Integrand size = 23, antiderivative size = 59 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^2(e+f x) \, dx=-a^2 x+\frac {a^2 \tan (e+f x)}{f}+\frac {b (2 a+b) \tan ^3(e+f x)}{3 f}+\frac {b^2 \tan ^5(e+f x)}{5 f} \] Output:
-a^2*x+a^2*tan(f*x+e)/f+1/3*b*(2*a+b)*tan(f*x+e)^3/f+1/5*b^2*tan(f*x+e)^5/ f
Leaf count is larger than twice the leaf count of optimal. \(281\) vs. \(2(59)=118\).
Time = 1.10 (sec) , antiderivative size = 281, normalized size of antiderivative = 4.76 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^2(e+f x) \, dx=-\frac {\sec (e) \sec ^5(e+f x) \left (150 a^2 f x \cos (f x)+150 a^2 f x \cos (2 e+f x)+75 a^2 f x \cos (2 e+3 f x)+75 a^2 f x \cos (4 e+3 f x)+15 a^2 f x \cos (4 e+5 f x)+15 a^2 f x \cos (6 e+5 f x)-180 a^2 \sin (f x)+80 a b \sin (f x)-20 b^2 \sin (f x)+120 a^2 \sin (2 e+f x)-120 a b \sin (2 e+f x)-60 b^2 \sin (2 e+f x)-120 a^2 \sin (2 e+3 f x)+40 a b \sin (2 e+3 f x)+20 b^2 \sin (2 e+3 f x)+30 a^2 \sin (4 e+3 f x)-60 a b \sin (4 e+3 f x)-30 a^2 \sin (4 e+5 f x)+20 a b \sin (4 e+5 f x)+4 b^2 \sin (4 e+5 f x)\right )}{480 f} \] Input:
Integrate[(a + b*Sec[e + f*x]^2)^2*Tan[e + f*x]^2,x]
Output:
-1/480*(Sec[e]*Sec[e + f*x]^5*(150*a^2*f*x*Cos[f*x] + 150*a^2*f*x*Cos[2*e + f*x] + 75*a^2*f*x*Cos[2*e + 3*f*x] + 75*a^2*f*x*Cos[4*e + 3*f*x] + 15*a^ 2*f*x*Cos[4*e + 5*f*x] + 15*a^2*f*x*Cos[6*e + 5*f*x] - 180*a^2*Sin[f*x] + 80*a*b*Sin[f*x] - 20*b^2*Sin[f*x] + 120*a^2*Sin[2*e + f*x] - 120*a*b*Sin[2 *e + f*x] - 60*b^2*Sin[2*e + f*x] - 120*a^2*Sin[2*e + 3*f*x] + 40*a*b*Sin[ 2*e + 3*f*x] + 20*b^2*Sin[2*e + 3*f*x] + 30*a^2*Sin[4*e + 3*f*x] - 60*a*b* Sin[4*e + 3*f*x] - 30*a^2*Sin[4*e + 5*f*x] + 20*a*b*Sin[4*e + 5*f*x] + 4*b ^2*Sin[4*e + 5*f*x]))/f
Time = 0.31 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4629, 2075, 364, 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.
\(\displaystyle \int \tan ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^2 \left (a+b \sec (e+f x)^2\right )^2dx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle \frac {\int \frac {\tan ^2(e+f x) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^2}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {\tan ^2(e+f x) \left (b \tan ^2(e+f x)+a+b\right )^2}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 364 |
\(\displaystyle \frac {\int \left (b^2 \tan ^4(e+f x)+b (2 a+b) \tan ^2(e+f x)+a^2-\frac {a^2}{\tan ^2(e+f x)+1}\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-a^2 \arctan (\tan (e+f x))+a^2 \tan (e+f x)+\frac {1}{3} b (2 a+b) \tan ^3(e+f x)+\frac {1}{5} b^2 \tan ^5(e+f x)}{f}\) |
Input:
Int[(a + b*Sec[e + f*x]^2)^2*Tan[e + f*x]^2,x]
Output:
(-(a^2*ArcTan[Tan[e + f*x]]) + a^2*Tan[e + f*x] + (b*(2*a + b)*Tan[e + f*x ]^3)/3 + (b^2*Tan[e + f*x]^5)/5)/f
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f _.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/f Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 )), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
Time = 2.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.15
method | result | size |
parts | \(\frac {a^{2} \left (\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {b^{2} \left (\frac {\tan \left (f x +e \right )^{5}}{5}+\frac {\tan \left (f x +e \right )^{3}}{3}\right )}{f}+\frac {2 a b \tan \left (f x +e \right )^{3}}{3 f}\) | \(68\) |
derivativedivides | \(\frac {a^{2} \left (\tan \left (f x +e \right )-f x -e \right )+\frac {2 a b \sin \left (f x +e \right )^{3}}{3 \cos \left (f x +e \right )^{3}}+b^{2} \left (\frac {\sin \left (f x +e \right )^{3}}{5 \cos \left (f x +e \right )^{5}}+\frac {2 \sin \left (f x +e \right )^{3}}{15 \cos \left (f x +e \right )^{3}}\right )}{f}\) | \(85\) |
default | \(\frac {a^{2} \left (\tan \left (f x +e \right )-f x -e \right )+\frac {2 a b \sin \left (f x +e \right )^{3}}{3 \cos \left (f x +e \right )^{3}}+b^{2} \left (\frac {\sin \left (f x +e \right )^{3}}{5 \cos \left (f x +e \right )^{5}}+\frac {2 \sin \left (f x +e \right )^{3}}{15 \cos \left (f x +e \right )^{3}}\right )}{f}\) | \(85\) |
risch | \(-a^{2} x -\frac {2 i \left (-15 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}+30 \,{\mathrm e}^{8 i \left (f x +e \right )} a b -60 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+60 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+30 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}-90 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+40 a b \,{\mathrm e}^{4 i \left (f x +e \right )}-10 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-60 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+20 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+10 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}-15 a^{2}+10 a b +2 b^{2}\right )}{15 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}\) | \(192\) |
