2}{d^2}\right )-\frac {7 d^{11} \, _2F_1\left (-\frac {5}{2},-\frac {5}{2};-\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{x^5 \sqrt {1-\frac {e^2 x^2}{d^2}}}-\frac {35 d^9 e^2 \, _2F_1\left (-\frac {5}{2},-\frac {3}{2};-\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{x^3 \sqrt {1-\frac {e^2 x^2}{d^2}}}\right )}{35 d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^3*(d^2 - e^2*x^2)^(5/2))/x^6,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*((-7*d^11*Hypergeometric2F1[-5/2, -5/2, -3/2, (e^2*x^2)/d^2])/(x^5*Sqrt[1 - (e^2*x^2)/d^2
]) - (35*d^9*e^2*Hypergeometric2F1[-5/2, -3/2, -1/2, (e^2*x^2)/d^2])/(x^3*Sqrt[1 - (e^2*x^2)/d^2]) + 5*e^5*(-d
^2 + e^2*x^2)^3*Hypergeometric2F1[2, 7/2, 9/2, 1 - (e^2*x^2)/d^2] + 15*e^5*(-d^2 + e^2*x^2)^3*Hypergeometric2F
1[3, 7/2, 9/2, 1 - (e^2*x^2)/d^2]))/(35*d^4)

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fricas [A]  time = 0.95, size = 180, normalized size = 0.83 \[ -\frac {1560 \, d^{3} e^{5} x^{5} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - 375 \, d^{3} e^{5} x^{5} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - 80 \, d^{3} e^{5} x^{5} - {\left (40 \, e^{7} x^{7} + 180 \, d e^{6} x^{6} + 80 \, d^{2} e^{5} x^{5} + 656 \, d^{3} e^{4} x^{4} + 345 \, d^{4} e^{3} x^{3} - 32 \, d^{5} e^{2} x^{2} - 90 \, d^{6} e x - 24 \, d^{7}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{120 \, x^{5}} \]

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^6,x, algorithm="fricas")

[Out]

-1/120*(1560*d^3*e^5*x^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - 375*d^3*e^5*x^5*log(-(d - sqrt(-e^2*x^2 +
d^2))/x) - 80*d^3*e^5*x^5 - (40*e^7*x^7 + 180*d*e^6*x^6 + 80*d^2*e^5*x^5 + 656*d^3*e^4*x^4 + 345*d^4*e^3*x^3
- 32*d^5*e^2*x^2 - 90*d^6*e*x - 24*d^7)*sqrt(-e^2*x^2 + d^2))/x^5

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giac [B]  time = 0.28, size = 430, normalized size = 1.99 \[ \frac {13}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{5} \mathrm {sgn}\relax (d) - \frac {25}{8} \, d^{3} e^{5} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right ) + \frac {{\left (6 \, d^{3} e^{12} + \frac {45 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{3} e^{10}}{x} + \frac {50 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{3} e^{8}}{x^{2}} - \frac {600 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{3} e^{6}}{x^{3}} - \frac {2580 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{3} e^{4}}{x^{4}}\right )} x^{5} e^{3}}{960 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5}} + \frac {1}{960} \, {\left (\frac {2580 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{3} e^{38}}{x} + \frac {600 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{3} e^{36}}{x^{2}} - \frac {50 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{3} e^{34}}{x^{3}} - \frac {45 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{3} e^{32}}{x^{4}} - \frac {6 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} d^{3} e^{30}}{x^{5}}\right )} e^{\left (-35\right )} + \frac {1}{6} \, {\left (4 \, d^{2} e^{5} + {\left (2 \, x e^{7} + 9 \, d e^{6}\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \]

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^6,x, algorithm="giac")

[Out]

13/2*d^3*arcsin(x*e/d)*e^5*sgn(d) - 25/8*d^3*e^5*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))
+ 1/960*(6*d^3*e^12 + 45*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^3*e^10/x + 50*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^3*
e^8/x^2 - 600*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d^3*e^6/x^3 - 2580*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d^3*e^4/x^4
)*x^5*e^3/(d*e + sqrt(-x^2*e^2 + d^2)*e)^5 + 1/960*(2580*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^3*e^38/x + 600*(d*e
+ sqrt(-x^2*e^2 + d^2)*e)^2*d^3*e^36/x^2 - 50*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d^3*e^34/x^3 - 45*(d*e + sqrt(-
x^2*e^2 + d^2)*e)^4*d^3*e^32/x^4 - 6*(d*e + sqrt(-x^2*e^2 + d^2)*e)^5*d^3*e^30/x^5)*e^(-35) + 1/6*(4*d^2*e^5 +
(2*x*e^7 + 9*d*e^6)*x)*sqrt(-x^2*e^2 + d^2)

