Integrand size = 27, antiderivative size = 189 \[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=-\frac {\left (15 a^2+48 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {\left (15 a^2-48 a b+35 b^2\right ) \log (1+\sin (c+d x))}{16 d}-\frac {\left (a^2+3 b^2\right ) \sin (c+d x)}{d}-\frac {a b \sin ^2(c+d x)}{d}-\frac {b^2 \sin ^3(c+d x)}{3 d}-\frac {\sec ^2(c+d x) (11 b+9 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d} \]
-1/16*(15*a^2+48*a*b+35*b^2)*ln(1-sin(d*x+c))/d+1/16*(15*a^2-48*a*b+35*b^2 )*ln(1+sin(d*x+c))/d-(a^2+3*b^2)*sin(d*x+c)/d-a*b*sin(d*x+c)^2/d-1/3*b^2*s in(d*x+c)^3/d-1/8*sec(d*x+c)^2*(11*b+9*a*sin(d*x+c))*(a+b*sin(d*x+c))/d+1/ 4*sec(d*x+c)^3*(a+b*sin(d*x+c))^2*tan(d*x+c)/d
Time = 1.20 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.98 \[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=\frac {-3 \left (15 a^2+48 a b+35 b^2\right ) \log (1-\sin (c+d x))+3 \left (15 a^2-48 a b+35 b^2\right ) \log (1+\sin (c+d x))+\frac {3 (a+b)^2}{(-1+\sin (c+d x))^2}+\frac {3 (a+b) (9 a+13 b)}{-1+\sin (c+d x)}-48 \left (a^2+3 b^2\right ) \sin (c+d x)-48 a b \sin ^2(c+d x)-16 b^2 \sin ^3(c+d x)-\frac {3 (a-b)^2}{(1+\sin (c+d x))^2}+\frac {3 (9 a-13 b) (a-b)}{1+\sin (c+d x)}}{48 d} \]
(-3*(15*a^2 + 48*a*b + 35*b^2)*Log[1 - Sin[c + d*x]] + 3*(15*a^2 - 48*a*b + 35*b^2)*Log[1 + Sin[c + d*x]] + (3*(a + b)^2)/(-1 + Sin[c + d*x])^2 + (3 *(a + b)*(9*a + 13*b))/(-1 + Sin[c + d*x]) - 48*(a^2 + 3*b^2)*Sin[c + d*x] - 48*a*b*Sin[c + d*x]^2 - 16*b^2*Sin[c + d*x]^3 - (3*(a - b)^2)/(1 + Sin[ c + d*x])^2 + (3*(9*a - 13*b)*(a - b))/(1 + Sin[c + d*x]))/(48*d)
Time = 0.62 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.14, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3316, 27, 531, 25, 2176, 25, 2341, 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.
| \(\displaystyle \int \sin (c+d x) \tan ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^6 (a+b \sin (c+d x))^2}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\sin ^6(c+d x) (a+b \sin (c+d x))^2}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {b^6 \sin ^6(c+d x) (a+b \sin (c+d x))^2}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 531 |
| \(\displaystyle \frac {\frac {\int -\frac {(a+b \sin (c+d x)) \left (4 \sin ^5(c+d x) b^7+4 \sin ^3(c+d x) b^7+3 \sin (c+d x) b^7+4 a \sin ^4(c+d x) b^6+4 a \sin ^2(c+d x) b^6+a b^6\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {b^5 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{b d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {b^5 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int \frac {(a+b \sin (c+d x)) \left (4 \sin ^5(c+d x) b^7+4 \sin ^3(c+d x) b^7+3 \sin (c+d x) b^7+4 a \sin ^4(c+d x) b^6+4 a \sin ^2(c+d x) b^6+a b^6\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}}{b d}\) |
| \(\Big \downarrow \) 2176 |
| \(\displaystyle \frac {\frac {b^5 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {\int -\frac {8 \sin ^4(c+d x) b^8+16 a \sin ^3(c+d x) b^7+32 a \sin (c+d x) b^7+8 \left (a^2+2 b^2\right ) \sin ^2(c+d x) b^6+\left (7 a^2+11 b^2\right ) b^6}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}+\frac {b^4 (a+b \sin (c+d x)) \left (9 a b \sin (c+d x)+11 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}}{b d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {b^5 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (a+b \sin (c+d x)) \left (9 a b \sin (c+d x)+11 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \frac {8 \sin ^4(c+d x) b^8+16 a \sin ^3(c+d x) b^7+32 a \sin (c+d x) b^7+8 \left (a^2+2 b^2\right ) \sin ^2(c+d x) b^6+\left (7 a^2+11 b^2\right ) b^6}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}}{4 b^2}}{b d}\) |
| \(\Big \downarrow \) 2341 |
