3.1.3 \(\int x^4 (a+b \sec ^{-1}(c x)) \, dx\) [3]
Optimal. Leaf size=89 \[ -\frac {3 b \sqrt {1-\frac {1}{c^2 x^2}} x^2}{40 c^3}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^4}{20 c}+\frac {1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )-\frac {3 b \tanh ^{-1}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{40 c^5} \]
[Out]
1/5*x^5*(a+b*arcsec(c*x))-3/40*b*arctanh((1-1/c^2/x^2)^(1/2))/c^5-3/40*b*x^2*(1-1/c^2/x^2)^(1/2)/c^3-1/20*b*x^
4*(1-1/c^2/x^2)^(1/2)/c
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Rubi [A]
time = 0.04, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5328, 272, 44,
65, 214} \begin {gather*} \frac {1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )-\frac {b x^4 \sqrt {1-\frac {1}{c^2 x^2}}}{20 c}-\frac {3 b \tanh ^{-1}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{40 c^5}-\frac {3 b x^2 \sqrt {1-\frac {1}{c^2 x^2}}}{40 c^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[x^4*(a + b*ArcSec[c*x]),x]
[Out]
(-3*b*Sqrt[1 - 1/(c^2*x^2)]*x^2)/(40*c^3) - (b*Sqrt[1 - 1/(c^2*x^2)]*x^4)/(20*c) + (x^5*(a + b*ArcSec[c*x]))/5
- (3*b*ArcTanh[Sqrt[1 - 1/(c^2*x^2)]])/(40*c^5)
Rule 44
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] && !IntegerQ[n] && LtQ[n, 0]
Rule 65
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 214
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]
Rule 272
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 5328
Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSec[c*x]
)/(d*(m + 1))), x] - Dist[b*(d/(c*(m + 1))), Int[(d*x)^(m - 1)/Sqrt[1 - 1/(c^2*x^2)], x], x] /; FreeQ[{a, b, c
, d, m}, x] && NeQ[m, -1]
Rubi steps
\begin {align*} \int x^4 \left (a+b \sec ^{-1}(c x)\right ) \, dx &=\frac {1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )-\frac {b \int \frac {x^3}{\sqrt {1-\frac {1}{c^2 x^2}}} \, dx}{5 c}\\ &=\frac {1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )+\frac {b \text {Subst}\left (\int \frac {1}{x^3 \sqrt {1-\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{10 c}\\ &=-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^4}{20 c}+\frac {1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )+\frac {(3 b) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1-\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{40 c^3}\\ &=-\frac {3 b \sqrt {1-\frac {1}{c^2 x^2}} x^2}{40 c^3}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^4}{20 c}+\frac {1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )+\frac {(3 b) \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{80 c^5}\\ &=-\frac {3 b \sqrt {1-\frac {1}{c^2 x^2}} x^2}{40 c^3}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^4}{20 c}+\frac {1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )-\frac {(3 b) \text {Subst}\left (\int \frac {1}{c^2-c^2 x^2} \, dx,x,\sqrt {1-\frac {1}{c^2 x^2}}\right )}{40 c^3}\\ &=-\frac {3 b \sqrt {1-\frac {1}{c^2 x^2}} x^2}{40 c^3}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^4}{20 c}+\frac {1}{5} x^5 \left (a+b \sec ^{-1}(c x)\right )-\frac {3 b \tanh ^{-1}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{40 c^5}\\ \end {align*}
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Mathematica [A]
time = 0.05, size = 97, normalized size = 1.09 \begin {gather*} \frac {a x^5}{5}+b \sqrt {\frac {-1+c^2 x^2}{c^2 x^2}} \left (-\frac {3 x^2}{40 c^3}-\frac {x^4}{20 c}\right )+\frac {1}{5} b x^5 \sec ^{-1}(c x)-\frac {3 b \log \left (x \left (1+\sqrt {\frac {-1+c^2 x^2}{c^2 x^2}}\right )\right )}{40 c^5} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[x^4*(a + b*ArcSec[c*x]),x]
