3.50 \(\int \frac {\sec ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx\)

Optimal. Leaf size=69 \[ \frac {i \text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )}{2 d}+\frac {i \sec ^{-1}(a+b x)^2}{2 d}-\frac {\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{d} \]

[Out]

1/2*I*arcsec(b*x+a)^2/d-arcsec(b*x+a)*ln(1+(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^2)/d+1/2*I*polylog(2,-(1/(b*x+a
)+I*(1-1/(b*x+a)^2)^(1/2))^2)/d

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Rubi [A]  time = 0.09, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {5256, 12, 5218, 4626, 3719, 2190, 2279, 2391} \[ \frac {i \text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )}{2 d}+\frac {i \sec ^{-1}(a+b x)^2}{2 d}-\frac {\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[a + b*x]/((a*d)/b + d*x),x]

[Out]

((I/2)*ArcSec[a + b*x]^2)/d - (ArcSec[a + b*x]*Log[1 + E^((2*I)*ArcSec[a + b*x])])/d + ((I/2)*PolyLog[2, -E^((
2*I)*ArcSec[a + b*x])])/d

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4626

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> -Subst[Int[(a + b*x)^n/Cot[x], x], x, ArcCos[c
*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 5218

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcCos[x/c])/x, x], x, 1/x] /; Fre
eQ[{a, b, c}, x]

Rule 5256

Int[((a_.) + ArcSec[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((f*x)/d)^m*(a + b*ArcSec[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0
] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\sec ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {b \sec ^{-1}(x)}{d x} \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\sec ^{-1}(x)}{x} \, dx,x,a+b x\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\cos ^{-1}(x)}{x} \, dx,x,\frac {1}{a+b x}\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int x \tan (x) \, dx,x,\cos ^{-1}\left (\frac {1}{a+b x}\right )\right )}{d}\\ &=\frac {i \cos ^{-1}\left (\frac {1}{a+b x}\right )^2}{2 d}-\frac {(2 i) \operatorname {Subst}\left (\int \frac {e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\cos ^{-1}\left (\frac {1}{a+b x}\right )\right )}{d}\\ &=\frac {i \cos ^{-1}\left (\frac {1}{a+b x}\right )^2}{2 d}-\frac {\cos ^{-1}\left (\frac {1}{a+b x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {1}{a+b x}\right )}\right )}{d}+\frac {\operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}\left (\frac {1}{a+b x}\right )\right )}{d}\\ &=\frac {i \cos ^{-1}\left (\frac {1}{a+b x}\right )^2}{2 d}-\frac {\cos ^{-1}\left (\frac {1}{a+b x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {1}{a+b x}\right )}\right )}{d}-\frac {i \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \cos ^{-1}\left (\frac {1}{a+b x}\right )}\right )}{2 d}\\ &=\frac {i \cos ^{-1}\left (\frac {1}{a+b x}\right )^2}{2 d}-\frac {\cos ^{-1}\left (\frac {1}{a+b x}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac {1}{a+b x}\right )}\right )}{d}+\frac {i \text {Li}_2\left (-e^{2 i \cos ^{-1}\left (\frac {1}{a+b x}\right )}\right )}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 59, normalized size = 0.86 \[ \frac {i \left (\text {Li}_2\left (-e^{2 i \sec ^{-1}(a+b x)}\right )+\sec ^{-1}(a+b x) \left (\sec ^{-1}(a+b x)+2 i \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*(ArcSec[a + b*x]*(ArcSec[a + b*x] + (2*I)*Log[1 + E^((2*I)*ArcSec[a + b*x])]) + PolyLog[2, -E^((2*I)*Ar
cSec[a + b*x])]))/d

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {arcsec}\left (b x + a\right )}{b d x + a d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b*arcsec(b*x + a)/(b*d*x + a*d), x)

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giac [A]  time = 0.59, size = 115, normalized size = 1.67 \[ -\frac {1}{4} \, b^{2} {\left (\frac {2 \, {\left (b x + a\right )}^{2} \arccos \left (\frac {1}{{\left ({\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a\right )} {\left (\frac {a}{b x + a} - 1\right )} + a}\right )}{b^{3} d} - \frac {{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} - \frac {1}{{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}}}{b^{3} d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.35, size = 92, normalized size = 1.33 \[ \frac {i \mathrm {arcsec}\left (b x +a \right )^{2}}{2 d}-\frac {\mathrm {arcsec}\left (b x +a \right ) \ln \left (1+\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )}{d}+\frac {i \polylog \left (2, -\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )^{2}\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(b*x+a)/(a*d/b+d*x),x)

[Out]

1/2*I*arcsec(b*x+a)^2/d-arcsec(b*x+a)*ln(1+(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^2)/d+1/2*I*polylog(2,-(1/(b*x+a
)+I*(1-1/(b*x+a)^2)^(1/2))^2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {-i \, {\left (b {\left (\frac {\log \left (b x + a + 1\right ) \log \left (b x + a\right ) - \log \left (b x + a\right )^{2} + \log \left (b x + a\right ) \log \left (b x + a - 1\right )}{b^{2} d} - \frac {\log \left (b x + a + 1\right ) \log \left (b x + a\right ) + {\rm Li}_2\left (-b x - a\right )}{b^{2} d} - \frac {\log \left (b x + a\right ) \log \left (-b x - a + 1\right ) + {\rm Li}_2\left (b x + a\right )}{b^{2} d}\right )} - {\left (\frac {\log \left (b x + a + 1\right )}{b d} - \frac {2 \, \log \left (b x + a\right )}{b d} + \frac {\log \left (b x + a - 1\right )}{b d}\right )} \log \left (b x + a\right )\right )} b d + 2 \, b \int \frac {\log \left (b x + a\right )}{\sqrt {b x + a + 1} \sqrt {b x + a - 1} b x + \sqrt {b x + a + 1} \sqrt {b x + a - 1} a}\,{d x} - 2 \, \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right ) \log \left (b x + a\right ) + i \, \log \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right ) \log \left (b x + a\right ) - i \, \log \left (b x + a + 1\right ) \log \left (b x + a\right ) - i \, \log \left (b x + a\right )^{2} - i \, \log \left (b x + a\right ) \log \left (-b x - a + 1\right ) - i \, {\rm Li}_2\left (b x + a\right ) - i \, {\rm Li}_2\left (-b x - a\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*d*integrate(sqrt(b*x + a + 1)*sqrt(b*x + a - 1)*log(b*x + a)/(b^3*d*x^3 + 3*a*b^2*d*x^2 + (3*a^2 - 1
)*b*d*x + (a^3 - a)*d), x) + 2*I*b*d*integrate(log(b*x + a)/(b^3*d*x^3 + 3*a*b^2*d*x^2 + (3*a^2 - 1)*b*d*x + (
a^3 - a)*d), x) - 2*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))*log(b*x + a) + I*log(b^2*x^2 + 2*a*b*x + a^2)*
log(b*x + a) - I*log(b*x + a + 1)*log(b*x + a) - I*log(b*x + a)^2 - I*log(b*x + a)*log(-b*x - a + 1) - I*dilog
(b*x + a) - I*dilog(-b*x - a))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}{d\,x+\frac {a\,d}{b}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {b \int \frac {\operatorname {asec}{\left (a + b x \right )}}{a + b x}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*Integral(asec(a + b*x)/(a + b*x), x)/d

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