Integrand size = 23, antiderivative size = 153 \[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {\left (48 a^2+80 a b+35 b^2\right ) \text {arctanh}(\sin (e+f x))}{128 f}+\frac {\left (48 a^2+80 a b+35 b^2\right ) \sec (e+f x) \tan (e+f x)}{128 f}+\frac {\left (48 a^2+80 a b+35 b^2\right ) \sec ^3(e+f x) \tan (e+f x)}{192 f}+\frac {b (16 a+7 b) \sec ^5(e+f x) \tan (e+f x)}{48 f}+\frac {b^2 \sec ^7(e+f x) \tan (e+f x)}{8 f} \] Output:
1/128*(48*a^2+80*a*b+35*b^2)*arctanh(sin(f*x+e))/f+1/128*(48*a^2+80*a*b+35 *b^2)*sec(f*x+e)*tan(f*x+e)/f+1/192*(48*a^2+80*a*b+35*b^2)*sec(f*x+e)^3*ta n(f*x+e)/f+1/48*b*(16*a+7*b)*sec(f*x+e)^5*tan(f*x+e)/f+1/8*b^2*sec(f*x+e)^ 7*tan(f*x+e)/f
Time = 0.04 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.69 \[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 a^2 \text {arctanh}(\sin (e+f x))}{8 f}+\frac {5 a b \text {arctanh}(\sin (e+f x))}{8 f}+\frac {35 b^2 \text {arctanh}(\sin (e+f x))}{128 f}+\frac {3 a^2 \sec (e+f x) \tan (e+f x)}{8 f}+\frac {5 a b \sec (e+f x) \tan (e+f x)}{8 f}+\frac {35 b^2 \sec (e+f x) \tan (e+f x)}{128 f}+\frac {a^2 \sec ^3(e+f x) \tan (e+f x)}{4 f}+\frac {5 a b \sec ^3(e+f x) \tan (e+f x)}{12 f}+\frac {35 b^2 \sec ^3(e+f x) \tan (e+f x)}{192 f}+\frac {a b \sec ^5(e+f x) \tan (e+f x)}{3 f}+\frac {7 b^2 \sec ^5(e+f x) \tan (e+f x)}{48 f}+\frac {b^2 \sec ^7(e+f x) \tan (e+f x)}{8 f} \] Input:
Integrate[Sec[e + f*x]^5*(a + b*Sec[e + f*x]^2)^2,x]
Output:
(3*a^2*ArcTanh[Sin[e + f*x]])/(8*f) + (5*a*b*ArcTanh[Sin[e + f*x]])/(8*f) + (35*b^2*ArcTanh[Sin[e + f*x]])/(128*f) + (3*a^2*Sec[e + f*x]*Tan[e + f*x ])/(8*f) + (5*a*b*Sec[e + f*x]*Tan[e + f*x])/(8*f) + (35*b^2*Sec[e + f*x]* Tan[e + f*x])/(128*f) + (a^2*Sec[e + f*x]^3*Tan[e + f*x])/(4*f) + (5*a*b*S ec[e + f*x]^3*Tan[e + f*x])/(12*f) + (35*b^2*Sec[e + f*x]^3*Tan[e + f*x])/ (192*f) + (a*b*Sec[e + f*x]^5*Tan[e + f*x])/(3*f) + (7*b^2*Sec[e + f*x]^5* Tan[e + f*x])/(48*f) + (b^2*Sec[e + f*x]^7*Tan[e + f*x])/(8*f)
Time = 0.32 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4635, 315, 25, 298, 215, 215, 219}
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.
