Integrand size = 21, antiderivative size = 117 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^5}{12 d (a-a \sin (c+d x))^3}+\frac {a^4}{8 d (a-a \sin (c+d x))^2}+\frac {3 a^7}{16 d \left (a^5-a^5 \sin (c+d x)\right )}-\frac {a^7}{16 d \left (a^5+a^5 \sin (c+d x)\right )} \] Output:
1/4*a^2*arctanh(sin(d*x+c))/d+1/12*a^5/d/(a-a*sin(d*x+c))^3+1/8*a^4/d/(a-a *sin(d*x+c))^2+3/16*a^7/d/(a^5-a^5*sin(d*x+c))-1/16*a^7/d/(a^5+a^5*sin(d*x +c))
Time = 0.16 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.73 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \sec ^6(c+d x) (1+\sin (c+d x))^2 \left (-4-\sin (c+d x)+6 \sin ^2(c+d x)-3 \sin ^3(c+d x)+3 \text {arctanh}(\sin (c+d x)) (-1+\sin (c+d x))^3 (1+\sin (c+d x))\right )}{12 d} \] Input:
Integrate[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2,x]
Output:
-1/12*(a^2*Sec[c + d*x]^6*(1 + Sin[c + d*x])^2*(-4 - Sin[c + d*x] + 6*Sin[ c + d*x]^2 - 3*Sin[c + d*x]^3 + 3*ArcTanh[Sin[c + d*x]]*(-1 + Sin[c + d*x] )^3*(1 + Sin[c + d*x])))/d
Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3146, 54, 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.
| \(\displaystyle \int \sec ^7(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^2}{\cos (c+d x)^7}dx\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {a^7 \int \frac {1}{(a-a \sin (c+d x))^4 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {a^7 \int \left (\frac {3}{16 a^4 (a-a \sin (c+d x))^2}+\frac {1}{16 a^4 (\sin (c+d x) a+a)^2}+\frac {1}{4 a^3 (a-a \sin (c+d x))^3}+\frac {1}{4 a^2 (a-a \sin (c+d x))^4}+\frac {1}{4 a^4 \left (a^2-a^2 \sin ^2(c+d x)\right )}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^7 \left (\frac {\text {arctanh}(\sin (c+d x))}{4 a^5}+\frac {3}{16 a^4 (a-a \sin (c+d x))}-\frac {1}{16 a^4 (a \sin (c+d x)+a)}+\frac {1}{8 a^3 (a-a \sin (c+d x))^2}+\frac {1}{12 a^2 (a-a \sin (c+d x))^3}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2,x]
Output:
(a^7*(ArcTanh[Sin[c + d*x]]/(4*a^5) + 1/(12*a^2*(a - a*Sin[c + d*x])^3) + 1/(8*a^3*(a - a*Sin[c + d*x])^2) + 3/(16*a^4*(a - a*Sin[c + d*x])) - 1/(16 *a^4*(a + a*Sin[c + d*x]))))/d
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 0.62 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.37
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {a^{2}}{3 \cos \left (d x +c \right )^{6}}+a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\) | \(160\) |
| default | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {a^{2}}{3 \cos \left (d x +c \right )^{6}}+a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\) | \(160\) |
| risch | \(-\frac {i a^{2} \left (-12 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}-8 i {\mathrm e}^{4 i \left (d x +c \right )}-13 \,{\mathrm e}^{5 i \left (d x +c \right )}-12 i {\mathrm e}^{2 i \left (d x +c \right )}+13 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{6 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}\) | \(162\) |
| parallelrisch | \(-\frac {\left (\left (\frac {5}{4}-\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-\frac {5}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}-\cos \left (2 d x +2 c \right )+\sin \left (d x +c \right )+\sin \left (3 d x +3 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 \cos \left (2 d x +2 c \right )}{3}-\frac {\cos \left (4 d x +4 c \right )}{3}-\frac {7 \sin \left (d x +c \right )}{2}-\frac {5 \sin \left (3 d x +3 c \right )}{6}+1\right ) a^{2}}{d \left (-\cos \left (4 d x +4 c \right )+5+4 \cos \left (2 d x +2 c \right )-4 \sin \left (3 d x +3 c \right )-4 \sin \left (d x +c \right )\right )}\) | \(203\) |
| norman | \(\frac {\frac {3 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {31 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}+\frac {77 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 d}+\frac {139 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {139 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 d}+\frac {77 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 d}+\frac {31 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{6 d}+\frac {3 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{2 d}+\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{d}+\frac {8 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {8 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {80 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d}+\frac {52 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}+\frac {52 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{6} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\) | \(357\) |
Input:
int(sec(d*x+c)^7*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(1/6*sin(d*x+c)^3/cos(d*x+c)^6+1/8*sin(d*x+c)^3/cos(d*x+c)^4+1/16 *sin(d*x+c)^3/cos(d*x+c)^2+1/16*sin(d*x+c)-1/16*ln(sec(d*x+c)+tan(d*x+c))) +1/3*a^2/cos(d*x+c)^6+a^2*(-(-1/6*sec(d*x+c)^5-5/24*sec(d*x+c)^3-5/16*sec( d*x+c))*tan(d*x+c)+5/16*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.74 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {12 \, a^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{2} - 3 \, {\left (a^{2} \cos \left (d x + c\right )^{4} + 2 \, a^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (a^{2} \cos \left (d x + c\right )^{4} + 2 \, a^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{2}\right )} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right )^{4} + 2 \, d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="fricas")
Output:
-1/24*(12*a^2*cos(d*x + c)^2 - 4*a^2 - 3*(a^2*cos(d*x + c)^4 + 2*a^2*cos(d *x + c)^2*sin(d*x + c) - 2*a^2*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + 3*( a^2*cos(d*x + c)^4 + 2*a^2*cos(d*x + c)^2*sin(d*x + c) - 2*a^2*cos(d*x + c )^2)*log(-sin(d*x + c) + 1) - 2*(3*a^2*cos(d*x + c)^2 - 4*a^2)*sin(d*x + c ))/(d*cos(d*x + c)^4 + 2*d*cos(d*x + c)^2*sin(d*x + c) - 2*d*cos(d*x + c)^ 2)
Timed out. \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**7*(a+a*sin(d*x+c))**2,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 \, a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, a^{2} \sin \left (d x + c\right )^{3} - 6 \, a^{2} \sin \left (d x + c\right )^{2} + a^{2} \sin \left (d x + c\right ) + 4 \, a^{2}\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{3} + 2 \, \sin \left (d x + c\right ) - 1}}{24 \, d} \] Input:
integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="maxima")
Output:
1/24*(3*a^2*log(sin(d*x + c) + 1) - 3*a^2*log(sin(d*x + c) - 1) - 2*(3*a^2 *sin(d*x + c)^3 - 6*a^2*sin(d*x + c)^2 + a^2*sin(d*x + c) + 4*a^2)/(sin(d* x + c)^4 - 2*sin(d*x + c)^3 + 2*sin(d*x + c) - 1))/d
Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.76 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {1}{24} \, a^{2} {\left (\frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{d} - \frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 6 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right ) + 4\right )}}{d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}^{3}}\right )} \] Input:
integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
1/24*a^2*(3*log(abs(sin(d*x + c) + 1))/d - 3*log(abs(sin(d*x + c) - 1))/d - 2*(3*sin(d*x + c)^3 - 6*sin(d*x + c)^2 + sin(d*x + c) + 4)/(d*(sin(d*x + c) + 1)*(sin(d*x + c) - 1)^3))
Time = 25.65 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.80 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{4\,d}-\frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^3}{4}-\frac {a^2\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {a^2\,\sin \left (c+d\,x\right )}{12}+\frac {a^2}{3}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^3+2\,\sin \left (c+d\,x\right )-1\right )} \] Input:
int((a + a*sin(c + d*x))^2/cos(c + d*x)^7,x)
Output:
(a^2*atanh(sin(c + d*x)))/(4*d) - ((a^2*sin(c + d*x))/12 + a^2/3 - (a^2*si n(c + d*x)^2)/2 + (a^2*sin(c + d*x)^3)/4)/(d*(2*sin(c + d*x) - 2*sin(c + d *x)^3 + sin(c + d*x)^4 - 1))
Time = 0.17 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.91 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \left (-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\sin \left (d x +c \right )^{4}-8 \sin \left (d x +c \right )^{3}+12 \sin \left (d x +c \right )^{2}-9\right )}{24 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )-1\right )} \] Input:
int(sec(d*x+c)^7*(a+a*sin(d*x+c))^2,x)
Output:
(a**2*( - 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 12*log(tan((c + d* x)/2) - 1)*sin(c + d*x)**3 - 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x) + 6 *log(tan((c + d*x)/2) - 1) + 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 + 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x) - 6*log(tan((c + d*x)/2) + 1) + sin(c + d*x)**4 - 8*sin(c + d*x)**3 + 12*sin(c + d*x)**2 - 9))/(24*d*(sin(c + d*x)**4 - 2*sin(c + d* x)**3 + 2*sin(c + d*x) - 1))