3.1.80 \(\int \frac {(d+e x)^3 (d^2-e^2 x^2)^{5/2}}{x^{10}} \, dx\)

Optimal. Leaf size=187 \[ -\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{9 x^9}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{8 x^8}-\frac {29 e^2 \left (d^2-e^2 x^2\right )^{7/2}}{63 d x^7}+\frac {55 e^9 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{128 d}-\frac {55 e^7 \sqrt {d^2-e^2 x^2}}{128 x^2}+\frac {55 e^5 \left (d^2-e^2 x^2\right )^{3/2}}{192 x^4}-\frac {11 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{48 x^6} \]

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Rubi [A]  time = 0.26, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1807, 807, 266, 47, 63, 208} \begin {gather*} -\frac {55 e^7 \sqrt {d^2-e^2 x^2}}{128 x^2}+\frac {55 e^5 \left (d^2-e^2 x^2\right )^{3/2}}{192 x^4}-\frac {11 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{48 x^6}-\frac {29 e^2 \left (d^2-e^2 x^2\right )^{7/2}}{63 d x^7}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{8 x^8}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{9 x^9}+\frac {55 e^9 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{128 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^3*(d^2 - e^2*x^2)^(5/2))/x^10,x]

[Out]

(-55*e^7*Sqrt[d^2 - e^2*x^2])/(128*x^2) + (55*e^5*(d^2 - e^2*x^2)^(3/2))/(192*x^4) - (11*e^3*(d^2 - e^2*x^2)^(
5/2))/(48*x^6) - (d*(d^2 - e^2*x^2)^(7/2))/(9*x^9) - (3*e*(d^2 - e^2*x^2)^(7/2))/(8*x^8) - (29*e^2*(d^2 - e^2*
x^2)^(7/2))/(63*d*x^7) + (55*e^9*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(128*d)

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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]

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 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(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]
&& EqQ[Simplify[m + 2*p + 3], 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^{10}} \, dx &=-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{9 x^9}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{5/2} \left (-27 d^4 e-29 d^3 e^2 x-9 d^2 e^3 x^2\right )}{x^9} \, dx}{9 d^2}\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{9 x^9}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{8 x^8}+\frac {\int \frac {\left (232 d^5 e^2+99 d^4 e^3 x\right ) \left (d^2-e^2 x^2\right )^{5/2}}{x^8} \, dx}{72 d^4}\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{9 x^9}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{8 x^8}-\frac {29 e^2 \left (d^2-e^2 x^2\right )^{7/2}}{63 d x^7}+\frac {1}{8} \left (11 e^3\right ) \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^7} \, dx\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{9 x^9}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{8 x^8}-\frac {29 e^2 \left (d^2-e^2 x^2\right )^{7/2}}{63 d x^7}+\frac {1}{16} \left (11 e^3\right ) \operatorname {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{5/2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac {11 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{48 x^6}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{9 x^9}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{8 x^8}-\frac {29 e^2 \left (d^2-e^2 x^2\right )^{7/2}}{63 d x^7}-\frac {1}{96} \left (55 e^5\right ) \operatorname {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=\frac {55 e^5 \left (d^2-e^2 x^2\right )^{3/2}}{192 x^4}-\frac {11 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{48 x^6}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{9 x^9}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{8 x^8}-\frac {29 e^2 \left (d^2-e^2 x^2\right )^{7/2}}{63 d x^7}+\frac {1}{128} \left (55 e^7\right ) \operatorname {Subst}\left (\int \frac {\sqrt {d^2-e^2 x}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {55 e^7 \sqrt {d^2-e^2 x^2}}{128 x^2}+\frac {55 e^5 \left (d^2-e^2 x^2\right )^{3/2}}{192 x^4}-\frac {11 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{48 x^6}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{9 x^9}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{8 x^8}-\frac {29 e^2 \left (d^2-e^2 x^2\right )^{7/2}}{63 d x^7}-\frac {1}{256} \left (55 e^9\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )\\ &=-\frac {55 e^7 \sqrt {d^2-e^2 x^2}}{128 x^2}+\frac {55 e^5 \left (d^2-e^2 x^2\right )^{3/2}}{192 x^4}-\frac {11 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{48 x^6}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{9 x^9}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{8 x^8}-\frac {29 e^2 \left (d^2-e^2 x^2\right )^{7/2}}{63 d x^7}+\frac {1}{128} \left (55 e^7\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 {55 e^7 \sqrt {d^2-e^2 x^2}}{128 x^2}+\frac {55 e^5 \left (d^2-e^2 x^2\right )^{3/2}}{192 x^4}-\frac {11 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{48 x^6}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{9 x^9}-\frac {3 e \left (d^2-e^2 x^2\right )^{7/2}}{8 x^8}-\frac {29 e^2 \left (d^2-e^2 x^2\right )^{7/2}}{63 d x^7}+\frac {55 e^9 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{128 d}\\ \end {align*}