| \(\displaystyle \frac {\frac {b^5 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (a+b \sin (c+d x)) \left (9 a b \sin (c+d x)+11 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \left (-8 \sin ^2(c+d x) b^6-16 a \sin (c+d x) b^5-8 \left (a^2+3 b^2\right ) b^4+\frac {48 a \sin (c+d x) b^7+5 \left (3 a^2+7 b^2\right ) b^6}{b^2-b^2 \sin ^2(c+d x)}\right )d(b \sin (c+d x))}{2 b^2}}{4 b^2}}{b d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^5 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (a+b \sin (c+d x)) \left (9 a b \sin (c+d x)+11 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {5 b^5 \left (3 a^2+7 b^2\right ) \text {arctanh}(\sin (c+d x))-8 b^5 \left (a^2+3 b^2\right ) \sin (c+d x)-8 a b^6 \sin ^2(c+d x)-24 a b^6 \log \left (b^2-b^2 \sin ^2(c+d x)\right )-\frac {8}{3} b^7 \sin ^3(c+d x)}{2 b^2}}{4 b^2}}{b d}\) |
((b^5*Sin[c + d*x]*(a + b*Sin[c + d*x])^2)/(4*(b^2 - b^2*Sin[c + d*x]^2)^2 ) - ((b^4*(a + b*Sin[c + d*x])*(11*b^2 + 9*a*b*Sin[c + d*x]))/(2*(b^2 - b^ 2*Sin[c + d*x]^2)) - (5*b^5*(3*a^2 + 7*b^2)*ArcTanh[Sin[c + d*x]] - 24*a*b ^6*Log[b^2 - b^2*Sin[c + d*x]^2] - 8*b^5*(a^2 + 3*b^2)*Sin[c + d*x] - 8*a* b^6*Sin[c + d*x]^2 - (8*b^7*Sin[c + d*x]^3)/3)/(2*b^2))/(4*b^2))/(b*d)
3.15.92.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb ol] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomi alRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a + b*x^2, x], x, 1]}, Simp[(c + d*x)^n*(a*f - b*e*x)*((a + b*x^2)^(p + 1)/( 2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b *x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(c + d*x)*Qx - a*d*f*n + b*c*e*(2*p + 3) + b*d*e*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGt Q[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && GtQ[n, 1] && IntegerQ[2*p]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[Pq, a + b*x^2, x], R = Coeff[PolynomialRema inder[Pq, a + b*x^2, x], x, 0], S = Coeff[PolynomialRemainder[Pq, a + b*x^2 , x], x, 1]}, Simp[(d + e*x)^m*(a + b*x^2)^(p + 1)*((a*S - b*R*x)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(d + e*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(d + e*x)*Qx - a*e*S*m + b*d*R*(2*p + 3) + b*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x ] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && !(IGtQ[m, 0] && R ationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 1.01 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.41
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a b \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin ^{9}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \left (\sin ^{9}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \left (\sin ^{7}\left (d x +c \right )\right )}{8}-\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(266\) |
| default | \(\frac {a^{2} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a b \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin ^{9}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \left (\sin ^{9}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \left (\sin ^{7}\left (d x +c \right )\right )}{8}-\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(266\) |
| parallelrisch | \(\frac {576 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-180 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+\frac {16}{5} a b +\frac {7}{3} b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+180 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}-\frac {16}{5} a b +\frac {7}{3} b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-54 a b \cos \left (2 d x +2 c \right )+72 \cos \left (4 d x +4 c \right ) a b +\left (-90 a^{2}-189 b^{2}\right ) \sin \left (3 d x +3 c \right )+\left (-12 a^{2}-35 b^{2}\right ) \sin \left (5 d x +5 c \right )+6 a b \cos \left (6 d x +6 c \right )+b^{2} \sin \left (7 d x +7 c \right )+\left (-30 a^{2}-105 b^{2}\right ) \sin \left (d x +c \right )-24 a b}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(283\) |