| \(\displaystyle \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (e+f x)^5 \left (a+b \sec (e+f x)^2\right )^2dx\) |
| \(\Big \downarrow \) 4635 |
| \(\displaystyle \frac {\int \frac {\left (-a \sin ^2(e+f x)+a+b\right )^2}{\left (1-\sin ^2(e+f x)\right )^5}d\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 315 |
| \(\displaystyle \frac {\frac {b \sin (e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{8 \left (1-\sin ^2(e+f x)\right )^4}-\frac {1}{8} \int -\frac {(a+b) (8 a+7 b)-a (8 a+5 b) \sin ^2(e+f x)}{\left (1-\sin ^2(e+f x)\right )^4}d\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{8} \int \frac {(a+b) (8 a+7 b)-a (8 a+5 b) \sin ^2(e+f x)}{\left (1-\sin ^2(e+f x)\right )^4}d\sin (e+f x)+\frac {b \sin (e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{8 \left (1-\sin ^2(e+f x)\right )^4}}{f}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (48 a^2+80 a b+35 b^2\right ) \int \frac {1}{\left (1-\sin ^2(e+f x)\right )^3}d\sin (e+f x)+\frac {b (10 a+7 b) \sin (e+f x)}{6 \left (1-\sin ^2(e+f x)\right )^3}\right )+\frac {b \sin (e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{8 \left (1-\sin ^2(e+f x)\right )^4}}{f}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (48 a^2+80 a b+35 b^2\right ) \left (\frac {3}{4} \int \frac {1}{\left (1-\sin ^2(e+f x)\right )^2}d\sin (e+f x)+\frac {\sin (e+f x)}{4 \left (1-\sin ^2(e+f x)\right )^2}\right )+\frac {b (10 a+7 b) \sin (e+f x)}{6 \left (1-\sin ^2(e+f x)\right )^3}\right )+\frac {b \sin (e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{8 \left (1-\sin ^2(e+f x)\right )^4}}{f}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (48 a^2+80 a b+35 b^2\right ) \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{1-\sin ^2(e+f x)}d\sin (e+f x)+\frac {\sin (e+f x)}{2 \left (1-\sin ^2(e+f x)\right )}\right )+\frac {\sin (e+f x)}{4 \left (1-\sin ^2(e+f x)\right )^2}\right )+\frac {b (10 a+7 b) \sin (e+f x)}{6 \left (1-\sin ^2(e+f x)\right )^3}\right )+\frac {b \sin (e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{8 \left (1-\sin ^2(e+f x)\right )^4}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (48 a^2+80 a b+35 b^2\right ) \left (\frac {3}{4} \left (\frac {1}{2} \text {arctanh}(\sin (e+f x))+\frac {\sin (e+f x)}{2 \left (1-\sin ^2(e+f x)\right )}\right )+\frac {\sin (e+f x)}{4 \left (1-\sin ^2(e+f x)\right )^2}\right )+\frac {b (10 a+7 b) \sin (e+f x)}{6 \left (1-\sin ^2(e+f x)\right )^3}\right )+\frac {b \sin (e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{8 \left (1-\sin ^2(e+f x)\right )^4}}{f}\) |
Input:
Int[Sec[e + f*x]^5*(a + b*Sec[e + f*x]^2)^2,x]
Output:
((b*Sin[e + f*x]*(a + b - a*Sin[e + f*x]^2))/(8*(1 - Sin[e + f*x]^2)^4) + ((b*(10*a + 7*b)*Sin[e + f*x])/(6*(1 - Sin[e + f*x]^2)^3) + ((48*a^2 + 80* a*b + 35*b^2)*(Sin[e + f*x]/(4*(1 - Sin[e + f*x]^2)^2) + (3*(ArcTanh[Sin[e + f*x]]/2 + Sin[e + f*x]/(2*(1 - Sin[e + f*x]^2))))/4))/6)/8)/f
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_ ))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && In tegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Time = 2.61 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.18
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )+2 a b \left (-\left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )+\frac {5 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16}\right )+b^{2} \left (-\left (-\frac {\sec \left (f x +e \right )^{7}}{8}-\frac {7 \sec \left (f x +e \right )^{5}}{48}-\frac {35 \sec \left (f x +e \right )^{3}}{192}-\frac {35 \sec \left (f x +e \right )}{128}\right ) \tan \left (f x +e \right )+\frac {35 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{128}\right )}{f}\) | \(180\) |
| default | \(\frac {a^{2} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )+2 a b \left (-\left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )+\frac {5 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16}\right )+b^{2} \left (-\left (-\frac {\sec \left (f x +e \right )^{7}}{8}-\frac {7 \sec \left (f x +e \right )^{5}}{48}-\frac {35 \sec \left (f x +e \right )^{3}}{192}-\frac {35 \sec \left (f x +e \right )}{128}\right ) \tan \left (f x +e \right )+\frac {35 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{128}\right )}{f}\) | \(180\) |
