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

Optimal. Leaf size=84 \[ \frac {(d+e x)^2 \left (a+b \sec ^{-1}(c x)\right )}{2 e}-\frac {b d \tanh ^{-1}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{c}-\frac {b e x \sqrt {1-\frac {1}{c^2 x^2}}}{2 c}+\frac {b d^2 \csc ^{-1}(c x)}{2 e} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {5226, 1568, 1396, 1807, 844, 216, 266, 63, 208} \[ \frac {(d+e x)^2 \left (a+b \sec ^{-1}(c x)\right )}{2 e}-\frac {b d \tanh ^{-1}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{c}-\frac {b e x \sqrt {1-\frac {1}{c^2 x^2}}}{2 c}+\frac {b d^2 \csc ^{-1}(c x)}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + b*ArcSec[c*x]),x]

[Out]

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

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 208

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

Rule 216

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

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 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1396

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> -Subst[Int[((d + e/x^n)^q*(a +
c/x^(2*n))^p)/x^2, x], x, 1/x] /; FreeQ[{a, c, d, e, p, q}, x] && EqQ[n2, 2*n] && ILtQ[n, 0]

Rule 1568

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Int[x^(m + mn*q
)*(e + d/x^mn)^q*(a + c*x^n2)^p, x] /; FreeQ[{a, c, d, e, m, mn, p}, x] && EqQ[n2, -2*mn] && IntegerQ[q] && (P
osQ[n2] ||  !IntegerQ[p])

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 5226

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + b
*ArcSec[c*x]))/(e*(m + 1)), x] - Dist[b/(c*e*(m + 1)), Int[(d + e*x)^(m + 1)/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x],
x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 114, normalized size = 1.36 \[ a d x+\frac {1}{2} a e x^2-\frac {b d x \sqrt {1-\frac {1}{c^2 x^2}} \tanh ^{-1}\left (\frac {c x}{\sqrt {c^2 x^2-1}}\right )}{\sqrt {c^2 x^2-1}}-\frac {b e x \sqrt {\frac {c^2 x^2-1}{c^2 x^2}}}{2 c}+b d x \sec ^{-1}(c x)+\frac {1}{2} b e x^2 \sec ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 130, normalized size = 1.55 \[ \frac {a c^{2} e x^{2} + 2 \, a c^{2} d x + 2 \, b c d \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) - \sqrt {c^{2} x^{2} - 1} b e + {\left (b c^{2} e x^{2} + 2 \, b c^{2} d x - 2 \, b c^{2} d - b c^{2} e\right )} \operatorname {arcsec}\left (c x\right ) + 2 \, {\left (2 \, b c^{2} d + b c^{2} e\right )} \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right )}{2 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.68, size = 1555, normalized size = 18.51 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*c*d*arccos(1/(c*x))/(c^3 + 2*c^3*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + c^3*(1/(c^2*x^2) - 1)^2/(1/(c*x)
+ 1)^4) - 2*b*c*d*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1))/(c^3 + 2*c^3*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)
^2 + c^3*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4) + 2*b*c*d*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) - 1))/(c^3 +
2*c^3*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + c^3*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4) + 2*a*c*d/(c^3 + 2*c^3*(1/(
c^2*x^2) - 1)/(1/(c*x) + 1)^2 + c^3*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4) + b*arccos(1/(c*x))*e/(c^3 + 2*c^3*(1
/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + c^3*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4) - 4*b*c*d*(1/(c^2*x^2) - 1)*log(abs
(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1))/((c^3 + 2*c^3*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + c^3*(1/(c^2*x^2) - 1
)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^2) + 4*b*c*d*(1/(c^2*x^2) - 1)*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) - 1
))/((c^3 + 2*c^3*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + c^3*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^2)
- 2*b*c*d*(1/(c^2*x^2) - 1)^2*arccos(1/(c*x))/((c^3 + 2*c^3*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + c^3*(1/(c^2*x
^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^4) + a*e/(c^3 + 2*c^3*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + c^3*(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^3 + 2*c^3*(1/(c^2*x^2) - 1)/(1/(c
*x) + 1)^2 + c^3*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^2) - 2*b*c*d*(1/(c^2*x^2) - 1)^2*log(abs(s
qrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1))/((c^3 + 2*c^3*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + c^3*(1/(c^2*x^2) - 1)^
2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^4) + 2*b*c*d*(1/(c^2*x^2) - 1)^2*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) - 1
))/((c^3 + 2*c^3*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + c^3*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^4)
- 2*b*sqrt(-1/(c^2*x^2) + 1)*e/((c^3 + 2*c^3*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + c^3*(1/(c^2*x^2) - 1)^2/(1/(
c*x) + 1)^4)*(1/(c*x) + 1)) - 2*a*c*d*(1/(c^2*x^2) - 1)^2/((c^3 + 2*c^3*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + c^
3*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^4) - 2*a*(1/(c^2*x^2) - 1)*e/((c^3 + 2*c^3*(1/(c^2*x^2) -
1)/(1/(c*x) + 1)^2 + c^3*(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^3 + 2*c^3*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + c^3*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4)*(1/(c*x
) + 1)^4) + 2*b*(-1/(c^2*x^2) + 1)^(3/2)*e/((c^3 + 2*c^3*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + c^3*(1/(c^2*x^2)
- 1)^2/(1/(c*x) + 1)^4)*(1/(c*x) + 1)^3) + a*(1/(c^2*x^2) - 1)^2*e/((c^3 + 2*c^3*(1/(c^2*x^2) - 1)/(1/(c*x) +
1)^2 + c^3*(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 = 0.05, size = 141, normalized size = 1.68 \[ \frac {a \,x^{2} e}{2}+a d x +\frac {b \,\mathrm {arcsec}\left (c x \right ) x^{2} e}{2}+b \,\mathrm {arcsec}\left (c x \right ) x d -\frac {b \sqrt {c^{2} x^{2}-1}\, d \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{c^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x}-\frac {b x e}{2 c \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}+\frac {b e}{2 c^{3} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2*e+a*d*x+1/2*b*arcsec(c*x)*x^2*e+b*arcsec(c*x)*x*d-1/c^2*b/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*(c^2*x^2-1)^
(1/2)*d*ln(c*x+(c^2*x^2-1)^(1/2))-1/2/c*b/((c^2*x^2-1)/c^2/x^2)^(1/2)*x*e+1/2/c^3*b/((c^2*x^2-1)/c^2/x^2)^(1/2
)/x*e

