Integrand size = 23, antiderivative size = 100 \[ \int \sqrt {a+b \sec (c+d x)} \tan ^3(c+d x) \, dx=\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 \sqrt {a+b \sec (c+d x)}}{d}-\frac {2 a (a+b \sec (c+d x))^{3/2}}{3 b^2 d}+\frac {2 (a+b \sec (c+d x))^{5/2}}{5 b^2 d} \]
-2/3*a*(a+b*sec(d*x+c))^(3/2)/b^2/d+2/5*(a+b*sec(d*x+c))^(5/2)/b^2/d+2*arc tanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))*a^(1/2)/d-2*(a+b*sec(d*x+c))^(1/2)/d
Time = 0.70 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int \sqrt {a+b \sec (c+d x)} \tan ^3(c+d x) \, dx=\frac {2 \left (15 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )+\frac {\sqrt {a+b \sec (c+d x)} \left (-2 a^2-15 b^2+a b \sec (c+d x)+3 b^2 \sec ^2(c+d x)\right )}{b^2}\right )}{15 d} \]
(2*(15*Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]] + (Sqrt[a + b*Sec [c + d*x]]*(-2*a^2 - 15*b^2 + a*b*Sec[c + d*x] + 3*b^2*Sec[c + d*x]^2))/b^ 2))/(15*d)
Time = 0.30 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.81, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 25, 4373, 517, 1584, 2009}
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.
| \(\displaystyle \int \tan ^3(c+d x) \sqrt {a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \sqrt {a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx\) |
| \(\Big \downarrow \) 4373 |
| \(\displaystyle -\frac {\int \frac {\cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (b^2-b^2 \sec ^2(c+d x)\right )}{b}d(b \sec (c+d x))}{b^2 d}\) |
| \(\Big \downarrow \) 517 |
| \(\displaystyle -\frac {2 \int \frac {b^2 \sec ^2(c+d x) \left (b^4 \sec ^4(c+d x)-2 a b^2 \sec ^2(c+d x)+a^2-b^2\right )}{a-b^2 \sec ^2(c+d x)}d\sqrt {a+b \sec (c+d x)}}{b^2 d}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle -\frac {2 \int \left (-b^4 \sec ^4(c+d x)+a b^2 \sec ^2(c+d x)+b^2-\frac {a b^2}{a-b^2 \sec ^2(c+d x)}\right )d\sqrt {a+b \sec (c+d x)}}{b^2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \left (-\sqrt {a} b^2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )+\frac {1}{3} a b^3 \sec ^3(c+d x)-\frac {1}{5} b^5 \sec ^5(c+d x)+b^3 \sec (c+d x)\right )}{b^2 d}\) |
(-2*(-(Sqrt[a]*b^2*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]]) + b^3*Sec[c + d*x] + (a*b^3*Sec[c + d*x]^3)/3 - (b^5*Sec[c + d*x]^5)/5))/(b^2*d)
3.4.19.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1)) Subst[Int[x^(2*n + 1)*(-c + x^ 2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-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]
Leaf count of result is larger than twice the leaf count of optimal. \(464\) vs. \(2(84)=168\).