| norman | \(\frac {\frac {\left (11 a^{2}-17 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {44 a b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {44 a b \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {32 a b}{3 d}-\frac {5 \left (3 a^{2}+7 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {5 \left (3 a^{2}+7 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {47 \left (3 a^{2}+7 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {47 \left (3 a^{2}+7 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {5 \left (3 a^{2}+7 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {5 \left (3 a^{2}+7 b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {4 a b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 a b \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {32 a b \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {\left (15 a^{2}-48 a b +35 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (15 a^{2}+48 a b +35 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {6 a b \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(402\) |
| risch | \(\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (9 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+13 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+5 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+48 i a b \,{\mathrm e}^{5 i \left (d x +c \right )}-a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+64 i a b \,{\mathrm e}^{3 i \left (d x +c \right )}-9 a^{2}-13 b^{2}+48 i a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {13 i {\mathrm e}^{i \left (d x +c \right )} b^{2}}{8 d}+\frac {a b \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 d}-\frac {13 i {\mathrm e}^{-i \left (d x +c \right )} b^{2}}{8 d}+\frac {12 i a b c}{d}-\frac {i b^{2} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {i b^{2} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {a b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 d}-\frac {i a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+6 i x a b +\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {15 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{d}+\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 d}-\frac {15 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{d}-\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 d}\) | \(475\) |
1/d*(a^2*(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* sin(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 )))+2*a*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)))+b^2*(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*sin(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 = 0.30 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.05 \[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=\frac {48 \, a b \cos \left (d x + c\right )^{6} - 24 \, a b \cos \left (d x + c\right )^{4} + 3 \, {\left (15 \, a^{2} - 48 \, a b + 35 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (15 \, a^{2} + 48 \, a b + 35 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 144 \, a b \cos \left (d x + c\right )^{2} + 24 \, a b + 2 \, {\left (8 \, b^{2} \cos \left (d x + c\right )^{6} - 8 \, {\left (3 \, a^{2} + 10 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 3 \, {\left (9 \, a^{2} + 13 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, a^{2} + 6 \, b^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
1/48*(48*a*b*cos(d*x + c)^6 - 24*a*b*cos(d*x + c)^4 + 3*(15*a^2 - 48*a*b + 35*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(15*a^2 + 48*a*b + 35*b^ 2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 144*a*b*cos(d*x + c)^2 + 24*a*b + 2*(8*b^2*cos(d*x + c)^6 - 8*(3*a^2 + 10*b^2)*cos(d*x + c)^4 - 3*(9*a^2 + 13*b^2)*cos(d*x + c)^2 + 6*a^2 + 6*b^2)*sin(d*x + c))/(d*cos(d*x + c)^4)
