\(\int \frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\) [173]

Optimal result

Integrand size = 21, antiderivative size = 31 \[ \int \frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3965, 65, 213} \[ \int \frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{\sqrt {a} d} \]

[In]

Int[Tan[c + d*x]/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(-2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d)

Rule 65

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 213

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

Rule 3965

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[-(d*b^(m - 1)
)^(-1), Subst[Int[(-a + b*x)^((m - 1)/2)*((a + b*x)^((m - 1)/2 + n)/x), x], x, Csc[c + d*x]], x] /; FreeQ[{a,
b, c, d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {2 \text {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{a d} \\ & = -\frac {2 \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {2 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right ) \sqrt {1+\sec (c+d x)}}{d \sqrt {a (1+\sec (c+d x))}} \]

[In]

Integrate[Tan[c + d*x]/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

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

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{d \sqrt {a}}\)	\(26\)
default	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{d \sqrt {a}}\)	\(26\)

[In]

int(tan(d*x+c)/(a+a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (25) = 50\).

Time = 0.32 (sec) , antiderivative size = 137, normalized size of antiderivative = 4.42 \[ \int \frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {\log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right )}{2 \, \sqrt {a} d}, \frac {\sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right )}{a d}\right ] \]

[In]

integrate(tan(d*x+c)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/2*log(-8*a*cos(d*x + c)^2 + 4*(2*cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c)) - 8*a*cos(d*x + c) - a)/(sqrt(a)*d), sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*c
os(d*x + c)/(2*a*cos(d*x + c) + a))/(a*d)]

Sympy [F]

\[ \int \frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\tan {\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

[In]

integrate(tan(d*x+c)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral(tan(c + d*x)/sqrt(a*(sec(c + d*x) + 1)), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{\sqrt {a} d} \]

[In]

integrate(tan(d*x+c)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos(d*x + c)) + sqrt(a)))/(sqrt(a)*d)

Giac [A] (verification not implemented)

none

Time = 0.78 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.55 \[ \int \frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {2 \, \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \]

[In]

integrate(tan(d*x+c)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

2*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*d*sgn(cos(d*x + c)))

Mupad [B] (verification not implemented)

Time = 15.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{\sqrt {a}}\right )}{\sqrt {a}\,d} \]

[In]

int(tan(c + d*x)/(a + a/cos(c + d*x))^(1/2),x)

[Out]

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