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Mathematica [C]  time = 0.17, size = 218, normalized size = 1.17 \begin {gather*} \frac {-112 d^{10}-16 d^8 e^2 x^2-168 d^7 e^3 x^3+1184 d^6 e^4 x^4+714 d^5 e^5 x^5-2336 d^4 e^6 x^6-1239 d^3 e^7 x^7+1744 d^2 e^8 x^8+315 d e^9 x^9 \sqrt {1-\frac {e^2 x^2}{d^2}} \tanh ^{-1}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )+693 d e^9 x^9-464 e^{10} x^{10}}{1008 d x^9 \sqrt {d^2-e^2 x^2}}-\frac {3 e^9 \left (d^2-e^2 x^2\right )^{7/2} \, _2F_1\left (\frac {7}{2},5;\frac {9}{2};1-\frac {e^2 x^2}{d^2}\right )}{7 d^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^3*(d^2 - e^2*x^2)^(5/2))/x^10,x]

[Out]

(-112*d^10 - 16*d^8*e^2*x^2 - 168*d^7*e^3*x^3 + 1184*d^6*e^4*x^4 + 714*d^5*e^5*x^5 - 2336*d^4*e^6*x^6 - 1239*d
^3*e^7*x^7 + 1744*d^2*e^8*x^8 + 693*d*e^9*x^9 - 464*e^10*x^10 + 315*d*e^9*x^9*Sqrt[1 - (e^2*x^2)/d^2]*ArcTanh[
Sqrt[1 - (e^2*x^2)/d^2]])/(1008*d*x^9*Sqrt[d^2 - e^2*x^2]) - (3*e^9*(d^2 - e^2*x^2)^(7/2)*Hypergeometric2F1[7/
2, 5, 9/2, 1 - (e^2*x^2)/d^2])/(7*d^8)

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IntegrateAlgebraic [A]  time = 0.88, size = 159, normalized size = 0.85 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-896 d^8-3024 d^7 e x-1024 d^6 e^2 x^2+7224 d^5 e^3 x^3+8448 d^4 e^4 x^4-3066 d^3 e^5 x^5-10240 d^2 e^6 x^6-4599 d e^7 x^7+3712 e^8 x^8\right )}{8064 d x^9}-\frac {55 e^9 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{64 d} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((d + e*x)^3*(d^2 - e^2*x^2)^(5/2))/x^10,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-896*d^8 - 3024*d^7*e*x - 1024*d^6*e^2*x^2 + 7224*d^5*e^3*x^3 + 8448*d^4*e^4*x^4 - 3066*
d^3*e^5*x^5 - 10240*d^2*e^6*x^6 - 4599*d*e^7*x^7 + 3712*e^8*x^8))/(8064*d*x^9) - (55*e^9*ArcTanh[(Sqrt[-e^2]*x
)/d - Sqrt[d^2 - e^2*x^2]/d])/(64*d)