Input:
int((a+b*sec(f*x+e)^2)^2*tan(f*x+e)^2,x,method=_RETURNVERBOSE)
Output:
a^2/f*(tan(f*x+e)-arctan(tan(f*x+e)))+b^2/f*(1/5*tan(f*x+e)^5+1/3*tan(f*x+ e)^3)+2/3*a*b/f*tan(f*x+e)^3
Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.46 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^2(e+f x) \, dx=-\frac {15 \, a^{2} f x \cos \left (f x + e\right )^{5} - {\left ({\left (15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + {\left (10 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, b^{2}\right )} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )^{5}} \] Input:
integrate((a+b*sec(f*x+e)^2)^2*tan(f*x+e)^2,x, algorithm="fricas")
Output:
-1/15*(15*a^2*f*x*cos(f*x + e)^5 - ((15*a^2 - 10*a*b - 2*b^2)*cos(f*x + e) ^4 + (10*a*b - b^2)*cos(f*x + e)^2 + 3*b^2)*sin(f*x + e))/(f*cos(f*x + e)^ 5)
\[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^2(e+f x) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \tan ^{2}{\left (e + f x \right )}\, dx \] Input:
integrate((a+b*sec(f*x+e)**2)**2*tan(f*x+e)**2,x)
Output:
Integral((a + b*sec(e + f*x)**2)**2*tan(e + f*x)**2, x)
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.98 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^2(e+f x) \, dx=\frac {3 \, b^{2} \tan \left (f x + e\right )^{5} + 5 \, {\left (2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{3} - 15 \, {\left (f x + e\right )} a^{2} + 15 \, a^{2} \tan \left (f x + e\right )}{15 \, f} \] Input:
integrate((a+b*sec(f*x+e)^2)^2*tan(f*x+e)^2,x, algorithm="maxima")
Output:
1/15*(3*b^2*tan(f*x + e)^5 + 5*(2*a*b + b^2)*tan(f*x + e)^3 - 15*(f*x + e) *a^2 + 15*a^2*tan(f*x + e))/f
Time = 0.38 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.37 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^2(e+f x) \, dx=-\frac {{\left (f x + e\right )} a^{2}}{f} + \frac {3 \, b^{2} f^{4} \tan \left (f x + e\right )^{5} + 10 \, a b f^{4} \tan \left (f x + e\right )^{3} + 5 \, b^{2} f^{4} \tan \left (f x + e\right )^{3} + 15 \, a^{2} f^{4} \tan \left (f x + e\right )}{15 \, f^{5}} \] Input:
integrate((a+b*sec(f*x+e)^2)^2*tan(f*x+e)^2,x, algorithm="giac")
Output:
-(f*x + e)*a^2/f + 1/15*(3*b^2*f^4*tan(f*x + e)^5 + 10*a*b*f^4*tan(f*x + e )^3 + 5*b^2*f^4*tan(f*x + e)^3 + 15*a^2*f^4*tan(f*x + e))/f^5
Time = 15.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.17 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^2(e+f x) \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )\,\left ({\left (a+b\right )}^2+b^2-2\,b\,\left (a+b\right )\right )-{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {b^2}{3}-\frac {2\,b\,\left (a+b\right )}{3}\right )+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^5}{5}-a^2\,f\,x}{f} \] Input:
int(tan(e + f*x)^2*(a + b/cos(e + f*x)^2)^2,x)
Output:
(tan(e + f*x)*((a + b)^2 + b^2 - 2*b*(a + b)) - tan(e + f*x)^3*(b^2/3 - (2 *b*(a + b))/3) + (b^2*tan(e + f*x)^5)/5 - a^2*f*x)/f
Time = 0.15 (sec) , antiderivative size = 208, normalized size of antiderivative = 3.53 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^2(e+f x) \, dx=\frac {15 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} \tan \left (f x +e \right ) a^{2}-15 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a^{2} f x -30 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} \tan \left (f x +e \right ) a^{2}+30 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a^{2} f x +15 \cos \left (f x +e \right ) \tan \left (f x +e \right ) a^{2}-15 \cos \left (f x +e \right ) a^{2} f x -10 \sin \left (f x +e \right )^{5} a b -2 \sin \left (f x +e \right )^{5} b^{2}+10 \sin \left (f x +e \right )^{3} a b +5 \sin \left (f x +e \right )^{3} b^{2}}{15 \cos \left (f x +e \right ) f \left (\sin \left (f x +e \right )^{4}-2 \sin \left (f x +e \right )^{2}+1\right )} \] Input:
int((a+b*sec(f*x+e)^2)^2*tan(f*x+e)^2,x)
Output:
(15*cos(e + f*x)*sin(e + f*x)**4*tan(e + f*x)*a**2 - 15*cos(e + f*x)*sin(e + f*x)**4*a**2*f*x - 30*cos(e + f*x)*sin(e + f*x)**2*tan(e + f*x)*a**2 + 30*cos(e + f*x)*sin(e + f*x)**2*a**2*f*x + 15*cos(e + f*x)*tan(e + f*x)*a* *2 - 15*cos(e + f*x)*a**2*f*x - 10*sin(e + f*x)**5*a*b - 2*sin(e + f*x)**5 *b**2 + 10*sin(e + f*x)**3*a*b + 5*sin(e + f*x)**3*b**2)/(15*cos(e + f*x)* f*(sin(e + f*x)**4 - 2*sin(e + f*x)**2 + 1))