Integrand size = 23, antiderivative size = 127 \[ \int \frac {\sec ^8(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {(a-b)^2 (5 a+b) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{7/2} d}-\frac {(2 a-3 b) \tan (c+d x)}{b^3 d}+\frac {\tan ^3(c+d x)}{3 b^2 d}-\frac {(a-b)^3 \tan (c+d x)}{2 a b^3 d \left (a+b \tan ^2(c+d x)\right )} \]
1/2*(a-b)^2*(5*a+b)*arctan(b^(1/2)*tan(d*x+c)/a^(1/2))/a^(3/2)/b^(7/2)/d-( 2*a-3*b)*tan(d*x+c)/b^3/d+1/3*tan(d*x+c)^3/b^2/d-1/2*(a-b)^3*tan(d*x+c)/a/ b^3/d/(a+b*tan(d*x+c)^2)
Time = 6.08 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^8(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\frac {3 (a-b)^2 (5 a+b) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{3/2}}+\frac {3 \sqrt {b} (-a+b)^3 \sin (2 (c+d x))}{a (a+b+(a-b) \cos (2 (c+d x)))}+4 \sqrt {b} (-3 a+4 b) \tan (c+d x)+2 b^{3/2} \sec ^2(c+d x) \tan (c+d x)}{6 b^{7/2} d} \]
((3*(a - b)^2*(5*a + b)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/a^(3/2) + (3*Sqrt[b]*(-a + b)^3*Sin[2*(c + d*x)])/(a*(a + b + (a - b)*Cos[2*(c + d*x )])) + 4*Sqrt[b]*(-3*a + 4*b)*Tan[c + d*x] + 2*b^(3/2)*Sec[c + d*x]^2*Tan[ c + d*x])/(6*b^(7/2)*d)
Time = 0.33 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.94, 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.
| \(\displaystyle \int \frac {\sec ^8(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^8}{\left (a+b \tan (c+d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4158 |
| \(\displaystyle \frac {\int \frac {\left (\tan ^2(c+d x)+1\right )^3}{\left (b \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 300 |
| \(\displaystyle \frac {\int \left (\frac {\tan ^2(c+d x)}{b^2}+\frac {3 b \tan ^2(c+d x) (a-b)^2+(2 a+b) (a-b)^2}{b^3 \left (b \tan ^2(c+d x)+a\right )^2}-\frac {2 a-3 b}{b^3}\right )d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {(5 a+b) (a-b)^2 \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{7/2}}-\frac {(a-b)^3 \tan (c+d x)}{2 a b^3 \left (a+b \tan ^2(c+d x)\right )}-\frac {(2 a-3 b) \tan (c+d x)}{b^3}+\frac {\tan ^3(c+d x)}{3 b^2}}{d}\) |
(((a - b)^2*(5*a + b)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(2*a^(3/2)*b ^(7/2)) - ((2*a - 3*b)*Tan[c + d*x])/b^3 + Tan[c + d*x]^3/(3*b^2) - ((a - b)^3*Tan[c + d*x])/(2*a*b^3*(a + b*Tan[c + d*x]^2)))/d
3.5.68.3.1 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 = 87.74 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.08
| method | result | size |
| derivativedivides | \(\frac {-\frac {-\frac {b \tan \left (d x +c \right )^{3}}{3}+2 a \tan \left (d x +c \right )-3 b \tan \left (d x +c \right )}{b^{3}}+\frac {-\frac {\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \tan \left (d x +c \right )}{2 a \left (a +b \tan \left (d x +c \right )^{2}\right )}+\frac {\left (5 a^{3}-9 a^{2} b +3 a \,b^{2}+b^{3}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}}{b^{3}}}{d}\) | \(137\) |
| default | \(\frac {-\frac {-\frac {b \tan \left (d x +c \right )^{3}}{3}+2 a \tan \left (d x +c \right )-3 b \tan \left (d x +c \right )}{b^{3}}+\frac {-\frac {\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \tan \left (d x +c \right )}{2 a \left (a +b \tan \left (d x +c \right )^{2}\right )}+\frac {\left (5 a^{3}-9 a^{2} b +3 a \,b^{2}+b^{3}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}}{b^{3}}}{d}\) | \(137\) |