[Out]
(a*x^5)/5 + b*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)]*((-3*x^2)/(40*c^3) - x^4/(20*c)) + (b*x^5*ArcSec[c*x])/5 - (3*b*L
og[x*(1 + Sqrt[(-1 + c^2*x^2)/(c^2*x^2)])])/(40*c^5)
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Maple [A]
time = 0.09, size = 148, normalized size = 1.66
| | |
method |
result |
size |
| | |
derivativedivides |
\(\frac {\frac {c^{5} x^{5} a}{5}+\frac {b \,c^{5} x^{5} \mathrm {arcsec}\left (c x \right )}{5}-\frac {b \left (c^{2} x^{2}-1\right ) c^{2} x^{2}}{20 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}-\frac {3 b \left (c^{2} x^{2}-1\right )}{40 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}-\frac {3 b \sqrt {c^{2} x^{2}-1}\, \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{40 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}}{c^{5}}\) |
\(148\) |
default |
\(\frac {\frac {c^{5} x^{5} a}{5}+\frac {b \,c^{5} x^{5} \mathrm {arcsec}\left (c x \right )}{5}-\frac {b \left (c^{2} x^{2}-1\right ) c^{2} x^{2}}{20 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}-\frac {3 b \left (c^{2} x^{2}-1\right )}{40 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}-\frac {3 b \sqrt {c^{2} x^{2}-1}\, \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{40 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}}{c^{5}}\) |
\(148\) |
| | |
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^4*(a+b*arcsec(c*x)),x,method=_RETURNVERBOSE)
[Out]
1/c^5*(1/5*c^5*x^5*a+1/5*b*c^5*x^5*arcsec(c*x)-1/20*b*(c^2*x^2-1)/((c^2*x^2-1)/c^2/x^2)^(1/2)*c^2*x^2-3/40*b*(
c^2*x^2-1)/((c^2*x^2-1)/c^2/x^2)^(1/2)-3/40*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/c/x*ln(c*x+(c^2*x^
2-1)^(1/2)))
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Maxima [A]
time = 0.27, size = 131, normalized size = 1.47 \begin {gather*} \frac {1}{5} \, a x^{5} + \frac {1}{80} \, {\left (16 \, x^{5} \operatorname {arcsec}\left (c x\right ) + \frac {\frac {2 \, {\left (3 \, {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 5 \, \sqrt {-\frac {1}{c^{2} x^{2}} + 1}\right )}}{c^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{2} + 2 \, c^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} + c^{4}} - \frac {3 \, \log \left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}{c^{4}} + \frac {3 \, \log \left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} - 1\right )}{c^{4}}}{c}\right )} b \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4*(a+b*arcsec(c*x)),x, algorithm="maxima")
[Out]
1/5*a*x^5 + 1/80*(16*x^5*arcsec(c*x) + (2*(3*(-1/(c^2*x^2) + 1)^(3/2) - 5*sqrt(-1/(c^2*x^2) + 1))/(c^4*(1/(c^2
*x^2) - 1)^2 + 2*c^4*(1/(c^2*x^2) - 1) + c^4) - 3*log(sqrt(-1/(c^2*x^2) + 1) + 1)/c^4 + 3*log(sqrt(-1/(c^2*x^2
) + 1) - 1)/c^4)/c)*b
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Fricas [A]
time = 2.60, size = 107, normalized size = 1.20 \begin {gather*} \frac {8 \, a c^{5} x^{5} + 16 \, b c^{5} \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + 8 \, {\left (b c^{5} x^{5} - b c^{5}\right )} \operatorname {arcsec}\left (c x\right ) + 3 \, b \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) - {\left (2 \, b c^{3} x^{3} + 3 \, b c x\right )} \sqrt {c^{2} x^{2} - 1}}{40 \, c^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4*(a+b*arcsec(c*x)),x, algorithm="fricas")
[Out]
1/40*(8*a*c^5*x^5 + 16*b*c^5*arctan(-c*x + sqrt(c^2*x^2 - 1)) + 8*(b*c^5*x^5 - b*c^5)*arcsec(c*x) + 3*b*log(-c
