Integrand size = 27, antiderivative size = 155 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {35 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \cos ^2(c+d x)}{2 d}-\frac {3 a \log (\cos (c+d x))}{d}-\frac {3 a \sec ^2(c+d x)}{2 d}+\frac {a \sec ^4(c+d x)}{4 d}-\frac {35 b \sin (c+d x)}{8 d}-\frac {35 b \sin ^3(c+d x)}{24 d}-\frac {7 b \sin ^3(c+d x) \tan ^2(c+d x)}{8 d}+\frac {b \sin ^3(c+d x) \tan ^4(c+d x)}{4 d} \]
35/8*b*arctanh(sin(d*x+c))/d+1/2*a*cos(d*x+c)^2/d-3*a*ln(cos(d*x+c))/d-3/2 *a*sec(d*x+c)^2/d+1/4*a*sec(d*x+c)^4/d-35/8*b*sin(d*x+c)/d-35/24*b*sin(d*x +c)^3/d-7/8*b*sin(d*x+c)^3*tan(d*x+c)^2/d+1/4*b*sin(d*x+c)^3*tan(d*x+c)^4/ d
Time = 0.43 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.12 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {35 b \text {arctanh}(\sin (c+d x))}{8 d}-\frac {a \left (12 \log (\cos (c+d x))+6 \sec ^2(c+d x)-\sec ^4(c+d x)+2 \sin ^2(c+d x)\right )}{4 d}+\frac {35 b \sec (c+d x) \tan (c+d x)}{8 d}-\frac {35 b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {35 b \sec (c+d x) \tan ^3(c+d x)}{3 d}-\frac {7 b \sin (c+d x) \tan ^4(c+d x)}{3 d}-\frac {b \sin ^3(c+d x) \tan ^4(c+d x)}{3 d} \]
(35*b*ArcTanh[Sin[c + d*x]])/(8*d) - (a*(12*Log[Cos[c + d*x]] + 6*Sec[c + d*x]^2 - Sec[c + d*x]^4 + 2*Sin[c + d*x]^2))/(4*d) + (35*b*Sec[c + d*x]*Ta n[c + d*x])/(8*d) - (35*b*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (35*b*Sec[c + d*x]*Tan[c + d*x]^3)/(3*d) - (7*b*Sin[c + d*x]*Tan[c + d*x]^4)/(3*d) - (b*Sin[c + d*x]^3*Tan[c + d*x]^4)/(3*d)
Time = 0.50 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.94, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 3313, 3042, 3070, 243, 49, 2009, 3072, 252, 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.
| \(\displaystyle \int \sin ^2(c+d x) \tan ^5(c+d x) (a+b \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^7 (a+b \sin (c+d x))}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3313 |
| \(\displaystyle a \int \sin ^2(c+d x) \tan ^5(c+d x)dx+b \int \sin ^3(c+d x) \tan ^5(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \sin (c+d x)^2 \tan (c+d x)^5dx+b \int \sin (c+d x)^3 \tan (c+d x)^5dx\) |
| \(\Big \downarrow \) 3070 |
| \(\displaystyle b \int \sin (c+d x)^3 \tan (c+d x)^5dx-\frac {a \int \left (1-\cos ^2(c+d x)\right )^3 \sec ^5(c+d x)d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle b \int \sin (c+d x)^3 \tan (c+d x)^5dx-\frac {a \int \left (1-\cos ^2(c+d x)\right )^3 \sec ^3(c+d x)d\cos ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle b \int \sin (c+d x)^3 \tan (c+d x)^5dx-\frac {a \int \left (\sec ^3(c+d x)-3 \sec ^2(c+d x)+3 \sec (c+d x)-1\right )d\cos ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle b \int \sin (c+d x)^3 \tan (c+d x)^5dx-\frac {a \left (-\cos ^2(c+d x)-\frac {1}{2} \sec ^2(c+d x)+3 \sec (c+d x)+3 \log \left (\cos ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 3072 |
| \(\displaystyle \frac {b \int \frac {\sin ^8(c+d x)}{\left (1-\sin ^2(c+d x)\right )^3}d\sin (c+d x)}{d}-\frac {a \left (-\cos ^2(c+d x)-\frac {1}{2} \sec ^2(c+d x)+3 \sec (c+d x)+3 \log \left (\cos ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {b \left (\frac {\sin ^7(c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}-\frac {7}{4} \int \frac {\sin ^6(c+d x)}{\left (1-\sin ^2(c+d x)\right )^2}d\sin (c+d x)\right )}{d}-\frac {a \left (-\cos ^2(c+d x)-\frac {1}{2} \sec ^2(c+d x)+3 \sec (c+d x)+3 \log \left (\cos ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {b \left (\frac {\sin ^7(c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}-\frac {7}{4} \left (\frac {\sin ^5(c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}-\frac {5}{2} \int \frac {\sin ^4(c+d x)}{1-\sin ^2(c+d x)}d\sin (c+d x)\right )\right )}{d}-\frac {a \left (-\cos ^2(c+d x)-\frac {1}{2} \sec ^2(c+d x)+3 \sec (c+d x)+3 \log \left (\cos ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle \frac {b \left (\frac {\sin ^7(c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}-\frac {7}{4} \left (\frac {\sin ^5(c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}-\frac {5}{2} \int \left (-\sin ^2(c+d x)+\frac {1}{1-\sin ^2(c+d x)}-1\right )d\sin (c+d x)\right )\right )}{d}-\frac {a \left (-\cos ^2(c+d x)-\frac {1}{2} \sec ^2(c+d x)+3 \sec (c+d x)+3 \log \left (\cos ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \left (\frac {\sin ^7(c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}-\frac {7}{4} \left (\frac {\sin ^5(c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}-\frac {5}{2} \left (\text {arctanh}(\sin (c+d x))-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )\right )\right )}{d}-\frac {a \left (-\cos ^2(c+d x)-\frac {1}{2} \sec ^2(c+d x)+3 \sec (c+d x)+3 \log \left (\cos ^2(c+d x)\right )\right )}{2 d}\) |
