Integrand size = 21, antiderivative size = 73 \[ \int \frac {\sin ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\cos ^3(c+d x)}{3 a d}-\frac {2 \cos ^5(c+d x)}{5 a d}+\frac {\cos ^7(c+d x)}{7 a d}+\frac {\sin ^6(c+d x)}{6 a d} \] Output:
1/3*cos(d*x+c)^3/a/d-2/5*cos(d*x+c)^5/a/d+1/7*cos(d*x+c)^7/a/d+1/6*sin(d*x +c)^6/a/d
Time = 0.76 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.71 \[ \int \frac {\sin ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {4 (123+197 \cos (c+d x)+85 \cos (2 (c+d x))+15 \cos (3 (c+d x))) \sin ^8\left (\frac {1}{2} (c+d x)\right )}{105 a d} \] Input:
Integrate[Sin[c + d*x]^7/(a + a*Sec[c + d*x]),x]
Output:
(4*(123 + 197*Cos[c + d*x] + 85*Cos[2*(c + d*x)] + 15*Cos[3*(c + d*x)])*Si n[(c + d*x)/2]^8)/(105*a*d)
Time = 0.51 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4360, 25, 25, 3042, 25, 3314, 25, 3042, 3044, 15, 3045, 244, 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.
\(\displaystyle \int \frac {\sin ^7(c+d x)}{a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos \left (c+d x-\frac {\pi }{2}\right )^7}{a-a \csc \left (c+d x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -\frac {\sin ^7(c+d x) \cos (c+d x)}{a (-\cos (c+d x))-a}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\frac {\cos (c+d x) \sin ^7(c+d x)}{\cos (c+d x) a+a}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\sin ^7(c+d x) \cos (c+d x)}{a \cos (c+d x)+a}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \cos \left (c+d x+\frac {\pi }{2}\right )^7}{a \sin \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^7 \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sin \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a}dx\) |
\(\Big \downarrow \) 3314 |
\(\displaystyle \frac {\int -\cos ^2(c+d x) \sin ^5(c+d x)dx}{a}-\frac {\int -\cos (c+d x) \sin ^5(c+d x)dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \cos (c+d x) \sin ^5(c+d x)dx}{a}-\frac {\int \cos ^2(c+d x) \sin ^5(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \cos (c+d x) \sin (c+d x)^5dx}{a}-\frac {\int \cos (c+d x)^2 \sin (c+d x)^5dx}{a}\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {\int \sin ^5(c+d x)d\sin (c+d x)}{a d}-\frac {\int \cos (c+d x)^2 \sin (c+d x)^5dx}{a}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\sin ^6(c+d x)}{6 a d}-\frac {\int \cos (c+d x)^2 \sin (c+d x)^5dx}{a}\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle \frac {\int \cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )^2d\cos (c+d x)}{a d}+\frac {\sin ^6(c+d x)}{6 a d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\cos ^6(c+d x)-2 \cos ^4(c+d x)+\cos ^2(c+d x)\right )d\cos (c+d x)}{a d}+\frac {\sin ^6(c+d x)}{6 a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sin ^6(c+d x)}{6 a d}+\frac {\frac {1}{7} \cos ^7(c+d x)-\frac {2}{5} \cos ^5(c+d x)+\frac {1}{3} \cos ^3(c+d x)}{a d}\) |
Input:
Int[Sin[c + d*x]^7/(a + a*Sec[c + d*x]),x]
Output:
(Cos[c + d*x]^3/3 - (2*Cos[c + d*x]^5)/5 + Cos[c + d*x]^7/7)/(a*d) + Sin[c + d*x]^6/(6*a*d)
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(a*f)^(-1) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/(( a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/a Int[Cos[e + f *x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[1/(b*d) Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] & & IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n, -p]))
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 0.92 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {\frac {\cos \left (d x +c \right )^{7}}{7}-\frac {\cos \left (d x +c \right )^{6}}{6}-\frac {2 \cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{4}}{2}+\frac {\cos \left (d x +c \right )^{3}}{3}-\frac {\cos \left (d x +c \right )^{2}}{2}}{d a}\) | \(69\) |
default | \(\frac {\frac {\cos \left (d x +c \right )^{7}}{7}-\frac {\cos \left (d x +c \right )^{6}}{6}-\frac {2 \cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{4}}{2}+\frac {\cos \left (d x +c \right )^{3}}{3}-\frac {\cos \left (d x +c \right )^{2}}{2}}{d a}\) | \(69\) |
