Integrand size = 23, antiderivative size = 95 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=-a^2 x+\frac {a^2 \tan (e+f x)}{f}-\frac {a^2 \tan ^3(e+f x)}{3 f}+\frac {a^2 \tan ^5(e+f x)}{5 f}+\frac {b (2 a+b) \tan ^7(e+f x)}{7 f}+\frac {b^2 \tan ^9(e+f x)}{9 f} \]
-a^2*x+a^2*tan(f*x+e)/f-1/3*a^2*tan(f*x+e)^3/f+1/5*a^2*tan(f*x+e)^5/f+1/7* b*(2*a+b)*tan(f*x+e)^7/f+1/9*b^2*tan(f*x+e)^9/f
Leaf count is larger than twice the leaf count of optimal. \(275\) vs. \(2(95)=190\).
Time = 3.71 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.89 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=-\frac {4 \left (b+a \cos ^2(e+f x)\right )^2 \sec ^9(e+f x) \left (315 a^2 f x \cos ^9(e+f x)-35 b^2 \sec (e) \sin (f x)-5 (18 a-19 b) b \cos ^2(e+f x) \sec (e) \sin (f x)-3 \left (21 a^2-90 a b+25 b^2\right ) \cos ^4(e+f x) \sec (e) \sin (f x)+\left (231 a^2-270 a b+5 b^2\right ) \cos ^6(e+f x) \sec (e) \sin (f x)-\left (483 a^2-90 a b-10 b^2\right ) \cos ^8(e+f x) \sec (e) \sin (f x)-35 b^2 \cos (e+f x) \tan (e)-5 (18 a-19 b) b \cos ^3(e+f x) \tan (e)-3 \left (21 a^2-90 a b+25 b^2\right ) \cos ^5(e+f x) \tan (e)+\left (231 a^2-270 a b+5 b^2\right ) \cos ^7(e+f x) \tan (e)\right )}{315 f (a+2 b+a \cos (2 (e+f x)))^2} \]
(-4*(b + a*Cos[e + f*x]^2)^2*Sec[e + f*x]^9*(315*a^2*f*x*Cos[e + f*x]^9 - 35*b^2*Sec[e]*Sin[f*x] - 5*(18*a - 19*b)*b*Cos[e + f*x]^2*Sec[e]*Sin[f*x] - 3*(21*a^2 - 90*a*b + 25*b^2)*Cos[e + f*x]^4*Sec[e]*Sin[f*x] + (231*a^2 - 270*a*b + 5*b^2)*Cos[e + f*x]^6*Sec[e]*Sin[f*x] - (483*a^2 - 90*a*b - 10* b^2)*Cos[e + f*x]^8*Sec[e]*Sin[f*x] - 35*b^2*Cos[e + f*x]*Tan[e] - 5*(18*a - 19*b)*b*Cos[e + f*x]^3*Tan[e] - 3*(21*a^2 - 90*a*b + 25*b^2)*Cos[e + f* x]^5*Tan[e] + (231*a^2 - 270*a*b + 5*b^2)*Cos[e + f*x]^7*Tan[e]))/(315*f*( a + 2*b + a*Cos[2*(e + f*x)])^2)
Time = 0.32 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.95, 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.
\(\displaystyle \int \tan ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^6 \left (a+b \sec (e+f x)^2\right )^2dx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle \frac {\int \frac {\tan ^6(e+f x) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^2}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {\tan ^6(e+f x) \left (b \tan ^2(e+f x)+a+b\right )^2}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 364 |
\(\displaystyle \frac {\int \left (b^2 \tan ^8(e+f x)+b (2 a+b) \tan ^6(e+f x)+a^2 \tan ^4(e+f x)-a^2 \tan ^2(e+f x)+a^2-\frac {a^2}{\tan ^2(e+f x)+1}\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-a^2 \arctan (\tan (e+f x))+\frac {1}{5} a^2 \tan ^5(e+f x)-\frac {1}{3} a^2 \tan ^3(e+f x)+a^2 \tan (e+f x)+\frac {1}{7} b (2 a+b) \tan ^7(e+f x)+\frac {1}{9} b^2 \tan ^9(e+f x)}{f}\) |
(-(a^2*ArcTan[Tan[e + f*x]]) + a^2*Tan[e + f*x] - (a^2*Tan[e + f*x]^3)/3 + (a^2*Tan[e + f*x]^5)/5 + (b*(2*a + b)*Tan[e + f*x]^7)/7 + (b^2*Tan[e + f* x]^9)/9)/f
3.4.30.3.1 Defintions of rubi rules used
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 = 12.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93
method | result | size |
parts | \(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{5}}{5}-\frac {\tan \left (f x +e \right )^{3}}{3}+\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {b^{2} \left (\frac {\tan \left (f x +e \right )^{9}}{9}+\frac {\tan \left (f x +e \right )^{7}}{7}\right )}{f}+\frac {2 a b \tan \left (f x +e \right )^{7}}{7 f}\) | \(88\) |
derivativedivides | \(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{5}}{5}-\frac {\tan \left (f x +e \right )^{3}}{3}+\tan \left (f x +e \right )-f x -e \right )+\frac {2 a b \sin \left (f x +e \right )^{7}}{7 \cos \left (f x +e \right )^{7}}+b^{2} \left (\frac {\sin \left (f x +e \right )^{7}}{9 \cos \left (f x +e \right )^{9}}+\frac {2 \sin \left (f x +e \right )^{7}}{63 \cos \left (f x +e \right )^{7}}\right )}{f}\) | \(105\) |
default | \(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{5}}{5}-\frac {\tan \left (f x +e \right )^{3}}{3}+\tan \left (f x +e \right )-f x -e \right )+\frac {2 a b \sin \left (f x +e \right )^{7}}{7 \cos \left (f x +e \right )^{7}}+b^{2} \left (\frac {\sin \left (f x +e \right )^{7}}{9 \cos \left (f x +e \right )^{9}}+\frac {2 \sin \left (f x +e \right )^{7}}{63 \cos \left (f x +e \right )^{7}}\right )}{f}\) | \(105\) |
