3.86 \(\int \frac {(d+e x^2)^2 (a+b \sec ^{-1}(c x))}{x^8} \, dx\)

Optimal. Leaf size=241 \[ -\frac {d^2 \left (a+b \sec ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}+\frac {b c d^2 \sqrt {c^2 x^2-1}}{49 x^6 \sqrt {c^2 x^2}}+\frac {2 b c d \sqrt {c^2 x^2-1} \left (15 c^2 d+49 e\right )}{1225 x^4 \sqrt {c^2 x^2}}+\frac {b c \sqrt {c^2 x^2-1} \left (360 c^4 d^2+1176 c^2 d e+1225 e^2\right )}{11025 x^2 \sqrt {c^2 x^2}}+\frac {2 b c^3 \sqrt {c^2 x^2-1} \left (360 c^4 d^2+1176 c^2 d e+1225 e^2\right )}{11025 \sqrt {c^2 x^2}} \]

[Out]

-1/7*d^2*(a+b*arcsec(c*x))/x^7-2/5*d*e*(a+b*arcsec(c*x))/x^5-1/3*e^2*(a+b*arcsec(c*x))/x^3+2/11025*b*c^3*(360*
c^4*d^2+1176*c^2*d*e+1225*e^2)*(c^2*x^2-1)^(1/2)/(c^2*x^2)^(1/2)+1/49*b*c*d^2*(c^2*x^2-1)^(1/2)/x^6/(c^2*x^2)^
(1/2)+2/1225*b*c*d*(15*c^2*d+49*e)*(c^2*x^2-1)^(1/2)/x^4/(c^2*x^2)^(1/2)+1/11025*b*c*(360*c^4*d^2+1176*c^2*d*e
+1225*e^2)*(c^2*x^2-1)^(1/2)/x^2/(c^2*x^2)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {270, 5238, 12, 1265, 453, 271, 264} \[ -\frac {d^2 \left (a+b \sec ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}+\frac {2 b c^3 \sqrt {c^2 x^2-1} \left (360 c^4 d^2+1176 c^2 d e+1225 e^2\right )}{11025 \sqrt {c^2 x^2}}+\frac {b c \sqrt {c^2 x^2-1} \left (360 c^4 d^2+1176 c^2 d e+1225 e^2\right )}{11025 x^2 \sqrt {c^2 x^2}}+\frac {b c d^2 \sqrt {c^2 x^2-1}}{49 x^6 \sqrt {c^2 x^2}}+\frac {2 b c d \sqrt {c^2 x^2-1} \left (15 c^2 d+49 e\right )}{1225 x^4 \sqrt {c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^2*(a + b*ArcSec[c*x]))/x^8,x]

[Out]

(2*b*c^3*(360*c^4*d^2 + 1176*c^2*d*e + 1225*e^2)*Sqrt[-1 + c^2*x^2])/(11025*Sqrt[c^2*x^2]) + (b*c*d^2*Sqrt[-1
+ c^2*x^2])/(49*x^6*Sqrt[c^2*x^2]) + (2*b*c*d*(15*c^2*d + 49*e)*Sqrt[-1 + c^2*x^2])/(1225*x^4*Sqrt[c^2*x^2]) +
(b*c*(360*c^4*d^2 + 1176*c^2*d*e + 1225*e^2)*Sqrt[-1 + c^2*x^2])/(11025*x^2*Sqrt[c^2*x^2]) - (d^2*(a + b*ArcS
ec[c*x]))/(7*x^7) - (2*d*e*(a + b*ArcSec[c*x]))/(5*x^5) - (e^2*(a + b*ArcSec[c*x]))/(3*x^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 1265

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wit
h[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x,
x]}, Simp[(R*(f*x)^(m + 1)*(d + e*x^2)^(q + 1))/(d*f*(m + 1)), x] + Dist[1/(d*f^2*(m + 1)), Int[(f*x)^(m + 2)
*(d + e*x^2)^q*ExpandToSum[(d*f*(m + 1)*Qx)/x - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q},
x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -1]