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fricas [A]  time = 0.46, size = 142, normalized size = 0.76 \begin {gather*} -\frac {3465 \, e^{9} x^{9} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (3712 \, e^{8} x^{8} - 4599 \, d e^{7} x^{7} - 10240 \, d^{2} e^{6} x^{6} - 3066 \, d^{3} e^{5} x^{5} + 8448 \, d^{4} e^{4} x^{4} + 7224 \, d^{5} e^{3} x^{3} - 1024 \, d^{6} e^{2} x^{2} - 3024 \, d^{7} e x - 896 \, d^{8}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{8064 \, d x^{9}} \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^10,x, algorithm="fricas")

[Out]

-1/8064*(3465*e^9*x^9*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (3712*e^8*x^8 - 4599*d*e^7*x^7 - 10240*d^2*e^6*x^6
- 3066*d^3*e^5*x^5 + 8448*d^4*e^4*x^4 + 7224*d^5*e^3*x^3 - 1024*d^6*e^2*x^2 - 3024*d^7*e*x - 896*d^8)*sqrt(-e^
2*x^2 + d^2))/(d*x^9)

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giac [B]  time = 0.47, size = 620, normalized size = 3.32 \begin {gather*} \frac {x^{9} {\left (\frac {189 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{18}}{x} + \frac {324 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{16}}{x^{2}} - \frac {672 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{14}}{x^{3}} - \frac {3024 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{12}}{x^{4}} - \frac {1512 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} e^{10}}{x^{5}} + \frac {9744 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{6} e^{8}}{x^{6}} + \frac {18144 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{7} e^{6}}{x^{7}} - \frac {16632 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{8} e^{4}}{x^{8}} + 28 \, e^{20}\right )} e^{7}}{129024 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{9} d} + \frac {55 \, e^{9} \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 )}{128 \, d} + \frac {{\left (\frac {16632 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{8} e^{106}}{x} - \frac {18144 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{8} e^{104}}{x^{2}} - \frac {9744 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{8} e^{102}}{x^{3}} + \frac {1512 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{8} e^{100}}{x^{4}} + \frac {3024 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} d^{8} e^{98}}{x^{5}} + \frac {672 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{6} d^{8} e^{96}}{x^{6}} - \frac {324 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{7} d^{8} e^{94}}{x^{7}} - \frac {189 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{8} d^{8} e^{92}}{x^{8}} - \frac {28 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{9} d^{8} e^{90}}{x^{9}}\right )} e^{\left (-99\right )}}{129024 \, d^{9}} \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^10,x, algorithm="giac")

[Out]

1/129024*x^9*(189*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^18/x + 324*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*e^16/x^2 - 672*
(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*e^14/x^3 - 3024*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^12/x^4 - 1512*(d*e + sqrt(
-x^2*e^2 + d^2)*e)^5*e^10/x^5 + 9744*(d*e + sqrt(-x^2*e^2 + d^2)*e)^6*e^8/x^6 + 18144*(d*e + sqrt(-x^2*e^2 + d
^2)*e)^7*e^6/x^7 - 16632*(d*e + sqrt(-x^2*e^2 + d^2)*e)^8*e^4/x^8 + 28*e^20)*e^7/((d*e + sqrt(-x^2*e^2 + d^2)*
e)^9*d) + 55/128*e^9*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d + 1/129024*(16632*(d*e +
sqrt(-x^2*e^2 + d^2)*e)*d^8*e^106/x - 18144*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^8*e^104/x^2 - 9744*(d*e + sqrt(
-x^2*e^2 + d^2)*e)^3*d^8*e^102/x^3 + 1512*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d^8*e^100/x^4 + 3024*(d*e + sqrt(-x
^2*e^2 + d^2)*e)^5*d^8*e^98/x^5 + 672*(d*e + sqrt(-x^2*e^2 + d^2)*e)^6*d^8*e^96/x^6 - 324*(d*e + sqrt(-x^2*e^2
+ d^2)*e)^7*d^8*e^94/x^7 - 189*(d*e + sqrt(-x^2*e^2 + d^2)*e)^8*d^8*e^92/x^8 - 28*(d*e + sqrt(-x^2*e^2 + d^2)
*e)^9*d^8*e^90/x^9)*e^(-99)/d^9

