Integrand size = 27, antiderivative size = 111 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {5 a x}{2}-\frac {3 a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}-\frac {3 a \sec (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 d}-\frac {2 a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d} \] Output:
5/2*a*x-3*a*cos(d*x+c)/d+1/3*a*cos(d*x+c)^3/d-3*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 = 0.89 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.76 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {a \left (30 c+30 d x-33 \cos (c+d x)+\cos (3 (c+d x))-36 \sec (c+d x)+4 \sec ^3(c+d x)-3 \sin (2 (c+d x))-28 \tan (c+d x)+4 \sec ^2(c+d x) \tan (c+d x)\right )}{12 d} \] Input:
Integrate[Sin[c + d*x]^2*(a + a*Sin[c + d*x])*Tan[c + d*x]^4,x]
Output:
(a*(30*c + 30*d*x - 33*Cos[c + d*x] + Cos[3*(c + d*x)] - 36*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]))/(12*d)
Time = 0.46 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.99, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, 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.
| \(\displaystyle \int \sin ^2(c+d x) \tan ^4(c+d x) (a \sin (c+d x)+a) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^6 (a \sin (c+d x)+a)}{\cos (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3317 |
| \(\displaystyle a \int \sin ^3(c+d x) \tan ^4(c+d x)dx+a \int \sin ^2(c+d x) \tan ^4(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \sin (c+d x)^2 \tan (c+d x)^4dx+a \int \sin (c+d x)^3 \tan (c+d x)^4dx\) |
| \(\Big \downarrow \) 3070 |
| \(\displaystyle a \int \sin (c+d x)^2 \tan (c+d x)^4dx-\frac {a \int \left (1-\cos ^2(c+d x)\right )^3 \sec ^4(c+d x)d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle a \int \sin (c+d x)^2 \tan (c+d x)^4dx-\frac {a \int \left (\sec ^4(c+d x)-3 \sec ^2(c+d x)-\cos ^2(c+d x)+3\right )d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a \int \sin (c+d x)^2 \tan (c+d x)^4dx-\frac {a \left (-\frac {1}{3} \cos ^3(c+d x)+3 \cos (c+d x)-\frac {1}{3} \sec ^3(c+d x)+3 \sec (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3071 |
| \(\displaystyle \frac {a \int \frac {\tan ^6(c+d x)}{\left (\tan ^2(c+d x)+1\right )^2}d\tan (c+d x)}{d}-\frac {a \left (-\frac {1}{3} \cos ^3(c+d x)+3 \cos (c+d x)-\frac {1}{3} \sec ^3(c+d x)+3 \sec (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {a \left (\frac {5}{2} \int \frac {\tan ^4(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)-\frac {\tan ^5(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d}-\frac {a \left (-\frac {1}{3} \cos ^3(c+d x)+3 \cos (c+d x)-\frac {1}{3} \sec ^3(c+d x)+3 \sec (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle \frac {a \left (\frac {5}{2} \int \left (\tan ^2(c+d x)+\frac {1}{\tan ^2(c+d x)+1}-1\right )d\tan (c+d x)-\frac {\tan ^5(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d}-\frac {a \left (-\frac {1}{3} \cos ^3(c+d x)+3 \cos (c+d x)-\frac {1}{3} \sec ^3(c+d x)+3 \sec (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \left (\frac {5}{2} \left (\arctan (\tan (c+d x))+\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )-\frac {\tan ^5(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d}-\frac {a \left (-\frac {1}{3} \cos ^3(c+d x)+3 \cos (c+d x)-\frac {1}{3} \sec ^3(c+d x)+3 \sec (c+d x)\right )}{d}\) |
Input:
Int[Sin[c + d*x]^2*(a + a*Sin[c + d*x])*Tan[c + d*x]^4,x]
Output:
-((a*(3*Cos[c + d*x] - Cos[c + d*x]^3/3 + 3*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]
Time = 1.96 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.48