Rule 5238

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSec[c*x], u, x] - Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&  !(ILtQ
[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right )}{x^8} \, dx &=-\frac {d^2 \left (a+b \sec ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac {(b c x) \int \frac {-15 d^2-42 d e x^2-35 e^2 x^4}{105 x^8 \sqrt {-1+c^2 x^2}} \, dx}{\sqrt {c^2 x^2}}\\ &=-\frac {d^2 \left (a+b \sec ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac {(b c x) \int \frac {-15 d^2-42 d e x^2-35 e^2 x^4}{x^8 \sqrt {-1+c^2 x^2}} \, dx}{105 \sqrt {c^2 x^2}}\\ &=\frac {b c d^2 \sqrt {-1+c^2 x^2}}{49 x^6 \sqrt {c^2 x^2}}-\frac {d^2 \left (a+b \sec ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac {(b c x) \int \frac {-6 d \left (15 c^2 d+49 e\right )-245 e^2 x^2}{x^6 \sqrt {-1+c^2 x^2}} \, dx}{735 \sqrt {c^2 x^2}}\\ &=\frac {b c d^2 \sqrt {-1+c^2 x^2}}{49 x^6 \sqrt {c^2 x^2}}+\frac {2 b c d \left (15 c^2 d+49 e\right ) \sqrt {-1+c^2 x^2}}{1225 x^4 \sqrt {c^2 x^2}}-\frac {d^2 \left (a+b \sec ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac {\left (b c \left (-1225 e^2-24 c^2 d \left (15 c^2 d+49 e\right )\right ) x\right ) \int \frac {1}{x^4 \sqrt {-1+c^2 x^2}} \, dx}{3675 \sqrt {c^2 x^2}}\\ &=\frac {b c d^2 \sqrt {-1+c^2 x^2}}{49 x^6 \sqrt {c^2 x^2}}+\frac {2 b c d \left (15 c^2 d+49 e\right ) \sqrt {-1+c^2 x^2}}{1225 x^4 \sqrt {c^2 x^2}}+\frac {b c \left (1225 e^2+24 c^2 d \left (15 c^2 d+49 e\right )\right ) \sqrt {-1+c^2 x^2}}{11025 x^2 \sqrt {c^2 x^2}}-\frac {d^2 \left (a+b \sec ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac {\left (2 b c^3 \left (-1225 e^2-24 c^2 d \left (15 c^2 d+49 e\right )\right ) x\right ) \int \frac {1}{x^2 \sqrt {-1+c^2 x^2}} \, dx}{11025 \sqrt {c^2 x^2}}\\ &=\frac {2 b c^3 \left (1225 e^2+24 c^2 d \left (15 c^2 d+49 e\right )\right ) \sqrt {-1+c^2 x^2}}{11025 \sqrt {c^2 x^2}}+\frac {b c d^2 \sqrt {-1+c^2 x^2}}{49 x^6 \sqrt {c^2 x^2}}+\frac {2 b c d \left (15 c^2 d+49 e\right ) \sqrt {-1+c^2 x^2}}{1225 x^4 \sqrt {c^2 x^2}}+\frac {b c \left (1225 e^2+24 c^2 d \left (15 c^2 d+49 e\right )\right ) \sqrt {-1+c^2 x^2}}{11025 x^2 \sqrt {c^2 x^2}}-\frac {d^2 \left (a+b \sec ^{-1}(c x)\right )}{7 x^7}-\frac {2 d e \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e^2 \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 153, normalized size = 0.63 \[ \frac {-105 a \left (15 d^2+42 d e x^2+35 e^2 x^4\right )+b c x \sqrt {1-\frac {1}{c^2 x^2}} \left (1225 e^2 x^4 \left (2 c^2 x^2+1\right )+294 d e x^2 \left (8 c^4 x^4+4 c^2 x^2+3\right )+45 d^2 \left (16 c^6 x^6+8 c^4 x^4+6 c^2 x^2+5\right )\right )-105 b \sec ^{-1}(c x) \left (15 d^2+42 d e x^2+35 e^2 x^4\right )}{11025 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^2*(a + b*ArcSec[c*x]))/x^8,x]

[Out]

(-105*a*(15*d^2 + 42*d*e*x^2 + 35*e^2*x^4) + b*c*Sqrt[1 - 1/(c^2*x^2)]*x*(1225*e^2*x^4*(1 + 2*c^2*x^2) + 294*d
*e*x^2*(3 + 4*c^2*x^2 + 8*c^4*x^4) + 45*d^2*(5 + 6*c^2*x^2 + 8*c^4*x^4 + 16*c^6*x^6)) - 105*b*(15*d^2 + 42*d*e
*x^2 + 35*e^2*x^4)*ArcSec[c*x])/(11025*x^7)

