3.1.71 \(\int (d+e x^2) (a+b \sec ^{-1}(c x)) \, dx\) [71]
Optimal. Leaf size=109 \[ -\frac {b e x^2 \sqrt {-1+c^2 x^2}}{6 c \sqrt {c^2 x^2}}+d x \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac {b \left (6 c^2 d+e\right ) x \tanh ^{-1}\left (\frac {c x}{\sqrt {-1+c^2 x^2}}\right )}{6 c^2 \sqrt {c^2 x^2}} \]
[Out]
d*x*(a+b*arcsec(c*x))+1/3*e*x^3*(a+b*arcsec(c*x))-1/6*b*(6*c^2*d+e)*x*arctanh(c*x/(c^2*x^2-1)^(1/2))/c^2/(c^2*
x^2)^(1/2)-1/6*b*e*x^2*(c^2*x^2-1)^(1/2)/c/(c^2*x^2)^(1/2)
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Rubi [A]
time = 0.04, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5336, 12, 396,
223, 212} \begin {gather*} d x \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac {b x \left (6 c^2 d+e\right ) \tanh ^{-1}\left (\frac {c x}{\sqrt {c^2 x^2-1}}\right )}{6 c^2 \sqrt {c^2 x^2}}-\frac {b e x^2 \sqrt {c^2 x^2-1}}{6 c \sqrt {c^2 x^2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(d + e*x^2)*(a + b*ArcSec[c*x]),x]
[Out]
-1/6*(b*e*x^2*Sqrt[-1 + c^2*x^2])/(c*Sqrt[c^2*x^2]) + d*x*(a + b*ArcSec[c*x]) + (e*x^3*(a + b*ArcSec[c*x]))/3
- (b*(6*c^2*d + e)*x*ArcTanh[(c*x)/Sqrt[-1 + c^2*x^2]])/(6*c^2*Sqrt[c^2*x^2])
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 223
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 396
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
Rule 5336
Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^p, x]}, Dist[a + b*ArcSec[c*x], u, x] - Dist[b*c*(x/Sqrt[c^2*x^2]), Int[SimplifyIntegrand[u/(x*Sqrt[c^2*x^2
- 1]), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])
Rubi steps
\begin {align*} \int \left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right ) \, dx &=d x \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac {(b c x) \int \frac {3 d+e x^2}{3 \sqrt {-1+c^2 x^2}} \, dx}{\sqrt {c^2 x^2}}\\ &=d x \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac {(b c x) \int \frac {3 d+e x^2}{\sqrt {-1+c^2 x^2}} \, dx}{3 \sqrt {c^2 x^2}}\\ &=-\frac {b e x^2 \sqrt {-1+c^2 x^2}}{6 c \sqrt {c^2 x^2}}+d x \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \sec ^{-1}(c x)\right )+\frac {\left (b \left (-6 c^2 d-e\right ) x\right ) \int \frac {1}{\sqrt {-1+c^2 x^2}} \, dx}{6 c \sqrt {c^2 x^2}}\\ &=-\frac {b e x^2 \sqrt {-1+c^2 x^2}}{6 c \sqrt {c^2 x^2}}+d x \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \sec ^{-1}(c x)\right )+\frac {\left (b \left (-6 c^2 d-e\right ) x\right ) \text {Subst}\left (\int \frac {1}{1-c^2 x^2} \, dx,x,\frac {x}{\sqrt {-1+c^2 x^2}}\right )}{6 c \sqrt {c^2 x^2}}\\ &=-\frac {b e x^2 \sqrt {-1+c^2 x^2}}{6 c \sqrt {c^2 x^2}}+d x \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac {b \left (6 c^2 d+e\right ) x \tanh ^{-1}\left (\frac {c x}{\sqrt {-1+c^2 x^2}}\right )}{6 c^2 \sqrt {c^2 x^2}}\\ \end {align*}
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Mathematica [A]
