Integrand size = 28, antiderivative size = 125 \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {a^2 \tan (c+d x)}{d}+\frac {a b \tan ^2(c+d x)}{d}+\frac {\left (2 a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {a b \tan ^4(c+d x)}{d}+\frac {\left (a^2+2 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {a b \tan ^6(c+d x)}{3 d}+\frac {b^2 \tan ^7(c+d x)}{7 d} \]
a^2*tan(d*x+c)/d+a*b*tan(d*x+c)^2/d+1/3*(2*a^2+b^2)*tan(d*x+c)^3/d+a*b*tan (d*x+c)^4/d+1/5*(a^2+2*b^2)*tan(d*x+c)^5/d+1/3*a*b*tan(d*x+c)^6/d+1/7*b^2* tan(d*x+c)^7/d
Time = 0.72 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.83 \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {\tan (c+d x) \left (105 a^2+105 a b \tan (c+d x)+35 \left (2 a^2+b^2\right ) \tan ^2(c+d x)+105 a b \tan ^3(c+d x)+21 \left (a^2+2 b^2\right ) \tan ^4(c+d x)+35 a b \tan ^5(c+d x)+15 b^2 \tan ^6(c+d x)\right )}{105 d} \]
(Tan[c + d*x]*(105*a^2 + 105*a*b*Tan[c + d*x] + 35*(2*a^2 + b^2)*Tan[c + d *x]^2 + 105*a*b*Tan[c + d*x]^3 + 21*(a^2 + 2*b^2)*Tan[c + d*x]^4 + 35*a*b* Tan[c + d*x]^5 + 15*b^2*Tan[c + d*x]^6))/(105*d)
Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3567, 522, 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.
\(\displaystyle \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \cos (c+d x)+b \sin (c+d x))^2}{\cos (c+d x)^8}dx\) |
\(\Big \downarrow \) 3567 |
\(\displaystyle -\frac {\int (b+a \cot (c+d x))^2 \left (\cot ^2(c+d x)+1\right )^2 \tan ^8(c+d x)d\cot (c+d x)}{d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle -\frac {\int \left (b^2 \tan ^8(c+d x)+2 a b \tan ^7(c+d x)+\left (a^2+2 b^2\right ) \tan ^6(c+d x)+4 a b \tan ^5(c+d x)+\left (2 a^2+b^2\right ) \tan ^4(c+d x)+2 a b \tan ^3(c+d x)+a^2 \tan ^2(c+d x)\right )d\cot (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{5} \left (a^2+2 b^2\right ) \tan ^5(c+d x)-\frac {1}{3} \left (2 a^2+b^2\right ) \tan ^3(c+d x)-a^2 \tan (c+d x)-\frac {1}{3} a b \tan ^6(c+d x)-a b \tan ^4(c+d x)-a b \tan ^2(c+d x)-\frac {1}{7} b^2 \tan ^7(c+d x)}{d}\) |
-((-(a^2*Tan[c + d*x]) - a*b*Tan[c + d*x]^2 - ((2*a^2 + b^2)*Tan[c + d*x]^ 3)/3 - a*b*Tan[c + d*x]^4 - ((a^2 + 2*b^2)*Tan[c + d*x]^5)/5 - (a*b*Tan[c + d*x]^6)/3 - (b^2*Tan[c + d*x]^7)/7)/d)
3.1.56.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*si n[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[x^m*((b + a*x)^n/(1 + x^2)^((m + n + 2)/2)), x], x, Cot[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] && !(GtQ[ n, 0] && GtQ[m, 1])
Time = 1.24 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {-a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+\frac {a b}{3 \cos \left (d x +c \right )^{6}}+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \sin \left (d x +c \right )^{3}}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \sin \left (d x +c \right )^{3}}{105 \cos \left (d x +c \right )^{3}}\right )}{d}\) | \(110\) |
default | \(\frac {-a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+\frac {a b}{3 \cos \left (d x +c \right )^{6}}+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \sin \left (d x +c \right )^{3}}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \sin \left (d x +c \right )^{3}}{105 \cos \left (d x +c \right )^{3}}\right )}{d}\) | \(110\) |
parts | \(-\frac {a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \sin \left (d x +c \right )^{3}}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \sin \left (d x +c \right )^{3}}{105 \cos \left (d x +c \right )^{3}}\right )}{d}+\frac {a b \sec \left (d x +c \right )^{6}}{3 d}\) | \(115\) |
risch | \(\frac {16 i \left (-140 i a b \,{\mathrm e}^{8 i \left (d x +c \right )}+70 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-70 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-140 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}+175 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+35 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+147 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-21 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+49 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-7 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+7 a^{2}-b^{2}\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}\) | \(171\) |