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin \left (d x +c \right )^{6}+\frac {6 \sin \left (d x +c \right )^{4}}{5}+\frac {8 \sin \left (d x +c \right )^{2}}{5}\right ) \cos \left (d x +c \right )}{3}\right )+a \left (\frac {\sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\) | \(164\) |
| default | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin \left (d x +c \right )^{6}+\frac {6 \sin \left (d x +c \right )^{4}}{5}+\frac {8 \sin \left (d x +c \right )^{2}}{5}\right ) \cos \left (d x +c \right )}{3}\right )+a \left (\frac {\sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\) | \(164\) |
| parts | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \sin \left (d x +c \right )^{7}}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}+\frac {a \left (\frac {\sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \sin \left (d x +c \right )^{8}}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin \left (d x +c \right )^{6}+\frac {6 \sin \left (d x +c \right )^{4}}{5}+\frac {8 \sin \left (d x +c \right )^{2}}{5}\right ) \cos \left (d x +c \right )}{3}\right )}{d}\) | \(166\) |
| risch | \(\frac {5 a x}{2}+\frac {a \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {11 a \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {11 a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {i a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {a \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {2 \left (5 a \,{\mathrm e}^{i \left (d x +c \right )}-9 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+9 a \,{\mathrm e}^{3 i \left (d x +c \right )}-7 i a \right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}\) | \(173\) |
Input:
int(sin(d*x+c)^2*(a+a*sin(d*x+c))*tan(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/3*sin(d*x+c)^8/cos(d*x+c)^3-5/3*sin(d*x+c)^8/cos(d*x+c)-5/3*(16/ 5+sin(d*x+c)^6+6/5*sin(d*x+c)^4+8/5*sin(d*x+c)^2)*cos(d*x+c))+a*(1/3*sin(d *x+c)^7/cos(d*x+c)^3-4/3*sin(d*x+c)^7/cos(d*x+c)-4/3*(sin(d*x+c)^5+5/4*sin (d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)+5/2*d*x+5/2*c))
Time = 0.08 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.97 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {a \cos \left (d x + c\right )^{4} - 15 \, a d x \cos \left (d x + c\right ) + 29 \, a \cos \left (d x + c\right )^{2} + {\left (2 \, a \cos \left (d x + c\right )^{4} + 15 \, a d x \cos \left (d x + c\right ) - 15 \, a \cos \left (d x + c\right )^{2} - 4 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{6 \, {\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \] Input:
integrate(sin(d*x+c)^2*(a+a*sin(d*x+c))*tan(d*x+c)^4,x, algorithm="fricas" )
Output:
1/6*(a*cos(d*x + c)^4 - 15*a*d*x*cos(d*x + c) + 29*a*cos(d*x + c)^2 + (2*a *cos(d*x + c)^4 + 15*a*d*x*cos(d*x + c) - 15*a*cos(d*x + c)^2 - 4*a)*sin(d *x + c) + 2*a)/(d*cos(d*x + c)*sin(d*x + c) - d*cos(d*x + c))
\[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=a \left (\int \sin ^{2}{\left (c + d x \right )} \tan ^{4}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \tan ^{4}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sin(d*x+c)**2*(a+a*sin(d*x+c))*tan(d*x+c)**4,x)
Output:
a*(Integral(sin(c + d*x)**2*tan(c + d*x)**4, x) + Integral(sin(c + d*x)**3 *tan(c + d*x)**4, x))
Time = 0.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.86 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {2 \, {\left (\cos \left (d x + c\right )^{3} - \frac {9 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} - 9 \, \cos \left (d x + c\right )\right )} a + {\left (2 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 15 \, c - \frac {3 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 12 \, \tan \left (d x + c\right )\right )} a}{6 \, d} \] Input:
integrate(sin(d*x+c)^2*(a+a*sin(d*x+c))*tan(d*x+c)^4,x, algorithm="maxima" )
Output:
1/6*(2*(cos(d*x + c)^3 - (9*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 - 9*cos(d*x + c))*a + (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)/d
Timed out. \[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)^2*(a+a*sin(d*x+c))*tan(d*x+c)^4,x, algorithm="giac")
Output:
Timed out
Time = 36.25 (sec) , antiderivative size = 360, normalized size of antiderivative = 3.24 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {5\,a\,x}{2}+\frac {\left (5\,a\,\left (c+d\,x\right )-\frac {a\,\left (30\,c+30\,d\,x-30\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {a\,\left (45\,c+45\,d\,x-60\right )}{6}-\frac {15\,a\,\left (c+d\,x\right )}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (10\,a\,\left (c+d\,x\right )-\frac {a\,\left (60\,c+60\,d\,x-80\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {a\,\left (30\,c+30\,d\,x-100\right )}{6}-5\,a\,\left (c+d\,x\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (5\,a\,\left (c+d\,x\right )-\frac {a\,\left (30\,c+30\,d\,x-28\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {a\,\left (60\,c+60\,d\,x-176\right )}{6}-10\,a\,\left (c+d\,x\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {15\,a\,\left (c+d\,x\right )}{2}-\frac {a\,\left (45\,c+45\,d\,x-132\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (\frac {a\,\left (30\,c+30\,d\,x-98\right )}{6}-5\,a\,\left (c+d\,x\right )\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a\,\left (15\,c+15\,d\,x-64\right )}{6}+\frac {5\,a\,\left (c+d\,x\right )}{2}}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3} \] Input:
int(sin(c + d*x)^2*tan(c + d*x)^4*(a + a*sin(c + d*x)),x)
Output:
(5*a*x)/2 + (tan(c/2 + (d*x)/2)*((a*(30*c + 30*d*x - 98))/6 - 5*a*(c + d*x )) - (a*(15*c + 15*d*x - 64))/6 + (5*a*(c + d*x))/2 - tan(c/2 + (d*x)/2)^4 *((a*(30*c + 30*d*x - 28))/6 - 5*a*(c + d*x)) - tan(c/2 + (d*x)/2)^9*((a*( 30*c + 30*d*x - 30))/6 - 5*a*(c + d*x)) + tan(c/2 + (d*x)/2)^6*((a*(30*c + 30*d*x - 100))/6 - 5*a*(c + d*x)) + tan(c/2 + (d*x)/2)^8*((a*(45*c + 45*d *x - 60))/6 - (15*a*(c + d*x))/2) - tan(c/2 + (d*x)/2)^7*((a*(60*c + 60*d* x - 80))/6 - 10*a*(c + d*x)) - tan(c/2 + (d*x)/2)^2*((a*(45*c + 45*d*x - 1 32))/6 - (15*a*(c + d*x))/2) + tan(c/2 + (d*x)/2)^3*((a*(60*c + 60*d*x - 1 76))/6 - 10*a*(c + d*x)) + 6*a*tan(c/2 + (d*x)/2)^5)/(d*(tan(c/2 + (d*x)/2 ) + 1)*(tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^3 - 1)^3)
Time = 0.21 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.30 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {a \left (15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c +15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) d x -17 \cos \left (d x +c \right ) \sin \left (d x +c \right )-15 \cos \left (d x +c \right ) c -15 \cos \left (d x +c \right ) d x +17 \cos \left (d x +c \right )+2 \sin \left (d x +c \right )^{5}+\sin \left (d x +c \right )^{4}+11 \sin \left (d x +c \right )^{3}-31 \sin \left (d x +c \right )^{2}-17 \sin \left (d x +c \right )+32\right )}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )-1\right )} \] Input:
int(sin(d*x+c)^2*(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 - 17 *cos(c + d*x)*sin(c + d*x) - 15*cos(c + d*x)*c - 15*cos(c + d*x)*d*x + 17* cos(c + d*x) + 2*sin(c + d*x)**5 + sin(c + d*x)**4 + 11*sin(c + d*x)**3 - 31*sin(c + d*x)**2 - 17*sin(c + d*x) + 32))/(6*cos(c + d*x)*d*(sin(c + d*x ) - 1))