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maple [A]  time = 0.10, size = 250, normalized size = 1.34 \begin {gather*} \frac {55 e^{9} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{128 \sqrt {d^{2}}}-\frac {55 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{9}}{128 d^{2}}-\frac {55 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{9}}{384 d^{4}}-\frac {11 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{9}}{128 d^{6}}-\frac {11 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{7}}{128 d^{6} x^{2}}+\frac {11 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{5}}{192 d^{4} x^{4}}-\frac {11 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{3}}{48 d^{2} x^{6}}-\frac {29 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{2}}{63 d \,x^{7}}-\frac {3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e}{8 x^{8}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} d}{9 x^{9}} \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^10,x)

[Out]

-29/63*e^2*(-e^2*x^2+d^2)^(7/2)/d/x^7-3/8*e*(-e^2*x^2+d^2)^(7/2)/x^8-11/48/d^2*e^3/x^6*(-e^2*x^2+d^2)^(7/2)+11
/192/d^4*e^5/x^4*(-e^2*x^2+d^2)^(7/2)-11/128/d^6*e^7/x^2*(-e^2*x^2+d^2)^(7/2)-11/128/d^6*e^9*(-e^2*x^2+d^2)^(5
/2)-55/384/d^4*e^9*(-e^2*x^2+d^2)^(3/2)-55/128/d^2*e^9*(-e^2*x^2+d^2)^(1/2)+55/128*e^9/(d^2)^(1/2)*ln((2*d^2+2
*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-1/9*d*(-e^2*x^2+d^2)^(7/2)/x^9

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maxima [A]  time = 1.00, size = 247, normalized size = 1.32 \begin {gather*} \frac {55 \, e^{9} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{128 \, d} - \frac {55 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{9}}{128 \, d^{2}} - \frac {55 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{9}}{384 \, d^{4}} - \frac {11 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{9}}{128 \, d^{6}} - \frac {11 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e^{7}}{128 \, d^{6} x^{2}} + \frac {11 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e^{5}}{192 \, d^{4} x^{4}} - \frac {11 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e^{3}}{48 \, d^{2} x^{6}} - \frac {29 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e^{2}}{63 \, d x^{7}} - \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e}{8 \, x^{8}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} d}{9 \, x^{9}} \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^10,x, algorithm="maxima")

[Out]

55/128*e^9*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d - 55/128*sqrt(-e^2*x^2 + d^2)*e^9/d^2 - 55/38
4*(-e^2*x^2 + d^2)^(3/2)*e^9/d^4 - 11/128*(-e^2*x^2 + d^2)^(5/2)*e^9/d^6 - 11/128*(-e^2*x^2 + d^2)^(7/2)*e^7/(
d^6*x^2) + 11/192*(-e^2*x^2 + d^2)^(7/2)*e^5/(d^4*x^4) - 11/48*(-e^2*x^2 + d^2)^(7/2)*e^3/(d^2*x^6) - 29/63*(-
e^2*x^2 + d^2)^(7/2)*e^2/(d*x^7) - 3/8*(-e^2*x^2 + d^2)^(7/2)*e/x^8 - 1/9*(-e^2*x^2 + d^2)^(7/2)*d/x^9

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}\,{\left (d+e\,x\right )}^3}{x^{10}} \,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^10,x)

[Out]

int(((d^2 - e^2*x^2)^(5/2)*(d + e*x)^3)/x^10, x)

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sympy [C]  time = 36.71, size = 1889, normalized size = 10.10

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**10,x)