*x + sqrt(c^2*x^2 - 1)) - (2*b*c^3*x^3 + 3*b*c*x)*sqrt(c^2*x^2 - 1))/c^5
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Sympy [A]
time = 4.50, size = 175, normalized size = 1.97 \begin {gather*} \frac {a x^{5}}{5} + \frac {b x^{5} \operatorname {asec}{\left (c x \right )}}{5} - \frac {b \left (\begin {cases} \frac {c x^{5}}{4 \sqrt {c^{2} x^{2} - 1}} + \frac {x^{3}}{8 c \sqrt {c^{2} x^{2} - 1}} - \frac {3 x}{8 c^{3} \sqrt {c^{2} x^{2} - 1}} + \frac {3 \operatorname {acosh}{\left (c x \right )}}{8 c^{4}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\- \frac {i c x^{5}}{4 \sqrt {- c^{2} x^{2} + 1}} - \frac {i x^{3}}{8 c \sqrt {- c^{2} x^{2} + 1}} + \frac {3 i x}{8 c^{3} \sqrt {- c^{2} x^{2} + 1}} - \frac {3 i \operatorname {asin}{\left (c x \right )}}{8 c^{4}} & \text {otherwise} \end {cases}\right )}{5 c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**4*(a+b*asec(c*x)),x)
[Out]
a*x**5/5 + b*x**5*asec(c*x)/5 - b*Piecewise((c*x**5/(4*sqrt(c**2*x**2 - 1)) + x**3/(8*c*sqrt(c**2*x**2 - 1)) -
3*x/(8*c**3*sqrt(c**2*x**2 - 1)) + 3*acosh(c*x)/(8*c**4), Abs(c**2*x**2) > 1), (-I*c*x**5/(4*sqrt(-c**2*x**2
+ 1)) - I*x**3/(8*c*sqrt(-c**2*x**2 + 1)) + 3*I*x/(8*c**3*sqrt(-c**2*x**2 + 1)) - 3*I*asin(c*x)/(8*c**4), True
))/(5*c)
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 4828 vs.
\(2 (75) = 150\).
time = 0.96, size = 4828, normalized size = 54.25 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4*(a+b*arcsec(c*x)),x, algorithm="giac")
[Out]
1/40*c*(8*b*arccos(1/(c*x))/(c^6 + 5*c^6*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x^2) - 1)^2/(1/(c*
x) + 1)^4 + 10*c^6*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + c^6*(1/(c
^2*x^2) - 1)^5/(1/(c*x) + 1)^10) - 3*b*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1))/(c^6 + 5*c^6*(1/(c^2*x^2
) - 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 10*c^6*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)
^6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + c^6*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10) + 3*b*log(abs(sqrt(
-1/(c^2*x^2) + 1) - 1/(c*x) - 1))/(c^6 + 5*c^6*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x^2) - 1)^2/
(1/(c*x) + 1)^4 + 10*c^6*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + c^6
*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10) + 8*a/(c^6 + 5*c^6*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x
^2) - 1)^2/(1/(c*x) + 1)^4 + 10*c^6*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(c*x) +
1)^8 + c^6*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10) - 40*b*(1/(c^2*x^2) - 1)*arccos(1/(c*x))/((c^6 + 5*c^6*(1/(c
^2*x^2) - 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 10*c^6*(1/(c^2*x^2) - 1)^3/(1/(c*x
) + 1)^6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + c^6*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10)*(1/(c*x) + 1)
^2) - 15*b*(1/(c^2*x^2) - 1)*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1))/((c^6 + 5*c^6*(1/(c^2*x^2) - 1)/(1
/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 10*c^6*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 5*c^
6*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + c^6*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10)*(1/(c*x) + 1)^2) + 15*b*(1/(
c^2*x^2) - 1)*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) - 1))/((c^6 + 5*c^6*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 +
10*c^6*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 10*c^6*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 5*c^6*(1/(c^2*x^2)
