\(\int \frac {\sec ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx\) [50]

Optimal result

Integrand size = 19, antiderivative size = 69 \[ \int \frac {\sec ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {i \sec ^{-1}(a+b x)^2}{2 d}-\frac {\sec ^{-1}(a+b x) \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )}{d}+\frac {i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )}{2 d} \] Output:

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
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.86 \[ \int \frac {\sec ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {i \left (\sec ^{-1}(a+b x) \left (\sec ^{-1}(a+b x)+2 i \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )\right )+\operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )\right )}{2 d} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.49 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {5779, 27, 5741, 5137, 3042, 4202, 2620, 2715, 2838}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 5779

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5741

\(\displaystyle -\frac {\int (a+b x) \arccos \left (\frac {1}{a+b x}\right )d\frac {1}{a+b x}}{d}\)

\(\Big \downarrow \) 5137

\(\displaystyle \frac {\int (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \arccos \left (\frac {1}{a+b x}\right )d\arccos \left (\frac {1}{a+b x}\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \arccos \left (\frac {1}{a+b x}\right ) \tan \left (\arccos \left (\frac {1}{a+b x}\right )\right )d\arccos \left (\frac {1}{a+b x}\right )}{d}\)

\(\Big \downarrow \) 4202

\(\displaystyle \frac {\frac {i}{2 (a+b x)^2}-2 i \int \frac {e^{2 i \arccos \left (\frac {1}{a+b x}\right )} \arccos \left (\frac {1}{a+b x}\right )}{1+e^{2 i \arccos \left (\frac {1}{a+b x}\right )}}d\arccos \left (\frac {1}{a+b x}\right )}{d}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {\frac {i}{2 (a+b x)^2}-2 i \left (\frac {1}{2} i \int \log \left (1+e^{2 i \arccos \left (\frac {1}{a+b x}\right )}\right )d\arccos \left (\frac {1}{a+b x}\right )-\frac {1}{2} i \arccos \left (\frac {1}{a+b x}\right ) \log \left (1+e^{2 i \arccos \left (\frac {1}{a+b x}\right )}\right )\right )}{d}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {\frac {i}{2 (a+b x)^2}-2 i \left (\frac {1}{4} \int (a+b x) \log \left (1+e^{2 i \arccos \left (\frac {1}{a+b x}\right )}\right )de^{2 i \arccos \left (\frac {1}{a+b x}\right )}-\frac {1}{2} i \arccos \left (\frac {1}{a+b x}\right ) \log \left (1+e^{2 i \arccos \left (\frac {1}{a+b x}\right )}\right )\right )}{d}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\frac {i}{2 (a+b x)^2}-2 i \left (-\frac {1}{4} \operatorname {PolyLog}(2,-a-b x)-\frac {1}{2} i \arccos \left (\frac {1}{a+b x}\right ) \log \left (1+e^{2 i \arccos \left (\frac {1}{a+b x}\right )}\right )\right )}{d}\)

Input:

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

Output:

((I/2)/(a + b*x)^2 - (2*I)*((-1/2*I)*ArcCos[(a + b*x)^(-1)]*Log[1 + E^((2* 
I)*ArcCos[(a + b*x)^(-1)])] - PolyLog[2, -a - b*x]/4))/d
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2620

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[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 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[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 2838

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[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] && IGt 
Q[m, 0]
 

rule 5137

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

rule 5741

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

rule 5779

Int[((a_.) + ArcSec[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m 
_.), x_Symbol] :> Simp[1/d   Subst[Int[(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]
 

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.43

method	result	size
derivativedivides	\(\frac {\frac {i b \operatorname {arcsec}\left (b x +a \right )^{2}}{2 d}-\frac {b \,\operatorname {arcsec}\left (b x +a \right ) \ln \left (1+{\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}^{2}\right )}{d}+\frac {i b \operatorname {polylog}\left (2, -{\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}^{2}\right )}{2 d}}{b}\)	\(99\)
default	\(\frac {\frac {i b \operatorname {arcsec}\left (b x +a \right )^{2}}{2 d}-\frac {b \,\operatorname {arcsec}\left (b x +a \right ) \ln \left (1+{\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}^{2}\right )}{d}+\frac {i b \operatorname {polylog}\left (2, -{\left (\frac {1}{b x +a}+i \sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}^{2}\right )}{2 d}}{b}\)	\(99\)

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {\sec ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int { \frac {\operatorname {arcsec}\left (b x + a\right )}{d x + \frac {a d}{b}} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {\sec ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {b \int \frac {\operatorname {asec}{\left (a + b x \right )}}{a + b x}\, dx}{d} \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {\sec ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int { \frac {\operatorname {arcsec}\left (b x + a\right )}{d x + \frac {a d}{b}} \,d x } \] Input:

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

Output:

-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*i 
ntegrate(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
 

Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.67 \[ \int \frac {\sec ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=-\frac {1}{4} \, b^{2} {\left (\frac {2 \, {\left (b x + a\right )}^{2} \arccos \left (\frac {1}{{\left ({\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a\right )} {\left (\frac {a}{b x + a} - 1\right )} + a}\right )}{b^{3} d} - \frac {{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} - \frac {1}{{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}}}{b^{3} d}\right )} \] Input:

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

Output:

-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))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int \frac {\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}{d\,x+\frac {a\,d}{b}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\sec ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {\left (\int \frac {\mathit {asec} \left (b x +a \right )}{b x +a}d x \right ) b}{d} \] Input:

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

Output:

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