parallelrisch | \(\frac {210 \cos \left (4 d x +4 c \right )+862-525 \cos \left (2 d x +2 c \right )-35 \cos \left (6 d x +6 c \right )+525 \cos \left (d x +c \right )+35 \cos \left (3 d x +3 c \right )-63 \cos \left (5 d x +5 c \right )+15 \cos \left (7 d x +7 c \right )}{6720 d a}\) | \(85\) |
norman | \(\frac {\frac {16}{105 a d}+\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{15 d a}+\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5 d a}+\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d a}+\frac {64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7}}\) | \(102\) |
risch | \(\frac {5 \cos \left (d x +c \right )}{64 a d}+\frac {\cos \left (7 d x +7 c \right )}{448 a d}-\frac {\cos \left (6 d x +6 c \right )}{192 a d}-\frac {3 \cos \left (5 d x +5 c \right )}{320 a d}+\frac {\cos \left (4 d x +4 c \right )}{32 a d}+\frac {\cos \left (3 d x +3 c \right )}{192 a d}-\frac {5 \cos \left (2 d x +2 c \right )}{64 a d}\) | \(118\) |
Input:
int(sin(d*x+c)^7/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/7*cos(d*x+c)^7-1/6*cos(d*x+c)^6-2/5*cos(d*x+c)^5+1/2*cos(d*x+c)^4 +1/3*cos(d*x+c)^3-1/2*cos(d*x+c)^2)
Time = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {30 \, \cos \left (d x + c\right )^{7} - 35 \, \cos \left (d x + c\right )^{6} - 84 \, \cos \left (d x + c\right )^{5} + 105 \, \cos \left (d x + c\right )^{4} + 70 \, \cos \left (d x + c\right )^{3} - 105 \, \cos \left (d x + c\right )^{2}}{210 \, a d} \] Input:
integrate(sin(d*x+c)^7/(a+a*sec(d*x+c)),x, algorithm="fricas")
Output:
1/210*(30*cos(d*x + c)^7 - 35*cos(d*x + c)^6 - 84*cos(d*x + c)^5 + 105*cos (d*x + c)^4 + 70*cos(d*x + c)^3 - 105*cos(d*x + c)^2)/(a*d)
Timed out. \[ \int \frac {\sin ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)**7/(a+a*sec(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {30 \, \cos \left (d x + c\right )^{7} - 35 \, \cos \left (d x + c\right )^{6} - 84 \, \cos \left (d x + c\right )^{5} + 105 \, \cos \left (d x + c\right )^{4} + 70 \, \cos \left (d x + c\right )^{3} - 105 \, \cos \left (d x + c\right )^{2}}{210 \, a d} \] Input:
integrate(sin(d*x+c)^7/(a+a*sec(d*x+c)),x, algorithm="maxima")
Output:
1/210*(30*cos(d*x + c)^7 - 35*cos(d*x + c)^6 - 84*cos(d*x + c)^5 + 105*cos (d*x + c)^4 + 70*cos(d*x + c)^3 - 105*cos(d*x + c)^2)/(a*d)
Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {30 \, \cos \left (d x + c\right )^{7} - 35 \, \cos \left (d x + c\right )^{6} - 84 \, \cos \left (d x + c\right )^{5} + 105 \, \cos \left (d x + c\right )^{4} + 70 \, \cos \left (d x + c\right )^{3} - 105 \, \cos \left (d x + c\right )^{2}}{210 \, a d} \] Input:
integrate(sin(d*x+c)^7/(a+a*sec(d*x+c)),x, algorithm="giac")
Output:
1/210*(30*cos(d*x + c)^7 - 35*cos(d*x + c)^6 - 84*cos(d*x + c)^5 + 105*cos (d*x + c)^4 + 70*cos(d*x + c)^3 - 105*cos(d*x + c)^2)/(a*d)
Time = 10.18 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.15 \[ \int \frac {\sin ^7(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {{\cos \left (c+d\,x\right )}^2}{2\,a}-\frac {{\cos \left (c+d\,x\right )}^3}{3\,a}-\frac {{\cos \left (c+d\,x\right )}^4}{2\,a}+\frac {2\,{\cos \left (c+d\,x\right )}^5}{5\,a}+\frac {{\cos \left (c+d\,x\right )}^6}{6\,a}-\frac {{\cos \left (c+d\,x\right )}^7}{7\,a}}{d} \] Input:
int(sin(c + d*x)^7/(a + a/cos(c + d*x)),x)
Output:
-(cos(c + d*x)^2/(2*a) - cos(c + d*x)^3/(3*a) - cos(c + d*x)^4/(2*a) + (2* cos(c + d*x)^5)/(5*a) + cos(c + d*x)^6/(6*a) - cos(c + d*x)^7/(7*a))/d
Time = 0.18 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int \frac {\sin ^7(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+16 \cos \left (d x +c \right )+35 \sin \left (d x +c \right )^{6}-16}{210 a d} \] Input:
int(sin(d*x+c)^7/(a+a*sec(d*x+c)),x)
Output:
( - 30*cos(c + d*x)*sin(c + d*x)**6 + 6*cos(c + d*x)*sin(c + d*x)**4 + 8*c os(c + d*x)*sin(c + d*x)**2 + 16*cos(c + d*x) + 35*sin(c + d*x)**6 - 16)/( 210*a*d)