Integrand size = 25, antiderivative size = 131 \[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {15 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {b \cos ^2(c+d x)}{2 d}-\frac {3 b \log (\cos (c+d x))}{d}-\frac {3 b \sec ^2(c+d x)}{2 d}+\frac {b \sec ^4(c+d x)}{4 d}-\frac {a \sin (c+d x)}{d}-\frac {9 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Output:
15/8*a*arctanh(sin(d*x+c))/d+1/2*b*cos(d*x+c)^2/d-3*b*ln(cos(d*x+c))/d-3/2 *b*sec(d*x+c)^2/d+1/4*b*sec(d*x+c)^4/d-a*sin(d*x+c)/d-9/8*a*sec(d*x+c)*tan (d*x+c)/d+1/4*a*sec(d*x+c)^3*tan(d*x+c)/d
Time = 0.41 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.11 \[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {15 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {b \left (12 \log (\cos (c+d x))+6 \sec ^2(c+d x)-\sec ^4(c+d x)+2 \sin ^2(c+d x)\right )}{4 d}+\frac {15 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {15 a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {5 a \sec (c+d x) \tan ^3(c+d x)}{d}-\frac {a \sin (c+d x) \tan ^4(c+d x)}{d} \] Input:
Integrate[Sin[c + d*x]*(a + b*Sin[c + d*x])*Tan[c + d*x]^5,x]
Output:
(15*a*ArcTanh[Sin[c + d*x]])/(8*d) - (b*(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) + (15*a*Sec[c + d*x]*Ta n[c + d*x])/(8*d) - (15*a*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*a*Sec[c + d*x]*Tan[c + d*x]^3)/d - (a*Sin[c + d*x]*Tan[c + d*x]^4)/d
Time = 0.44 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3313, 3042, 3070, 243, 49, 2009, 3072, 252, 252, 262, 219}
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.
| \(\displaystyle \int \sin (c+d x) \tan ^5(c+d x) (a+b \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^6 (a+b \sin (c+d x))}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3313 |
| \(\displaystyle a \int \sin (c+d x) \tan ^5(c+d x)dx+b \int \sin ^2(c+d x) \tan ^5(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \sin (c+d x) \tan (c+d x)^5dx+b \int \sin (c+d x)^2 \tan (c+d x)^5dx\) |
| \(\Big \downarrow \) 3070 |
| \(\displaystyle a \int \sin (c+d x) \tan (c+d x)^5dx-\frac {b \int \left (1-\cos ^2(c+d x)\right )^3 \sec ^5(c+d x)d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle a \int \sin (c+d x) \tan (c+d x)^5dx-\frac {b \int \left (1-\cos ^2(c+d x)\right )^3 \sec ^3(c+d x)d\cos ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle a \int \sin (c+d x) \tan (c+d x)^5dx-\frac {b \int \left (\sec ^3(c+d x)-3 \sec ^2(c+d x)+3 \sec (c+d x)-1\right )d\cos ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a \int \sin (c+d x) \tan (c+d x)^5dx-\frac {b \left (-\cos ^2(c+d x)-\frac {1}{2} \sec ^2(c+d x)+3 \sec (c+d x)+3 \log \left (\cos ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 3072 |
| \(\displaystyle \frac {a \int \frac {\sin ^6(c+d x)}{\left (1-\sin ^2(c+d x)\right )^3}d\sin (c+d x)}{d}-\frac {b \left (-\cos ^2(c+d x)-\frac {1}{2} \sec ^2(c+d x)+3 \sec (c+d x)+3 \log \left (\cos ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {a \left (\frac {\sin ^5(c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}-\frac {5}{4} \int \frac {\sin ^4(c+d x)}{\left (1-\sin ^2(c+d x)\right )^2}d\sin (c+d x)\right )}{d}-\frac {b \left (-\cos ^2(c+d x)-\frac {1}{2} \sec ^2(c+d x)+3 \sec (c+d x)+3 \log \left (\cos ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {a \left (\frac {\sin ^5(c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}-\frac {5}{4} \left (\frac {\sin ^3(c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}-\frac {3}{2} \int \frac {\sin ^2(c+d x)}{1-\sin ^2(c+d x)}d\sin (c+d x)\right )\right )}{d}-\frac {b \left (-\cos ^2(c+d x)-\frac {1}{2} \sec ^2(c+d x)+3 \sec (c+d x)+3 \log \left (\cos ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {a \left (\frac {\sin ^5(c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}-\frac {5}{4} \left (\frac {\sin ^3(c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}-\frac {3}{2} \left (\int \frac {1}{1-\sin ^2(c+d x)}d\sin (c+d x)-\sin (c+d x)\right )\right )\right )}{d}-\frac {b \left (-\cos ^2(c+d x)-\frac {1}{2} \sec ^2(c+d x)+3 \sec (c+d x)+3 \log \left (\cos ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {a \left (\frac {\sin ^5(c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}-\frac {5}{4} \left (\frac {\sin ^3(c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}-\frac {3}{2} (\text {arctanh}(\sin (c+d x))-\sin (c+d x))\right )\right )}{d}-\frac {b \left (-\cos ^2(c+d x)-\frac {1}{2} \sec ^2(c+d x)+3 \sec (c+d x)+3 \log \left (\cos ^2(c+d x)\right )\right )}{2 d}\) |
