3.1.4 \(\int x^3 (a+b \sec ^{-1}(c x)) \, dx\) [4]
Optimal. Leaf size=64 \[ -\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x}{6 c^3}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^3}{12 c}+\frac {1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right ) \]
[Out]
1/4*x^4*(a+b*arcsec(c*x))-1/6*b*x*(1-1/c^2/x^2)^(1/2)/c^3-1/12*b*x^3*(1-1/c^2/x^2)^(1/2)/c
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Rubi [A]
time = 0.02, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5328, 277, 197}
\begin {gather*} \frac {1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )-\frac {b x^3 \sqrt {1-\frac {1}{c^2 x^2}}}{12 c}-\frac {b x \sqrt {1-\frac {1}{c^2 x^2}}}{6 c^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[x^3*(a + b*ArcSec[c*x]),x]
[Out]
-1/6*(b*Sqrt[1 - 1/(c^2*x^2)]*x)/c^3 - (b*Sqrt[1 - 1/(c^2*x^2)]*x^3)/(12*c) + (x^4*(a + b*ArcSec[c*x]))/4
Rule 197
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]
Rule 277
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 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^3 \left (a+b \sec ^{-1}(c x)\right ) \, dx &=\frac {1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )-\frac {b \int \frac {x^2}{\sqrt {1-\frac {1}{c^2 x^2}}} \, dx}{4 c}\\ &=-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^3}{12 c}+\frac {1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )-\frac {b \int \frac {1}{\sqrt {1-\frac {1}{c^2 x^2}}} \, dx}{6 c^3}\\ &=-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x}{6 c^3}-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^3}{12 c}+\frac {1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )\\ \end {align*}
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Mathematica [A]
time = 0.07, size = 62, normalized size = 0.97 \begin {gather*} \frac {a x^4}{4}+b \sqrt {\frac {-1+c^2 x^2}{c^2 x^2}} \left (-\frac {x}{6 c^3}-\frac {x^3}{12 c}\right )+\frac {1}{4} b x^4 \sec ^{-1}(c x) \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[x^3*(a + b*ArcSec[c*x]),x]
[Out]
(a*x^4)/4 + b*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)]*(-1/6*x/c^3 - x^3/(12*c)) + (b*x^4*ArcSec[c*x])/4
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Maple [A]
time = 0.09, size = 74, normalized size = 1.16
| | |
| method |
result |
size |
| | |
| derivativedivides |
\(\frac {\frac {c^{4} x^{4} a}{4}+b \left (\frac {c^{4} x^{4} \mathrm {arcsec}\left (c x \right )}{4}-\frac {\left (c^{2} x^{2}-1\right ) \left (c^{2} x^{2}+2\right )}{12 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{4}}\) |
\(74\) |
| default |
\(\frac {\frac {c^{4} x^{4} a}{4}+b \left (\frac {c^{4} x^{4} \mathrm {arcsec}\left (c x \right )}{4}-\frac {\left (c^{2} x^{2}-1\right ) \left (c^{2} x^{2}+2\right )}{12 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{4}}\) |
\(74\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*(a+b*arcsec(c*x)),x,method=_RETURNVERBOSE)
[Out]
1/c^4*(1/4*c^4*x^4*a+b*(1/4*c^4*x^4*arcsec(c*x)-1/12*(c^2*x^2-1)*(c^2*x^2+2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/c/x))
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Maxima [A]
time = 0.27, size = 60, normalized size = 0.94 \begin {gather*} \frac {1}{4} \, a x^{4} + \frac {1}{12} \, {\left (3 \, x^{4} \operatorname {arcsec}\left (c x\right ) - \frac {c^{2} x^{3} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 3 \, x \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c^{3}}\right )} b \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(a+b*arcsec(c*x)),x, algorithm="maxima")
[Out]
1/4*a*x^4 + 1/12*(3*x^4*arcsec(c*x) - (c^2*x^3*(-1/(c^2*x^2) + 1)^(3/2) + 3*x*sqrt(-1/(c^2*x^2) + 1))/c^3)*b
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Fricas [A]
time = 2.33, size = 53, normalized size = 0.83 \begin {gather*} \frac {3 \, b c^{4} x^{4} \operatorname {arcsec}\left (c x\right ) + 3 \, a c^{4} x^{4} - {\left (b c^{2} x^{2} + 2 \, b\right )} \sqrt {c^{2} x^{2} - 1}}{12 \, c^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(a+b*arcsec(c*x)),x, algorithm="fricas")
