3.1.15 \(\int \frac {(d+e x) (d^2-e^2 x^2)^{3/2}}{x^9} \, dx\)

Optimal. Leaf size=201 \[ -\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{7 d^2 x^7}-\frac {3 e^8 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{128 d^4}-\frac {2 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^4 x^5}-\frac {e^2 \left (d^2-e^2 x^2\right )^{5/2}}{16 d^3 x^6}+\frac {3 e^6 \sqrt {d^2-e^2 x^2}}{128 d^3 x^2}-\frac {e^4 \left (d^2-e^2 x^2\right )^{3/2}}{64 d^3 x^4} \]

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Rubi [A]  time = 0.16, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {835, 807, 266, 47, 63, 208} \begin {gather*} \frac {3 e^6 \sqrt {d^2-e^2 x^2}}{128 d^3 x^2}-\frac {e^4 \left (d^2-e^2 x^2\right )^{3/2}}{64 d^3 x^4}-\frac {2 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^4 x^5}-\frac {e^2 \left (d^2-e^2 x^2\right )^{5/2}}{16 d^3 x^6}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{7 d^2 x^7}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}-\frac {3 e^8 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{128 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^9,x]

[Out]

(3*e^6*Sqrt[d^2 - e^2*x^2])/(128*d^3*x^2) - (e^4*(d^2 - e^2*x^2)^(3/2))/(64*d^3*x^4) - (d^2 - e^2*x^2)^(5/2)/(
8*d*x^8) - (e*(d^2 - e^2*x^2)^(5/2))/(7*d^2*x^7) - (e^2*(d^2 - e^2*x^2)^(5/2))/(16*d^3*x^6) - (2*e^3*(d^2 - e^
2*x^2)^(5/2))/(35*d^4*x^5) - (3*e^8*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(128*d^4)

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 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^9} \, dx &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}-\frac {\int \frac {\left (-8 d^2 e-3 d e^2 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x^8} \, dx}{8 d^2}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{7 d^2 x^7}+\frac {\int \frac {\left (21 d^3 e^2+16 d^2 e^3 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x^7} \, dx}{56 d^4}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{7 d^2 x^7}-\frac {e^2 \left (d^2-e^2 x^2\right )^{5/2}}{16 d^3 x^6}-\frac {\int \frac {\left (-96 d^4 e^3-21 d^3 e^4 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x^6} \, dx}{336 d^6}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{7 d^2 x^7}-\frac {e^2 \left (d^2-e^2 x^2\right )^{5/2}}{16 d^3 x^6}-\frac {2 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^4 x^5}+\frac {e^4 \int \frac {\left (d^2-e^2 x^2\right )^{3/2}}{x^5} \, dx}{16 d^3}\\ &=-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{7 d^2 x^7}-\frac {e^2 \left (d^2-e^2 x^2\right )^{5/2}}{16 d^3 x^6}-\frac {2 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^4 x^5}+\frac {e^4 \operatorname {Subst}\left (\int \frac {\left (d^2-e^2 x\right )^{3/2}}{x^3} \, dx,x,x^2\right )}{32 d^3}\\ &=-\frac {e^4 \left (d^2-e^2 x^2\right )^{3/2}}{64 d^3 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{7 d^2 x^7}-\frac {e^2 \left (d^2-e^2 x^2\right )^{5/2}}{16 d^3 x^6}-\frac {2 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^4 x^5}-\frac {\left (3 e^6\right ) \operatorname {Subst}\left (\int \frac {\sqrt {d^2-e^2 x}}{x^2} \, dx,x,x^2\right )}{128 d^3}\\ &=\frac {3 e^6 \sqrt {d^2-e^2 x^2}}{128 d^3 x^2}-\frac {e^4 \left (d^2-e^2 x^2\right )^{3/2}}{64 d^3 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{7 d^2 x^7}-\frac {e^2 \left (d^2-e^2 x^2\right )^{5/2}}{16 d^3 x^6}-\frac {2 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^4 x^5}+\frac {\left (3 e^8\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{256 d^3}\\ &=\frac {3 e^6 \sqrt {d^2-e^2 x^2}}{128 d^3 x^2}-\frac {e^4 \left (d^2-e^2 x^2\right )^{3/2}}{64 d^3 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{7 d^2 x^7}-\frac {e^2 \left (d^2-e^2 x^2\right )^{5/2}}{16 d^3 x^6}-\frac {2 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^4 x^5}-\frac {\left (3 e^6\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 )}{128 d^3}\\ &=\frac {3 e^6 \sqrt {d^2-e^2 x^2}}{128 d^3 x^2}-\frac {e^4 \left (d^2-e^2 x^2\right )^{3/2}}{64 d^3 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{7 d^2 x^7}-\frac {e^2 \left (d^2-e^2 x^2\right )^{5/2}}{16 d^3 x^6}-\frac {2 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^4 x^5}-\frac {3 e^8 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{128 d^4}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 73, normalized size = 0.36 \begin {gather*} -\frac {e \left (d^2-e^2 x^2\right )^{5/2} \left (5 d^7+2 d^5 e^2 x^2+7 e^7 x^7 \, _2F_1\left (\frac {5}{2},5;\frac {7}{2};1-\frac {e^2 x^2}{d^2}\right )\right )}{35 d^9 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^9,x]