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fricas [A]  time = 1.55, size = 159, normalized size = 0.66 \[ -\frac {3675 \, a e^{2} x^{4} + 4410 \, a d e x^{2} + 1575 \, a d^{2} + 105 \, {\left (35 \, b e^{2} x^{4} + 42 \, b d e x^{2} + 15 \, b d^{2}\right )} \operatorname {arcsec}\left (c x\right ) - {\left (2 \, {\left (360 \, b c^{6} d^{2} + 1176 \, b c^{4} d e + 1225 \, b c^{2} e^{2}\right )} x^{6} + {\left (360 \, b c^{4} d^{2} + 1176 \, b c^{2} d e + 1225 \, b e^{2}\right )} x^{4} + 225 \, b d^{2} + 18 \, {\left (15 \, b c^{2} d^{2} + 49 \, b d e\right )} x^{2}\right )} \sqrt {c^{2} x^{2} - 1}}{11025 \, x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arcsec(c*x))/x^8,x, algorithm="fricas")

[Out]

-1/11025*(3675*a*e^2*x^4 + 4410*a*d*e*x^2 + 1575*a*d^2 + 105*(35*b*e^2*x^4 + 42*b*d*e*x^2 + 15*b*d^2)*arcsec(c
*x) - (2*(360*b*c^6*d^2 + 1176*b*c^4*d*e + 1225*b*c^2*e^2)*x^6 + (360*b*c^4*d^2 + 1176*b*c^2*d*e + 1225*b*e^2)
*x^4 + 225*b*d^2 + 18*(15*b*c^2*d^2 + 49*b*d*e)*x^2)*sqrt(c^2*x^2 - 1))/x^7

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giac [A]  time = 0.16, size = 294, normalized size = 1.22 \[ \frac {1}{11025} \, {\left (720 \, b c^{6} d^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 2352 \, b c^{4} d \sqrt {-\frac {1}{c^{2} x^{2}} + 1} e + \frac {360 \, b c^{4} d^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x^{2}} + 2450 \, b c^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} e^{2} + \frac {1176 \, b c^{2} d \sqrt {-\frac {1}{c^{2} x^{2}} + 1} e}{x^{2}} + \frac {270 \, b c^{2} d^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x^{4}} + \frac {1225 \, b \sqrt {-\frac {1}{c^{2} x^{2}} + 1} e^{2}}{x^{2}} + \frac {882 \, b d \sqrt {-\frac {1}{c^{2} x^{2}} + 1} e}{x^{4}} - \frac {3675 \, b \arccos \left (\frac {1}{c x}\right ) e^{2}}{c x^{3}} - \frac {3675 \, a e^{2}}{c x^{3}} - \frac {4410 \, b d \arccos \left (\frac {1}{c x}\right ) e}{c x^{5}} + \frac {225 \, b d^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x^{6}} - \frac {4410 \, a d e}{c x^{5}} - \frac {1575 \, b d^{2} \arccos \left (\frac {1}{c x}\right )}{c x^{7}} - \frac {1575 \, a d^{2}}{c x^{7}}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arcsec(c*x))/x^8,x, algorithm="giac")

[Out]

1/11025*(720*b*c^6*d^2*sqrt(-1/(c^2*x^2) + 1) + 2352*b*c^4*d*sqrt(-1/(c^2*x^2) + 1)*e + 360*b*c^4*d^2*sqrt(-1/
(c^2*x^2) + 1)/x^2 + 2450*b*c^2*sqrt(-1/(c^2*x^2) + 1)*e^2 + 1176*b*c^2*d*sqrt(-1/(c^2*x^2) + 1)*e/x^2 + 270*b
*c^2*d^2*sqrt(-1/(c^2*x^2) + 1)/x^4 + 1225*b*sqrt(-1/(c^2*x^2) + 1)*e^2/x^2 + 882*b*d*sqrt(-1/(c^2*x^2) + 1)*e
/x^4 - 3675*b*arccos(1/(c*x))*e^2/(c*x^3) - 3675*a*e^2/(c*x^3) - 4410*b*d*arccos(1/(c*x))*e/(c*x^5) + 225*b*d^
2*sqrt(-1/(c^2*x^2) + 1)/x^6 - 4410*a*d*e/(c*x^5) - 1575*b*d^2*arccos(1/(c*x))/(c*x^7) - 1575*a*d^2/(c*x^7))*c

