Integrand size = 29, antiderivative size = 142 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=-\frac {3 b (a+b) (3 a+5 b) \log (1-\sin (c+d x))}{16 d}+\frac {3 (3 a-5 b) (a-b) b \log (1+\sin (c+d x))}{16 d}-\frac {15 b^3 \sin (c+d x)}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{8 d} \]
-3/16*b*(a+b)*(3*a+5*b)*ln(1-sin(d*x+c))/d+3/16*(3*a-5*b)*(a-b)*b*ln(1+sin (d*x+c))/d-15/8*b^3*sin(d*x+c)/d+1/4*sec(d*x+c)^4*(a+b*sin(d*x+c))^3/d-1/8 *sec(d*x+c)^2*(a+b*sin(d*x+c))^2*(4*a+7*b*sin(d*x+c))/d
Time = 0.32 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.04 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {-3 b (a+b) (3 a+5 b) \log (1-\sin (c+d x))+3 (3 a-5 b) (a-b) b \log (1+\sin (c+d x))+\frac {(a+b)^3}{(-1+\sin (c+d x))^2}+\frac {3 (a+b)^2 (a+3 b)}{-1+\sin (c+d x)}-16 b^3 \sin (c+d x)+\frac {(a-b)^3}{(1+\sin (c+d x))^2}-\frac {3 (a-3 b) (a-b)^2}{1+\sin (c+d x)}}{16 d} \]
(-3*b*(a + b)*(3*a + 5*b)*Log[1 - Sin[c + d*x]] + 3*(3*a - 5*b)*(a - b)*b* Log[1 + Sin[c + d*x]] + (a + b)^3/(-1 + Sin[c + d*x])^2 + (3*(a + b)^2*(a + 3*b))/(-1 + Sin[c + d*x]) - 16*b^3*Sin[c + d*x] + (a - b)^3/(1 + Sin[c + d*x])^2 - (3*(a - 3*b)*(a - b)^2)/(1 + Sin[c + d*x]))/(16*d)
Time = 0.49 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.16, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3316, 27, 531, 25, 2176, 27, 657, 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.
| \(\displaystyle \int \tan ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3 (a+b \sin (c+d x))^3}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\sin ^3(c+d x) (a+b \sin (c+d x))^3}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^2 \int \frac {b^3 \sin ^3(c+d x) (a+b \sin (c+d x))^3}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 531 |
| \(\displaystyle \frac {b^2 \left (\frac {\int -\frac {(a+b \sin (c+d x))^2 \left (4 \sin ^2(c+d x) b^4+3 b^4+4 a \sin (c+d x) b^3\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {b^2 (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^2 \left (\frac {b^2 (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int \frac {(a+b \sin (c+d x))^2 \left (4 \sin ^2(c+d x) b^4+3 b^4+4 a \sin (c+d x) b^3\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 2176 |
| \(\displaystyle \frac {b^2 \left (\frac {b^2 (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {\int -\frac {3 b^4 (a+b \sin (c+d x)) (3 a+5 b \sin (c+d x))}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}+\frac {b^2 (4 a+7 b \sin (c+d x)) (a+b \sin (c+d x))^2}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^2 \left (\frac {b^2 (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^2 (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {3}{2} b^2 \int \frac {(a+b \sin (c+d x)) (3 a+5 b \sin (c+d x))}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle \frac {b^2 \left (\frac {b^2 (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^2 (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {3}{2} b^2 \int \left (\frac {3 a^2+8 b \sin (c+d x) a+5 b^2}{b^2-b^2 \sin ^2(c+d x)}-5\right )d(b \sin (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^2 \left (\frac {b^2 (a+b \sin (c+d x))^3}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^2 (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {3}{2} b^2 \left (\frac {\left (3 a^2+5 b^2\right ) \text {arctanh}(\sin (c+d x))}{b}-4 a \log \left (b^2-b^2 \sin ^2(c+d x)\right )-5 b \sin (c+d x)\right )}{4 b^2}\right )}{d}\) |
(b^2*((b^2*(a + b*Sin[c + d*x])^3)/(4*(b^2 - b^2*Sin[c + d*x]^2)^2) - ((-3 *b^2*(((3*a^2 + 5*b^2)*ArcTanh[Sin[c + d*x]])/b - 4*a*Log[b^2 - b^2*Sin[c + d*x]^2] - 5*b*Sin[c + d*x]))/2 + (b^2*(a + b*Sin[c + d*x])^2*(4*a + 7*b* Sin[c + d*x]))/(2*(b^2 - b^2*Sin[c + d*x]^2)))/(4*b^2)))/d
3.16.3.3.1 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[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
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[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 = 0.84 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.58
| method | result | size |
| derivativedivides | \(\frac {\frac {a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+3 a^{2} b \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a \,b^{2} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(225\) |
| default | \(\frac {\frac {a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+3 a^{2} b \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a \,b^{2} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(225\) |
