Integrand size = 25, antiderivative size = 95 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {5 a x}{2}-\frac {a \cos (c+d x)}{d}-\frac {2 a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 d}-\frac {2 a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d} \] Output:
5/2*a*x-a*cos(d*x+c)/d-2*a*sec(d*x+c)/d+1/3*a*sec(d*x+c)^3/d-1/2*a*cos(d*x +c)*sin(d*x+c)/d-2*a*tan(d*x+c)/d+1/3*a*tan(d*x+c)^3/d
Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.80 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {a \left (30 c+30 d x-12 \cos (c+d x)-24 \sec (c+d x)+4 \sec ^3(c+d x)-3 \sin (2 (c+d x))-28 \tan (c+d x)+4 \sec ^2(c+d x) \tan (c+d x)\right )}{12 d} \] Input:
Integrate[Sin[c + d*x]*(a + a*Sin[c + d*x])*Tan[c + d*x]^4,x]
Output:
(a*(30*c + 30*d*x - 12*Cos[c + d*x] - 24*Sec[c + d*x] + 4*Sec[c + d*x]^3 - 3*Sin[2*(c + d*x)] - 28*Tan[c + d*x] + 4*Sec[c + d*x]^2*Tan[c + d*x]))/(1 2*d)
Time = 0.44 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3317, 3042, 3070, 244, 2009, 3071, 252, 254, 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.
| \(\displaystyle \int \sin (c+d x) \tan ^4(c+d x) (a \sin (c+d x)+a) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^5 (a \sin (c+d x)+a)}{\cos (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3317 |
| \(\displaystyle a \int \sin ^2(c+d x) \tan ^4(c+d x)dx+a \int \sin (c+d x) \tan ^4(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \sin (c+d x) \tan (c+d x)^4dx+a \int \sin (c+d x)^2 \tan (c+d x)^4dx\) |
| \(\Big \downarrow \) 3070 |
| \(\displaystyle a \int \sin (c+d x)^2 \tan (c+d x)^4dx-\frac {a \int \left (1-\cos ^2(c+d x)\right )^2 \sec ^4(c+d x)d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle a \int \sin (c+d x)^2 \tan (c+d x)^4dx-\frac {a \int \left (\sec ^4(c+d x)-2 \sec ^2(c+d x)+1\right )d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a \int \sin (c+d x)^2 \tan (c+d x)^4dx-\frac {a \left (\cos (c+d x)-\frac {1}{3} \sec ^3(c+d x)+2 \sec (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3071 |
| \(\displaystyle \frac {a \int \frac {\tan ^6(c+d x)}{\left (\tan ^2(c+d x)+1\right )^2}d\tan (c+d x)}{d}-\frac {a \left (\cos (c+d x)-\frac {1}{3} \sec ^3(c+d x)+2 \sec (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {a \left (\frac {5}{2} \int \frac {\tan ^4(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)-\frac {\tan ^5(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d}-\frac {a \left (\cos (c+d x)-\frac {1}{3} \sec ^3(c+d x)+2 \sec (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle \frac {a \left (\frac {5}{2} \int \left (\tan ^2(c+d x)+\frac {1}{\tan ^2(c+d x)+1}-1\right )d\tan (c+d x)-\frac {\tan ^5(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d}-\frac {a \left (\cos (c+d x)-\frac {1}{3} \sec ^3(c+d x)+2 \sec (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \left (\frac {5}{2} \left (\arctan (\tan (c+d x))+\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )-\frac {\tan ^5(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d}-\frac {a \left (\cos (c+d x)-\frac {1}{3} \sec ^3(c+d x)+2 \sec (c+d x)\right )}{d}\) |
Input:
Int[Sin[c + d*x]*(a + a*Sin[c + d*x])*Tan[c + d*x]^4,x]
Output:
-((a*(Cos[c + d*x] + 2*Sec[c + d*x] - Sec[c + d*x]^3/3))/d) + (a*(-1/2*Tan [c + d*x]^5/(1 + Tan[c + d*x]^2) + (5*(ArcTan[Tan[c + d*x]] - Tan[c + d*x] + Tan[c + d*x]^3/3))/2))/d
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Simp[-f^(-1) Subst[Int[(1 - x^2)^((m + n - 1)/2)/x^n, x], x, Cos[e + f *x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[b*(ff/f) Subst[I nt[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[(g*Co s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[(g*Cos[e + f*x])^ p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
Result contains complex when optimal does not.