Input:
Int[Sin[c + d*x]*(a + b*Sin[c + d*x])*Tan[c + d*x]^5,x]
Output:
-1/2*(b*(-Cos[c + d*x]^2 + 3*Log[Cos[c + d*x]^2] + 3*Sec[c + d*x] - Sec[c + d*x]^2/2))/d + (a*(Sin[c + d*x]^5/(4*(1 - Sin[c + d*x]^2)^2) - (5*((-3*( ArcTanh[Sin[c + d*x]] - Sin[c + d*x]))/2 + Sin[c + d*x]^3/(2*(1 - Sin[c + d*x]^2))))/4))/d
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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
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 = 2.35 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.27
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(167\) |
| default | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(167\) |
| parts | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {b \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(169\) |
| risch | \(3 i x b +\frac {b \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {6 i b c}{d}+\frac {i \left (9 a \,{\mathrm e}^{7 i \left (d x +c \right )}+a \,{\mathrm e}^{5 i \left (d x +c \right )}+24 i b \,{\mathrm e}^{6 i \left (d x +c \right )}-a \,{\mathrm e}^{3 i \left (d x +c \right )}+32 i b \,{\mathrm e}^{4 i \left (d x +c \right )}-9 a \,{\mathrm e}^{i \left (d x +c \right )}+24 i b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {15 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{d}-\frac {15 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{d}\) | \(259\) |
Input:
int(sin(d*x+c)*(a+b*sin(d*x+c))*tan(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/4*sin(d*x+c)^7/cos(d*x+c)^4-3/8*sin(d*x+c)^7/cos(d*x+c)^2-3/8*si n(d*x+c)^5-5/8*sin(d*x+c)^3-15/8*sin(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c)) )+b*(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*sin(d *x+c)^6-3/4*sin(d*x+c)^4-3/2*sin(d*x+c)^2-3*ln(cos(d*x+c))))
Time = 0.11 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.05 \[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {8 \, b \cos \left (d x + c\right )^{6} + 3 \, {\left (5 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 4 \, b \cos \left (d x + c\right )^{4} - 24 \, b \cos \left (d x + c\right )^{2} - 2 \, {\left (8 \, a \cos \left (d x + c\right )^{4} + 9 \, a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right ) + 4 \, b}{16 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sin(d*x+c)*(a+b*sin(d*x+c))*tan(d*x+c)^5,x, algorithm="fricas")
Output:
1/16*(8*b*cos(d*x + c)^6 + 3*(5*a - 8*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(5*a + 8*b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 4*b*cos(d*x + c)^4 - 24*b*cos(d*x + c)^2 - 2*(8*a*cos(d*x + c)^4 + 9*a*cos(d*x + c)^2 - 2*a)*sin(d*x + c) + 4*b)/(d*cos(d*x + c)^4)
\[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \sin {\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx \] Input:
integrate(sin(d*x+c)*(a+b*sin(d*x+c))*tan(d*x+c)**5,x)
Output:
Integral((a + b*sin(c + d*x))*sin(c + d*x)*tan(c + d*x)**5, x)
Time = 0.03 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.92 \[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {8 \, b \sin \left (d x + c\right )^{2} - 3 \, {\left (5 \, a - 8 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (5 \, a + 8 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, a \sin \left (d x + c\right ) - \frac {2 \, {\left (9 \, a \sin \left (d x + c\right )^{3} + 12 \, b \sin \left (d x + c\right )^{2} - 7 \, a \sin \left (d x + c\right ) - 10 \, b\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \] Input:
integrate(sin(d*x+c)*(a+b*sin(d*x+c))*tan(d*x+c)^5,x, algorithm="maxima")
Output:
-1/16*(8*b*sin(d*x + c)^2 - 3*(5*a - 8*b)*log(sin(d*x + c) + 1) + 3*(5*a + 8*b)*log(sin(d*x + c) - 1) + 16*a*sin(d*x + c) - 2*(9*a*sin(d*x + c)^3 + 12*b*sin(d*x + c)^2 - 7*a*sin(d*x + c) - 10*b)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.48 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.01 \[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {3 \, {\left (5 \, a - 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} - \frac {3 \, {\left (5 \, a + 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} - \frac {b d \sin \left (d x + c\right )^{2} + 2 \, a d \sin \left (d x + c\right )}{2 \, d^{2}} + \frac {9 \, a \sin \left (d x + c\right )^{3} + 12 \, b \sin \left (d x + c\right )^{2} - 7 \, a \sin \left (d x + c\right ) - 10 \, b}{8 \, d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(sin(d*x+c)*(a+b*sin(d*x+c))*tan(d*x+c)^5,x, algorithm="giac")
Output:
3/16*(5*a - 8*b)*log(abs(sin(d*x + c) + 1))/d - 3/16*(5*a + 8*b)*log(abs(s in(d*x + c) - 1))/d - 1/2*(b*d*sin(d*x + c)^2 + 2*a*d*sin(d*x + c))/d^2 + 1/8*(9*a*sin(d*x + c)^3 + 12*b*sin(d*x + c)^2 - 7*a*sin(d*x + c) - 10*b)/( d*(sin(d*x + c) + 1)^2*(sin(d*x + c) - 1)^2)
Time = 19.69 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.32 \[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {3\,b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {15\,a}{8}+3\,b\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {15\,a}{8}-3\,b\right )}{d}-\frac {-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{4}-6\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {25\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+12\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}+12\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {25\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}-6\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {15\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(sin(c + d*x)*tan(c + d*x)^5*(a + b*sin(c + d*x)),x)
Output:
(3*b*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (log(tan(c/2 + (d*x)/2) - 1)*((15* a)/8 + 3*b))/d + (log(tan(c/2 + (d*x)/2) + 1)*((15*a)/8 - 3*b))/d - ((25*a *tan(c/2 + (d*x)/2)^3)/4 - (15*a*tan(c/2 + (d*x)/2))/4 + (11*a*tan(c/2 + ( d*x)/2)^5)/2 + (11*a*tan(c/2 + (d*x)/2)^7)/2 + (25*a*tan(c/2 + (d*x)/2)^9) /4 - (15*a*tan(c/2 + (d*x)/2)^11)/4 - 6*b*tan(c/2 + (d*x)/2)^2 + 12*b*tan( c/2 + (d*x)/2)^4 + 4*b*tan(c/2 + (d*x)/2)^6 + 12*b*tan(c/2 + (d*x)/2)^8 - 6*b*tan(c/2 + (d*x)/2)^10)/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2) ^4 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 2*tan(c/2 + (d*x)/2)^ 10 - tan(c/2 + (d*x)/2)^12 - 1))
Time = 0.20 (sec) , antiderivative size = 403, normalized size of antiderivative = 3.08 \[ \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{4} b -48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} b +24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b -15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a -24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b +30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b -15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a -24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b -30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b +15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -4 \sin \left (d x +c \right )^{6} b -8 \sin \left (d x +c \right )^{5} a +18 \sin \left (d x +c \right )^{4} b +25 \sin \left (d x +c \right )^{3} a -12 \sin \left (d x +c \right )^{2} b -15 \sin \left (d x +c \right ) a}{8 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sin(d*x+c)*(a+b*sin(d*x+c))*tan(d*x+c)^5,x)
Output:
(24*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4*b - 48*log(tan((c + d*x)/ 2)**2 + 1)*sin(c + d*x)**2*b + 24*log(tan((c + d*x)/2)**2 + 1)*b - 15*log( tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 24*log(tan((c + d*x)/2) - 1)*sin (c + d*x)**4*b + 30*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 48*log(t an((c + d*x)/2) - 1)*sin(c + d*x)**2*b - 15*log(tan((c + d*x)/2) - 1)*a - 24*log(tan((c + d*x)/2) - 1)*b + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x) **4*a - 24*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b - 30*log(tan((c + d *x)/2) + 1)*sin(c + d*x)**2*a + 48*log(tan((c + d*x)/2) + 1)*sin(c + d*x)* *2*b + 15*log(tan((c + d*x)/2) + 1)*a - 24*log(tan((c + d*x)/2) + 1)*b - 4 *sin(c + d*x)**6*b - 8*sin(c + d*x)**5*a + 18*sin(c + d*x)**4*b + 25*sin(c + d*x)**3*a - 12*sin(c + d*x)**2*b - 15*sin(c + d*x)*a)/(8*d*(sin(c + d*x )**4 - 2*sin(c + d*x)**2 + 1))