time = 0.20, size = 150, normalized size = 1.38 \begin {gather*} a d x+\frac {1}{3} a e x^3-\frac {b e x^2 \sqrt {\frac {-1+c^2 x^2}{c^2 x^2}}}{6 c}+b d x \sec ^{-1}(c x)+\frac {1}{3} b e x^3 \sec ^{-1}(c x)-\frac {b d \sqrt {1-\frac {1}{c^2 x^2}} x \tanh ^{-1}\left (\frac {c x}{\sqrt {-1+c^2 x^2}}\right )}{\sqrt {-1+c^2 x^2}}-\frac {b e \log \left (x \left (1+\sqrt {\frac {-1+c^2 x^2}{c^2 x^2}}\right )\right )}{6 c^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(d + e*x^2)*(a + b*ArcSec[c*x]),x]
[Out]
a*d*x + (a*e*x^3)/3 - (b*e*x^2*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)])/(6*c) + b*d*x*ArcSec[c*x] + (b*e*x^3*ArcSec[c*x
])/3 - (b*d*Sqrt[1 - 1/(c^2*x^2)]*x*ArcTanh[(c*x)/Sqrt[-1 + c^2*x^2]])/Sqrt[-1 + c^2*x^2] - (b*e*Log[x*(1 + Sq
rt[(-1 + c^2*x^2)/(c^2*x^2)])])/(6*c^3)
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(191\) vs.
\(2(95)=190\).
time = 0.12, size = 192, normalized size = 1.76
| | |
| method |
result |
size |
| | |
| derivativedivides |
\(\frac {\frac {a \left (d \,c^{3} x +\frac {1}{3} e \,c^{3} x^{3}\right )}{c^{2}}+b \,\mathrm {arcsec}\left (c x \right ) d c x +\frac {b c \,\mathrm {arcsec}\left (c x \right ) e \,x^{3}}{3}-\frac {b \sqrt {c^{2} x^{2}-1}\, d \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{\sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}-\frac {b \left (c^{2} x^{2}-1\right ) e}{6 c^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}-\frac {b \sqrt {c^{2} x^{2}-1}\, e \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{6 c^{3} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x}}{c}\) |
\(192\) |
| default |
\(\frac {\frac {a \left (d \,c^{3} x +\frac {1}{3} e \,c^{3} x^{3}\right )}{c^{2}}+b \,\mathrm {arcsec}\left (c x \right ) d c x +\frac {b c \,\mathrm {arcsec}\left (c x \right ) e \,x^{3}}{3}-\frac {b \sqrt {c^{2} x^{2}-1}\, d \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{\sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}-\frac {b \left (c^{2} x^{2}-1\right ) e}{6 c^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}-\frac {b \sqrt {c^{2} x^{2}-1}\, e \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{6 c^{3} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x}}{c}\) |
\(192\) |
| | |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*x^2+d)*(a+b*arcsec(c*x)),x,method=_RETURNVERBOSE)
[Out]
1/c*(a/c^2*(d*c^3*x+1/3*e*c^3*x^3)+b*arcsec(c*x)*d*c*x+1/3*b*c*arcsec(c*x)*e*x^3-b*(c^2*x^2-1)^(1/2)/((c^2*x^2
-1)/c^2/x^2)^(1/2)/c/x*d*ln(c*x+(c^2*x^2-1)^(1/2))-1/6*b/c^2*(c^2*x^2-1)/((c^2*x^2-1)/c^2/x^2)^(1/2)*e-1/6*b/c
^3*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*e*ln(c*x+(c^2*x^2-1)^(1/2)))
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Maxima [A]
time = 0.28, size = 156, normalized size = 1.43 \begin {gather*} \frac {1}{3} \, a x^{3} e + a d x + \frac {1}{12} \, {\left (4 \, x^{3} \operatorname {arcsec}\left (c x\right ) - \frac {\frac {2 \, \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c^{2} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} + c^{2}} + \frac {\log \left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}{c^{2}} - \frac {\log \left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} - 1\right )}{c^{2}}}{c}\right )} b e + \frac {{\left (2 \, c x \operatorname {arcsec}\left (c x\right ) - \log \left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right ) + \log \left (-\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )\right )} b d}{2 \, c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x^2+d)*(a+b*arcsec(c*x)),x, algorithm="maxima")