-1/2*(a*(-Cos[c + d*x]^2 + 3*Log[Cos[c + d*x]^2] + 3*Sec[c + d*x] - Sec[c + d*x]^2/2))/d + (b*(Sin[c + d*x]^7/(4*(1 - Sin[c + d*x]^2)^2) - (7*(Sin[c + d*x]^5/(2*(1 - Sin[c + d*x]^2)) - (5*(ArcTanh[Sin[c + d*x]] - Sin[c + d *x] - Sin[c + d*x]^3/3))/2))/4))/d
3.15.81.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ (ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x ]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_ ) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[Cos[e + f*x]^ p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[Cos[e + f*x]^p*(d*Sin[e + f*x ])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2 ] && IntegerQ[n] && ((LtQ[p, 0] && NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] | | LtQ[p + 1, -n, 2*p + 1])
Time = 1.10 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.14
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+b \left (\frac {\sin ^{9}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \left (\sin ^{9}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \left (\sin ^{7}\left (d x +c \right )\right )}{8}-\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(177\) |
| default | \(\frac {a \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+b \left (\frac {\sin ^{9}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \left (\sin ^{9}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \left (\sin ^{7}\left (d x +c \right )\right )}{8}-\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(177\) |
| parallelrisch | \(\frac {288 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-288 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +\frac {35 b}{24}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-288 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -\frac {35 b}{24}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-603 a \cos \left (2 d x +2 c \right )-108 \cos \left (4 d x +4 c \right ) a +3 a \cos \left (6 d x +6 c \right )-189 b \sin \left (3 d x +3 c \right )-35 b \sin \left (5 d x +5 c \right )+b \sin \left (7 d x +7 c \right )-105 b \sin \left (d x +c \right )-444 a}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(233\) |
| risch | \(3 i a x -\frac {i b \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {13 i b \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {13 i b \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {i b \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {6 i a c}{d}+\frac {i \left (24 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+13 b \,{\mathrm e}^{7 i \left (d x +c \right )}+32 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+5 b \,{\mathrm e}^{5 i \left (d x +c \right )}+24 i a \,{\mathrm e}^{2 i \left (d x +c \right )}-5 b \,{\mathrm e}^{3 i \left (d x +c \right )}-13 b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{8 d}\) | \(292\) |
| norman | \(\frac {\frac {22 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {22 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {16 a}{3 d}-\frac {35 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {35 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {329 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {17 b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {329 b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {35 b \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {35 b \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {16 a \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {3 a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (24 a -35 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (24 a +35 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\) | \(310\) |
1/d*(a*(1/4*sin(d*x+c)^8/cos(d*x+c)^4-1/2*sin(d*x+c)^8/cos(d*x+c)^2-1/2*si n(d*x+c)^6-3/4*sin(d*x+c)^4-3/2*sin(d*x+c)^2-3*ln(cos(d*x+c)))+b*(1/4*sin( d*x+c)^9/cos(d*x+c)^4-5/8*sin(d*x+c)^9/cos(d*x+c)^2-5/8*sin(d*x+c)^7-7/8*s in(d*x+c)^5-35/24*sin(d*x+c)^3-35/8*sin(d*x+c)+35/8*ln(sec(d*x+c)+tan(d*x+ c))))