| parts | \(\frac {a^{2} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )}{f}+\frac {b^{2} \left (-\left (-\frac {\sec \left (f x +e \right )^{7}}{8}-\frac {7 \sec \left (f x +e \right )^{5}}{48}-\frac {35 \sec \left (f x +e \right )^{3}}{192}-\frac {35 \sec \left (f x +e \right )}{128}\right ) \tan \left (f x +e \right )+\frac {35 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{128}\right )}{f}+\frac {2 a b \left (-\left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )+\frac {5 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16}\right )}{f}\) | \(185\) |
| parallelrisch | \(\frac {-8064 \left (a^{2}+\frac {5}{3} a b +\frac {35}{48} b^{2}\right ) \left (\frac {5}{8}+\frac {\cos \left (8 f x +8 e \right )}{56}+\frac {\cos \left (6 f x +6 e \right )}{7}+\frac {\cos \left (4 f x +4 e \right )}{2}+\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+8064 \left (a^{2}+\frac {5}{3} a b +\frac {35}{48} b^{2}\right ) \left (\frac {5}{8}+\frac {\cos \left (8 f x +8 e \right )}{56}+\frac {\cos \left (6 f x +6 e \right )}{7}+\frac {\cos \left (4 f x +4 e \right )}{2}+\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )+\left (4896 a^{2}+12256 a b +5362 b^{2}\right ) \sin \left (3 f x +3 e \right )+\left (2208 a^{2}+3680 a b +1610 b^{2}\right ) \sin \left (5 f x +5 e \right )+\left (288 a^{2}+480 a b +210 b^{2}\right ) \sin \left (7 f x +7 e \right )+2976 \left (a^{2}+\frac {283}{93} a b +\frac {163}{48} b^{2}\right ) \sin \left (f x +e \right )}{384 f \left (35+\cos \left (8 f x +8 e \right )+8 \cos \left (6 f x +6 e \right )+28 \cos \left (4 f x +4 e \right )+56 \cos \left (2 f x +2 e \right )\right )}\) | \(291\) |
| norman | \(\frac {\frac {\left (80 a^{2}+176 a b +93 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{64 f}+\frac {\left (80 a^{2}+176 a b +93 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{15}}{64 f}-\frac {\left (432 a^{2}+1360 a b -1085 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{192 f}-\frac {\left (432 a^{2}+1360 a b -1085 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{192 f}-\frac {\left (816 a^{2}+976 a b +91 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{192 f}-\frac {\left (816 a^{2}+976 a b +91 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{13}}{192 f}+\frac {\left (1008 a^{2}+1808 a b +1799 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{192 f}+\frac {\left (1008 a^{2}+1808 a b +1799 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{192 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{8}}-\frac {\left (48 a^{2}+80 a b +35 b^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{128 f}+\frac {\left (48 a^{2}+80 a b +35 b^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{128 f}\) | \(329\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (f x +e \right )} \left (144 a^{2} {\mathrm e}^{14 i \left (f x +e \right )}+240 a b \,{\mathrm e}^{14 i \left (f x +e \right )}+105 b^{2} {\mathrm e}^{14 i \left (f x +e \right )}+1104 \,{\mathrm e}^{12 i \left (f x +e \right )} a^{2}+1840 a b \,{\mathrm e}^{12 i \left (f x +e \right )}+805 b^{2} {\mathrm e}^{12 i \left (f x +e \right )}+2448 \,{\mathrm e}^{10 i \left (f x +e \right )} a^{2}+6128 \,{\mathrm e}^{10 i \left (f x +e \right )} a b +2681 b^{2} {\mathrm e}^{10 i \left (f x +e \right )}+1488 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}+4528 \,{\mathrm e}^{8 i \left (f x +e \right )} a b +5053 \,{\mathrm e}^{8 i \left (f x +e \right )} b^{2}-1488 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}-4528 a b \,{\mathrm e}^{6 i \left (f x +e \right )}-5053 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}-2448 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-6128 a b \,{\mathrm e}^{4 i \left (f x +e \right )}-2681 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-1104 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-1840 a b \,{\mathrm e}^{2 i \left (f x +e \right )}-805 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}-144 a^{2}-240 a b -105 b^{2}\right )}{192 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{8}}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) a^{2}}{8 f}+\frac {5 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) a b}{8 f}+\frac {35 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) b^{2}}{128 f}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) a^{2}}{8 f}-\frac {5 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) a b}{8 f}-\frac {35 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) b^{2}}{128 f}\) | \(500\) |