[Out]

-1/35*(e*(d^2 - e^2*x^2)^(5/2)*(5*d^7 + 2*d^5*e^2*x^2 + 7*e^7*x^7*Hypergeometric2F1[5/2, 5, 7/2, 1 - (e^2*x^2)
/d^2]))/(d^9*x^7)

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IntegrateAlgebraic [A]  time = 0.74, size = 148, normalized size = 0.74 \begin {gather*} \frac {3 e^8 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{64 d^4}+\frac {\sqrt {d^2-e^2 x^2} \left (-560 d^7-640 d^6 e x+840 d^5 e^2 x^2+1024 d^4 e^3 x^3-70 d^3 e^4 x^4-128 d^2 e^5 x^5-105 d e^6 x^6-256 e^7 x^7\right )}{4480 d^4 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((d + e*x)*(d^2 - e^2*x^2)^(3/2))/x^9,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-560*d^7 - 640*d^6*e*x + 840*d^5*e^2*x^2 + 1024*d^4*e^3*x^3 - 70*d^3*e^4*x^4 - 128*d^2*e
^5*x^5 - 105*d*e^6*x^6 - 256*e^7*x^7))/(4480*d^4*x^8) + (3*e^8*ArcTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/
d])/(64*d^4)

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fricas [A]  time = 0.45, size = 131, normalized size = 0.65 \begin {gather*} \frac {105 \, e^{8} x^{8} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (256 \, e^{7} x^{7} + 105 \, d e^{6} x^{6} + 128 \, d^{2} e^{5} x^{5} + 70 \, d^{3} e^{4} x^{4} - 1024 \, d^{4} e^{3} x^{3} - 840 \, d^{5} e^{2} x^{2} + 640 \, d^{6} e x + 560 \, d^{7}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{4480 \, d^{4} x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^9,x, algorithm="fricas")

[Out]

1/4480*(105*e^8*x^8*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (256*e^7*x^7 + 105*d*e^6*x^6 + 128*d^2*e^5*x^5 + 70*d
^3*e^4*x^4 - 1024*d^4*e^3*x^3 - 840*d^5*e^2*x^2 + 640*d^6*e*x + 560*d^7)*sqrt(-e^2*x^2 + d^2))/(d^4*x^8)

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giac [B]  time = 0.26, size = 431, normalized size = 2.14 \begin {gather*} \frac {x^{8} {\left (\frac {80 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{16}}{x} - \frac {112 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} e^{12}}{x^{3}} - \frac {280 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} e^{10}}{x^{4}} - \frac {560 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} e^{8}}{x^{5}} + \frac {1680 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{7} e^{4}}{x^{7}} + 35 \, e^{18}\right )} e^{6}}{71680 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{8} d^{4}} - \frac {3 \, e^{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 )}{128 \, d^{4}} - \frac {{\left (\frac {1680 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{28} e^{86}}{x} - \frac {560 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{28} e^{82}}{x^{3}} - \frac {280 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d^{28} e^{80}}{x^{4}} - \frac {112 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} d^{28} e^{78}}{x^{5}} + \frac {80 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{7} d^{28} e^{74}}{x^{7}} + \frac {35 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{8} d^{28} e^{72}}{x^{8}}\right )} e^{\left (-80\right )}}{71680 \, d^{32}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^9,x, algorithm="giac")

[Out]

1/71680*x^8*(80*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^16/x - 112*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*e^12/x^3 - 280*(d
*e + sqrt(-x^2*e^2 + d^2)*e)^4*e^10/x^4 - 560*(d*e + sqrt(-x^2*e^2 + d^2)*e)^5*e^8/x^5 + 1680*(d*e + sqrt(-x^2
*e^2 + d^2)*e)^7*e^4/x^7 + 35*e^18)*e^6/((d*e + sqrt(-x^2*e^2 + d^2)*e)^8*d^4) - 3/128*e^8*log(1/2*abs(-2*d*e
- 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^4 - 1/71680*(1680*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^28*e^86/x - 56
0*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d^28*e^82/x^3 - 280*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d^28*e^80/x^4 - 112*(d
*e + sqrt(-x^2*e^2 + d^2)*e)^5*d^28*e^78/x^5 + 80*(d*e + sqrt(-x^2*e^2 + d^2)*e)^7*d^28*e^74/x^7 + 35*(d*e + s
qrt(-x^2*e^2 + d^2)*e)^8*d^28*e^72/x^8)*e^(-80)/d^32

