Integrand size = 21, antiderivative size = 87 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {7 x}{8 a^2}-\frac {2 \sin (c+d x)}{a^2 d}+\frac {7 \cos (c+d x) \sin (c+d x)}{8 a^2 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a^2 d}+\frac {2 \sin ^3(c+d x)}{3 a^2 d} \] Output:
7/8*x/a^2-2*sin(d*x+c)/a^2/d+7/8*cos(d*x+c)*sin(d*x+c)/a^2/d+1/4*cos(d*x+c )^3*sin(d*x+c)/a^2/d+2/3*sin(d*x+c)^3/a^2/d
Time = 0.35 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (84 d x-144 \sin (c+d x)+48 \sin (2 (c+d x))-16 \sin (3 (c+d x))+3 \sin (4 (c+d x))+2 \tan \left (\frac {c}{2}\right )\right )}{24 a^2 d (1+\sec (c+d x))^2} \] Input:
Integrate[Sin[c + d*x]^4/(a + a*Sec[c + d*x])^2,x]
Output:
(Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*(84*d*x - 144*Sin[c + d*x] + 48*Sin[2*( c + d*x)] - 16*Sin[3*(c + d*x)] + 3*Sin[4*(c + d*x)] + 2*Tan[c/2]))/(24*a^ 2*d*(1 + Sec[c + d*x])^2)
Time = 0.53 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4360, 3042, 3348, 3042, 3236, 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.
\(\displaystyle \int \frac {\sin ^4(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos \left (c+d x-\frac {\pi }{2}\right )^4}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \frac {\sin ^4(c+d x) \cos ^2(c+d x)}{(a (-\cos (c+d x))-a)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \cos \left (c+d x+\frac {\pi }{2}\right )^4}{\left (a \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )-a\right )^2}dx\) |
\(\Big \downarrow \) 3348 |
\(\displaystyle \frac {\int \cos ^2(c+d x) (a-a \cos (c+d x))^2dx}{a^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a-a \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2dx}{a^4}\) |
\(\Big \downarrow \) 3236 |
\(\displaystyle \frac {\int \left (a^2 \cos ^4(c+d x)-2 a^2 \cos ^3(c+d x)+a^2 \cos ^2(c+d x)\right )dx}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {7 a^2 x}{8}}{a^4}\) |
Input:
Int[Sin[c + d*x]^4/(a + a*Sec[c + d*x])^2,x]
Output:
((7*a^2*x)/8 - (2*a^2*Sin[c + d*x])/d + (7*a^2*Cos[c + d*x]*Sin[c + d*x])/ (8*d) + (a^2*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (2*a^2*Sin[c + d*x]^3)/( 3*d))/a^4
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IGt Q[m, 0] && RationalQ[n]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[a^(2*m) Int[(d* Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]
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 = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.63
method | result | size |
parallelrisch | \(\frac {84 d x -16 \sin \left (3 d x +3 c \right )-144 \sin \left (d x +c \right )+3 \sin \left (4 d x +4 c \right )+48 \sin \left (2 d x +2 c \right )}{96 d \,a^{2}}\) | \(55\) |
risch | \(\frac {7 x}{8 a^{2}}-\frac {3 \sin \left (d x +c \right )}{2 a^{2} d}+\frac {\sin \left (4 d x +4 c \right )}{32 d \,a^{2}}-\frac {\sin \left (3 d x +3 c \right )}{6 d \,a^{2}}+\frac {\sin \left (2 d x +2 c \right )}{2 d \,a^{2}}\) | \(73\) |
derivativedivides | \(\frac {\frac {8 \left (-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{32}-\frac {83 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{96}-\frac {77 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{96}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {7 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d \,a^{2}}\) | \(89\) |
default | \(\frac {\frac {8 \left (-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{32}-\frac {83 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{96}-\frac {77 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{96}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {7 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{d \,a^{2}}\) | \(89\) |
norman | \(\frac {\frac {7 x}{8 a}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {77 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}-\frac {83 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 a d}-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 a d}+\frac {7 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}+\frac {21 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 a}+\frac {7 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 a}+\frac {7 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8 a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a}\) | \(169\) |
Input:
int(sin(d*x+c)^4/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/96*(84*d*x-16*sin(3*d*x+3*c)-144*sin(d*x+c)+3*sin(4*d*x+4*c)+48*sin(2*d* x+2*c))/d/a^2
Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.57 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {21 \, d x + {\left (6 \, \cos \left (d x + c\right )^{3} - 16 \, \cos \left (d x + c\right )^{2} + 21 \, \cos \left (d x + c\right ) - 32\right )} \sin \left (d x + c\right )}{24 \, a^{2} d} \] Input:
integrate(sin(d*x+c)^4/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
Output:
1/24*(21*d*x + (6*cos(d*x + c)^3 - 16*cos(d*x + c)^2 + 21*cos(d*x + c) - 3 2)*sin(d*x + c))/(a^2*d)
\[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\sin ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(sin(d*x+c)**4/(a+a*sec(d*x+c))**2,x)
Output:
Integral(sin(c + d*x)**4/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2
Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (79) = 158\).
Time = 0.11 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.37 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {83 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {75 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {21 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{12 \, d} \] Input:
integrate(sin(d*x+c)^4/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
Output:
-1/12*((21*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 83*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 75*sin(d*x + c)^7/(co s(d*x + c) + 1)^7)/(a^2 + 4*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a^ 2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) - 21*arctan(sin(d*x + c )/(cos(d*x + c) + 1))/a^2)/d
Time = 0.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {21 \, {\left (d x + c\right )}}{a^{2}} - \frac {2 \, {\left (75 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 83 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 77 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{2}}}{24 \, d} \] Input:
integrate(sin(d*x+c)^4/(a+a*sec(d*x+c))^2,x, algorithm="giac")
Output:
1/24*(21*(d*x + c)/a^2 - 2*(75*tan(1/2*d*x + 1/2*c)^7 + 83*tan(1/2*d*x + 1 /2*c)^5 + 77*tan(1/2*d*x + 1/2*c)^3 + 21*tan(1/2*d*x + 1/2*c))/((tan(1/2*d *x + 1/2*c)^2 + 1)^4*a^2))/d
Time = 14.47 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {7\,x}{8\,a^2}-\frac {\frac {25\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {83\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {77\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}+\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{a^2\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \] Input:
int(sin(c + d*x)^4/(a + a/cos(c + d*x))^2,x)
Output:
(7*x)/(8*a^2) - ((7*tan(c/2 + (d*x)/2))/4 + (77*tan(c/2 + (d*x)/2)^3)/12 + (83*tan(c/2 + (d*x)/2)^5)/12 + (25*tan(c/2 + (d*x)/2)^7)/4)/(a^2*d*(tan(c /2 + (d*x)/2)^2 + 1)^4)
Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.70 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+27 \cos \left (d x +c \right ) \sin \left (d x +c \right )+16 \sin \left (d x +c \right )^{3}-48 \sin \left (d x +c \right )+21 d x}{24 a^{2} d} \] Input:
int(sin(d*x+c)^4/(a+a*sec(d*x+c))^2,x)
Output:
( - 6*cos(c + d*x)*sin(c + d*x)**3 + 27*cos(c + d*x)*sin(c + d*x) + 16*sin (c + d*x)**3 - 48*sin(c + d*x) + 21*d*x)/(24*a**2*d)