| risch | \(\frac {i \left (15 a^{3} {\mathrm e}^{8 i \left (d x +c \right )}-27 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+9 a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+60 a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-78 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+12 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+6 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+90 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-112 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-26 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+60 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-98 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+28 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+15 a^{3}-37 a^{2} b +25 a \,b^{2}-3 b^{3}\right )}{3 d \,b^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} a \left (-a \,{\mathrm e}^{4 i \left (d x +c \right )}+b \,{\mathrm e}^{4 i \left (d x +c \right )}-2 a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a +b \right )}-\frac {5 a^{2} \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 )}{4 \sqrt {-a b}\, d \,b^{3}}+\frac {9 a \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 )}{4 \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 )}{4 \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 )}{4 \sqrt {-a b}\, d a}+\frac {5 a^{2} \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 )}{4 \sqrt {-a b}\, d \,b^{3}}-\frac {9 a \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 )}{4 \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 )}{4 \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 )}{4 \sqrt {-a b}\, d a}\) | \(818\) |
1/d*(-1/b^3*(-1/3*b*tan(d*x+c)^3+2*a*tan(d*x+c)-3*b*tan(d*x+c))+1/b^3*(-1/ 2*(a^3-3*a^2*b+3*a*b^2-b^3)/a*tan(d*x+c)/(a+b*tan(d*x+c)^2)+1/2*(5*a^3-9*a ^2*b+3*a*b^2+b^3)/a/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (113) = 226\).
Time = 0.33 (sec) , antiderivative size = 597, normalized size of antiderivative = 4.70 \[ \int \frac {\sec ^8(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\left [-\frac {3 \, {\left ({\left (5 \, a^{4} - 14 \, a^{3} b + 12 \, a^{2} b^{2} - 2 \, a b^{3} - b^{4}\right )} \cos \left (d x + c\right )^{5} + {\left (5 \, a^{3} b - 9 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {-a b} \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 (2 \, a^{2} b^{3} - {\left (15 \, a^{4} b - 37 \, a^{3} b^{2} + 25 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (5 \, a^{3} b^{2} - 7 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{2} b^{5} d \cos \left (d x + c\right )^{3} + {\left (a^{3} b^{4} - a^{2} b^{5}\right )} d \cos \left (d x + c\right )^{5}\right )}}, -\frac {3 \, {\left ({\left (5 \, a^{4} - 14 \, a^{3} b + 12 \, a^{2} b^{2} - 2 \, a b^{3} - b^{4}\right )} \cos \left (d x + c\right )^{5} + {\left (5 \, a^{3} b - 9 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{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 ) - 2 \, {\left (2 \, a^{2} b^{3} - {\left (15 \, a^{4} b - 37 \, a^{3} b^{2} + 25 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (5 \, a^{3} b^{2} - 7 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} b^{5} d \cos \left (d x + c\right )^{3} + {\left (a^{3} b^{4} - a^{2} b^{5}\right )} d \cos \left (d x + c\right )^{5}\right )}}\right ] \]