Time = 0.29 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.96 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {24 \, a \cos \left (d x + c\right )^{6} - 3 \, {\left (24 \, a - 35 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (24 \, a + 35 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 12 \, a \cos \left (d x + c\right )^{4} - 72 \, a \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, b \cos \left (d x + c\right )^{6} - 80 \, b \cos \left (d x + c\right )^{4} - 39 \, b \cos \left (d x + c\right )^{2} + 6 \, b\right )} \sin \left (d x + c\right ) + 12 \, a}{48 \, d \cos \left (d x + c\right )^{4}} \]
1/48*(24*a*cos(d*x + c)^6 - 3*(24*a - 35*b)*cos(d*x + c)^4*log(sin(d*x + c ) + 1) - 3*(24*a + 35*b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 12*a*cos( d*x + c)^4 - 72*a*cos(d*x + c)^2 + 2*(8*b*cos(d*x + c)^6 - 80*b*cos(d*x + c)^4 - 39*b*cos(d*x + c)^2 + 6*b)*sin(d*x + c) + 12*a)/(d*cos(d*x + c)^4)
Timed out. \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.85 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {16 \, b \sin \left (d x + c\right )^{3} + 24 \, a \sin \left (d x + c\right )^{2} + 3 \, {\left (24 \, a - 35 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (24 \, a + 35 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 144 \, b \sin \left (d x + c\right ) - \frac {6 \, {\left (13 \, b \sin \left (d x + c\right )^{3} + 12 \, a \sin \left (d x + c\right )^{2} - 11 \, b \sin \left (d x + c\right ) - 10 \, a\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{48 \, d} \]
-1/48*(16*b*sin(d*x + c)^3 + 24*a*sin(d*x + c)^2 + 3*(24*a - 35*b)*log(sin (d*x + c) + 1) + 3*(24*a + 35*b)*log(sin(d*x + c) - 1) + 144*b*sin(d*x + c ) - 6*(13*b*sin(d*x + c)^3 + 12*a*sin(d*x + c)^2 - 11*b*sin(d*x + c) - 10* a)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.37 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.87 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {16 \, b \sin \left (d x + c\right )^{3} + 24 \, a \sin \left (d x + c\right )^{2} + 3 \, {\left (24 \, a - 35 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \, {\left (24 \, a + 35 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 144 \, b \sin \left (d x + c\right ) - \frac {6 \, {\left (18 \, a \sin \left (d x + c\right )^{4} + 13 \, b \sin \left (d x + c\right )^{3} - 24 \, a \sin \left (d x + c\right )^{2} - 11 \, b \sin \left (d x + c\right ) + 8 \, a\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{48 \, d} \]
-1/48*(16*b*sin(d*x + c)^3 + 24*a*sin(d*x + c)^2 + 3*(24*a - 35*b)*log(abs (sin(d*x + c) + 1)) + 3*(24*a + 35*b)*log(abs(sin(d*x + c) - 1)) + 144*b*s in(d*x + c) - 6*(18*a*sin(d*x + c)^4 + 13*b*sin(d*x + c)^3 - 24*a*sin(d*x + c)^2 - 11*b*sin(d*x + c) + 8*a)/(sin(d*x + c)^2 - 1)^2)/d
Time = 12.62 (sec) , antiderivative size = 346, normalized size of antiderivative = 2.23 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {3\,a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {-\frac {35\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}-6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\frac {35\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{6}+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {329\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{12}+16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-17\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {329\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {35\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}-6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {35\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (3\,a+\frac {35\,b}{8}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (3\,a-\frac {35\,b}{8}\right )}{d} \]
(3*a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (6*a*tan(c/2 + (d*x)/2)^4 - 6*a*ta n(c/2 + (d*x)/2)^2 - (35*b*tan(c/2 + (d*x)/2))/4 + 16*a*tan(c/2 + (d*x)/2) ^6 + 16*a*tan(c/2 + (d*x)/2)^8 + 6*a*tan(c/2 + (d*x)/2)^10 - 6*a*tan(c/2 + (d*x)/2)^12 + (35*b*tan(c/2 + (d*x)/2)^3)/6 + (329*b*tan(c/2 + (d*x)/2)^5 )/12 - 17*b*tan(c/2 + (d*x)/2)^7 + (329*b*tan(c/2 + (d*x)/2)^9)/12 + (35*b *tan(c/2 + (d*x)/2)^11)/6 - (35*b*tan(c/2 + (d*x)/2)^13)/4)/(d*(tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 + 3*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 - tan(c/2 + (d*x)/2)^14 - 1)) - (log(tan(c/2 + (d*x)/2) - 1)*(3*a + (35*b)/8))/d - (lo g(tan(c/2 + (d*x)/2) + 1)*(3*a - (35*b)/8))/d