3.1.78 \(\int \frac {(d+e x)^3 (d^2-e^2 x^2)^{5/2}}{x^8} \, dx\)

Optimal. Leaf size=206 \[ -\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{7 x^7}-\frac {e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^6}-\frac {e^2 (24 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{40 x^5}-3 d e^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {15}{16} d e^7 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )-\frac {3 e^6 (16 d-5 e x) \sqrt {d^2-e^2 x^2}}{16 x}+\frac {e^4 (16 d+5 e x) \left (d^2-e^2 x^2\right )^{3/2}}{16 x^3} \]

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Rubi [A]  time = 0.31, antiderivative size = 206, normalized size of antiderivative = 1.00, 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, 811, 813, 844, 217, 203, 266, 63, 208} \begin {gather*} -\frac {3 e^6 (16 d-5 e x) \sqrt {d^2-e^2 x^2}}{16 x}+\frac {e^4 (16 d+5 e x) \left (d^2-e^2 x^2\right )^{3/2}}{16 x^3}-\frac {e^2 (24 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{40 x^5}-\frac {e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^6}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{7 x^7}-3 d e^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {15}{16} d e^7 \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^8,x]

[Out]

(-3*e^6*(16*d - 5*e*x)*Sqrt[d^2 - e^2*x^2])/(16*x) + (e^4*(16*d + 5*e*x)*(d^2 - e^2*x^2)^(3/2))/(16*x^3) - (e^
2*(24*d + 5*e*x)*(d^2 - e^2*x^2)^(5/2))/(40*x^5) - (d*(d^2 - e^2*x^2)^(7/2))/(7*x^7) - (e*(d^2 - e^2*x^2)^(7/2
))/(2*x^6) - 3*d*e^7*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]] - (15*d*e^7*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/16

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 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

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^8} \, dx &=-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{7 x^7}-\frac {\int \frac {\left (d^2-e^2 x^2\right )^{5/2} \left (-21 d^4 e-21 d^3 e^2 x-7 d^2 e^3 x^2\right )}{x^7} \, dx}{7 d^2}\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{7 x^7}-\frac {e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^6}+\frac {\int \frac {\left (126 d^5 e^2+21 d^4 e^3 x\right ) \left (d^2-e^2 x^2\right )^{5/2}}{x^6} \, dx}{42 d^4}\\ &=-\frac {e^2 (24 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{40 x^5}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{7 x^7}-\frac {e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^6}-\frac {\int \frac {\left (1008 d^7 e^4+210 d^6 e^5 x\right ) \left (d^2-e^2 x^2\right )^{3/2}}{x^4} \, dx}{336 d^6}\\ &=\frac {e^4 (16 d+5 e x) \left (d^2-e^2 x^2\right )^{3/2}}{16 x^3}-\frac {e^2 (24 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{40 x^5}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{7 x^7}-\frac {e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^6}+\frac {\int \frac {\left (4032 d^9 e^6+1260 d^8 e^7 x\right ) \sqrt {d^2-e^2 x^2}}{x^2} \, dx}{1344 d^8}\\ &=-\frac {3 e^6 (16 d-5 e x) \sqrt {d^2-e^2 x^2}}{16 x}+\frac {e^4 (16 d+5 e x) \left (d^2-e^2 x^2\right )^{3/2}}{16 x^3}-\frac {e^2 (24 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{40 x^5}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{7 x^7}-\frac {e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^6}-\frac {\int \frac {-2520 d^{10} e^7+8064 d^9 e^8 x}{x \sqrt {d^2-e^2 x^2}} \, dx}{2688 d^8}\\ &=-\frac {3 e^6 (16 d-5 e x) \sqrt {d^2-e^2 x^2}}{16 x}+\frac {e^4 (16 d+5 e x) \left (d^2-e^2 x^2\right )^{3/2}}{16 x^3}-\frac {e^2 (24 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{40 x^5}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{7 x^7}-\frac {e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^6}+\frac {1}{16} \left (15 d^2 e^7\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx-\left (3 d e^8\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {3 e^6 (16 d-5 e x) \sqrt {d^2-e^2 x^2}}{16 x}+\frac {e^4 (16 d+5 e x) \left (d^2-e^2 x^2\right )^{3/2}}{16 x^3}-\frac {e^2 (24 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{40 x^5}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{7 x^7}-\frac {e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^6}+\frac {1}{32} \left (15 d^2 e^7\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )-\left (3 d e^8\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 {3 e^6 (16 d-5 e x) \sqrt {d^2-e^2 x^2}}{16 x}+\frac {e^4 (16 d+5 e x) \left (d^2-e^2 x^2\right )^{3/2}}{16 x^3}-\frac {e^2 (24 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{40 x^5}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{7 x^7}-\frac {e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^6}-3 d e^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {1}{16} \left (15 d^2 e^5\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 {3 e^6 (16 d-5 e x) \sqrt {d^2-e^2 x^2}}{16 x}+\frac {e^4 (16 d+5 e x) \left (d^2-e^2 x^2\right )^{3/2}}{16 x^3}-\frac {e^2 (24 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{40 x^5}-\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{7 x^7}-\frac {e \left (d^2-e^2 x^2\right )^{7/2}}{2 x^6}-3 d e^7 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {15}{16} d e^7 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\\ \end {align*}