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maxima [A]  time = 0.32, size = 93, normalized size = 1.11 \[ \frac {1}{2} \, a e x^{2} + \frac {1}{2} \, {\left (x^{2} \operatorname {arcsec}\left (c x\right ) - \frac {x \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c}\right )} b e + a d x + \frac {{\left (2 \, c x \operatorname {arcsec}\left (c x\right ) - \log \left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right ) + \log \left (-\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )\right )} b d}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.90, size = 77, normalized size = 0.92 \[ \frac {a\,x\,\left (2\,d+e\,x\right )}{2}-\frac {b\,d\,\mathrm {atanh}\left (\frac {1}{\sqrt {1-\frac {1}{c^2\,x^2}}}\right )}{c}+b\,d\,x\,\mathrm {acos}\left (\frac {1}{c\,x}\right )-\frac {b\,e\,x\,\left (\sqrt {1-\frac {1}{c^2\,x^2}}-c\,x\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{2\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 4.42, size = 104, normalized size = 1.24 \[ a d x + \frac {a e x^{2}}{2} + b d x \operatorname {asec}{\left (c x \right )} + \frac {b e x^{2} \operatorname {asec}{\left (c x \right )}}{2} - \frac {b d \left (\begin {cases} \operatorname {acosh}{\left (c x \right )} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\- i \operatorname {asin}{\left (c x \right )} & \text {otherwise} \end {cases}\right )}{c} - \frac {b e \left (\begin {cases} \frac {\sqrt {c^{2} x^{2} - 1}}{c} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {i \sqrt {- c^{2} x^{2} + 1}}{c} & \text {otherwise} \end {cases}\right )}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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