[Out]
1/3*a*x^3*e + a*d*x + 1/12*(4*x^3*arcsec(c*x) - (2*sqrt(-1/(c^2*x^2) + 1)/(c^2*(1/(c^2*x^2) - 1) + c^2) + log(
sqrt(-1/(c^2*x^2) + 1) + 1)/c^2 - log(sqrt(-1/(c^2*x^2) + 1) - 1)/c^2)/c)*b*e + 1/2*(2*c*x*arcsec(c*x) - log(s
qrt(-1/(c^2*x^2) + 1) + 1) + log(-sqrt(-1/(c^2*x^2) + 1) + 1))*b*d/c
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Fricas [A]
time = 4.81, size = 147, normalized size = 1.35 \begin {gather*} \frac {2 \, a c^{3} x^{3} e + 6 \, a c^{3} d x - \sqrt {c^{2} x^{2} - 1} b c x e + 2 \, {\left (3 \, b c^{3} d x - 3 \, b c^{3} d + {\left (b c^{3} x^{3} - b c^{3}\right )} e\right )} \operatorname {arcsec}\left (c x\right ) + 4 \, {\left (3 \, b c^{3} d + b c^{3} e\right )} \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + {\left (6 \, b c^{2} d + b e\right )} \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right )}{6 \, c^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x^2+d)*(a+b*arcsec(c*x)),x, algorithm="fricas")
[Out]
1/6*(2*a*c^3*x^3*e + 6*a*c^3*d*x - sqrt(c^2*x^2 - 1)*b*c*x*e + 2*(3*b*c^3*d*x - 3*b*c^3*d + (b*c^3*x^3 - b*c^3
)*e)*arcsec(c*x) + 4*(3*b*c^3*d + b*c^3*e)*arctan(-c*x + sqrt(c^2*x^2 - 1)) + (6*b*c^2*d + b*e)*log(-c*x + sqr
t(c^2*x^2 - 1)))/c^3
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Sympy [A]
time = 4.35, size = 153, normalized size = 1.40 \begin {gather*} a d x + \frac {a e x^{3}}{3} + b d x \operatorname {asec}{\left (c x \right )} + \frac {b e x^{3} \operatorname {asec}{\left (c x \right )}}{3} - \frac {b d \left (\begin {cases} \operatorname {acosh}{\left (c x \right )} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\- i \operatorname {asin}{\left (c x \right )} & \text {otherwise} \end {cases}\right )}{c} - \frac {b e \left (\begin {cases} \frac {x \sqrt {c^{2} x^{2} - 1}}{2 c} + \frac {\operatorname {acosh}{\left (c x \right )}}{2 c^{2}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\- \frac {i c x^{3}}{2 \sqrt {- c^{2} x^{2} + 1}} + \frac {i x}{2 c \sqrt {- c^{2} x^{2} + 1}} - \frac {i \operatorname {asin}{\left (c x \right )}}{2 c^{2}} & \text {otherwise} \end {cases}\right )}{3 c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x**2+d)*(a+b*asec(c*x)),x)
[Out]
a*d*x + a*e*x**3/3 + b*d*x*asec(c*x) + b*e*x**3*asec(c*x)/3 - b*d*Piecewise((acosh(c*x), Abs(c**2*x**2) > 1),
(-I*asin(c*x), True))/c - b*e*Piecewise((x*sqrt(c**2*x**2 - 1)/(2*c) + acosh(c*x)/(2*c**2), Abs(c**2*x**2) > 1
), (-I*c*x**3/(2*sqrt(-c**2*x**2 + 1)) + I*x/(2*c*sqrt(-c**2*x**2 + 1)) - I*asin(c*x)/(2*c**2), True))/(3*c)
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 4051 vs.