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Mathematica [C]  time = 0.15, size = 247, normalized size = 1.20 \begin {gather*} -\frac {d \left (d^2-e^2 x^2\right )^{7/2}}{7 x^7}-\frac {e^7 \left (d^2-e^2 x^2\right )^{7/2} \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};1-\frac {e^2 x^2}{d^2}\right )}{7 d^6}-\frac {3 d^5 e^2 \sqrt {d^2-e^2 x^2} \, _2F_1\left (-\frac {5}{2},-\frac {5}{2};-\frac {3}{2};\frac {e^2 x^2}{d^2}\right )}{5 x^5 \sqrt {1-\frac {e^2 x^2}{d^2}}}+\frac {-8 d^8 e+34 d^6 e^3 x^2-59 d^4 e^5 x^4+33 d^2 e^7 x^6+15 d^2 e^7 x^6 \sqrt {1-\frac {e^2 x^2}{d^2}} \tanh ^{-1}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )}{16 x^6 \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^8,x]

[Out]

-1/7*(d*(d^2 - e^2*x^2)^(7/2))/x^7 + (-8*d^8*e + 34*d^6*e^3*x^2 - 59*d^4*e^5*x^4 + 33*d^2*e^7*x^6 + 15*d^2*e^7
*x^6*Sqrt[1 - (e^2*x^2)/d^2]*ArcTanh[Sqrt[1 - (e^2*x^2)/d^2]])/(16*x^6*Sqrt[d^2 - e^2*x^2]) - (3*d^5*e^2*Sqrt[
d^2 - e^2*x^2]*Hypergeometric2F1[-5/2, -5/2, -3/2, (e^2*x^2)/d^2])/(5*x^5*Sqrt[1 - (e^2*x^2)/d^2]) - (e^7*(d^2
- e^2*x^2)^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, 1 - (e^2*x^2)/d^2])/(7*d^6)

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IntegrateAlgebraic [A]  time = 0.80, size = 188, normalized size = 0.91 \begin {gather*} \frac {15}{8} d e^7 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )-3 d \sqrt {-e^2} e^6 \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )+\frac {\sqrt {d^2-e^2 x^2} \left (-80 d^7-280 d^6 e x-96 d^5 e^2 x^2+770 d^4 e^3 x^3+992 d^3 e^4 x^4-525 d^2 e^5 x^5-2496 d e^6 x^6+560 e^7 x^7\right )}{560 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((d + e*x)^3*(d^2 - e^2*x^2)^(5/2))/x^8,x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-80*d^7 - 280*d^6*e*x - 96*d^5*e^2*x^2 + 770*d^4*e^3*x^3 + 992*d^3*e^4*x^4 - 525*d^2*e^5
*x^5 - 2496*d*e^6*x^6 + 560*e^7*x^7))/(560*x^7) + (15*d*e^7*ArcTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d])
/8 - 3*d*e^6*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]]

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fricas [A]  time = 0.41, size = 173, normalized size = 0.84 \begin {gather*} \frac {3360 \, d e^{7} x^{7} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + 525 \, d e^{7} x^{7} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + 560 \, d e^{7} x^{7} + {\left (560 \, e^{7} x^{7} - 2496 \, d e^{6} x^{6} - 525 \, d^{2} e^{5} x^{5} + 992 \, d^{3} e^{4} x^{4} + 770 \, d^{4} e^{3} x^{3} - 96 \, d^{5} e^{2} x^{2} - 280 \, d^{6} e x - 80 \, d^{7}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{560 \, x^{7}} \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^8,x, algorithm="fricas")

[Out]

1/560*(3360*d*e^7*x^7*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + 525*d*e^7*x^7*log(-(d - sqrt(-e^2*x^2 + d^2)
)/x) + 560*d*e^7*x^7 + (560*e^7*x^7 - 2496*d*e^6*x^6 - 525*d^2*e^5*x^5 + 992*d^3*e^4*x^4 + 770*d^4*e^3*x^3 - 9
6*d^5*e^2*x^2 - 280*d^6*e*x - 80*d^7)*sqrt(-e^2*x^2 + d^2))/x^7

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giac [B]  time = 0.31, size = 510, normalized size = 2.48 \begin {gather*} -3 \, d \arcsin \left (\frac {x e}{d}\right ) e^{7} \mathrm {sgn}\relax (d) - \frac {15}{16} \, d e^{7} \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 (5 \, d e^{16} + \frac {35 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d e^{14}}{x} + \frac {49 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d e^{12}}{x^{2}} - \frac {245 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d e^{10}}{x^{3}} - \frac {875 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d e^{8}}{x^{4}} + \frac {455 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} d e^{6}}{x^{5}} + \frac {9065 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{6} d e^{4}}{x^{6}}\right )} x^{7} e^{5}}{4480 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{7}} - \frac {1}{4480} \, {\left (\frac {9065 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d e^{68}}{x} + \frac {455 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d e^{66}}{x^{2}} - \frac {875 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d e^{64}}{x^{3}} - \frac {245 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{4} d e^{62}}{x^{4}} + \frac {49 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{5} d e^{60}}{x^{5}} + \frac {35 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{6} d e^{58}}{x^{6}} + \frac {5 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{7} d e^{56}}{x^{7}}\right )} e^{\left (-63\right )} + \sqrt {-x^{2} e^{2} + d^{2}} e^{7} \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^8,x, algorithm="giac")

[Out]

-3*d*arcsin(x*e/d)*e^7*sgn(d) - 15/16*d*e^7*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x)) + 1/
4480*(5*d*e^16 + 35*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d*e^14/x + 49*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d*e^12/x^2 -
245*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d*e^10/x^3 - 875*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d*e^8/x^4 + 455*(d*e +
sqrt(-x^2*e^2 + d^2)*e)^5*d*e^6/x^5 + 9065*(d*e + sqrt(-x^2*e^2 + d^2)*e)^6*d*e^4/x^6)*x^7*e^5/(d*e + sqrt(-x
^2*e^2 + d^2)*e)^7 - 1/4480*(9065*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d*e^68/x + 455*(d*e + sqrt(-x^2*e^2 + d^2)*e)
^2*d*e^66/x^2 - 875*(d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d*e^64/x^3 - 245*(d*e + sqrt(-x^2*e^2 + d^2)*e)^4*d*e^62/
x^4 + 49*(d*e + sqrt(-x^2*e^2 + d^2)*e)^5*d*e^60/x^5 + 35*(d*e + sqrt(-x^2*e^2 + d^2)*e)^6*d*e^58/x^6 + 5*(d*e
+ sqrt(-x^2*e^2 + d^2)*e)^7*d*e^56/x^7)*e^(-63) + sqrt(-x^2*e^2 + d^2)*e^7

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maple [B]  time = 0.05, size = 377, normalized size = 1.83 \begin {gather*} -\frac {15 d^{2} e^{7} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{16 \sqrt {d^{2}}}-\frac {3 d \,e^{8} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{8} x}{d}+\frac {15 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{7}}{16}-\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{8} x}{d^{3}}+\frac {5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e^{7}}{16 d^{2}}-\frac {8 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{8} x}{5 d^{5}}+\frac {3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} e^{7}}{16 d^{4}}-\frac {8 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{6}}{5 d^{5} x}+\frac {3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{5}}{16 d^{4} x^{2}}+\frac {2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{4}}{5 d^{3} x^{3}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{3}}{8 d^{2} x^{4}}-\frac {3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e^{2}}{5 d \,x^{5}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} e}{2 x^{6}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}} d}{7 x^{7}} \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^8,x)

