Integrand size = 23, antiderivative size = 108 \[ \int \frac {\sec ^8(c+d x)}{a+b \tan ^2(c+d x)} \, dx=-\frac {(a-b)^3 \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2} d}+\frac {\left (a^2-3 a b+3 b^2\right ) \tan (c+d x)}{b^3 d}-\frac {(a-3 b) \tan ^3(c+d x)}{3 b^2 d}+\frac {\tan ^5(c+d x)}{5 b d} \]
-(a-b)^3*arctan(b^(1/2)*tan(d*x+c)/a^(1/2))/b^(7/2)/d/a^(1/2)+(a^2-3*a*b+3 *b^2)*tan(d*x+c)/b^3/d-1/3*(a-3*b)*tan(d*x+c)^3/b^2/d+1/5*tan(d*x+c)^5/b/d
Time = 6.12 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^8(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {-\frac {15 (a-b)^3 \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a}}+\sqrt {b} \left (15 a^2-40 a b+33 b^2-(5 a-9 b) b \sec ^2(c+d x)+3 b^2 \sec ^4(c+d x)\right ) \tan (c+d x)}{15 b^{7/2} d} \]
((-15*(a - b)^3*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/Sqrt[a] + Sqrt[b]* (15*a^2 - 40*a*b + 33*b^2 - (5*a - 9*b)*b*Sec[c + d*x]^2 + 3*b^2*Sec[c + d *x]^4)*Tan[c + d*x])/(15*b^(7/2)*d)
Time = 0.31 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4158, 300, 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.
| \(\displaystyle \int \frac {\sec ^8(c+d x)}{a+b \tan ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^8}{a+b \tan (c+d x)^2}dx\) |
| \(\Big \downarrow \) 4158 |
| \(\displaystyle \frac {\int \frac {\left (\tan ^2(c+d x)+1\right )^3}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 300 |
| \(\displaystyle \frac {\int \left (\frac {\tan ^4(c+d x)}{b}-\frac {(a-3 b) \tan ^2(c+d x)}{b^2}+\frac {a^2-3 b a+3 b^2}{b^3}+\frac {-a^3+3 b a^2-3 b^2 a+b^3}{b^3 \left (b \tan ^2(c+d x)+a\right )}\right )d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\left (a^2-3 a b+3 b^2\right ) \tan (c+d x)}{b^3}-\frac {(a-b)^3 \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2}}-\frac {(a-3 b) \tan ^3(c+d x)}{3 b^2}+\frac {\tan ^5(c+d x)}{5 b}}{d}\) |
(-(((a - b)^3*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(Sqrt[a]*b^(7/2))) + ((a^2 - 3*a*b + 3*b^2)*Tan[c + d*x])/b^3 - ((a - 3*b)*Tan[c + d*x]^3)/(3* b^2) + Tan[c + d*x]^5/(5*b))/d
3.5.56.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/(c^(m - 1)*f) Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)^n)^ p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && I ntegerQ[m/2] && (IntegersQ[n, p] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Time = 40.24 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.14
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {b^{2} \tan \left (d x +c \right )^{5}}{5}-\frac {\tan \left (d x +c \right )^{3} a b}{3}+b^{2} \tan \left (d x +c \right )^{3}+a^{2} \tan \left (d x +c \right )-3 a b \tan \left (d x +c \right )+3 b^{2} \tan \left (d x +c \right )}{b^{3}}+\frac {\left (-a^{3}+3 a^{2} b -3 a \,b^{2}+b^{3}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}}{d}\) | \(123\) |
| default | \(\frac {\frac {\frac {b^{2} \tan \left (d x +c \right )^{5}}{5}-\frac {\tan \left (d x +c \right )^{3} a b}{3}+b^{2} \tan \left (d x +c \right )^{3}+a^{2} \tan \left (d x +c \right )-3 a b \tan \left (d x +c \right )+3 b^{2} \tan \left (d x +c \right )}{b^{3}}+\frac {\left (-a^{3}+3 a^{2} b -3 a \,b^{2}+b^{3}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}}{d}\) | \(123\) |
| risch | \(\frac {2 i \left (15 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-30 a b \,{\mathrm e}^{8 i \left (d x +c \right )}+15 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+60 \,{\mathrm e}^{6 i \left (d x +c \right )} a^{2}-150 a b \,{\mathrm e}^{6 i \left (d x +c \right )}+90 \,{\mathrm e}^{6 i \left (d x +c \right )} b^{2}+90 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-250 a b \,{\mathrm e}^{4 i \left (d x +c \right )}+240 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+60 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}-170 a b \,{\mathrm e}^{2 i \left (d x +c \right )}+150 \,{\mathrm e}^{2 i \left (d x +c \right )} b^{2}+15 a^{2}-40 a b +33 b^{2}\right )}{15 d \,b^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a^{3}}{2 \sqrt {-a b}\, d \,b^{3}}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a^{2}}{2 \sqrt {-a b}\, d \,b^{2}}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a}{2 \sqrt {-a b}\, d b}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {-2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{2 \sqrt {-a b}\, d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a^{3}}{2 \sqrt {-a b}\, d \,b^{3}}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a^{2}}{2 \sqrt {-a b}\, d \,b^{2}}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right ) a}{2 \sqrt {-a b}\, d b}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a b +\sqrt {-a b}\, a +\sqrt {-a b}\, b}{\left (a -b \right ) \sqrt {-a b}}\right )}{2 \sqrt {-a b}\, d}\) | \(699\) |
