Integrand size = 21, antiderivative size = 72 \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {a \tan (e+f x)}{f}+\frac {(3 a+b) \tan ^3(e+f x)}{3 f}+\frac {(3 a+2 b) \tan ^5(e+f x)}{5 f}+\frac {(a+b) \tan ^7(e+f x)}{7 f} \]
a*tan(f*x+e)/f+1/3*(3*a+b)*tan(f*x+e)^3/f+1/5*(3*a+2*b)*tan(f*x+e)^5/f+1/7 *(a+b)*tan(f*x+e)^7/f
Time = 0.49 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.19 \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {\tan (e+f x) \left (105 a-8 b-4 b \sec ^2(e+f x)-3 b \sec ^4(e+f x)+15 b \sec ^6(e+f x)+105 a \tan ^2(e+f x)+63 a \tan ^4(e+f x)+15 a \tan ^6(e+f x)\right )}{105 f} \]
(Tan[e + f*x]*(105*a - 8*b - 4*b*Sec[e + f*x]^2 - 3*b*Sec[e + f*x]^4 + 15* b*Sec[e + f*x]^6 + 105*a*Tan[e + f*x]^2 + 63*a*Tan[e + f*x]^4 + 15*a*Tan[e + f*x]^6))/(105*f)
Time = 0.25 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3670, 290, 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.
\(\displaystyle \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \sin (e+f x)^2}{\cos (e+f x)^8}dx\) |
\(\Big \downarrow \) 3670 |
\(\displaystyle \frac {\int \left (\tan ^2(e+f x)+1\right )^2 \left ((a+b) \tan ^2(e+f x)+a\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 290 |
\(\displaystyle \frac {\int \left ((a+b) \tan ^6(e+f x)+(3 a+2 b) \tan ^4(e+f x)+(3 a+b) \tan ^2(e+f x)+a\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{7} (a+b) \tan ^7(e+f x)+\frac {1}{5} (3 a+2 b) \tan ^5(e+f x)+\frac {1}{3} (3 a+b) \tan ^3(e+f x)+a \tan (e+f x)}{f}\) |
(a*Tan[e + f*x] + ((3*a + b)*Tan[e + f*x]^3)/3 + ((3*a + 2*b)*Tan[e + f*x] ^5)/5 + ((a + b)*Tan[e + f*x]^7)/7)/f
3.3.92.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I nt[ExpandIntegrand[(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] && IGtQ[q, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Su bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Time = 1.14 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.44
method | result | size |
derivativedivides | \(\frac {-a \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (f x +e \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (f x +e \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (f x +e \right )\right )}{35}\right ) \tan \left (f x +e \right )+b \left (\frac {\sin ^{3}\left (f x +e \right )}{7 \cos \left (f x +e \right )^{7}}+\frac {4 \left (\sin ^{3}\left (f x +e \right )\right )}{35 \cos \left (f x +e \right )^{5}}+\frac {8 \left (\sin ^{3}\left (f x +e \right )\right )}{105 \cos \left (f x +e \right )^{3}}\right )}{f}\) | \(104\) |
default | \(\frac {-a \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (f x +e \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (f x +e \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (f x +e \right )\right )}{35}\right ) \tan \left (f x +e \right )+b \left (\frac {\sin ^{3}\left (f x +e \right )}{7 \cos \left (f x +e \right )^{7}}+\frac {4 \left (\sin ^{3}\left (f x +e \right )\right )}{35 \cos \left (f x +e \right )^{5}}+\frac {8 \left (\sin ^{3}\left (f x +e \right )\right )}{105 \cos \left (f x +e \right )^{3}}\right )}{f}\) | \(104\) |
risch | \(-\frac {16 i \left (70 b \,{\mathrm e}^{8 i \left (f x +e \right )}-210 a \,{\mathrm e}^{6 i \left (f x +e \right )}-35 b \,{\mathrm e}^{6 i \left (f x +e \right )}-126 a \,{\mathrm e}^{4 i \left (f x +e \right )}+21 \,{\mathrm e}^{4 i \left (f x +e \right )} b -42 a \,{\mathrm e}^{2 i \left (f x +e \right )}+7 \,{\mathrm e}^{2 i \left (f x +e \right )} b -6 a +b \right )}{105 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{7}}\) | \(109\) |
parallelrisch | \(-\frac {2 \left (a \left (\tan ^{12}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-2 a +\frac {4 b}{3}\right ) \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {43 a}{5}+\frac {16 b}{15}\right ) \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-\frac {212 a}{35}+\frac {152 b}{35}\right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {43 a}{5}+\frac {16 b}{15}\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-2 a +\frac {4 b}{3}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}\) | \(140\) |
1/f*(-a*(-16/35-1/7*sec(f*x+e)^6-6/35*sec(f*x+e)^4-8/35*sec(f*x+e)^2)*tan( f*x+e)+b*(1/7*sin(f*x+e)^3/cos(f*x+e)^7+4/35*sin(f*x+e)^3/cos(f*x+e)^5+8/1 05*sin(f*x+e)^3/cos(f*x+e)^3))
Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.07 \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {{\left (8 \, {\left (6 \, a - b\right )} \cos \left (f x + e\right )^{6} + 4 \, {\left (6 \, a - b\right )} \cos \left (f x + e\right )^{4} + 3 \, {\left (6 \, a - b\right )} \cos \left (f x + e\right )^{2} + 15 \, a + 15 \, b\right )} \sin \left (f x + e\right )}{105 \, f \cos \left (f x + e\right )^{7}} \]
1/105*(8*(6*a - b)*cos(f*x + e)^6 + 4*(6*a - b)*cos(f*x + e)^4 + 3*(6*a - b)*cos(f*x + e)^2 + 15*a + 15*b)*sin(f*x + e)/(f*cos(f*x + e)^7)
Timed out. \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\text {Timed out} \]
Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.83 \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {15 \, {\left (a + b\right )} \tan \left (f x + e\right )^{7} + 21 \, {\left (3 \, a + 2 \, b\right )} \tan \left (f x + e\right )^{5} + 35 \, {\left (3 \, a + b\right )} \tan \left (f x + e\right )^{3} + 105 \, a \tan \left (f x + e\right )}{105 \, f} \]
1/105*(15*(a + b)*tan(f*x + e)^7 + 21*(3*a + 2*b)*tan(f*x + e)^5 + 35*(3*a + b)*tan(f*x + e)^3 + 105*a*tan(f*x + e))/f
Time = 0.37 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.12 \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {15 \, a \tan \left (f x + e\right )^{7} + 15 \, b \tan \left (f x + e\right )^{7} + 63 \, a \tan \left (f x + e\right )^{5} + 42 \, b \tan \left (f x + e\right )^{5} + 105 \, a \tan \left (f x + e\right )^{3} + 35 \, b \tan \left (f x + e\right )^{3} + 105 \, a \tan \left (f x + e\right )}{105 \, f} \]
1/105*(15*a*tan(f*x + e)^7 + 15*b*tan(f*x + e)^7 + 63*a*tan(f*x + e)^5 + 4 2*b*tan(f*x + e)^5 + 105*a*tan(f*x + e)^3 + 35*b*tan(f*x + e)^3 + 105*a*ta n(f*x + e))/f
Time = 13.65 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.82 \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {\left (\frac {a}{7}+\frac {b}{7}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^7+\left (\frac {3\,a}{5}+\frac {2\,b}{5}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (a+\frac {b}{3}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+a\,\mathrm {tan}\left (e+f\,x\right )}{f} \]