\(2 (95) = 190\).
time = 1.27, size = 4051, normalized size = 37.17 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x^2+d)*(a+b*arcsec(c*x)),x, algorithm="giac")
[Out]
1/6*(6*b*c^2*d*arccos(1/(c*x))/(c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(
c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6) - 6*b*c^2*d*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) +
1))/(c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x
^2) - 1)^3/(1/(c*x) + 1)^6) + 6*b*c^2*d*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) - 1))/(c^4 + 3*c^4*(1/(c^2*x^
2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)
+ 6*a*c^2*d/(c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*
(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6) + 6*b*c^2*d*(1/(c^2*x^2) - 1)*arccos(1/(c*x))/((c^4 + 3*c^4*(1/(c^2*x^2)
- 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1
/(c*x) + 1)^2) - 18*b*c^2*d*(1/(c^2*x^2) - 1)*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1))/((c^4 + 3*c^4*(1/
(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x)
+ 1)^6)*(1/(c*x) + 1)^2) + 18*b*c^2*d*(1/(c^2*x^2) - 1)*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) - 1))/((c^4 +
3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3
/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^2) + 6*a*c^2*d*(1/(c^2*x^2) - 1)/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)
^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^2) + 2
*b*e*arccos(1/(c*x))/(c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^
4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6) - 6*b*c^2*d*(1/(c^2*x^2) - 1)^2*arccos(1/(c*x))/((c^4 + 3*c^4*(1/
(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x)
+ 1)^6)*(1/(c*x) + 1)^4) - b*e*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1))/(c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(
1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6) - 18*b*c
^2*d*(1/(c^2*x^2) - 1)^2*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1))/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*
x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)
^4) + b*e*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) - 1))/(c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^
4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6) + 18*b*c^2*d*(1/(c^2*x^2) - 1
)^2*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) - 1))/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/
(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^4) + 2*a*e/(c^4 + 3*
c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1
/(c*x) + 1)^6) - 6*a*c^2*d*(1/(c^2*x^2) - 1)^2/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2
*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^4) - 6*b*e*(1/(c^2*x^2)
- 1)*arccos(1/(c*x))/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)
^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^2) - 6*b*c^2*d*(1/(c^2*x^2) - 1)^3*arccos(1/(c*x))
/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2
) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^6) - 3*b*e*(1/(c^2*x^2) - 1)*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x)
+ 1))/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^
2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^2) - 6*b*c^2*d*(1/(c^2*x^2) - 1)^3*log(abs(sqrt(-1/(c^2*x^2) + 1)
+ 1/(c*x) + 1))/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 +
c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^6) + 3*b*e*(1/(c^2*x^2) - 1)*log(abs(sqrt(-1/(c^2*x^2)
+ 1) - 1/(c*x) - 1))/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1
)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^2) + 6*b*c^2*d*(1/(c^2*x^2) - 1)^3*log(abs(sqrt(-
1/(c^2*x^2) + 1) - 1/(c*x) - 1))/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(
1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^6) - 2*b*e*sqrt(-1/(c^2*x^2) + 1)/((c^
4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1
)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)) - 6*a*e*(1/(c^2*x^2) - 1)/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2
+ 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^2) - 6*a*
c^2*d*(1/(c^2*x^2) - 1)^3/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x)
+ 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^6) + 6*b*e*(1/(c^2*x^2) - 1)^2*arccos(1/(c*x)
)/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)...
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (e\,x^2+d\right )\,\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((d + e*x^2)*(a + b*acos(1/(c*x))),x)
[Out]
int((d + e*x^2)*(a + b*acos(1/(c*x))), x)
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