[-1/24*(3*((5*a^4 - 14*a^3*b + 12*a^2*b^2 - 2*a*b^3 - b^4)*cos(d*x + c)^5 + (5*a^3*b - 9*a^2*b^2 + 3*a*b^3 + b^4)*cos(d*x + c)^3)*sqrt(-a*b)*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* (2*a^2*b^3 - (15*a^4*b - 37*a^3*b^2 + 25*a^2*b^3 - 3*a*b^4)*cos(d*x + c)^4 - 2*(5*a^3*b^2 - 7*a^2*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a^2*b^5*d*cos( d*x + c)^3 + (a^3*b^4 - a^2*b^5)*d*cos(d*x + c)^5), -1/12*(3*((5*a^4 - 14* a^3*b + 12*a^2*b^2 - 2*a*b^3 - b^4)*cos(d*x + c)^5 + (5*a^3*b - 9*a^2*b^2 + 3*a*b^3 + b^4)*cos(d*x + c)^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))) - 2*(2*a^2*b^3 - (15*a ^4*b - 37*a^3*b^2 + 25*a^2*b^3 - 3*a*b^4)*cos(d*x + c)^4 - 2*(5*a^3*b^2 - 7*a^2*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a^2*b^5*d*cos(d*x + c)^3 + (a^3* b^4 - a^2*b^5)*d*cos(d*x + c)^5)]
\[ \int \frac {\sec ^8(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\int \frac {\sec ^{8}{\left (c + d x \right )}}{\left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^8(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=-\frac {\frac {3 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \tan \left (d x + c\right )}{a b^{4} \tan \left (d x + c\right )^{2} + a^{2} b^{3}} - \frac {2 \, {\left (b \tan \left (d x + c\right )^{3} - 3 \, {\left (2 \, a - 3 \, b\right )} \tan \left (d x + c\right )\right )}}{b^{3}} - \frac {3 \, {\left (5 \, a^{3} - 9 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a b^{3}}}{6 \, d} \]
-1/6*(3*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*tan(d*x + c)/(a*b^4*tan(d*x + c)^2 + a^2*b^3) - 2*(b*tan(d*x + c)^3 - 3*(2*a - 3*b)*tan(d*x + c))/b^3 - 3*(5 *a^3 - 9*a^2*b + 3*a*b^2 + b^3)*arctan(b*tan(d*x + c)/sqrt(a*b))/(sqrt(a*b )*a*b^3))/d
Time = 0.65 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.42 \[ \int \frac {\sec ^8(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\frac {3 \, {\left (5 \, a^{3} - 9 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )}}{\sqrt {a b} a b^{3}} - \frac {3 \, {\left (a^{3} \tan \left (d x + c\right ) - 3 \, a^{2} b \tan \left (d x + c\right ) + 3 \, a b^{2} \tan \left (d x + c\right ) - b^{3} \tan \left (d x + c\right )\right )}}{{\left (b \tan \left (d x + c\right )^{2} + a\right )} a b^{3}} + \frac {2 \, {\left (b^{4} \tan \left (d x + c\right )^{3} - 6 \, a b^{3} \tan \left (d x + c\right ) + 9 \, b^{4} \tan \left (d x + c\right )\right )}}{b^{6}}}{6 \, d} \]
1/6*(3*(5*a^3 - 9*a^2*b + 3*a*b^2 + b^3)*(pi*floor((d*x + c)/pi + 1/2)*sgn (b) + arctan(b*tan(d*x + c)/sqrt(a*b)))/(sqrt(a*b)*a*b^3) - 3*(a^3*tan(d*x + c) - 3*a^2*b*tan(d*x + c) + 3*a*b^2*tan(d*x + c) - b^3*tan(d*x + c))/(( b*tan(d*x + c)^2 + a)*a*b^3) + 2*(b^4*tan(d*x + c)^3 - 6*a*b^3*tan(d*x + c ) + 9*b^4*tan(d*x + c))/b^6)/d
Time = 11.64 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.31 \[ \int \frac {\sec ^8(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,b^2\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {2\,a}{b^3}-\frac {3}{b^2}\right )}{d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,a\,d\,\left (b^4\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\,b^3\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (c+d\,x\right )\,{\left (a-b\right )}^2\,\left (5\,a+b\right )}{\sqrt {a}\,\left (5\,a^3-9\,a^2\,b+3\,a\,b^2+b^3\right )}\right )\,{\left (a-b\right )}^2\,\left (5\,a+b\right )}{2\,a^{3/2}\,b^{7/2}\,d} \]