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

Optimal. Leaf size=87 \[ -\frac {d \left (a+b \sec ^{-1}(c x)\right )}{x}+e x \left (a+b \sec ^{-1}(c x)\right )+\frac {b c d \sqrt {c^2 x^2-1}}{\sqrt {c^2 x^2}}-\frac {b e x \tanh ^{-1}\left (\frac {c x}{\sqrt {c^2 x^2-1}}\right )}{\sqrt {c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {14, 5238, 451, 217, 206} \[ -\frac {d \left (a+b \sec ^{-1}(c x)\right )}{x}+e x \left (a+b \sec ^{-1}(c x)\right )+\frac {b c d \sqrt {c^2 x^2-1}}{\sqrt {c^2 x^2}}-\frac {b e x \tanh ^{-1}\left (\frac {c x}{\sqrt {c^2 x^2-1}}\right )}{\sqrt {c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 217

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 451

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[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -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 ) \left (a+b \sec ^{-1}(c x)\right )}{x^2} \, dx &=-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{x}+e x \left (a+b \sec ^{-1}(c x)\right )-\frac {(b c x) \int \frac {-d+e x^2}{x^2 \sqrt {-1+c^2 x^2}} \, dx}{\sqrt {c^2 x^2}}\\ &=\frac {b c d \sqrt {-1+c^2 x^2}}{\sqrt {c^2 x^2}}-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{x}+e x \left (a+b \sec ^{-1}(c x)\right )-\frac {(b c e x) \int \frac {1}{\sqrt {-1+c^2 x^2}} \, dx}{\sqrt {c^2 x^2}}\\ &=\frac {b c d \sqrt {-1+c^2 x^2}}{\sqrt {c^2 x^2}}-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{x}+e x \left (a+b \sec ^{-1}(c x)\right )-\frac {(b c e x) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x^2} \, dx,x,\frac {x}{\sqrt {-1+c^2 x^2}}\right )}{\sqrt {c^2 x^2}}\\ &=\frac {b c d \sqrt {-1+c^2 x^2}}{\sqrt {c^2 x^2}}-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{x}+e x \left (a+b \sec ^{-1}(c x)\right )-\frac {b e x \tanh ^{-1}\left (\frac {c x}{\sqrt {-1+c^2 x^2}}\right )}{\sqrt {c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 104, normalized size = 1.20 \[ -\frac {a d}{x}+a e x+b c d \sqrt {\frac {c^2 x^2-1}{c^2 x^2}}-\frac {b e x \sqrt {1-\frac {1}{c^2 x^2}} \tanh ^{-1}\left (\frac {c x}{\sqrt {c^2 x^2-1}}\right )}{\sqrt {c^2 x^2-1}}-\frac {b d \sec ^{-1}(c x)}{x}+b e x \sec ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 123, normalized size = 1.41 \[ \frac {b c^{2} d x + a c e x^{2} + b e x \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + \sqrt {c^{2} x^{2} - 1} b c d - a c d - 2 \, {\left (b c d - b c e\right )} x \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + {\left (b c e x^{2} - b c d + {\left (b c d - b c e\right )} x\right )} \operatorname {arcsec}\left (c x\right )}{c x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.62, size = 1098, normalized size = 12.62 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*c^2*d*arccos(1/(c*x))/(c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4) + a*c^2*d/(c^2 - c^2*(1/(c^2*x^2) -
1)^2/(1/(c*x) + 1)^4) + 2*b*c^2*d*(1/(c^2*x^2) - 1)*arccos(1/(c*x))/((c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) +
1)^4)*(1/(c*x) + 1)^2) - 2*b*c^2*d*sqrt(-1/(c^2*x^2) + 1)/((c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4)*(1
/(c*x) + 1)) + 2*a*c^2*d*(1/(c^2*x^2) - 1)/((c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^2) +
b*c^2*d*(1/(c^2*x^2) - 1)^2*arccos(1/(c*x))/((c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^4)
- b*arccos(1/(c*x))*e/(c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4) + b*e*log(abs(sqrt(-1/(c^2*x^2) + 1) +
1/(c*x) + 1))/(c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4) - b*e*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) -
1))/(c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4) + 2*b*c^2*d*(-1/(c^2*x^2) + 1)^(3/2)/((c^2 - c^2*(1/(c^2*
x^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^3) + a*c^2*d*(1/(c^2*x^2) - 1)^2/((c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1
/(c*x) + 1)^4)*(1/(c*x) + 1)^4) - a*e/(c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4) + 2*b*(1/(c^2*x^2) - 1)*
arccos(1/(c*x))*e/((c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^2) + 2*a*(1/(c^2*x^2) - 1)*e/
((c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^2) - b*(1/(c^2*x^2) - 1)^2*arccos(1/(c*x))*e/((
c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^4) - b*(1/(c^2*x^2) - 1)^2*e*log(abs(sqrt(-1/(c^2
*x^2) + 1) + 1/(c*x) + 1))/((c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^4) + b*(1/(c^2*x^2)
- 1)^2*e*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) - 1))/((c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c
*x) + 1)^4) - a*(1/(c^2*x^2) - 1)^2*e/((c^2 - c^2*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^4))*c

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maple [A]  time = 0.06, size = 137, normalized size = 1.57 \[ a e x -\frac {a d}{x}+b \,\mathrm {arcsec}\left (c x \right ) e x -\frac {b \,\mathrm {arcsec}\left (c x \right ) d}{x}+\frac {c b d}{\sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}-\frac {b d}{c \,x^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}-\frac {b \sqrt {c^{2} x^{2}-1}\, e \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{c^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*e*x-a*d/x+b*arcsec(c*x)*e*x-b*arcsec(c*x)*d/x+c*b/((c^2*x^2-1)/c^2/x^2)^(1/2)*d-b/c/x^2/((c^2*x^2-1)/c^2/x^2
)^(1/2)*d-b/c^2*(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 = 0.32, size = 89, normalized size = 1.02 \[ {\left (c \sqrt {-\frac {1}{c^{2} x^{2}} + 1} - \frac {\operatorname {arcsec}\left (c x\right )}{x}\right )} b d + a e x + \frac {{\left (2 \, c x \operatorname {arcsec}\left (c x\right ) - \log \left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right ) + \log \left (-\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )\right )} b e}{2 \, c} - \frac {a d}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.79, size = 72, normalized size = 0.83 \[ a\,e\,x-\frac {d\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )-b\,c\,x\,\sqrt {1-\frac {1}{c^2\,x^2}}\right )}{x}-\frac {b\,e\,\mathrm {atanh}\left (\frac {1}{\sqrt {1-\frac {1}{c^2\,x^2}}}\right )}{c}+b\,e\,x\,\mathrm {acos}\left (\frac {1}{c\,x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 5.11, size = 73, normalized size = 0.84 \[ - \frac {a d}{x} + a e x + b c d \sqrt {1 - \frac {1}{c^{2} x^{2}}} - \frac {b d \operatorname {asec}{\left (c x \right )}}{x} + b e x \operatorname {asec}{\left (c x \right )} - \frac {b e \left (\begin {cases} \operatorname {acosh}{\left (c x \right )} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\- i \operatorname {asin}{\left (c x \right )} & \text {otherwise} \end {cases}\right )}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*d/x + a*e*x + b*c*d*sqrt(1 - 1/(c**2*x**2)) - b*d*asec(c*x)/x + b*e*x*asec(c*x) - b*e*Piecewise((acosh(c*x)
, Abs(c**2*x**2) > 1), (-I*asin(c*x), True))/c

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