risch | \(-a^{2} x -\frac {2 i \left (90 a b -483 a^{2}+10 b^{2}-28350 a^{2} {\mathrm e}^{10 i \left (f x +e \right )}+5040 a b \,{\mathrm e}^{8 i \left (f x +e \right )}+3780 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+1980 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+180 a b \,{\mathrm e}^{2 i \left (f x +e \right )}-32508 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}-1890 b^{2} {\mathrm e}^{8 i \left (f x +e \right )}-24402 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+1890 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}-11718 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-270 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-3402 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+90 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}-5670 a^{2} {\mathrm e}^{14 i \left (f x +e \right )}-16170 a^{2} {\mathrm e}^{12 i \left (f x +e \right )}-1050 b^{2} {\mathrm e}^{12 i \left (f x +e \right )}+630 a b \,{\mathrm e}^{16 i \left (f x +e \right )}+1260 a b \,{\mathrm e}^{14 i \left (f x +e \right )}+3780 a b \,{\mathrm e}^{12 i \left (f x +e \right )}-945 a^{2} {\mathrm e}^{16 i \left (f x +e \right )}+630 b^{2} {\mathrm e}^{14 i \left (f x +e \right )}+6300 a b \,{\mathrm e}^{10 i \left (f x +e \right )}+3150 b^{2} {\mathrm e}^{10 i \left (f x +e \right )}\right )}{315 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{9}}\) | \(356\) |
a^2/f*(1/5*tan(f*x+e)^5-1/3*tan(f*x+e)^3+tan(f*x+e)-arctan(tan(f*x+e)))+b^ 2/f*(1/9*tan(f*x+e)^9+1/7*tan(f*x+e)^7)+2/7*a*b/f*tan(f*x+e)^7
Time = 0.27 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.44 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=-\frac {315 \, a^{2} f x \cos \left (f x + e\right )^{9} - {\left ({\left (483 \, a^{2} - 90 \, a b - 10 \, b^{2}\right )} \cos \left (f x + e\right )^{8} - {\left (231 \, a^{2} - 270 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + 3 \, {\left (21 \, a^{2} - 90 \, a b + 25 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 5 \, {\left (18 \, a b - 19 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 35 \, b^{2}\right )} \sin \left (f x + e\right )}{315 \, f \cos \left (f x + e\right )^{9}} \]
-1/315*(315*a^2*f*x*cos(f*x + e)^9 - ((483*a^2 - 90*a*b - 10*b^2)*cos(f*x + e)^8 - (231*a^2 - 270*a*b + 5*b^2)*cos(f*x + e)^6 + 3*(21*a^2 - 90*a*b + 25*b^2)*cos(f*x + e)^4 + 5*(18*a*b - 19*b^2)*cos(f*x + e)^2 + 35*b^2)*sin (f*x + e))/(f*cos(f*x + e)^9)
\[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \tan ^{6}{\left (e + f x \right )}\, dx \]
Time = 0.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=\frac {35 \, b^{2} \tan \left (f x + e\right )^{9} + 45 \, {\left (2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{7} + 63 \, a^{2} \tan \left (f x + e\right )^{5} - 105 \, a^{2} \tan \left (f x + e\right )^{3} - 315 \, {\left (f x + e\right )} a^{2} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, f} \]
1/315*(35*b^2*tan(f*x + e)^9 + 45*(2*a*b + b^2)*tan(f*x + e)^7 + 63*a^2*ta n(f*x + e)^5 - 105*a^2*tan(f*x + e)^3 - 315*(f*x + e)*a^2 + 315*a^2*tan(f* x + e))/f
Time = 2.52 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.96 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=\frac {35 \, b^{2} \tan \left (f x + e\right )^{9} + 90 \, a b \tan \left (f x + e\right )^{7} + 45 \, b^{2} \tan \left (f x + e\right )^{7} + 63 \, a^{2} \tan \left (f x + e\right )^{5} - 105 \, a^{2} \tan \left (f x + e\right )^{3} - 315 \, {\left (f x + e\right )} a^{2} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, f} \]
1/315*(35*b^2*tan(f*x + e)^9 + 90*a*b*tan(f*x + e)^7 + 45*b^2*tan(f*x + e) ^7 + 63*a^2*tan(f*x + e)^5 - 105*a^2*tan(f*x + e)^3 - 315*(f*x + e)*a^2 + 315*a^2*tan(f*x + e))/f
Time = 19.69 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.33 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )\,\left ({\left (a+b\right )}^2+b^2-2\,b\,\left (a+b\right )\right )-{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {{\left (a+b\right )}^2}{3}+\frac {b^2}{3}-\frac {2\,b\,\left (a+b\right )}{3}\right )+{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (\frac {{\left (a+b\right )}^2}{5}+\frac {b^2}{5}-\frac {2\,b\,\left (a+b\right )}{5}\right )-{\mathrm {tan}\left (e+f\,x\right )}^7\,\left (\frac {b^2}{7}-\frac {2\,b\,\left (a+b\right )}{7}\right )+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^9}{9}-a^2\,f\,x}{f} \]