3.38 \(\int x (a+b \sec ^{-1}(c+d x^2)) \, dx\)

Optimal. Leaf size=58 \[ \frac {a x^2}{2}+\frac {b \left (c+d x^2\right ) \sec ^{-1}\left (c+d x^2\right )}{2 d}-\frac {b \tanh ^{-1}\left (\sqrt {1-\frac {1}{\left (c+d x^2\right )^2}}\right )}{2 d} \]

[Out]

1/2*a*x^2+1/2*b*(d*x^2+c)*arcsec(d*x^2+c)/d-1/2*b*arctanh((1-1/(d*x^2+c)^2)^(1/2))/d

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Rubi [A]  time = 0.07, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6715, 5250, 372, 266, 63, 206} \[ \frac {a x^2}{2}+\frac {b \left (c+d x^2\right ) \sec ^{-1}\left (c+d x^2\right )}{2 d}-\frac {b \tanh ^{-1}\left (\sqrt {1-\frac {1}{\left (c+d x^2\right )^2}}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcSec[c + d*x^2]),x]

[Out]

(a*x^2)/2 + (b*(c + d*x^2)*ArcSec[c + d*x^2])/(2*d) - (b*ArcTanh[Sqrt[1 - (c + d*x^2)^(-2)]])/(2*d)

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 372

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m), Subst[Int[x^m
*(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, m, n, p}, x] && LinearPairQ[u, v, x]

Rule 5250

Int[ArcSec[(c_) + (d_.)*(x_)], x_Symbol] :> Simp[((c + d*x)*ArcSec[c + d*x])/d, x] - Int[1/((c + d*x)*Sqrt[1 -
1/(c + d*x)^2]), x] /; FreeQ[{c, d}, x]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

\begin {align*} \int x \left (a+b \sec ^{-1}\left (c+d x^2\right )\right ) \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \left (a+b \sec ^{-1}(c+d x)\right ) \, dx,x,x^2\right )\\ &=\frac {a x^2}{2}+\frac {1}{2} b \operatorname {Subst}\left (\int \sec ^{-1}(c+d x) \, dx,x,x^2\right )\\ &=\frac {a x^2}{2}+\frac {b \left (c+d x^2\right ) \sec ^{-1}\left (c+d x^2\right )}{2 d}-\frac {1}{2} b \operatorname {Subst}\left (\int \frac {1}{(c+d x) \sqrt {1-\frac {1}{(c+d x)^2}}} \, dx,x,x^2\right )\\ &=\frac {a x^2}{2}+\frac {b \left (c+d x^2\right ) \sec ^{-1}\left (c+d x^2\right )}{2 d}-\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {1}{x^2}} x} \, dx,x,c+d x^2\right )}{2 d}\\ &=\frac {a x^2}{2}+\frac {b \left (c+d x^2\right ) \sec ^{-1}\left (c+d x^2\right )}{2 d}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,\frac {1}{\left (c+d x^2\right )^2}\right )}{4 d}\\ &=\frac {a x^2}{2}+\frac {b \left (c+d x^2\right ) \sec ^{-1}\left (c+d x^2\right )}{2 d}-\frac {b \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-\frac {1}{\left (c+d x^2\right )^2}}\right )}{2 d}\\ &=\frac {a x^2}{2}+\frac {b \left (c+d x^2\right ) \sec ^{-1}\left (c+d x^2\right )}{2 d}-\frac {b \tanh ^{-1}\left (\sqrt {1-\frac {1}{\left (c+d x^2\right )^2}}\right )}{2 d}\\ \end {align*}

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Mathematica [B]  time = 0.24, size = 148, normalized size = 2.55 \[ \frac {a x^2}{2}-\frac {b \left (c+d x^2\right ) \sqrt {\frac {c^2+2 c d x^2+d^2 x^4-1}{\left (c+d x^2\right )^2}} \left (\tanh ^{-1}\left (\frac {c+d x^2}{\sqrt {c^2+2 c d x^2+d^2 x^4-1}}\right )-c \tan ^{-1}\left (\sqrt {\left (c+d x^2\right )^2-1}\right )\right )}{2 d \sqrt {c^2+2 c d x^2+d^2 x^4-1}}+\frac {1}{2} b x^2 \sec ^{-1}\left (c+d x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x^2)/2 + (b*x^2*ArcSec[c + d*x^2])/2 - (b*(c + d*x^2)*Sqrt[(-1 + c^2 + 2*c*d*x^2 + d^2*x^4)/(c + d*x^2)^2]*
(-(c*ArcTan[Sqrt[-1 + (c + d*x^2)^2]]) + ArcTanh[(c + d*x^2)/Sqrt[-1 + c^2 + 2*c*d*x^2 + d^2*x^4]]))/(2*d*Sqrt
[-1 + c^2 + 2*c*d*x^2 + d^2*x^4])

