Integrand size = 23, antiderivative size = 99 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 (a+a \sec (c+d x))^{3/2}}{3 a d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a^2 d} \]
-2/3*(a+a*sec(d*x+c))^(3/2)/a/d+2/5*(a+a*sec(d*x+c))^(5/2)/a^2/d+2*arctanh ((a+a*sec(d*x+c))^(1/2)/a^(1/2))*a^(1/2)/d-2*(a+a*sec(d*x+c))^(1/2)/d
Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.81 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=\frac {2 \sqrt {a (1+\sec (c+d x))} \left (15 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )+\sqrt {1+\sec (c+d x)} \left (-17+\sec (c+d x)+3 \sec ^2(c+d x)\right )\right )}{15 d \sqrt {1+\sec (c+d x)}} \]
(2*Sqrt[a*(1 + Sec[c + d*x])]*(15*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + Sqrt[1 + Sec[c + d*x]]*(-17 + Sec[c + d*x] + 3*Sec[c + d*x]^2)))/(15*d*Sqrt[1 + Sec[c + d*x]])
Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.96, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 25, 4368, 25, 27, 90, 60, 60, 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.
\(\displaystyle \int \tan ^3(c+d x) \sqrt {a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \sqrt {\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a}dx\) |
\(\Big \downarrow \) 4368 |
\(\displaystyle \frac {\int -a \cos (c+d x) (1-\sec (c+d x)) (\sec (c+d x) a+a)^{3/2}d\sec (c+d x)}{a^2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int a \cos (c+d x) (1-\sec (c+d x)) (\sec (c+d x) a+a)^{3/2}d\sec (c+d x)}{a^2 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \cos (c+d x) (1-\sec (c+d x)) (\sec (c+d x) a+a)^{3/2}d\sec (c+d x)}{a d}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle -\frac {\int \cos (c+d x) (\sec (c+d x) a+a)^{3/2}d\sec (c+d x)-\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 a}}{a d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {a \int \cos (c+d x) \sqrt {\sec (c+d x) a+a}d\sec (c+d x)-\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 a}+\frac {2}{3} (a \sec (c+d x)+a)^{3/2}}{a d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {a \left (a \int \frac {\cos (c+d x)}{\sqrt {\sec (c+d x) a+a}}d\sec (c+d x)+2 \sqrt {a \sec (c+d x)+a}\right )-\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 a}+\frac {2}{3} (a \sec (c+d x)+a)^{3/2}}{a d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {a \left (2 \int \frac {1}{\frac {\sec (c+d x) a+a}{a}-1}d\sqrt {\sec (c+d x) a+a}+2 \sqrt {a \sec (c+d x)+a}\right )-\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 a}+\frac {2}{3} (a \sec (c+d x)+a)^{3/2}}{a d}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {a \left (2 \sqrt {a \sec (c+d x)+a}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )\right )-\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 a}+\frac {2}{3} (a \sec (c+d x)+a)^{3/2}}{a d}\) |
-(((2*(a + a*Sec[c + d*x])^(3/2))/3 - (2*(a + a*Sec[c + d*x])^(5/2))/(5*a) + a*(-2*Sqrt[a]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]] + 2*Sqrt[a + a* Sec[c + d*x]]))/(a*d))
3.2.36.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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 = 4.74 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.82
method | result | size |
default | \(-\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (15 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+17-\sec \left (d x +c \right )-3 \sec \left (d x +c \right )^{2}\right )}{15 d}\) | \(81\) |
-2/15/d*(a*(1+sec(d*x+c)))^(1/2)*(15*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^( 1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+17-sec(d*x+c)-3*sec(d*x+c)^2)
Time = 0.33 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.62 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=\left [\frac {15 \, \sqrt {a} \cos \left (d x + c\right )^{2} \log \left (-8 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) - 4 \, {\left (17 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 3\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{30 \, d \cos \left (d x + c\right )^{2}}, -\frac {15 \, \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{2} + 2 \, {\left (17 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 3\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}}\right ] \]
[1/30*(15*sqrt(a)*cos(d*x + c)^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) - 4*(17*cos(d*x + c)^2 - cos(d*x + c) - 3)*sqrt((a*cos( d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^2), -1/15*(15*sqrt(-a)*arctan (2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(2*a*cos( d*x + c) + a))*cos(d*x + c)^2 + 2*(17*cos(d*x + c)^2 - cos(d*x + c) - 3)*s qrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^2)]
\[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \tan ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08 \[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=-\frac {15 \, \sqrt {a} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 30 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \frac {6 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{a^{2}} + \frac {10 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}}}{a}}{15 \, d} \]
-1/15*(15*sqrt(a)*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos (d*x + c)) + sqrt(a))) + 30*sqrt(a + a/cos(d*x + c)) - 6*(a + a/cos(d*x + c))^(5/2)/a^2 + 10*(a + a/cos(d*x + c))^(3/2)/a)/d
\[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \tan \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sqrt {a+a \sec (c+d x)} \tan ^3(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^3\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \]