Integrand size = 23, antiderivative size = 80 \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {a^2 \tan (e+f x)}{f}+\frac {a (3 a+2 b) \tan ^3(e+f x)}{3 f}+\frac {(a+b) (3 a+b) \tan ^5(e+f x)}{5 f}+\frac {(a+b)^2 \tan ^7(e+f x)}{7 f} \] Output:
a^2*tan(f*x+e)/f+1/3*a*(3*a+2*b)*tan(f*x+e)^3/f+1/5*(a+b)*(3*a+b)*tan(f*x+ e)^5/f+1/7*(a+b)^2*tan(f*x+e)^7/f
Time = 3.53 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.15 \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {\left (48 a^2-16 a b+6 b^2+\left (24 a^2-8 a b+3 b^2\right ) \sec ^2(e+f x)+6 \left (3 a^2-a b-4 b^2\right ) \sec ^4(e+f x)+15 (a+b)^2 \sec ^6(e+f x)\right ) \tan (e+f x)}{105 f} \] Input:
Integrate[Sec[e + f*x]^8*(a + b*Sin[e + f*x]^2)^2,x]
Output:
((48*a^2 - 16*a*b + 6*b^2 + (24*a^2 - 8*a*b + 3*b^2)*Sec[e + f*x]^2 + 6*(3 *a^2 - a*b - 4*b^2)*Sec[e + f*x]^4 + 15*(a + b)^2*Sec[e + f*x]^6)*Tan[e + f*x])/(105*f)
Time = 0.28 (sec) , antiderivative size = 72, 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.
\(\displaystyle \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin (e+f x)^2\right )^2}{\cos (e+f x)^8}dx\) |
\(\Big \downarrow \) 3670 |
\(\displaystyle \frac {\int \left (\tan ^2(e+f x)+1\right ) \left ((a+b) \tan ^2(e+f x)+a\right )^2d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 290 |
\(\displaystyle \frac {\int \left ((a+b)^2 \tan ^6(e+f x)+(a+b) (3 a+b) \tan ^4(e+f x)+a (3 a+2 b) \tan ^2(e+f x)+a^2\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \tan (e+f x)+\frac {1}{7} (a+b)^2 \tan ^7(e+f x)+\frac {1}{5} (a+b) (3 a+b) \tan ^5(e+f x)+\frac {1}{3} a (3 a+2 b) \tan ^3(e+f x)}{f}\) |
Input:
Int[Sec[e + f*x]^8*(a + b*Sin[e + f*x]^2)^2,x]
Output:
(a^2*Tan[e + f*x] + (a*(3*a + 2*b)*Tan[e + f*x]^3)/3 + ((a + b)*(3*a + b)* Tan[e + f*x]^5)/5 + ((a + b)^2*Tan[e + f*x]^7)/7)/f
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 = 1.19 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.86
method | result | size |
derivativedivides | \(\frac {-a^{2} \left (-\frac {16}{35}-\frac {\sec \left (f x +e \right )^{6}}{7}-\frac {6 \sec \left (f x +e \right )^{4}}{35}-\frac {8 \sec \left (f x +e \right )^{2}}{35}\right ) \tan \left (f x +e \right )+2 a b \left (\frac {\sin \left (f x +e \right )^{3}}{7 \cos \left (f x +e \right )^{7}}+\frac {4 \sin \left (f x +e \right )^{3}}{35 \cos \left (f x +e \right )^{5}}+\frac {8 \sin \left (f x +e \right )^{3}}{105 \cos \left (f x +e \right )^{3}}\right )+b^{2} \left (\frac {\sin \left (f x +e \right )^{5}}{7 \cos \left (f x +e \right )^{7}}+\frac {2 \sin \left (f x +e \right )^{5}}{35 \cos \left (f x +e \right )^{5}}\right )}{f}\) | \(149\) |
default | \(\frac {-a^{2} \left (-\frac {16}{35}-\frac {\sec \left (f x +e \right )^{6}}{7}-\frac {6 \sec \left (f x +e \right )^{4}}{35}-\frac {8 \sec \left (f x +e \right )^{2}}{35}\right ) \tan \left (f x +e \right )+2 a b \left (\frac {\sin \left (f x +e \right )^{3}}{7 \cos \left (f x +e \right )^{7}}+\frac {4 \sin \left (f x +e \right )^{3}}{35 \cos \left (f x +e \right )^{5}}+\frac {8 \sin \left (f x +e \right )^{3}}{105 \cos \left (f x +e \right )^{3}}\right )+b^{2} \left (\frac {\sin \left (f x +e \right )^{5}}{7 \cos \left (f x +e \right )^{7}}+\frac {2 \sin \left (f x +e \right )^{5}}{35 \cos \left (f x +e \right )^{5}}\right )}{f}\) | \(149\) |
parallelrisch | \(-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{12} a^{2}-2 a \left (a -\frac {4 b}{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}+\frac {\left (43 a^{2}+\frac {32}{3} a b +16 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{5}+\frac {4 \left (-53 a^{2}+76 a b +24 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{35}+\frac {\left (43 a^{2}+\frac {32}{3} a b +16 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{5}-2 a \left (a -\frac {4 b}{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+a^{2}\right )}{f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{7}}\) | \(171\) |
risch | \(\frac {4 i \left (105 b^{2} {\mathrm e}^{10 i \left (f x +e \right )}-560 a b \,{\mathrm e}^{8 i \left (f x +e \right )}-105 b^{2} {\mathrm e}^{8 i \left (f x +e \right )}+840 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+280 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+210 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+504 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-168 a b \,{\mathrm e}^{4 i \left (f x +e \right )}-42 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+168 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-56 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+21 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+24 a^{2}-8 a b +3 b^{2}\right )}{105 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{7}}\) | \(199\) |
Input:
int(sec(f*x+e)^8*(a+b*sin(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(-a^2*(-16/35-1/7*sec(f*x+e)^6-6/35*sec(f*x+e)^4-8/35*sec(f*x+e)^2)*ta n(f*x+e)+2*a*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+8/105*sin(f*x+e)^3/cos(f*x+e)^3)+b^2*(1/7*sin(f*x+e)^5/cos(f*x+e)^7+2/3 5*sin(f*x+e)^5/cos(f*x+e)^5))
Time = 0.08 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.35 \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {{\left (2 \, {\left (24 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + {\left (24 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 6 \, {\left (3 \, a^{2} - a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 15 \, a^{2} + 30 \, a b + 15 \, b^{2}\right )} \sin \left (f x + e\right )}{105 \, f \cos \left (f x + e\right )^{7}} \] Input:
integrate(sec(f*x+e)^8*(a+b*sin(f*x+e)^2)^2,x, algorithm="fricas")
Output:
1/105*(2*(24*a^2 - 8*a*b + 3*b^2)*cos(f*x + e)^6 + (24*a^2 - 8*a*b + 3*b^2 )*cos(f*x + e)^4 + 6*(3*a^2 - a*b - 4*b^2)*cos(f*x + e)^2 + 15*a^2 + 30*a* b + 15*b^2)*sin(f*x + e)/(f*cos(f*x + e)^7)
Timed out. \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\text {Timed out} \] Input:
integrate(sec(f*x+e)**8*(a+b*sin(f*x+e)**2)**2,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.01 \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{7} + 21 \, {\left (3 \, a^{2} + 4 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + 35 \, {\left (3 \, a^{2} + 2 \, a b\right )} \tan \left (f x + e\right )^{3} + 105 \, a^{2} \tan \left (f x + e\right )}{105 \, f} \] Input:
integrate(sec(f*x+e)^8*(a+b*sin(f*x+e)^2)^2,x, algorithm="maxima")
Output:
1/105*(15*(a^2 + 2*a*b + b^2)*tan(f*x + e)^7 + 21*(3*a^2 + 4*a*b + b^2)*ta n(f*x + e)^5 + 35*(3*a^2 + 2*a*b)*tan(f*x + e)^3 + 105*a^2*tan(f*x + e))/f
Time = 0.50 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.48 \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {15 \, a^{2} \tan \left (f x + e\right )^{7} + 30 \, a b \tan \left (f x + e\right )^{7} + 15 \, b^{2} \tan \left (f x + e\right )^{7} + 63 \, a^{2} \tan \left (f x + e\right )^{5} + 84 \, a b \tan \left (f x + e\right )^{5} + 21 \, b^{2} \tan \left (f x + e\right )^{5} + 105 \, a^{2} \tan \left (f x + e\right )^{3} + 70 \, a b \tan \left (f x + e\right )^{3} + 105 \, a^{2} \tan \left (f x + e\right )}{105 \, f} \] Input:
integrate(sec(f*x+e)^8*(a+b*sin(f*x+e)^2)^2,x, algorithm="giac")
Output:
1/105*(15*a^2*tan(f*x + e)^7 + 30*a*b*tan(f*x + e)^7 + 15*b^2*tan(f*x + e) ^7 + 63*a^2*tan(f*x + e)^5 + 84*a*b*tan(f*x + e)^5 + 21*b^2*tan(f*x + e)^5 + 105*a^2*tan(f*x + e)^3 + 70*a*b*tan(f*x + e)^3 + 105*a^2*tan(f*x + e))/ f
Time = 36.66 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90 \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {a^2\,\mathrm {tan}\left (e+f\,x\right )+\frac {{\mathrm {tan}\left (e+f\,x\right )}^7\,{\left (a+b\right )}^2}{7}+{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (\frac {3\,a^2}{5}+\frac {4\,a\,b}{5}+\frac {b^2}{5}\right )+\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (3\,a+2\,b\right )}{3}}{f} \] Input:
int((a + b*sin(e + f*x)^2)^2/cos(e + f*x)^8,x)
Output:
(a^2*tan(e + f*x) + (tan(e + f*x)^7*(a + b)^2)/7 + tan(e + f*x)^5*((4*a*b) /5 + (3*a^2)/5 + b^2/5) + (a*tan(e + f*x)^3*(3*a + 2*b))/3)/f
Time = 0.17 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.98 \[ \int \sec ^8(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {\sin \left (f x +e \right ) \left (48 \sin \left (f x +e \right )^{6} a^{2}-16 \sin \left (f x +e \right )^{6} a b +6 \sin \left (f x +e \right )^{6} b^{2}-168 \sin \left (f x +e \right )^{4} a^{2}+56 \sin \left (f x +e \right )^{4} a b -21 \sin \left (f x +e \right )^{4} b^{2}+210 \sin \left (f x +e \right )^{2} a^{2}-70 \sin \left (f x +e \right )^{2} a b -105 a^{2}\right )}{105 \cos \left (f x +e \right ) f \left (\sin \left (f x +e \right )^{6}-3 \sin \left (f x +e \right )^{4}+3 \sin \left (f x +e \right )^{2}-1\right )} \] Input:
int(sec(f*x+e)^8*(a+b*sin(f*x+e)^2)^2,x)
Output:
(sin(e + f*x)*(48*sin(e + f*x)**6*a**2 - 16*sin(e + f*x)**6*a*b + 6*sin(e + f*x)**6*b**2 - 168*sin(e + f*x)**4*a**2 + 56*sin(e + f*x)**4*a*b - 21*si n(e + f*x)**4*b**2 + 210*sin(e + f*x)**2*a**2 - 70*sin(e + f*x)**2*a*b - 1 05*a**2))/(105*cos(e + f*x)*f*(sin(e + f*x)**6 - 3*sin(e + f*x)**4 + 3*sin (e + f*x)**2 - 1))