1/d*(1/b^3*(1/5*b^2*tan(d*x+c)^5-1/3*tan(d*x+c)^3*a*b+b^2*tan(d*x+c)^3+a^2 *tan(d*x+c)-3*a*b*tan(d*x+c)+3*b^2*tan(d*x+c))+(-a^3+3*a^2*b-3*a*b^2+b^3)/ b^3/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2)))
Time = 0.32 (sec) , antiderivative size = 425, normalized size of antiderivative = 3.94 \[ \int \frac {\sec ^8(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\left [\frac {15 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {-a b} \cos \left (d x + c\right )^{5} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cos \left (d x + c\right )^{3} - b \cos \left (d x + c\right )\right )} \sqrt {-a b} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 4 \, {\left ({\left (15 \, a^{3} b - 40 \, a^{2} b^{2} + 33 \, a b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, a b^{3} - {\left (5 \, a^{2} b^{2} - 9 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, a b^{4} d \cos \left (d x + c\right )^{5}}, \frac {15 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{5} + 2 \, {\left ({\left (15 \, a^{3} b - 40 \, a^{2} b^{2} + 33 \, a b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, a b^{3} - {\left (5 \, a^{2} b^{2} - 9 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{30 \, a b^{4} d \cos \left (d x + c\right )^{5}}\right ] \]
[1/60*(15*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(-a*b)*cos(d*x + c)^5*log((( a^2 + 6*a*b + b^2)*cos(d*x + c)^4 - 2*(3*a*b + b^2)*cos(d*x + c)^2 + 4*((a + b)*cos(d*x + c)^3 - b*cos(d*x + c))*sqrt(-a*b)*sin(d*x + c) + b^2)/((a^ 2 - 2*a*b + b^2)*cos(d*x + c)^4 + 2*(a*b - b^2)*cos(d*x + c)^2 + b^2)) + 4 *((15*a^3*b - 40*a^2*b^2 + 33*a*b^3)*cos(d*x + c)^4 + 3*a*b^3 - (5*a^2*b^2 - 9*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a*b^4*d*cos(d*x + c)^5), 1/30*( 15*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b)*arctan(1/2*((a + b)*cos(d*x + c)^2 - b)*sqrt(a*b)/(a*b*cos(d*x + c)*sin(d*x + c)))*cos(d*x + c)^5 + 2*( (15*a^3*b - 40*a^2*b^2 + 33*a*b^3)*cos(d*x + c)^4 + 3*a*b^3 - (5*a^2*b^2 - 9*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a*b^4*d*cos(d*x + c)^5)]
\[ \int \frac {\sec ^8(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\int \frac {\sec ^{8}{\left (c + d x \right )}}{a + b \tan ^{2}{\left (c + d x \right )}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^8(c+d x)}{a+b \tan ^2(c+d x)} \, dx=-\frac {\frac {15 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {3 \, b^{2} \tan \left (d x + c\right )^{5} - 5 \, {\left (a b - 3 \, b^{2}\right )} \tan \left (d x + c\right )^{3} + 15 \, {\left (a^{2} - 3 \, a b + 3 \, b^{2}\right )} \tan \left (d x + c\right )}{b^{3}}}{15 \, d} \]
-1/15*(15*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*arctan(b*tan(d*x + c)/sqrt(a*b)) /(sqrt(a*b)*b^3) - (3*b^2*tan(d*x + c)^5 - 5*(a*b - 3*b^2)*tan(d*x + c)^3 + 15*(a^2 - 3*a*b + 3*b^2)*tan(d*x + c))/b^3)/d
Time = 0.50 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^8(c+d x)}{a+b \tan ^2(c+d x)} \, dx=-\frac {\frac {15 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )}}{\sqrt {a b} b^{3}} - \frac {3 \, b^{4} \tan \left (d x + c\right )^{5} - 5 \, a b^{3} \tan \left (d x + c\right )^{3} + 15 \, b^{4} \tan \left (d x + c\right )^{3} + 15 \, a^{2} b^{2} \tan \left (d x + c\right ) - 45 \, a b^{3} \tan \left (d x + c\right ) + 45 \, b^{4} \tan \left (d x + c\right )}{b^{5}}}{15 \, d} \]
-1/15*(15*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(pi*floor((d*x + c)/pi + 1/2)*sg n(b) + arctan(b*tan(d*x + c)/sqrt(a*b)))/(sqrt(a*b)*b^3) - (3*b^4*tan(d*x + c)^5 - 5*a*b^3*tan(d*x + c)^3 + 15*b^4*tan(d*x + c)^3 + 15*a^2*b^2*tan(d *x + c) - 45*a*b^3*tan(d*x + c) + 45*b^4*tan(d*x + c))/b^5)/d
Time = 12.33 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.26 \[ \int \frac {\sec ^8(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {3}{b}+\frac {a\,\left (\frac {a}{b^2}-\frac {3}{b}\right )}{b}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,b\,d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {a}{3\,b^2}-\frac {1}{b}\right )}{d}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (c+d\,x\right )\,{\left (a-b\right )}^3}{\sqrt {a}\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}\right )\,{\left (a-b\right )}^3}{\sqrt {a}\,b^{7/2}\,d} \]