parallelrisch | \(-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (105 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}-210 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11} a b -350 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10} a^{2}+140 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10} b^{2}+210 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} a b +791 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} a^{2}+112 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} b^{2}-700 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a b -1092 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a^{2}+456 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} b^{2}+700 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a b +791 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{2}+112 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b^{2}-210 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a b -350 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2}+140 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{2}+210 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a b +105 a^{2}\right )}{105 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{7}}\) | \(300\) |
1/d*(-a^2*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+1/3*a*b/co s(d*x+c)^6+b^2*(1/7*sin(d*x+c)^3/cos(d*x+c)^7+4/35*sin(d*x+c)^3/cos(d*x+c) ^5+8/105*sin(d*x+c)^3/cos(d*x+c)^3))
Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.80 \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {35 \, a b \cos \left (d x + c\right ) + {\left (8 \, {\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \]
1/105*(35*a*b*cos(d*x + c) + (8*(7*a^2 - b^2)*cos(d*x + c)^6 + 4*(7*a^2 - b^2)*cos(d*x + c)^4 + 3*(7*a^2 - b^2)*cos(d*x + c)^2 + 15*b^2)*sin(d*x + c ))/(d*cos(d*x + c)^7)
Timed out. \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\text {Timed out} \]
Time = 0.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.73 \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {7 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{2} + {\left (15 \, \tan \left (d x + c\right )^{7} + 42 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3}\right )} b^{2} - \frac {35 \, a b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{105 \, d} \]
1/105*(7*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^2 + (1 5*tan(d*x + c)^7 + 42*tan(d*x + c)^5 + 35*tan(d*x + c)^3)*b^2 - 35*a*b/(si n(d*x + c)^2 - 1)^3)/d
Time = 0.34 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.94 \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {15 \, b^{2} \tan \left (d x + c\right )^{7} + 35 \, a b \tan \left (d x + c\right )^{6} + 21 \, a^{2} \tan \left (d x + c\right )^{5} + 42 \, b^{2} \tan \left (d x + c\right )^{5} + 105 \, a b \tan \left (d x + c\right )^{4} + 70 \, a^{2} \tan \left (d x + c\right )^{3} + 35 \, b^{2} \tan \left (d x + c\right )^{3} + 105 \, a b \tan \left (d x + c\right )^{2} + 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \]
1/105*(15*b^2*tan(d*x + c)^7 + 35*a*b*tan(d*x + c)^6 + 21*a^2*tan(d*x + c) ^5 + 42*b^2*tan(d*x + c)^5 + 105*a*b*tan(d*x + c)^4 + 70*a^2*tan(d*x + c)^ 3 + 35*b^2*tan(d*x + c)^3 + 105*a*b*tan(d*x + c)^2 + 105*a^2*tan(d*x + c)) /d
Time = 22.70 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.04 \[ \int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {\frac {b^2\,\sin \left (c+d\,x\right )}{7}+{\cos \left (c+d\,x\right )}^2\,\left (\frac {a^2\,\sin \left (c+d\,x\right )}{5}-\frac {b^2\,\sin \left (c+d\,x\right )}{35}\right )+{\cos \left (c+d\,x\right )}^4\,\left (\frac {4\,a^2\,\sin \left (c+d\,x\right )}{15}-\frac {4\,b^2\,\sin \left (c+d\,x\right )}{105}\right )+{\cos \left (c+d\,x\right )}^6\,\left (\frac {8\,a^2\,\sin \left (c+d\,x\right )}{15}-\frac {8\,b^2\,\sin \left (c+d\,x\right )}{105}\right )+\frac {a\,b\,\cos \left (c+d\,x\right )}{3}}{d\,{\cos \left (c+d\,x\right )}^7} \]