[Out]
1/12*(3*b*c^4*x^4*arcsec(c*x) + 3*a*c^4*x^4 - (b*c^2*x^2 + 2*b)*sqrt(c^2*x^2 - 1))/c^4
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Sympy [A]
time = 2.49, size = 107, normalized size = 1.67 \begin {gather*} \frac {a x^{4}}{4} + \frac {b x^{4} \operatorname {asec}{\left (c x \right )}}{4} - \frac {b \left (\begin {cases} \frac {x^{2} \sqrt {c^{2} x^{2} - 1}}{3 c} + \frac {2 \sqrt {c^{2} x^{2} - 1}}{3 c^{3}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {i x^{2} \sqrt {- c^{2} x^{2} + 1}}{3 c} + \frac {2 i \sqrt {- c^{2} x^{2} + 1}}{3 c^{3}} & \text {otherwise} \end {cases}\right )}{4 c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**3*(a+b*asec(c*x)),x)
[Out]
a*x**4/4 + b*x**4*asec(c*x)/4 - b*Piecewise((x**2*sqrt(c**2*x**2 - 1)/(3*c) + 2*sqrt(c**2*x**2 - 1)/(3*c**3),
Abs(c**2*x**2) > 1), (I*x**2*sqrt(-c**2*x**2 + 1)/(3*c) + 2*I*sqrt(-c**2*x**2 + 1)/(3*c**3), True))/(4*c)
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1926 vs.
\(2 (54) = 108\).
time = 0.42, size = 1926, normalized size = 30.09 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(a+b*arcsec(c*x)),x, algorithm="giac")
[Out]
1/12*c*(3*b*arccos(1/(c*x))/(c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x
) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8) + 3*a/(c^5 + 4
*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3
/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8) - 12*b*(1/(c^2*x^2) - 1)*arccos(1/(c*x))/((c^5 + 4
*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3
/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^2) - 6*b*sqrt(-1/(c^2*x^2) + 1)/((c^
5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) -
1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)) - 12*a*(1/(c^2*x^2) - 1)/((c^5
+ 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) -
1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^2) + 18*b*(1/(c^2*x^2) - 1)^2*ar
ccos(1/(c*x))/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*
c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^4) + 10*b*(-1
/(c^2*x^2) + 1)^(3/2)/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1
)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^3) +
18*a*(1/(c^2*x^2) - 1)^2/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x)
+ 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^4)
- 12*b*(1/(c^2*x^2) - 1)^3*arccos(1/(c*x))/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^
2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^
8)*(1/(c*x) + 1)^6) - 10*b*(1/(c^2*x^2) - 1)^2*sqrt(-1/(c^2*x^2) + 1)/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x)
+ 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*
x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^5) - 12*a*(1/(c^2*x^2) - 1)^3/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c
*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c
^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^6) + 3*b*(1/(c^2*x^2) - 1)^4*arccos(1/(c*x))/((c^5 + 4*c^5*(1/(c
^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(c^2*x^2) - 1)^3/(1/(c*x)
+ 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^8) - 6*b*(1/(c^2*x^2) - 1)^3*sqrt(-1/(c^2*x^2)
+ 1)/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1/(
c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^7) + 3*a*(1/(c^2*x^2)
- 1)^4/((c^5 + 4*c^5*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 6*c^5*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + 4*c^5*(1
/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6 + c^5*(1/(c^2*x^2) - 1)^4/(1/(c*x) + 1)^8)*(1/(c*x) + 1)^8))
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^3\,\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^3*(a + b*acos(1/(c*x))),x)
[Out]
int(x^3*(a + b*acos(1/(c*x))), x)
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