Input:
int(sec(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(a^2*(-(-1/4*sec(f*x+e)^3-3/8*sec(f*x+e))*tan(f*x+e)+3/8*ln(sec(f*x+e) +tan(f*x+e)))+2*a*b*(-(-1/6*sec(f*x+e)^5-5/24*sec(f*x+e)^3-5/16*sec(f*x+e) )*tan(f*x+e)+5/16*ln(sec(f*x+e)+tan(f*x+e)))+b^2*(-(-1/8*sec(f*x+e)^7-7/48 *sec(f*x+e)^5-35/192*sec(f*x+e)^3-35/128*sec(f*x+e))*tan(f*x+e)+35/128*ln( sec(f*x+e)+tan(f*x+e))))
Time = 0.09 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.10 \[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, {\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \cos \left (f x + e\right )^{8} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, {\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \cos \left (f x + e\right )^{8} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (3 \, {\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + 2 \, {\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (16 \, a b + 7 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 48 \, b^{2}\right )} \sin \left (f x + e\right )}{768 \, f \cos \left (f x + e\right )^{8}} \] Input:
integrate(sec(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")
Output:
1/768*(3*(48*a^2 + 80*a*b + 35*b^2)*cos(f*x + e)^8*log(sin(f*x + e) + 1) - 3*(48*a^2 + 80*a*b + 35*b^2)*cos(f*x + e)^8*log(-sin(f*x + e) + 1) + 2*(3 *(48*a^2 + 80*a*b + 35*b^2)*cos(f*x + e)^6 + 2*(48*a^2 + 80*a*b + 35*b^2)* cos(f*x + e)^4 + 8*(16*a*b + 7*b^2)*cos(f*x + e)^2 + 48*b^2)*sin(f*x + e)) /(f*cos(f*x + e)^8)
\[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \sec ^{5}{\left (e + f x \right )}\, dx \] Input:
integrate(sec(f*x+e)**5*(a+b*sec(f*x+e)**2)**2,x)
Output:
Integral((a + b*sec(e + f*x)**2)**2*sec(e + f*x)**5, x)
Time = 0.04 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.31 \[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, {\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, {\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (f x + e\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \sin \left (f x + e\right )^{7} - 11 \, {\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \sin \left (f x + e\right )^{5} + {\left (624 \, a^{2} + 1168 \, a b + 511 \, b^{2}\right )} \sin \left (f x + e\right )^{3} - 3 \, {\left (80 \, a^{2} + 176 \, a b + 93 \, b^{2}\right )} \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{8} - 4 \, \sin \left (f x + e\right )^{6} + 6 \, \sin \left (f x + e\right )^{4} - 4 \, \sin \left (f x + e\right )^{2} + 1}}{768 \, f} \] Input:
integrate(sec(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")
Output:
1/768*(3*(48*a^2 + 80*a*b + 35*b^2)*log(sin(f*x + e) + 1) - 3*(48*a^2 + 80 *a*b + 35*b^2)*log(sin(f*x + e) - 1) - 2*(3*(48*a^2 + 80*a*b + 35*b^2)*sin (f*x + e)^7 - 11*(48*a^2 + 80*a*b + 35*b^2)*sin(f*x + e)^5 + (624*a^2 + 11 68*a*b + 511*b^2)*sin(f*x + e)^3 - 3*(80*a^2 + 176*a*b + 93*b^2)*sin(f*x + e))/(sin(f*x + e)^8 - 4*sin(f*x + e)^6 + 6*sin(f*x + e)^4 - 4*sin(f*x + e )^2 + 1))/f
Time = 0.17 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.44 \[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, {\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \log \left ({\left | \sin \left (f x + e\right ) + 1 \right |}\right ) - 3 \, {\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \log \left ({\left | \sin \left (f x + e\right ) - 1 \right |}\right ) - \frac {2 \, {\left (144 \, a^{2} \sin \left (f x + e\right )^{7} + 240 \, a b \sin \left (f x + e\right )^{7} + 105 \, b^{2} \sin \left (f x + e\right )^{7} - 528 \, a^{2} \sin \left (f x + e\right )^{5} - 880 \, a b \sin \left (f x + e\right )^{5} - 385 \, b^{2} \sin \left (f x + e\right )^{5} + 624 \, a^{2} \sin \left (f x + e\right )^{3} + 1168 \, a b \sin \left (f x + e\right )^{3} + 511 \, b^{2} \sin \left (f x + e\right )^{3} - 240 \, a^{2} \sin \left (f x + e\right ) - 528 \, a b \sin \left (f x + e\right ) - 279 \, b^{2} \sin \left (f x + e\right )\right )}}{{\left (\sin \left (f x + e\right )^{2} - 1\right )}^{4}}}{768 \, f} \] Input:
