Integrand size = 21, antiderivative size = 54 \[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2}{a d \sqrt {a+a \sec (c+d x)}} \] Output:
-2*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d+2/a/d/(a+a*sec(d*x+c) )^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.70 \[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\sec (c+d x)\right )}{a d \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[Tan[c + d*x]/(a + a*Sec[c + d*x])^(3/2),x]
Output:
(2*Hypergeometric2F1[-1/2, 1, 1/2, 1 + Sec[c + d*x]])/(a*d*Sqrt[a*(1 + Sec [c + d*x])])
Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 25, 4368, 61, 73, 220}
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.
\(\displaystyle \int \frac {\tan (c+d x)}{(a \sec (c+d x)+a)^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cot \left (c+d x+\frac {\pi }{2}\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4368 |
\(\displaystyle \frac {\int \frac {\cos (c+d x)}{(\sec (c+d x) a+a)^{3/2}}d\sec (c+d x)}{d}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {\frac {\int \frac {\cos (c+d x)}{\sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{a}+\frac {2}{a \sqrt {a \sec (c+d x)+a}}}{d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {2 \int \frac {1}{\frac {\sec (c+d x) a+a}{a}-1}d\sqrt {\sec (c+d x) a+a}}{a^2}+\frac {2}{a \sqrt {a \sec (c+d x)+a}}}{d}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {\frac {2}{a \sqrt {a \sec (c+d x)+a}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{a^{3/2}}}{d}\) |
Input:
Int[Tan[c + d*x]/(a + a*Sec[c + d*x])^(3/2),x]
Output:
((-2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/a^(3/2) + 2/(a*Sqrt[a + a* Sec[c + d*x]]))/d
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] - Simp[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] && 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]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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])
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(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]
Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83
method | result | size |
derivativedivides | \(\frac {\frac {2}{a \sqrt {a +a \sec \left (d x +c \right )}}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}}{d}\) | \(45\) |
default | \(\frac {\frac {2}{a \sqrt {a +a \sec \left (d x +c \right )}}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {a +a \sec \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}}{d}\) | \(45\) |
Input:
int(tan(d*x+c)/(a+a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/d*(2/a/(a+a*sec(d*x+c))^(1/2)-2/a^(3/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a ^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (46) = 92\).
Time = 0.13 (sec) , antiderivative size = 244, normalized size of antiderivative = 4.52 \[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {\sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, \frac {\sqrt {-a} {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) + 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a^{2} d \cos \left (d x + c\right ) + a^{2} d}\right ] \] Input:
integrate(tan(d*x+c)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")
Output:
[1/2*(sqrt(a)*(cos(d*x + c) + 1)*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) + 4*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c ))/(a^2*d*cos(d*x + c) + a^2*d), (sqrt(-a)*(cos(d*x + c) + 1)*arctan(2*sqr t(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + a)) + 2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c))/(a^2*d* cos(d*x + c) + a^2*d)]
\[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\tan {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(tan(d*x+c)/(a+a*sec(d*x+c))**(3/2),x)
Output:
Integral(tan(c + d*x)/(a*(sec(c + d*x) + 1))**(3/2), x)
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.30 \[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\frac {\log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a}}{d} \] Input:
integrate(tan(d*x+c)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")
Output:
(log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos(d*x + c)) + sqrt (a)))/a^(3/2) + 2/(sqrt(a + a/cos(d*x + c))*a))/d
Time = 0.63 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.61 \[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\frac {2 \, \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{d} \] Input:
integrate(tan(d*x+c)/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")
Output:
(2*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt( -a)*a*sgn(cos(d*x + c))) + sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/(a^ 2*sgn(cos(d*x + c))))/d
Time = 12.97 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.93 \[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {2}{a\,d\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{\sqrt {a}}\right )}{a^{3/2}\,d} \] Input:
int(tan(c + d*x)/(a + a/cos(c + d*x))^(3/2),x)
Output:
2/(a*d*(a + a/cos(c + d*x))^(1/2)) - (2*atanh((a + a/cos(c + d*x))^(1/2)/a ^(1/2)))/(a^(3/2)*d)
\[ \int \frac {\tan (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right )}{a^{2}} \] Input:
int(tan(d*x+c)/(a+a*sec(d*x+c))^(3/2),x)
Output:
(sqrt(a)*int((sqrt(sec(c + d*x) + 1)*tan(c + d*x))/(sec(c + d*x)**2 + 2*se c(c + d*x) + 1),x))/a**2