Integrand size = 27, antiderivative size = 147 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {35 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \cos ^2(c+d x)}{2 d}-\frac {3 a \log (\cos (c+d x))}{d}-\frac {3 a \sec ^2(c+d x)}{2 d}+\frac {a \sec ^4(c+d x)}{4 d}-\frac {3 b \sin (c+d x)}{d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {13 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Output:
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-3*b*sin(d*x+c)/d-1/3*b*sin(d*x+c)^3 /d-13/8*b*sec(d*x+c)*tan(d*x+c)/d+1/4*b*sec(d*x+c)^3*tan(d*x+c)/d
Time = 0.66 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.18 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {35 b \text {arctanh}(\sin (c+d x))}{8 d}-\frac {a \left (12 \log (\cos (c+d x))+6 \sec ^2(c+d x)-\sec ^4(c+d x)+2 \sin ^2(c+d x)\right )}{4 d}+\frac {35 b \sec (c+d x) \tan (c+d x)}{8 d}-\frac {35 b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {35 b \sec (c+d x) \tan ^3(c+d x)}{3 d}-\frac {7 b \sin (c+d x) \tan ^4(c+d x)}{3 d}-\frac {b \sin ^3(c+d x) \tan ^4(c+d x)}{3 d} \] Input:
Integrate[Sin[c + d*x]^2*(a + b*Sin[c + d*x])*Tan[c + d*x]^5,x]
Output:
(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 = 0.99, 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 |
| \(\displaystyle \frac {b \left (\frac {\sin ^7(c+d x)}{4 \left (1-\sin ^2(c+d x)\right )^2}-\frac {7}{4} \left (\frac {\sin ^5(c+d x)}{2 \left (1-\sin ^2(c+d x)\right )}-\frac {5}{2} \left (\text {arctanh}(\sin (c+d x))-\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )\right )\right )}{d}-\frac {a \left (-\cos ^2(c+d x)-\frac {1}{2} \sec ^2(c+d x)+3 \sec (c+d x)+3 \log \left (\cos ^2(c+d x)\right )\right )}{2 d}\) |
Input:
Int[Sin[c + d*x]^2*(a + b*Sin[c + d*x])*Tan[c + d*x]^5,x]
Output:
-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
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 = 3.52 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.20
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+b \left (\frac {\sin \left (d x +c \right )^{9}}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \sin \left (d x +c \right )^{9}}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \sin \left (d x +c \right )^{7}}{8}-\frac {7 \sin \left (d x +c \right )^{5}}{8}-\frac {35 \sin \left (d x +c \right )^{3}}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(177\) |
| default | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+b \left (\frac {\sin \left (d x +c \right )^{9}}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \sin \left (d x +c \right )^{9}}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \sin \left (d x +c \right )^{7}}{8}-\frac {7 \sin \left (d x +c \right )^{5}}{8}-\frac {35 \sin \left (d x +c \right )^{3}}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(177\) |
| parts | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}+\frac {b \left (\frac {\sin \left (d x +c \right )^{9}}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \sin \left (d x +c \right )^{9}}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \sin \left (d x +c \right )^{7}}{8}-\frac {7 \sin \left (d x +c \right )^{5}}{8}-\frac {35 \sin \left (d x +c \right )^{3}}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(179\) |
| risch | \(3 i a x -\frac {i b \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {13 i b \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {13 i b \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {i b \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {6 i a c}{d}+\frac {i \left (24 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+13 b \,{\mathrm e}^{7 i \left (d x +c \right )}+32 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+5 b \,{\mathrm e}^{5 i \left (d x +c \right )}+24 i a \,{\mathrm e}^{2 i \left (d x +c \right )}-5 b \,{\mathrm e}^{3 i \left (d x +c \right )}-13 b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{8 d}\) | \(292\) |
Input:
int(sin(d*x+c)^2*(a+b*sin(d*x+c))*tan(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
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 = 0.10 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.01 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {24 \, a \cos \left (d x + c\right )^{6} - 3 \, {\left (24 \, a - 35 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (24 \, a + 35 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 12 \, a \cos \left (d x + c\right )^{4} - 72 \, a \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, b \cos \left (d x + c\right )^{6} - 80 \, b \cos \left (d x + c\right )^{4} - 39 \, b \cos \left (d x + c\right )^{2} + 6 \, b\right )} \sin \left (d x + c\right ) + 12 \, a}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sin(d*x+c)^2*(a+b*sin(d*x+c))*tan(d*x+c)^5,x, algorithm="fricas" )
Output:
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)
Time = 17.33 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.29 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\begin {cases} \frac {a \left (- \frac {3 \log {\left (\csc ^{2}{\left (c + d x \right )} - 1 \right )}}{2} + \frac {3 \log {\left (\csc ^{2}{\left (c + d x \right )} \right )}}{2} - \frac {1}{2 \csc ^{2}{\left (c + d x \right )}} - \frac {1}{\csc ^{2}{\left (c + d x \right )} - 1} + \frac {1}{4 \left (\csc ^{2}{\left (c + d x \right )} - 1\right )^{2}}\right ) + b \left (- \frac {35 \log {\left (\csc {\left (c + d x \right )} - 1 \right )}}{16} + \frac {35 \log {\left (\csc {\left (c + d x \right )} + 1 \right )}}{16} - \frac {3}{\csc {\left (c + d x \right )}} - \frac {1}{3 \csc ^{3}{\left (c + d x \right )}} - \frac {11}{16 \left (\csc {\left (c + d x \right )} + 1\right )} - \frac {1}{16 \left (\csc {\left (c + d x \right )} + 1\right )^{2}} - \frac {11}{16 \left (\csc {\left (c + d x \right )} - 1\right )} + \frac {1}{16 \left (\csc {\left (c + d x \right )} - 1\right )^{2}}\right )}{d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right ) \sin ^{2}{\left (c \right )} \tan ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(sin(d*x+c)**2*(a+b*sin(d*x+c))*tan(d*x+c)**5,x)
Output:
Piecewise(((a*(-3*log(csc(c + d*x)**2 - 1)/2 + 3*log(csc(c + d*x)**2)/2 - 1/(2*csc(c + d*x)**2) - 1/(csc(c + d*x)**2 - 1) + 1/(4*(csc(c + d*x)**2 - 1)**2)) + b*(-35*log(csc(c + d*x) - 1)/16 + 35*log(csc(c + d*x) + 1)/16 - 3/csc(c + d*x) - 1/(3*csc(c + d*x)**3) - 11/(16*(csc(c + d*x) + 1)) - 1/(1 6*(csc(c + d*x) + 1)**2) - 11/(16*(csc(c + d*x) - 1)) + 1/(16*(csc(c + d*x ) - 1)**2)))/d, Ne(d, 0)), (x*(a + b*sin(c))*sin(c)**2*tan(c)**5, True))
Time = 0.03 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.90 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {16 \, b \sin \left (d x + c\right )^{3} + 24 \, a \sin \left (d x + c\right )^{2} + 3 \, {\left (24 \, a - 35 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (24 \, a + 35 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 144 \, b \sin \left (d x + c\right ) - \frac {6 \, {\left (13 \, b \sin \left (d x + c\right )^{3} + 12 \, a \sin \left (d x + c\right )^{2} - 11 \, b \sin \left (d x + c\right ) - 10 \, a\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{48 \, d} \] Input:
integrate(sin(d*x+c)^2*(a+b*sin(d*x+c))*tan(d*x+c)^5,x, algorithm="maxima" )
Output:
-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 = 0.48 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.03 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {{\left (24 \, a - 35 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} - \frac {{\left (24 \, a + 35 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} - \frac {2 \, b d^{2} \sin \left (d x + c\right )^{3} + 3 \, a d^{2} \sin \left (d x + c\right )^{2} + 18 \, b d^{2} \sin \left (d x + c\right )}{6 \, d^{3}} + \frac {13 \, b \sin \left (d x + c\right )^{3} + 12 \, a \sin \left (d x + c\right )^{2} - 11 \, b \sin \left (d x + c\right ) - 10 \, a}{8 \, d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(sin(d*x+c)^2*(a+b*sin(d*x+c))*tan(d*x+c)^5,x, algorithm="giac")
Output:
-1/16*(24*a - 35*b)*log(abs(sin(d*x + c) + 1))/d - 1/16*(24*a + 35*b)*log( abs(sin(d*x + c) - 1))/d - 1/6*(2*b*d^2*sin(d*x + c)^3 + 3*a*d^2*sin(d*x + c)^2 + 18*b*d^2*sin(d*x + c))/d^3 + 1/8*(13*b*sin(d*x + c)^3 + 12*a*sin(d *x + c)^2 - 11*b*sin(d*x + c) - 10*a)/(d*(sin(d*x + c) + 1)^2*(sin(d*x + c ) - 1)^2)
Time = 19.92 (sec) , antiderivative size = 346, normalized size of antiderivative = 2.35 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {3\,a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {-\frac {35\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{4}-6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\frac {35\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{6}+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {329\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{12}+16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-17\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {329\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {35\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}-6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {35\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (3\,a+\frac {35\,b}{8}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (3\,a-\frac {35\,b}{8}\right )}{d} \] Input:
int(sin(c + d*x)^2*tan(c + d*x)^5*(a + b*sin(c + d*x)),x)
Output:
(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
Time = 0.18 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.82 \[ \int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {72 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{4} a -144 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} a +72 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a -72 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a -105 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b +144 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +210 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b -72 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -105 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -72 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a +105 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b +144 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a -210 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -72 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +105 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -8 \sin \left (d x +c \right )^{7} b -12 \sin \left (d x +c \right )^{6} a -56 \sin \left (d x +c \right )^{5} b +54 \sin \left (d x +c \right )^{4} a +175 \sin \left (d x +c \right )^{3} b -36 \sin \left (d x +c \right )^{2} a -105 \sin \left (d x +c \right ) b}{24 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)^2*(a+b*sin(d*x+c))*tan(d*x+c)^5,x)
Output:
(72*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4*a - 144*log(tan((c + d*x) /2)**2 + 1)*sin(c + d*x)**2*a + 72*log(tan((c + d*x)/2)**2 + 1)*a - 72*log (tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 105*log(tan((c + d*x)/2) - 1)*s in(c + d*x)**4*b + 144*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 210*l og(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b - 72*log(tan((c + d*x)/2) - 1)* a - 105*log(tan((c + d*x)/2) - 1)*b - 72*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a + 105*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b + 144*log(tan ((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 210*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 72*log(tan((c + d*x)/2) + 1)*a + 105*log(tan((c + d*x)/2) + 1)*b - 8*sin(c + d*x)**7*b - 12*sin(c + d*x)**6*a - 56*sin(c + d*x)**5*b + 54*sin(c + d*x)**4*a + 175*sin(c + d*x)**3*b - 36*sin(c + d*x)**2*a - 10 5*sin(c + d*x)*b)/(24*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))