Timed out. \[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.95 \[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=-\frac {16 \, b^{2} \sin \left (d x + c\right )^{3} + 48 \, a b \sin \left (d x + c\right )^{2} - 3 \, {\left (15 \, a^{2} - 48 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (15 \, a^{2} + 48 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 48 \, {\left (a^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right ) - \frac {6 \, {\left (24 \, a b \sin \left (d x + c\right )^{2} + {\left (9 \, a^{2} + 13 \, b^{2}\right )} \sin \left (d x + c\right )^{3} - 20 \, a b - {\left (7 \, a^{2} + 11 \, b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{48 \, d} \]
-1/48*(16*b^2*sin(d*x + c)^3 + 48*a*b*sin(d*x + c)^2 - 3*(15*a^2 - 48*a*b + 35*b^2)*log(sin(d*x + c) + 1) + 3*(15*a^2 + 48*a*b + 35*b^2)*log(sin(d*x + c) - 1) + 48*(a^2 + 3*b^2)*sin(d*x + c) - 6*(24*a*b*sin(d*x + c)^2 + (9 *a^2 + 13*b^2)*sin(d*x + c)^3 - 20*a*b - (7*a^2 + 11*b^2)*sin(d*x + c))/(s in(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.50 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.05 \[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=-\frac {16 \, b^{2} \sin \left (d x + c\right )^{3} + 48 \, a b \sin \left (d x + c\right )^{2} + 48 \, a^{2} \sin \left (d x + c\right ) + 144 \, b^{2} \sin \left (d x + c\right ) - 3 \, {\left (15 \, a^{2} - 48 \, a b + 35 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \, {\left (15 \, a^{2} + 48 \, a b + 35 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (36 \, a b \sin \left (d x + c\right )^{4} + 9 \, a^{2} \sin \left (d x + c\right )^{3} + 13 \, b^{2} \sin \left (d x + c\right )^{3} - 48 \, a b \sin \left (d x + c\right )^{2} - 7 \, a^{2} \sin \left (d x + c\right ) - 11 \, b^{2} \sin \left (d x + c\right ) + 16 \, a b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{48 \, d} \]
-1/48*(16*b^2*sin(d*x + c)^3 + 48*a*b*sin(d*x + c)^2 + 48*a^2*sin(d*x + c) + 144*b^2*sin(d*x + c) - 3*(15*a^2 - 48*a*b + 35*b^2)*log(abs(sin(d*x + c ) + 1)) + 3*(15*a^2 + 48*a*b + 35*b^2)*log(abs(sin(d*x + c) - 1)) - 6*(36* a*b*sin(d*x + c)^4 + 9*a^2*sin(d*x + c)^3 + 13*b^2*sin(d*x + c)^3 - 48*a*b *sin(d*x + c)^2 - 7*a^2*sin(d*x + c) - 11*b^2*sin(d*x + c) + 16*a*b)/(sin( d*x + c)^2 - 1)^2)/d
Time = 11.05 (sec) , antiderivative size = 433, normalized size of antiderivative = 2.29 \[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {15\,a^2}{8}-6\,a\,b+\frac {35\,b^2}{8}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {15\,a^2}{8}+6\,a\,b+\frac {35\,b^2}{8}\right )}{d}-\frac {\left (-\frac {15\,a^2}{4}-\frac {35\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\left (\frac {5\,a^2}{2}+\frac {35\,b^2}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\left (\frac {47\,a^2}{4}+\frac {329\,b^2}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+32\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (11\,a^2-17\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {47\,a^2}{4}+\frac {329\,b^2}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {5\,a^2}{2}+\frac {35\,b^2}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (-\frac {15\,a^2}{4}-\frac {35\,b^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{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 {6\,a\,b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]
(log(tan(c/2 + (d*x)/2) + 1)*((15*a^2)/8 - 6*a*b + (35*b^2)/8))/d - (log(t an(c/2 + (d*x)/2) - 1)*(6*a*b + (15*a^2)/8 + (35*b^2)/8))/d - (tan(c/2 + ( d*x)/2)^7*(11*a^2 - 17*b^2) + tan(c/2 + (d*x)/2)^3*((5*a^2)/2 + (35*b^2)/6 ) + tan(c/2 + (d*x)/2)^11*((5*a^2)/2 + (35*b^2)/6) - tan(c/2 + (d*x)/2)^13 *((15*a^2)/4 + (35*b^2)/4) + tan(c/2 + (d*x)/2)^5*((47*a^2)/4 + (329*b^2)/ 12) + tan(c/2 + (d*x)/2)^9*((47*a^2)/4 + (329*b^2)/12) - tan(c/2 + (d*x)/2 )*((15*a^2)/4 + (35*b^2)/4) - 12*a*b*tan(c/2 + (d*x)/2)^2 + 12*a*b*tan(c/2 + (d*x)/2)^4 + 32*a*b*tan(c/2 + (d*x)/2)^6 + 32*a*b*tan(c/2 + (d*x)/2)^8 + 12*a*b*tan(c/2 + (d*x)/2)^10 - 12*a*b*tan(c/2 + (d*x)/2)^12)/(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)) + (6*a*b*log(tan(c/2 + (d*x)/2)^2 + 1))/d