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maple [A]  time = 0.06, size = 223, normalized size = 0.93 \[ c^{7} \left (\frac {a \left (-\frac {d^{2}}{7 c^{3} x^{7}}-\frac {e^{2}}{3 c^{3} x^{3}}-\frac {2 e d}{5 c^{3} x^{5}}\right )}{c^{4}}+\frac {b \left (-\frac {\mathrm {arcsec}\left (c x \right ) d^{2}}{7 c^{3} x^{7}}-\frac {\mathrm {arcsec}\left (c x \right ) e^{2}}{3 c^{3} x^{3}}-\frac {2 \,\mathrm {arcsec}\left (c x \right ) e d}{5 c^{3} x^{5}}+\frac {\left (c^{2} x^{2}-1\right ) \left (720 c^{10} d^{2} x^{6}+2352 c^{8} e d \,x^{6}+360 x^{4} c^{8} d^{2}+2450 e^{2} c^{6} x^{6}+1176 c^{6} e d \,x^{4}+270 x^{2} c^{6} d^{2}+1225 c^{4} e^{2} x^{4}+882 c^{4} d e \,x^{2}+225 d^{2} c^{4}\right )}{11025 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{8} x^{8}}\right )}{c^{4}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^2*(a+b*arcsec(c*x))/x^8,x)

[Out]

c^7*(a/c^4*(-1/7*d^2/c^3/x^7-1/3*e^2/c^3/x^3-2/5/c^3*e*d/x^5)+b/c^4*(-1/7*arcsec(c*x)*d^2/c^3/x^7-1/3*arcsec(c
*x)*e^2/c^3/x^3-2/5*arcsec(c*x)/c^3*e*d/x^5+1/11025*(c^2*x^2-1)*(720*c^10*d^2*x^6+2352*c^8*d*e*x^6+360*c^8*d^2
*x^4+2450*c^6*e^2*x^6+1176*c^6*d*e*x^4+270*c^6*d^2*x^2+1225*c^4*e^2*x^4+882*c^4*d*e*x^2+225*c^4*d^2)/((c^2*x^2
-1)/c^2/x^2)^(1/2)/c^8/x^8))

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maxima [A]  time = 0.33, size = 241, normalized size = 1.00 \[ -\frac {1}{245} \, b d^{2} {\left (\frac {5 \, c^{8} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {7}{2}} - 21 \, c^{8} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {5}{2}} + 35 \, c^{8} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 35 \, c^{8} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c} + \frac {35 \, \operatorname {arcsec}\left (c x\right )}{x^{7}}\right )} + \frac {2}{75} \, b d e {\left (\frac {3 \, c^{6} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {5}{2}} - 10 \, c^{6} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 15 \, c^{6} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c} - \frac {15 \, \operatorname {arcsec}\left (c x\right )}{x^{5}}\right )} - \frac {1}{9} \, b e^{2} {\left (\frac {c^{4} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 3 \, c^{4} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c} + \frac {3 \, \operatorname {arcsec}\left (c x\right )}{x^{3}}\right )} - \frac {a e^{2}}{3 \, x^{3}} - \frac {2 \, a d e}{5 \, x^{5}} - \frac {a d^{2}}{7 \, x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arcsec(c*x))/x^8,x, algorithm="maxima")

[Out]

-1/245*b*d^2*((5*c^8*(-1/(c^2*x^2) + 1)^(7/2) - 21*c^8*(-1/(c^2*x^2) + 1)^(5/2) + 35*c^8*(-1/(c^2*x^2) + 1)^(3
/2) - 35*c^8*sqrt(-1/(c^2*x^2) + 1))/c + 35*arcsec(c*x)/x^7) + 2/75*b*d*e*((3*c^6*(-1/(c^2*x^2) + 1)^(5/2) - 1
0*c^6*(-1/(c^2*x^2) + 1)^(3/2) + 15*c^6*sqrt(-1/(c^2*x^2) + 1))/c - 15*arcsec(c*x)/x^5) - 1/9*b*e^2*((c^4*(-1/
(c^2*x^2) + 1)^(3/2) - 3*c^4*sqrt(-1/(c^2*x^2) + 1))/c + 3*arcsec(c*x)/x^3) - 1/3*a*e^2/x^3 - 2/5*a*d*e/x^5 -
1/7*a*d^2/x^7

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e\,x^2+d\right )}^2\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{x^8} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*x^2)^2*(a + b*acos(1/(c*x))))/x^8,x)

[Out]

int(((d + e*x^2)^2*(a + b*acos(1/(c*x))))/x^8, x)