Time = 10.67 (sec) , antiderivative size = 465, normalized size of antiderivative = 4.65
| method | result | size |
| default | \(\frac {\sqrt {a +b \sec \left (d x +c \right )}\, \left (15 \sqrt {a}\, \cos \left (d x +c \right ) \ln \left (4 \cos \left (d x +c \right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \sqrt {a}+4 a \cos \left (d x +c \right )+4 \sqrt {a}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+2 b \right ) b^{2}-4 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a^{2} \cos \left (d x +c \right )-30 \cos \left (d x +c \right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, b^{2}-4 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a^{2}+2 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a b -30 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, b^{2}+2 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a b \sec \left (d x +c \right )+6 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, b^{2} \sec \left (d x +c \right )+6 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, b^{2} \sec \left (d x +c \right )^{2}\right )}{15 d \,b^{2} \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}}\) | \(465\) |
1/15/d/b^2*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)+1)/((b+a*cos(d*x+c))*cos(d*x +c)/(cos(d*x+c)+1)^2)^(1/2)*(15*a^(1/2)*cos(d*x+c)*ln(4*cos(d*x+c)*((b+a*c os(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^( 1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*b^2-4*((b+a *cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^2*cos(d*x+c)-30*cos(d*x+ c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*b^2-4*((b+a*cos(d* x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^2+2*((b+a*cos(d*x+c))*cos(d*x+c )/(cos(d*x+c)+1)^2)^(1/2)*a*b-30*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+ 1)^2)^(1/2)*b^2+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a*b *sec(d*x+c)+6*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*b^2*sec (d*x+c)+6*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*b^2*sec(d*x +c)^2)
Time = 0.44 (sec) , antiderivative size = 311, normalized size of antiderivative = 3.11 \[ \int \sqrt {a+b \sec (c+d x)} \tan ^3(c+d x) \, dx=\left [\frac {15 \, \sqrt {a} b^{2} \cos \left (d x + c\right )^{2} \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} - 4 \, {\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right ) + 4 \, {\left (a b \cos \left (d x + c\right ) - {\left (2 \, a^{2} + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{30 \, b^{2} d \cos \left (d x + c\right )^{2}}, -\frac {15 \, \sqrt {-a} b^{2} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) \cos \left (d x + c\right )^{2} - 2 \, {\left (a b \cos \left (d x + c\right ) - {\left (2 \, a^{2} + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{15 \, b^{2} d \cos \left (d x + c\right )^{2}}\right ] \]
[1/30*(15*sqrt(a)*b^2*cos(d*x + c)^2*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))) + 4*(a*b*cos(d*x + c) - (2*a^2 + 15*b^2)* cos(d*x + c)^2 + 3*b^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/(b^2*d*co s(d*x + c)^2), -1/15*(15*sqrt(-a)*b^2*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b))*cos(d*x + c)^2 - 2*(a*b*cos(d*x + c) - (2*a^2 + 15*b^2)*cos(d*x + c)^2 + 3*b^2)*sqrt((a*c os(d*x + c) + b)/cos(d*x + c)))/(b^2*d*cos(d*x + c)^2)]
\[ \int \sqrt {a+b \sec (c+d x)} \tan ^3(c+d x) \, dx=\int \sqrt {a + b \sec {\left (c + d x \right )}} \tan ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.08 \[ \int \sqrt {a+b \sec (c+d x)} \tan ^3(c+d x) \, dx=-\frac {15 \, \sqrt {a} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 30 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \frac {6 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{b^{2}} + \frac {10 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a}{b^{2}}}{15 \, d} \]
-1/15*(15*sqrt(a)*log((sqrt(a + b/cos(d*x + c)) - sqrt(a))/(sqrt(a + b/cos (d*x + c)) + sqrt(a))) + 30*sqrt(a + b/cos(d*x + c)) - 6*(a + b/cos(d*x + c))^(5/2)/b^2 + 10*(a + b/cos(d*x + c))^(3/2)*a/b^2)/d
Leaf count of result is larger than twice the leaf count of optimal. 539 vs. \(2 (84) = 168\).
Time = 0.70 (sec) , antiderivative size = 539, normalized size of antiderivative = 5.39 \[ \int \sqrt {a+b \sec (c+d x)} \tan ^3(c+d x) \, dx=-\frac {2 \, {\left (\frac {15 \, a \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}} - \frac {2 \, {\left (15 \, {\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 )}^{4} a - 30 \, {\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 )}^{3} {\left (a + 2 \, b\right )} \sqrt {a - b} + 20 \, {\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 )}^{2} {\left (4 \, a b - 3 \, b^{2}\right )} - 15 \, a^{3} - 10 \, a^{2} b - 35 \, a b^{2} + 12 \, b^{3} + 10 \, {\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 )} {\left (3 \, a^{2} - a b + 6 \, b^{2}\right )} \sqrt {a - b}\right )}}{{\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} - \sqrt {a - b}\right )}^{5}}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{15 \, d} \]
-2/15*(15*a*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) - 2*(15*(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))^4*a - 30*(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*t an(1/2*d*x + 1/2*c)^2 + a + b))^3*(a + 2*b)*sqrt(a - b) + 20*(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))^2*(4*a*b - 3*b^2) - 15*a^3 - 10*a^2*b - 35*a*b^2 + 12*b^3 + 10*(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))*(3*a^2 - a*b + 6*b^2)*sqrt(a - 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))^5)*sgn(cos(d*x + c))/d
Timed out. \[ \int \sqrt {a+b \sec (c+d x)} \tan ^3(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^3\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \]