Integrand size = 23, antiderivative size = 106 \[ \int \sec ^{10}(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {a^2 \tan (e+f x)}{f}+\frac {2 a (2 a+b) \tan ^3(e+f x)}{3 f}+\frac {\left (6 a^2+6 a b+b^2\right ) \tan ^5(e+f x)}{5 f}+\frac {2 (a+b) (2 a+b) \tan ^7(e+f x)}{7 f}+\frac {(a+b)^2 \tan ^9(e+f x)}{9 f} \]
a^2*tan(f*x+e)/f+2/3*a*(2*a+b)*tan(f*x+e)^3/f+1/5*(6*a^2+6*a*b+b^2)*tan(f* x+e)^5/f+2/7*(a+b)*(2*a+b)*tan(f*x+e)^7/f+1/9*(a+b)^2*tan(f*x+e)^9/f
Time = 6.06 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.01 \[ \int \sec ^{10}(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {\sec ^9(e+f x) \left (252 \left (8 a^2+8 a b+3 b^2\right ) \sin (e+f x)+336 \left (4 a^2-a b-b^2\right ) \sin (3 (e+f x))+\left (16 a^2-4 a b+b^2\right ) (36 \sin (5 (e+f x))+9 \sin (7 (e+f x))+\sin (9 (e+f x)))\right )}{10080 f} \]
(Sec[e + f*x]^9*(252*(8*a^2 + 8*a*b + 3*b^2)*Sin[e + f*x] + 336*(4*a^2 - a *b - b^2)*Sin[3*(e + f*x)] + (16*a^2 - 4*a*b + b^2)*(36*Sin[5*(e + f*x)] + 9*Sin[7*(e + f*x)] + Sin[9*(e + f*x)])))/(10080*f)
Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, 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.
| \(\displaystyle \int \sec ^{10}(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin (e+f x)^2\right )^2}{\cos (e+f x)^{10}}dx\) |
| \(\Big \downarrow \) 3670 |
| \(\displaystyle \frac {\int \left (\tan ^2(e+f x)+1\right )^2 \left ((a+b) \tan ^2(e+f x)+a\right )^2d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 290 |
| \(\displaystyle \frac {\int \left ((a+b)^2 \tan ^8(e+f x)+2 (a+b) (2 a+b) \tan ^6(e+f x)+\left (6 a^2+6 b a+b^2\right ) \tan ^4(e+f x)+2 a (2 a+b) \tan ^2(e+f x)+a^2\right )d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{5} \left (6 a^2+6 a b+b^2\right ) \tan ^5(e+f x)+a^2 \tan (e+f x)+\frac {1}{9} (a+b)^2 \tan ^9(e+f x)+\frac {2}{7} (a+b) (2 a+b) \tan ^7(e+f x)+\frac {2}{3} a (2 a+b) \tan ^3(e+f x)}{f}\) |
(a^2*Tan[e + f*x] + (2*a*(2*a + b)*Tan[e + f*x]^3)/3 + ((6*a^2 + 6*a*b + b ^2)*Tan[e + f*x]^5)/5 + (2*(a + b)*(2*a + b)*Tan[e + f*x]^7)/7 + ((a + b)^ 2*Tan[e + f*x]^9)/9)/f
3.3.100.3.1 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 = 1.69 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.84
| method | result | size |
| derivativedivides | \(\frac {-a^{2} \left (-\frac {128}{315}-\frac {\left (\sec ^{8}\left (f x +e \right )\right )}{9}-\frac {8 \left (\sec ^{6}\left (f x +e \right )\right )}{63}-\frac {16 \left (\sec ^{4}\left (f x +e \right )\right )}{105}-\frac {64 \left (\sec ^{2}\left (f x +e \right )\right )}{315}\right ) \tan \left (f x +e \right )+2 a b \left (\frac {\sin ^{3}\left (f x +e \right )}{9 \cos \left (f x +e \right )^{9}}+\frac {2 \left (\sin ^{3}\left (f x +e \right )\right )}{21 \cos \left (f x +e \right )^{7}}+\frac {8 \left (\sin ^{3}\left (f x +e \right )\right )}{105 \cos \left (f x +e \right )^{5}}+\frac {16 \left (\sin ^{3}\left (f x +e \right )\right )}{315 \cos \left (f x +e \right )^{3}}\right )+b^{2} \left (\frac {\sin ^{5}\left (f x +e \right )}{9 \cos \left (f x +e \right )^{9}}+\frac {4 \left (\sin ^{5}\left (f x +e \right )\right )}{63 \cos \left (f x +e \right )^{7}}+\frac {8 \left (\sin ^{5}\left (f x +e \right )\right )}{315 \cos \left (f x +e \right )^{5}}\right )}{f}\) | \(195\) |