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sympy [A]  time = 57.34, size = 508, normalized size = 2.11 \[ - \frac {a d^{2}}{7 x^{7}} - \frac {2 a d e}{5 x^{5}} - \frac {a e^{2}}{3 x^{3}} - \frac {b d^{2} \operatorname {asec}{\left (c x \right )}}{7 x^{7}} - \frac {2 b d e \operatorname {asec}{\left (c x \right )}}{5 x^{5}} - \frac {b e^{2} \operatorname {asec}{\left (c x \right )}}{3 x^{3}} + \frac {b d^{2} \left (\begin {cases} \frac {16 c^{7} \sqrt {c^{2} x^{2} - 1}}{35 x} + \frac {8 c^{5} \sqrt {c^{2} x^{2} - 1}}{35 x^{3}} + \frac {6 c^{3} \sqrt {c^{2} x^{2} - 1}}{35 x^{5}} + \frac {c \sqrt {c^{2} x^{2} - 1}}{7 x^{7}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {16 i c^{7} \sqrt {- c^{2} x^{2} + 1}}{35 x} + \frac {8 i c^{5} \sqrt {- c^{2} x^{2} + 1}}{35 x^{3}} + \frac {6 i c^{3} \sqrt {- c^{2} x^{2} + 1}}{35 x^{5}} + \frac {i c \sqrt {- c^{2} x^{2} + 1}}{7 x^{7}} & \text {otherwise} \end {cases}\right )}{7 c} + \frac {2 b d e \left (\begin {cases} \frac {8 c^{5} \sqrt {c^{2} x^{2} - 1}}{15 x} + \frac {4 c^{3} \sqrt {c^{2} x^{2} - 1}}{15 x^{3}} + \frac {c \sqrt {c^{2} x^{2} - 1}}{5 x^{5}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {8 i c^{5} \sqrt {- c^{2} x^{2} + 1}}{15 x} + \frac {4 i c^{3} \sqrt {- c^{2} x^{2} + 1}}{15 x^{3}} + \frac {i c \sqrt {- c^{2} x^{2} + 1}}{5 x^{5}} & \text {otherwise} \end {cases}\right )}{5 c} + \frac {b e^{2} \left (\begin {cases} \frac {2 c^{3} \sqrt {c^{2} x^{2} - 1}}{3 x} + \frac {c \sqrt {c^{2} x^{2} - 1}}{3 x^{3}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {2 i c^{3} \sqrt {- c^{2} x^{2} + 1}}{3 x} + \frac {i c \sqrt {- c^{2} x^{2} + 1}}{3 x^{3}} & \text {otherwise} \end {cases}\right )}{3 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**2*(a+b*asec(c*x))/x**8,x)

[Out]

-a*d**2/(7*x**7) - 2*a*d*e/(5*x**5) - a*e**2/(3*x**3) - b*d**2*asec(c*x)/(7*x**7) - 2*b*d*e*asec(c*x)/(5*x**5)
- b*e**2*asec(c*x)/(3*x**3) + b*d**2*Piecewise((16*c**7*sqrt(c**2*x**2 - 1)/(35*x) + 8*c**5*sqrt(c**2*x**2 -
1)/(35*x**3) + 6*c**3*sqrt(c**2*x**2 - 1)/(35*x**5) + c*sqrt(c**2*x**2 - 1)/(7*x**7), Abs(c**2*x**2) > 1), (16
*I*c**7*sqrt(-c**2*x**2 + 1)/(35*x) + 8*I*c**5*sqrt(-c**2*x**2 + 1)/(35*x**3) + 6*I*c**3*sqrt(-c**2*x**2 + 1)/
(35*x**5) + I*c*sqrt(-c**2*x**2 + 1)/(7*x**7), True))/(7*c) + 2*b*d*e*Piecewise((8*c**5*sqrt(c**2*x**2 - 1)/(1
5*x) + 4*c**3*sqrt(c**2*x**2 - 1)/(15*x**3) + c*sqrt(c**2*x**2 - 1)/(5*x**5), Abs(c**2*x**2) > 1), (8*I*c**5*s
qrt(-c**2*x**2 + 1)/(15*x) + 4*I*c**3*sqrt(-c**2*x**2 + 1)/(15*x**3) + I*c*sqrt(-c**2*x**2 + 1)/(5*x**5), True
))/(5*c) + b*e**2*Piecewise((2*c**3*sqrt(c**2*x**2 - 1)/(3*x) + c*sqrt(c**2*x**2 - 1)/(3*x**3), Abs(c**2*x**2)
> 1), (2*I*c**3*sqrt(-c**2*x**2 + 1)/(3*x) + I*c*sqrt(-c**2*x**2 + 1)/(3*x**3), True))/(3*c)

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