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fricas [A]  time = 0.60, size = 96, normalized size = 1.66 \[ \frac {b d x^{2} \operatorname {arcsec}\left (d x^{2} + c\right ) + a d x^{2} + 2 \, b c \arctan \left (-d x^{2} - c + \sqrt {d^{2} x^{4} + 2 \, c d x^{2} + c^{2} - 1}\right ) + b \log \left (-d x^{2} - c + \sqrt {d^{2} x^{4} + 2 \, c d x^{2} + c^{2} - 1}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*d*x^2*arcsec(d*x^2 + c) + a*d*x^2 + 2*b*c*arctan(-d*x^2 - c + sqrt(d^2*x^4 + 2*c*d*x^2 + c^2 - 1)) + b*
log(-d*x^2 - c + sqrt(d^2*x^4 + 2*c*d*x^2 + c^2 - 1)))/d

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giac [A]  time = 0.38, size = 100, normalized size = 1.72 \[ \frac {1}{2} \, a x^{2} + \frac {1}{4} \, b d {\left (\frac {2 \, {\left (d x^{2} + c\right )} \arccos \left (-\frac {1}{{\left (d x^{2} + c\right )} {\left (\frac {c}{d x^{2} + c} - 1\right )} - c}\right )}{d^{2}} - \frac {\log \left (\sqrt {-\frac {1}{{\left (d x^{2} + c\right )}^{2}} + 1} + 1\right ) - \log \left (-\sqrt {-\frac {1}{{\left (d x^{2} + c\right )}^{2}} + 1} + 1\right )}{d^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2 + 1/4*b*d*(2*(d*x^2 + c)*arccos(-1/((d*x^2 + c)*(c/(d*x^2 + c) - 1) - c))/d^2 - (log(sqrt(-1/(d*x^2
+ c)^2 + 1) + 1) - log(-sqrt(-1/(d*x^2 + c)^2 + 1) + 1))/d^2)

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maple [A]  time = 0.10, size = 81, normalized size = 1.40 \[ \frac {\mathrm {arcsec}\left (d \,x^{2}+c \right ) x^{2} b}{2}+\frac {a \,x^{2}}{2}+\frac {\mathrm {arcsec}\left (d \,x^{2}+c \right ) b c}{2 d}-\frac {\ln \left (d \,x^{2}+c +\left (d \,x^{2}+c \right ) \sqrt {1-\frac {1}{\left (d \,x^{2}+c \right )^{2}}}\right ) b}{2 d}+\frac {a c}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arcsec(d*x^2+c)*x^2*b+1/2*a*x^2+1/2/d*arcsec(d*x^2+c)*b*c-1/2/d*ln(d*x^2+c+(d*x^2+c)*(1-1/(d*x^2+c)^2)^(1/
2))*b+1/2/d*a*c

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maxima [A]  time = 0.35, size = 71, normalized size = 1.22 \[ \frac {1}{2} \, a x^{2} + \frac {{\left (2 \, {\left (d x^{2} + c\right )} \operatorname {arcsec}\left (d x^{2} + c\right ) - \log \left (\sqrt {-\frac {1}{{\left (d x^{2} + c\right )}^{2}} + 1} + 1\right ) + \log \left (-\sqrt {-\frac {1}{{\left (d x^{2} + c\right )}^{2}} + 1} + 1\right )\right )} b}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2 + 1/4*(2*(d*x^2 + c)*arcsec(d*x^2 + c) - log(sqrt(-1/(d*x^2 + c)^2 + 1) + 1) + log(-sqrt(-1/(d*x^2 +
c)^2 + 1) + 1))*b/d

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mupad [B]  time = 1.03, size = 52, normalized size = 0.90 \[ \frac {a\,x^2}{2}-\frac {b\,\mathrm {atanh}\left (\frac {1}{\sqrt {1-\frac {1}{{\left (d\,x^2+c\right )}^2}}}\right )}{2\,d}+\frac {b\,\mathrm {acos}\left (\frac {1}{d\,x^2+c}\right )\,\left (d\,x^2+c\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*acos(1/(c + d*x^2))),x)

[Out]

(a*x^2)/2 - (b*atanh(1/(1 - 1/(c + d*x^2)^2)^(1/2)))/(2*d) + (b*acos(1/(c + d*x^2))*(c + d*x^2))/(2*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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