Time = 1.48 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.46
| method | result | size |
| risch | \(\frac {5 a x}{2}+\frac {i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 a \left (-3 i {\mathrm e}^{2 i \left (d x +c \right )}+6 \,{\mathrm e}^{3 i \left (d x +c \right )}-7 i+8 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}\) | \(139\) |
| derivativedivides | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+a \left (\frac {\sin \left (d x +c \right )^{6}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(154\) |
| default | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+a \left (\frac {\sin \left (d x +c \right )^{6}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(154\) |
| parts | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{6}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}+\frac {a \left (\frac {\sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\) | \(156\) |
Input:
int(sin(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
5/2*a*x+1/8*I*a/d*exp(2*I*(d*x+c))-1/2*a/d*exp(I*(d*x+c))-1/2*a/d*exp(-I*( d*x+c))-1/8*I*a/d*exp(-2*I*(d*x+c))-2/3*a*(-3*I*exp(2*I*(d*x+c))+6*exp(3*I *(d*x+c))-7*I+8*exp(I*(d*x+c)))/(exp(I*(d*x+c))+I)/(exp(I*(d*x+c))-I)^3/d
Time = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.03 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {3 \, a \cos \left (d x + c\right )^{4} - 15 \, a d x \cos \left (d x + c\right ) + 17 \, a \cos \left (d x + c\right )^{2} + {\left (15 \, a d x \cos \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right ) - 4 \, a}{6 \, {\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^4,x, algorithm="fricas")
Output:
1/6*(3*a*cos(d*x + c)^4 - 15*a*d*x*cos(d*x + c) + 17*a*cos(d*x + c)^2 + (1 5*a*d*x*cos(d*x + c) - 3*a*cos(d*x + c)^2 + 2*a)*sin(d*x + c) - 4*a)/(d*co s(d*x + c)*sin(d*x + c) - d*cos(d*x + c))
\[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=a \left (\int \sin {\left (c + d x \right )} \tan ^{4}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \tan ^{4}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)**4,x)
Output:
a*(Integral(sin(c + d*x)*tan(c + d*x)**4, x) + Integral(sin(c + d*x)**2*ta n(c + d*x)**4, x))
Time = 0.11 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.92 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {{\left (2 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 15 \, c - \frac {3 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 12 \, \tan \left (d x + c\right )\right )} a - 2 \, a {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )}}{6 \, d} \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^4,x, algorithm="maxima")
Output:
1/6*((2*tan(d*x + c)^3 + 15*d*x + 15*c - 3*tan(d*x + c)/(tan(d*x + c)^2 + 1) - 12*tan(d*x + c))*a - 2*a*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*c os(d*x + c)))/d
Timed out. \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^4,x, algorithm="giac")
Output:
Timed out
Time = 35.50 (sec) , antiderivative size = 285, normalized size of antiderivative = 3.00 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {5\,a\,x}{2}-\frac {\left (\frac {a\,\left (30\,c+30\,d\,x-30\right )}{6}-5\,a\,\left (c+d\,x\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (5\,a\,\left (c+d\,x\right )-\frac {a\,\left (30\,c+30\,d\,x-60\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {a\,\left (30\,c+30\,d\,x-50\right )}{6}-5\,a\,\left (c+d\,x\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\left (5\,a\,\left (c+d\,x\right )-\frac {a\,\left (30\,c+30\,d\,x-14\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {a\,\left (30\,c+30\,d\,x-4\right )}{6}-5\,a\,\left (c+d\,x\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (5\,a\,\left (c+d\,x\right )-\frac {a\,\left (30\,c+30\,d\,x-34\right )}{6}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a\,\left (15\,c+15\,d\,x-32\right )}{6}-\frac {5\,a\,\left (c+d\,x\right )}{2}}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \] Input:
int(sin(c + d*x)*tan(c + d*x)^4*(a + a*sin(c + d*x)),x)
Output:
(5*a*x)/2 - ((a*(15*c + 15*d*x - 32))/6 - tan(c/2 + (d*x)/2)*((a*(30*c + 3 0*d*x - 34))/6 - 5*a*(c + d*x)) - (5*a*(c + d*x))/2 + tan(c/2 + (d*x)/2)^2 *((a*(30*c + 30*d*x - 4))/6 - 5*a*(c + d*x)) - tan(c/2 + (d*x)/2)^3*((a*(3 0*c + 30*d*x - 14))/6 - 5*a*(c + d*x)) + tan(c/2 + (d*x)/2)^7*((a*(30*c + 30*d*x - 30))/6 - 5*a*(c + d*x)) + tan(c/2 + (d*x)/2)^5*((a*(30*c + 30*d*x - 50))/6 - 5*a*(c + d*x)) - tan(c/2 + (d*x)/2)^6*((a*(30*c + 30*d*x - 60) )/6 - 5*a*(c + d*x)) + (20*a*tan(c/2 + (d*x)/2)^4)/3)/(d*(tan(c/2 + (d*x)/ 2) - 1)^3*(tan(c/2 + (d*x)/2) + 1)*(tan(c/2 + (d*x)/2)^2 + 1)^2)
Time = 0.22 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.41 \[ \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {a \left (15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c +15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) d x -\cos \left (d x +c \right ) \sin \left (d x +c \right )-15 \cos \left (d x +c \right ) c -15 \cos \left (d x +c \right ) d x +\cos \left (d x +c \right )+3 \sin \left (d x +c \right )^{4}+3 \sin \left (d x +c \right )^{3}-23 \sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )+16\right )}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )-1\right )} \] Input:
int(sin(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^4,x)
Output:
(a*(15*cos(c + d*x)*sin(c + d*x)*c + 15*cos(c + d*x)*sin(c + d*x)*d*x - co s(c + d*x)*sin(c + d*x) - 15*cos(c + d*x)*c - 15*cos(c + d*x)*d*x + cos(c + d*x) + 3*sin(c + d*x)**4 + 3*sin(c + d*x)**3 - 23*sin(c + d*x)**2 - sin( c + d*x) + 16))/(6*cos(c + d*x)*d*(sin(c + d*x) - 1))