[Out]

d**7*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/(9*x**8) + e**3*sqrt(d**2/(e**2*x**2) - 1)/(63*d**2*x**6) + 2*e*
*5*sqrt(d**2/(e**2*x**2) - 1)/(105*d**4*x**4) + 8*e**7*sqrt(d**2/(e**2*x**2) - 1)/(315*d**6*x**2) + 16*e**9*sq
rt(d**2/(e**2*x**2) - 1)/(315*d**8), Abs(d**2/(e**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(9*x**8) +
I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(63*d**2*x**6) + 2*I*e**5*sqrt(-d**2/(e**2*x**2) + 1)/(105*d**4*x**4) + 8*I
*e**7*sqrt(-d**2/(e**2*x**2) + 1)/(315*d**6*x**2) + 16*I*e**9*sqrt(-d**2/(e**2*x**2) + 1)/(315*d**8), True)) +
3*d**6*e*Piecewise((-d**2/(8*e*x**9*sqrt(d**2/(e**2*x**2) - 1)) + 7*e/(48*x**7*sqrt(d**2/(e**2*x**2) - 1)) +
e**3/(192*d**2*x**5*sqrt(d**2/(e**2*x**2) - 1)) + 5*e**5/(384*d**4*x**3*sqrt(d**2/(e**2*x**2) - 1)) - 5*e**7/(
128*d**6*x*sqrt(d**2/(e**2*x**2) - 1)) + 5*e**8*acosh(d/(e*x))/(128*d**7), Abs(d**2/(e**2*x**2)) > 1), (I*d**2
/(8*e*x**9*sqrt(-d**2/(e**2*x**2) + 1)) - 7*I*e/(48*x**7*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**3/(192*d**2*x**5*
sqrt(-d**2/(e**2*x**2) + 1)) - 5*I*e**5/(384*d**4*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + 5*I*e**7/(128*d**6*x*sqr
t(-d**2/(e**2*x**2) + 1)) - 5*I*e**8*asin(d/(e*x))/(128*d**7), True)) + d**5*e**2*Piecewise((-e*sqrt(d**2/(e**
2*x**2) - 1)/(7*x**6) + e**3*sqrt(d**2/(e**2*x**2) - 1)/(35*d**2*x**4) + 4*e**5*sqrt(d**2/(e**2*x**2) - 1)/(10
5*d**4*x**2) + 8*e**7*sqrt(d**2/(e**2*x**2) - 1)/(105*d**6), Abs(d**2/(e**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**
2*x**2) + 1)/(7*x**6) + I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(35*d**2*x**4) + 4*I*e**5*sqrt(-d**2/(e**2*x**2) +
1)/(105*d**4*x**2) + 8*I*e**7*sqrt(-d**2/(e**2*x**2) + 1)/(105*d**6), True)) - 5*d**4*e**3*Piecewise((-d**2/(6
*e*x**7*sqrt(d**2/(e**2*x**2) - 1)) + 5*e/(24*x**5*sqrt(d**2/(e**2*x**2) - 1)) + e**3/(48*d**2*x**3*sqrt(d**2/
(e**2*x**2) - 1)) - e**5/(16*d**4*x*sqrt(d**2/(e**2*x**2) - 1)) + e**6*acosh(d/(e*x))/(16*d**5), Abs(d**2/(e**
2*x**2)) > 1), (I*d**2/(6*e*x**7*sqrt(-d**2/(e**2*x**2) + 1)) - 5*I*e/(24*x**5*sqrt(-d**2/(e**2*x**2) + 1)) -
I*e**3/(48*d**2*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**5/(16*d**4*x*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**6*as
in(d/(e*x))/(16*d**5), True)) - 5*d**3*e**4*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
)) + d**2*e**5*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*sq
rt(-d**2/(e**2*x**2) + 1)) - I*e**4*asin(d/(e*x))/(8*d**3), True)) + 3*d*e**6*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)) + e**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))

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