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maple [A]  time = 0.03, size = 327, normalized size = 1.51 \[ -\frac {25 d^{4} e^{5} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{8 \sqrt {d^{2}}}+\frac {13 d^{3} e^{6} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}+\frac {13 \sqrt {-e^{2} x^{2}+d^{2}}\, d \,e^{6} x}{2}+\frac {25 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} e^{5}}{8}+\frac {13 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{6} x}{3 d}+\frac {25 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{5}}{24}+\frac {52 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{6} x}{15 d^{3}}+\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{5}}{8 d^{2}}+\frac {52 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{4}}{15 d^{3} x}+\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{3}}{8 d^{2} x^{2}}-\frac {13 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{2}}{15 d \,x^{3}}-\frac {3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e}{4 x^{4}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} d}{5 x^{5}} \]

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^6,x)

[Out]

-3/4*e*(-e^2*x^2+d^2)^(7/2)/x^4+5/8/d^2*e^3/x^2*(-e^2*x^2+d^2)^(7/2)+5/8/d^2*e^5*(-e^2*x^2+d^2)^(5/2)+25/24*e^
5*(-e^2*x^2+d^2)^(3/2)+25/8*d^2*e^5*(-e^2*x^2+d^2)^(1/2)-25/8*d^4*e^5/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^
2*x^2+d^2)^(1/2))/x)-13/15/d*e^2/x^3*(-e^2*x^2+d^2)^(7/2)+52/15/d^3*e^4/x*(-e^2*x^2+d^2)^(7/2)+52/15/d^3*e^6*x
*(-e^2*x^2+d^2)^(5/2)+13/3/d*e^6*x*(-e^2*x^2+d^2)^(3/2)+13/2*d*e^6*x*(-e^2*x^2+d^2)^(1/2)+13/2*d^3*e^6/(e^2)^(
1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-1/5*d*(-e^2*x^2+d^2)^(7/2)/x^5

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maxima [A]  time = 0.99, size = 278, normalized size = 1.29 \[ \frac {13}{2} \, d^{3} e^{5} \arcsin \left (\frac {e x}{d}\right ) - \frac {25}{8} \, d^{3} e^{5} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) + \frac {13}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{6} x + \frac {25}{8} \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e^{5} + \frac {13 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{6} x}{3 \, d} + \frac {25}{24} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{5} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{5}}{8 \, d^{2}} + \frac {52 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}}{15 \, d x} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e^{3}}{8 \, d^{2} x^{2}} - \frac {13 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e^{2}}{15 \, d x^{3}} - \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e}{4 \, x^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} d}{5 \, x^{5}} \]

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^6,x, algorithm="maxima")

[Out]

13/2*d^3*e^5*arcsin(e*x/d) - 25/8*d^3*e^5*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x)) + 13/2*sqrt(-e^2
*x^2 + d^2)*d*e^6*x + 25/8*sqrt(-e^2*x^2 + d^2)*d^2*e^5 + 13/3*(-e^2*x^2 + d^2)^(3/2)*e^6*x/d + 25/24*(-e^2*x^
2 + d^2)^(3/2)*e^5 + 5/8*(-e^2*x^2 + d^2)^(5/2)*e^5/d^2 + 52/15*(-e^2*x^2 + d^2)^(5/2)*e^4/(d*x) + 5/8*(-e^2*x
^2 + d^2)^(7/2)*e^3/(d^2*x^2) - 13/15*(-e^2*x^2 + d^2)^(7/2)*e^2/(d*x^3) - 3/4*(-e^2*x^2 + d^2)^(7/2)*e/x^4 -
1/5*(-e^2*x^2 + d^2)^(7/2)*d/x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}\,{\left (d+e\,x\right )}^3}{x^6} \,d x \]

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^6,x)

[Out]

int(((d^2 - e^2*x^2)^(5/2)*(d + e*x)^3)/x^6, x)

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sympy [C]  time = 20.70, size = 1178, normalized size = 5.45 \[ \text {result too large to display} \]

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**6,x)

[Out]

d**7*Piecewise((3*I*d**3*sqrt(-1 + e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) - 4*I*d*e**2*x**2*sqrt(-1 +
e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) + 2*I*e**6*x**6*sqrt(-1 + e**2*x**2/d**2)/(-15*d**5*x**5 + 15*d
**3*e**2*x**7) - I*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)/(-15*d**3*x**5 + 15*d*e**2*x**7), Abs(e**2*x**2/d**2) >
1), (3*d**3*sqrt(1 - e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e**2*x**7) - 4*d*e**2*x**2*sqrt(1 - e**2*x**2/d**2)/
(-15*d**2*x**5 + 15*e**2*x**7) + 2*e**6*x**6*sqrt(1 - e**2*x**2/d**2)/(-15*d**5*x**5 + 15*d**3*e**2*x**7) - e*
*4*x**4*sqrt(1 - e**2*x**2/d**2)/(-15*d**3*x**5 + 15*d*e**2*x**7), True)) + 3*d**6*e*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)) + d**5*e**2*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)) - 5*d**4*e**3*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)) - 5*d**3*e**4*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**2*e**5*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)) +
3*d*e**6*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
)) + e**7*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True))

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