- 1)^4/(1/(c*x) + 1)^8 + c^6*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10)*(1/(c*x) + 1)^2) - 10*b*sqrt(-1/(c^2*x^2) +
1)/((c^6 + 5*c^6*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 10*c^6*(1/(
c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + c^6*(1/(c^2*x^2) - 1)^5/(1/(c*x)
+ 1)^10)*(1/(c*x) + 1)) - 40*a*(1/(c^2*x^2) - 1)/((c^6 + 5*c^6*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/
(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 10*c^6*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(
c*x) + 1)^8 + c^6*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10)*(1/(c*x) + 1)^2) + 80*b*(1/(c^2*x^2) - 1)^2*arccos(1/(
c*x))/((c^6 + 5*c^6*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 10*c^6*(1
/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + c^6*(1/(c^2*x^2) - 1)^5/(1/(c*
x) + 1)^10)*(1/(c*x) + 1)^4) - 30*b*(1/(c^2*x^2) - 1)^2*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1))/((c^6 +
5*c^6*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 10*c^6*(1/(c^2*x^2) -
1)^3/(1/(c*x) + 1)^6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + c^6*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10)*(
1/(c*x) + 1)^4) + 30*b*(1/(c^2*x^2) - 1)^2*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) - 1))/((c^6 + 5*c^6*(1/(c^
2*x^2) - 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 10*c^6*(1/(c^2*x^2) - 1)^3/(1/(c*x)
+ 1)^6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + c^6*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10)*(1/(c*x) + 1)^
4) + 4*b*(-1/(c^2*x^2) + 1)^(3/2)/((c^6 + 5*c^6*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x^2) - 1)^2
/(1/(c*x) + 1)^4 + 10*c^6*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + c^
6*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10)*(1/(c*x) + 1)^3) + 80*a*(1/(c^2*x^2) - 1)^2/((c^6 + 5*c^6*(1/(c^2*x^2)
- 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 10*c^6*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^
6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + c^6*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10)*(1/(c*x) + 1)^4) - 8
0*b*(1/(c^2*x^2) - 1)^3*arccos(1/(c*x))/((c^6 + 5*c^6*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x^2)
- 1)^2/(1/(c*x) + 1)^4 + 10*c^6*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^
8 + c^6*(1/(c^2*x^2) - 1)^5/(1/(c*x) + 1)^10)*(1/(c*x) + 1)^6) - 30*b*(1/(c^2*x^2) - 1)^3*log(abs(sqrt(-1/(c^2
*x^2) + 1) + 1/(c*x) + 1))/((c^6 + 5*c^6*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x^2) - 1)^2/(1/(c*
x) + 1)^4 + 10*c^6*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + c^6*(1/(c
^2*x^2) - 1)^5/(1/(c*x) + 1)^10)*(1/(c*x) + 1)^6) + 30*b*(1/(c^2*x^2) - 1)^3*log(abs(sqrt(-1/(c^2*x^2) + 1) -
1/(c*x) - 1))/((c^6 + 5*c^6*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 10*c^6*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 1
0*c^6*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + 5*c^6*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8 + c^6*(1/(c^2*x^2) - 1)^
5/(1/(c*x) + 1)^10)*(1/(c*x) + 1)^6) - 80*a*(1/(c^2*x^2) - 1)^3/((c^6 + 5*c^6*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^
2 + 10*c^6*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 ...
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^4\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^4*(a + b*acos(1/(c*x))),x)
[Out]
int(x^4*(a + b*acos(1/(c*x))), x)
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