| default | \(\frac {-a^{2} \left (-\frac {128}{315}-\frac {\left (\sec ^{8}\left (f x +e \right )\right )}{9}-\frac {8 \left (\sec ^{6}\left (f x +e \right )\right )}{63}-\frac {16 \left (\sec ^{4}\left (f x +e \right )\right )}{105}-\frac {64 \left (\sec ^{2}\left (f x +e \right )\right )}{315}\right ) \tan \left (f x +e \right )+2 a b \left (\frac {\sin ^{3}\left (f x +e \right )}{9 \cos \left (f x +e \right )^{9}}+\frac {2 \left (\sin ^{3}\left (f x +e \right )\right )}{21 \cos \left (f x +e \right )^{7}}+\frac {8 \left (\sin ^{3}\left (f x +e \right )\right )}{105 \cos \left (f x +e \right )^{5}}+\frac {16 \left (\sin ^{3}\left (f x +e \right )\right )}{315 \cos \left (f x +e \right )^{3}}\right )+b^{2} \left (\frac {\sin ^{5}\left (f x +e \right )}{9 \cos \left (f x +e \right )^{9}}+\frac {4 \left (\sin ^{5}\left (f x +e \right )\right )}{63 \cos \left (f x +e \right )^{7}}+\frac {8 \left (\sin ^{5}\left (f x +e \right )\right )}{315 \cos \left (f x +e \right )^{5}}\right )}{f}\) | \(195\) |
| parallelrisch | \(-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (a^{2} \left (\tan ^{16}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {8 a \left (a -b \right ) \left (\tan ^{14}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+\frac {4 \left (19 a^{2}+4 a b +4 b^{2}\right ) \left (\tan ^{12}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}+\frac {8 \left (-71 a^{2}+79 a b +24 b^{2}\right ) \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{35}+\frac {2 \left (5329 a^{2}+1424 a b +1744 b^{2}\right ) \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{315}+\frac {8 \left (-71 a^{2}+79 a b +24 b^{2}\right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{35}+\frac {4 \left (19 a^{2}+4 a b +4 b^{2}\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5}-\frac {8 a \left (a -b \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+a^{2}\right )}{f \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{9}}\) | \(227\) |
| risch | \(\frac {16 i \left (210 b^{2} {\mathrm e}^{12 i \left (f x +e \right )}-1260 a b \,{\mathrm e}^{10 i \left (f x +e \right )}-315 b^{2} {\mathrm e}^{10 i \left (f x +e \right )}+2016 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}+756 a b \,{\mathrm e}^{8 i \left (f x +e \right )}+441 b^{2} {\mathrm e}^{8 i \left (f x +e \right )}+1344 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}-336 a b \,{\mathrm e}^{6 i \left (f x +e \right )}-126 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+576 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-144 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+36 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+144 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-36 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+9 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+16 a^{2}-4 a b +b^{2}\right )}{315 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{9}}\) | \(238\) |
1/f*(-a^2*(-128/315-1/9*sec(f*x+e)^8-8/63*sec(f*x+e)^6-16/105*sec(f*x+e)^4 -64/315*sec(f*x+e)^2)*tan(f*x+e)+2*a*b*(1/9*sin(f*x+e)^3/cos(f*x+e)^9+2/21 *sin(f*x+e)^3/cos(f*x+e)^7+8/105*sin(f*x+e)^3/cos(f*x+e)^5+16/315*sin(f*x+ e)^3/cos(f*x+e)^3)+b^2*(1/9*sin(f*x+e)^5/cos(f*x+e)^9+4/63*sin(f*x+e)^5/co s(f*x+e)^7+8/315*sin(f*x+e)^5/cos(f*x+e)^5))
Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.21 \[ \int \sec ^{10}(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {{\left (8 \, {\left (16 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{8} + 4 \, {\left (16 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{6} + 3 \, {\left (16 \, a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 10 \, {\left (4 \, a^{2} - a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 35 \, a^{2} + 70 \, a b + 35 \, b^{2}\right )} \sin \left (f x + e\right )}{315 \, f \cos \left (f x + e\right )^{9}} \]
1/315*(8*(16*a^2 - 4*a*b + b^2)*cos(f*x + e)^8 + 4*(16*a^2 - 4*a*b + b^2)* cos(f*x + e)^6 + 3*(16*a^2 - 4*a*b + b^2)*cos(f*x + e)^4 + 10*(4*a^2 - a*b - 5*b^2)*cos(f*x + e)^2 + 35*a^2 + 70*a*b + 35*b^2)*sin(f*x + e)/(f*cos(f *x + e)^9)
Timed out. \[ \int \sec ^{10}(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\text {Timed out} \]
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.97 \[ \int \sec ^{10}(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {35 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{9} + 90 \, {\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{7} + 63 \, {\left (6 \, a^{2} + 6 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + 210 \, {\left (2 \, a^{2} + a b\right )} \tan \left (f x + e\right )^{3} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, f} \]
1/315*(35*(a^2 + 2*a*b + b^2)*tan(f*x + e)^9 + 90*(2*a^2 + 3*a*b + b^2)*ta n(f*x + e)^7 + 63*(6*a^2 + 6*a*b + b^2)*tan(f*x + e)^5 + 210*(2*a^2 + a*b) *tan(f*x + e)^3 + 315*a^2*tan(f*x + e))/f
Time = 0.37 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.47 \[ \int \sec ^{10}(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {35 \, a^{2} \tan \left (f x + e\right )^{9} + 70 \, a b \tan \left (f x + e\right )^{9} + 35 \, b^{2} \tan \left (f x + e\right )^{9} + 180 \, a^{2} \tan \left (f x + e\right )^{7} + 270 \, a b \tan \left (f x + e\right )^{7} + 90 \, b^{2} \tan \left (f x + e\right )^{7} + 378 \, a^{2} \tan \left (f x + e\right )^{5} + 378 \, a b \tan \left (f x + e\right )^{5} + 63 \, b^{2} \tan \left (f x + e\right )^{5} + 420 \, a^{2} \tan \left (f x + e\right )^{3} + 210 \, a b \tan \left (f x + e\right )^{3} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, f} \]
1/315*(35*a^2*tan(f*x + e)^9 + 70*a*b*tan(f*x + e)^9 + 35*b^2*tan(f*x + e) ^9 + 180*a^2*tan(f*x + e)^7 + 270*a*b*tan(f*x + e)^7 + 90*b^2*tan(f*x + e) ^7 + 378*a^2*tan(f*x + e)^5 + 378*a*b*tan(f*x + e)^5 + 63*b^2*tan(f*x + e) ^5 + 420*a^2*tan(f*x + e)^3 + 210*a*b*tan(f*x + e)^3 + 315*a^2*tan(f*x + e ))/f
Time = 14.00 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.89 \[ \int \sec ^{10}(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {a^2\,\mathrm {tan}\left (e+f\,x\right )+\frac {{\mathrm {tan}\left (e+f\,x\right )}^9\,{\left (a+b\right )}^2}{9}+{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (\frac {6\,a^2}{5}+\frac {6\,a\,b}{5}+\frac {b^2}{5}\right )+{\mathrm {tan}\left (e+f\,x\right )}^7\,\left (\frac {4\,a^2}{7}+\frac {6\,a\,b}{7}+\frac {2\,b^2}{7}\right )+\frac {2\,a\,{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (2\,a+b\right )}{3}}{f} \]