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maple [A]  time = 0.07, size = 236, normalized size = 1.17 \begin {gather*} -\frac {3 e^{8} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{128 \sqrt {d^{2}}\, d^{3}}+\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{8}}{128 d^{5}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{8}}{128 d^{7}}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{6}}{128 d^{7} x^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{4}}{64 d^{5} x^{4}}-\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{3}}{35 d^{4} x^{5}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{2}}{16 d^{3} x^{6}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e}{7 d^{2} x^{7}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{8 d \,x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^9,x)

[Out]

-1/8*(-e^2*x^2+d^2)^(5/2)/d/x^8-1/16*e^2*(-e^2*x^2+d^2)^(5/2)/d^3/x^6-1/64*e^4/d^5/x^4*(-e^2*x^2+d^2)^(5/2)+1/
128*e^6/d^7/x^2*(-e^2*x^2+d^2)^(5/2)+1/128*e^8/d^7*(-e^2*x^2+d^2)^(3/2)+3/128*e^8/d^5*(-e^2*x^2+d^2)^(1/2)-3/1
28*e^8/d^3/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-1/7*e*(-e^2*x^2+d^2)^(5/2)/d^2/x^7-2/3
5*e^3*(-e^2*x^2+d^2)^(5/2)/d^4/x^5

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maxima [A]  time = 0.99, size = 230, normalized size = 1.14 \begin {gather*} -\frac {3 \, e^{8} \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^{4}} + \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{8}}{128 \, d^{5}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{8}}{128 \, d^{7}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}{128 \, d^{7} x^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}}{64 \, d^{5} x^{4}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}}{35 \, d^{4} x^{5}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}}{16 \, d^{3} x^{6}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e}{7 \, d^{2} x^{7}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{8 \, d x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x^9,x, algorithm="maxima")

[Out]

-3/128*e^8*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^4 + 3/128*sqrt(-e^2*x^2 + d^2)*e^8/d^5 + 1/12
8*(-e^2*x^2 + d^2)^(3/2)*e^8/d^7 + 1/128*(-e^2*x^2 + d^2)^(5/2)*e^6/(d^7*x^2) - 1/64*(-e^2*x^2 + d^2)^(5/2)*e^
4/(d^5*x^4) - 2/35*(-e^2*x^2 + d^2)^(5/2)*e^3/(d^4*x^5) - 1/16*(-e^2*x^2 + d^2)^(5/2)*e^2/(d^3*x^6) - 1/7*(-e^
2*x^2 + d^2)^(5/2)*e/(d^2*x^7) - 1/8*(-e^2*x^2 + d^2)^(5/2)/(d*x^8)

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mupad [B]  time = 6.04, size = 212, normalized size = 1.05 \begin {gather*} \frac {3\,d^3\,\sqrt {d^2-e^2\,x^2}}{128\,x^8}-\frac {11\,d\,{\left (d^2-e^2\,x^2\right )}^{3/2}}{128\,x^8}-\frac {11\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{128\,d\,x^8}+\frac {3\,{\left (d^2-e^2\,x^2\right )}^{7/2}}{128\,d^3\,x^8}+\frac {8\,e^3\,\sqrt {d^2-e^2\,x^2}}{35\,x^5}-\frac {e^5\,\sqrt {d^2-e^2\,x^2}}{35\,d^2\,x^3}-\frac {2\,e^7\,\sqrt {d^2-e^2\,x^2}}{35\,d^4\,x}-\frac {d^2\,e\,\sqrt {d^2-e^2\,x^2}}{7\,x^7}+\frac {e^8\,\mathrm {atan}\left (\frac {\sqrt {d^2-e^2\,x^2}\,1{}\mathrm {i}}{d}\right )\,3{}\mathrm {i}}{128\,d^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d^2 - e^2*x^2)^(3/2)*(d + e*x))/x^9,x)

[Out]

(3*d^3*(d^2 - e^2*x^2)^(1/2))/(128*x^8) - (11*d*(d^2 - e^2*x^2)^(3/2))/(128*x^8) - (11*(d^2 - e^2*x^2)^(5/2))/
(128*d*x^8) + (3*(d^2 - e^2*x^2)^(7/2))/(128*d^3*x^8) + (8*e^3*(d^2 - e^2*x^2)^(1/2))/(35*x^5) + (e^8*atan(((d
^2 - e^2*x^2)^(1/2)*1i)/d)*3i)/(128*d^4) - (e^5*(d^2 - e^2*x^2)^(1/2))/(35*d^2*x^3) - (2*e^7*(d^2 - e^2*x^2)^(
1/2))/(35*d^4*x) - (d^2*e*(d^2 - e^2*x^2)^(1/2))/(7*x^7)

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sympy [C]  time = 22.55, size = 1159, normalized size = 5.77

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e**2*x**2+d**2)**(3/2)/x**9,x)

[Out]

d**3*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*sqrt(-d*
*2/(e**2*x**2) + 1)) - 5*I*e**8*asin(d/(e*x))/(128*d**7), True)) + d**2*e*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)/(105*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)) - d*e**2*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*asin(d/(e*x))/(
16*d**5), True)) - e**3*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), Ab
s(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))

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