integrate(sec(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")
Output:
1/768*(3*(48*a^2 + 80*a*b + 35*b^2)*log(abs(sin(f*x + e) + 1)) - 3*(48*a^2 + 80*a*b + 35*b^2)*log(abs(sin(f*x + e) - 1)) - 2*(144*a^2*sin(f*x + e)^7 + 240*a*b*sin(f*x + e)^7 + 105*b^2*sin(f*x + e)^7 - 528*a^2*sin(f*x + e)^ 5 - 880*a*b*sin(f*x + e)^5 - 385*b^2*sin(f*x + e)^5 + 624*a^2*sin(f*x + e) ^3 + 1168*a*b*sin(f*x + e)^3 + 511*b^2*sin(f*x + e)^3 - 240*a^2*sin(f*x + e) - 528*a*b*sin(f*x + e) - 279*b^2*sin(f*x + e))/(sin(f*x + e)^2 - 1)^4)/ f
Time = 15.90 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.11 \[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {\left (-\frac {3\,a^2}{8}-\frac {5\,a\,b}{8}-\frac {35\,b^2}{128}\right )\,{\sin \left (e+f\,x\right )}^7+\left (\frac {11\,a^2}{8}+\frac {55\,a\,b}{24}+\frac {385\,b^2}{384}\right )\,{\sin \left (e+f\,x\right )}^5+\left (-\frac {13\,a^2}{8}-\frac {73\,a\,b}{24}-\frac {511\,b^2}{384}\right )\,{\sin \left (e+f\,x\right )}^3+\left (\frac {5\,a^2}{8}+\frac {11\,a\,b}{8}+\frac {93\,b^2}{128}\right )\,\sin \left (e+f\,x\right )}{f\,\left ({\sin \left (e+f\,x\right )}^8-4\,{\sin \left (e+f\,x\right )}^6+6\,{\sin \left (e+f\,x\right )}^4-4\,{\sin \left (e+f\,x\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\sin \left (e+f\,x\right )\right )\,\left (\frac {3\,a^2}{8}+\frac {5\,a\,b}{8}+\frac {35\,b^2}{128}\right )}{f} \] Input:
int((a + b/cos(e + f*x)^2)^2/cos(e + f*x)^5,x)
Output:
(sin(e + f*x)*((11*a*b)/8 + (5*a^2)/8 + (93*b^2)/128) - sin(e + f*x)^7*((5 *a*b)/8 + (3*a^2)/8 + (35*b^2)/128) + sin(e + f*x)^5*((55*a*b)/24 + (11*a^ 2)/8 + (385*b^2)/384) - sin(e + f*x)^3*((73*a*b)/24 + (13*a^2)/8 + (511*b^ 2)/384))/(f*(6*sin(e + f*x)^4 - 4*sin(e + f*x)^2 - 4*sin(e + f*x)^6 + sin( e + f*x)^8 + 1)) + (atanh(sin(e + f*x))*((5*a*b)/8 + (3*a^2)/8 + (35*b^2)/ 128))/f
Time = 0.17 (sec) , antiderivative size = 886, normalized size of antiderivative = 5.79 \[ \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx =\text {Too large to display} \] Input:
int(sec(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x)
Output:
( - 144*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**8*a**2 - 240*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**8*a*b - 105*log(tan((e + f*x)/2) - 1)*sin(e + f *x)**8*b**2 + 576*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**6*a**2 + 960*log (tan((e + f*x)/2) - 1)*sin(e + f*x)**6*a*b + 420*log(tan((e + f*x)/2) - 1) *sin(e + f*x)**6*b**2 - 864*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**4*a**2 - 1440*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**4*a*b - 630*log(tan((e + f *x)/2) - 1)*sin(e + f*x)**4*b**2 + 576*log(tan((e + f*x)/2) - 1)*sin(e + f *x)**2*a**2 + 960*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**2*a*b + 420*log( tan((e + f*x)/2) - 1)*sin(e + f*x)**2*b**2 - 144*log(tan((e + f*x)/2) - 1) *a**2 - 240*log(tan((e + f*x)/2) - 1)*a*b - 105*log(tan((e + f*x)/2) - 1)* b**2 + 144*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**8*a**2 + 240*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**8*a*b + 105*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**8*b**2 - 576*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**6*a**2 - 960* log(tan((e + f*x)/2) + 1)*sin(e + f*x)**6*a*b - 420*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**6*b**2 + 864*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**4*a **2 + 1440*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**4*a*b + 630*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**4*b**2 - 576*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**2*a**2 - 960*log(tan((e + f*x)/2) + 1)*sin(e + f*x)**2*a*b - 420*l og(tan((e + f*x)/2) + 1)*sin(e + f*x)**2*b**2 + 144*log(tan((e + f*x)/2) + 1)*a**2 + 240*log(tan((e + f*x)/2) + 1)*a*b + 105*log(tan((e + f*x)/2)...