\(\int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [337]

Optimal result

Integrand size = 21, antiderivative size = 54 \[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2}{a d \sqrt {a+b \sec (c+d x)}} \]

[Out]

-2*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d+2/a/d/(a+b*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3970, 53, 65, 213} \[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2}{a d \sqrt {a+b \sec (c+d x)}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d} \]

[In]

Int[Tan[c + d*x]/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(-2*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) + 2/(a*d*Sqrt[a + b*Sec[c + d*x]])

Rule 53

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

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 3970

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x (a+x)^{3/2}} \, dx,x,b \sec (c+d x)\right )}{d} \\ & = \frac {2}{a d \sqrt {a+b \sec (c+d x)}}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+x}} \, dx,x,b \sec (c+d x)\right )}{a d} \\ & = \frac {2}{a d \sqrt {a+b \sec (c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {1}{-a+x^2} \, dx,x,\sqrt {a+b \sec (c+d x)}\right )}{a d} \\ & = -\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2}{a d \sqrt {a+b \sec (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.10 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.80 \[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \sec (c+d x)}{a}\right )}{a d \sqrt {a+b \sec (c+d x)}} \]

[In]

Integrate[Tan[c + d*x]/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(2*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*Sec[c + d*x])/a])/(a*d*Sqrt[a + b*Sec[c + d*x]])

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}+\frac {2}{a \sqrt {a +b \sec \left (d x +c \right )}}}{d}\)	\(45\)
default	\(\frac {-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}+\frac {2}{a \sqrt {a +b \sec \left (d x +c \right )}}}{d}\)	\(45\)

[In]

int(tan(d*x+c)/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/d*(-2/a^(3/2)*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))+2/a/(a+b*sec(d*x+c))^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (46) = 92\).

Time = 0.42 (sec) , antiderivative size = 260, normalized size of antiderivative = 4.81 \[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\left [\frac {4 \, a \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right ) + b\right )} \sqrt {a} \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} + 4 \, {\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right )}{2 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{2} b d\right )}}, \frac {{\left (a \cos \left (d x + c\right ) + b\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) + 2 \, a \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a^{3} d \cos \left (d x + c\right ) + a^{2} b d}\right ] \]

[In]

integrate(tan(d*x+c)/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/2*(4*a*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c) + (a*cos(d*x + c) + b)*sqrt(a)*log(-8*a^2*cos(d
*x + c)^2 - 8*a*b*cos(d*x + c) - b^2 + 4*(2*a*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) +
b)/cos(d*x + c))))/(a^3*d*cos(d*x + c) + a^2*b*d), ((a*cos(d*x + c) + b)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*co
s(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b)) + 2*a*sqrt((a*cos(d*x + c) + b)/cos(d*x + c
))*cos(d*x + c))/(a^3*d*cos(d*x + c) + a^2*b*d)]

Sympy [F]

\[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\tan {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(tan(d*x+c)/(a+b*sec(d*x+c))**(3/2),x)

[Out]

Integral(tan(c + d*x)/(a + b*sec(c + d*x))**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.30 \[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {\frac {\log \left (\frac {\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} a}}{d} \]

[In]

integrate(tan(d*x+c)/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

(log((sqrt(a + b/cos(d*x + c)) - sqrt(a))/(sqrt(a + b/cos(d*x + c)) + sqrt(a)))/a^(3/2) + 2/(sqrt(a + b/cos(d*
x + c))*a))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (46) = 92\).

Time = 0.78 (sec) , antiderivative size = 214, normalized size of antiderivative = 3.96 \[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \, {\left (\frac {\arctan \left (-\frac {\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b} + \sqrt {a - b}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {2 \, b}{{\left ({\left (\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}\right )} \sqrt {a - b} - a - b\right )} \sqrt {a - b} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )}}{d} \]

[In]

integrate(tan(d*x+c)/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

2*(arctan(-1/2*(sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4
- 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b) + sqrt(a - b))/sqrt(-a))/(sqrt(-a)*a*sgn(cos(d*x + c))) + 2*b/(((sqrt(a
- b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2
*c)^2 + a + b))*sqrt(a - b) - a - b)*sqrt(a - b)*a*sgn(cos(d*x + c))))/d

Mupad [B] (verification not implemented)

Time = 15.59 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.93 \[ \int \frac {\tan (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2}{a\,d\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}}-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}}{\sqrt {a}}\right )}{a^{3/2}\,d} \]

[In]

int(tan(c + d*x)/(a + b/cos(c + d*x))^(3/2),x)

[Out]

2/(a*d*(a + b/cos(c + d*x))^(1/2)) - (2*atanh((a + b/cos(c + d*x))^(1/2)/a^(1/2)))/(a^(3/2)*d)