[Out]

-3/5/d*e^2/x^5*(-e^2*x^2+d^2)^(7/2)+2/5/d^3*e^4/x^3*(-e^2*x^2+d^2)^(7/2)-8/5/d^5*e^6/x*(-e^2*x^2+d^2)^(7/2)-8/
5/d^5*e^8*x*(-e^2*x^2+d^2)^(5/2)-2/d^3*e^8*x*(-e^2*x^2+d^2)^(3/2)-3/d*e^8*x*(-e^2*x^2+d^2)^(1/2)-3*d*e^8/(e^2)
^(1/2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-1/2*e*(-e^2*x^2+d^2)^(7/2)/x^6-1/8/d^2*e^3/x^4*(-e^2*x^2+d^2
)^(7/2)+3/16/d^4*e^5/x^2*(-e^2*x^2+d^2)^(7/2)+3/16/d^4*e^7*(-e^2*x^2+d^2)^(5/2)+5/16/d^2*e^7*(-e^2*x^2+d^2)^(3
/2)+15/16*e^7*(-e^2*x^2+d^2)^(1/2)-15/16*d^2*e^7/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-
1/7*d*(-e^2*x^2+d^2)^(7/2)/x^7

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maxima [A]  time = 1.01, size = 326, normalized size = 1.58 \begin {gather*} -3 \, d e^{7} \arcsin \left (\frac {e x}{d}\right ) - \frac {15}{16} \, d e^{7} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{8} x}{d} + \frac {15}{16} \, \sqrt {-e^{2} x^{2} + d^{2}} e^{7} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{8} x}{d^{3}} + \frac {5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{7}}{16 \, d^{2}} + \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{7}}{16 \, d^{4}} - \frac {8 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}{5 \, d^{3} x} + \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e^{5}}{16 \, d^{4} x^{2}} + \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e^{4}}{5 \, d^{3} x^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e^{3}}{8 \, d^{2} x^{4}} - \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e^{2}}{5 \, d x^{5}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} e}{2 \, x^{6}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} d}{7 \, x^{7}} \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^8,x, algorithm="maxima")

[Out]

-3*d*e^7*arcsin(e*x/d) - 15/16*d*e^7*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x)) - 3*sqrt(-e^2*x^2 + d
^2)*e^8*x/d + 15/16*sqrt(-e^2*x^2 + d^2)*e^7 - 2*(-e^2*x^2 + d^2)^(3/2)*e^8*x/d^3 + 5/16*(-e^2*x^2 + d^2)^(3/2
)*e^7/d^2 + 3/16*(-e^2*x^2 + d^2)^(5/2)*e^7/d^4 - 8/5*(-e^2*x^2 + d^2)^(5/2)*e^6/(d^3*x) + 3/16*(-e^2*x^2 + d^
2)^(7/2)*e^5/(d^4*x^2) + 2/5*(-e^2*x^2 + d^2)^(7/2)*e^4/(d^3*x^3) - 1/8*(-e^2*x^2 + d^2)^(7/2)*e^3/(d^2*x^4) -
3/5*(-e^2*x^2 + d^2)^(7/2)*e^2/(d*x^5) - 1/2*(-e^2*x^2 + d^2)^(7/2)*e/x^6 - 1/7*(-e^2*x^2 + d^2)^(7/2)*d/x^7

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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^8} \,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^8,x)

[Out]

int(((d^2 - e^2*x^2)^(5/2)*(d + e*x)^3)/x^8, x)

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sympy [C]  time = 22.29, size = 1513, normalized size = 7.34

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**8,x)

[Out]

d**7*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))
+ 3*d**6*e*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*s
qrt(-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)) + d**5*e**2*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)) - 5*d**4*e**3*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), A
bs(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)) - 5*d**3*e**4
*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))
+ d**2*e**5*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*e**6*Piecewise((I*d/(x*sqrt(-1 + e**2*x**2/d**2)) + I*e*acosh(e*x/d) - I*e**2*x/(d*sqr
t(-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)) + e**7*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))

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