| parallelrisch | \(\frac {48 b^{2} \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-18 b \left (a +\frac {5 b}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+18 b \left (a -\frac {5 b}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (-a^{3}-3 a \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (a^{3}+9 a \,b^{2}\right ) \cos \left (4 d x +4 c \right )+15 \left (-a^{2} b -b^{3}\right ) \sin \left (3 d x +3 c \right )-2 b^{3} \sin \left (5 d x +5 c \right )+\left (9 a^{2} b -5 b^{3}\right ) \sin \left (d x +c \right )+3 a^{3}+3 a \,b^{2}}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(273\) |
| risch | \(3 i a \,b^{2} x +\frac {i b^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {6 i b^{2} a c}{d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (8 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+48 i a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+15 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+48 i a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-9 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+8 i a^{3} {\mathrm e}^{i \left (d x +c \right )}+48 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}+9 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-15 a^{2} b -9 b^{3}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{2}}{d}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{3}}{8 d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{8 d}\) | \(413\) |
| norman | \(\frac {\frac {\left (4 a^{3}+6 a \,b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 a^{3}+6 a \,b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b \left (45 a^{2}+11 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a \,b^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a \,b^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (4 a^{3}+16 a \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (4 a^{3}+16 a \,b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 b \left (3 a^{2}+5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {b \left (3 a^{2}+5 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {b \left (3 a^{2}+5 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {3 b \left (3 a^{2}+5 b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {b \left (105 a^{2}+47 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {b \left (105 a^{2}+47 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {3 a \,b^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 b \left (3 a^{2}-8 a b +5 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {3 b \left (3 a^{2}+8 a b +5 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\) | \(466\) |
1/d*(1/4*a^3*sin(d*x+c)^4/cos(d*x+c)^4+3*a^2*b*(1/4*sin(d*x+c)^5/cos(d*x+c )^4-1/8*sin(d*x+c)^5/cos(d*x+c)^2-1/8*sin(d*x+c)^3-3/8*sin(d*x+c)+3/8*ln(s ec(d*x+c)+tan(d*x+c)))+3*a*b^2*(1/4*tan(d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d *x+c)))+b^3*(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))))
Time = 0.30 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.24 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {3 \, {\left (3 \, a^{2} b - 8 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, a^{2} b + 8 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{3} + 12 \, a b^{2} - 8 \, {\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (8 \, b^{3} \cos \left (d x + c\right )^{4} - 6 \, a^{2} b - 2 \, b^{3} + 3 \, {\left (5 \, a^{2} b + 3 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
1/16*(3*(3*a^2*b - 8*a*b^2 + 5*b^3)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(3*a^2*b + 8*a*b^2 + 5*b^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*a ^3 + 12*a*b^2 - 8*(a^3 + 6*a*b^2)*cos(d*x + c)^2 - 2*(8*b^3*cos(d*x + c)^4 - 6*a^2*b - 2*b^3 + 3*(5*a^2*b + 3*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d* cos(d*x + c)^4)
Timed out. \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.22 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=-\frac {16 \, b^{3} \sin \left (d x + c\right ) - 3 \, {\left (3 \, a^{2} b - 8 \, a b^{2} + 5 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (3 \, a^{2} b + 8 \, a b^{2} + 5 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (5 \, a^{2} b + 3 \, b^{3}\right )} \sin \left (d x + c\right )^{3} - 2 \, a^{3} - 18 \, a b^{2} + 4 \, {\left (a^{3} + 6 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} - {\left (9 \, a^{2} b + 7 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
-1/16*(16*b^3*sin(d*x + c) - 3*(3*a^2*b - 8*a*b^2 + 5*b^3)*log(sin(d*x + c ) + 1) + 3*(3*a^2*b + 8*a*b^2 + 5*b^3)*log(sin(d*x + c) - 1) - 2*(3*(5*a^2 *b + 3*b^3)*sin(d*x + c)^3 - 2*a^3 - 18*a*b^2 + 4*(a^3 + 6*a*b^2)*sin(d*x + c)^2 - (9*a^2*b + 7*b^3)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^ 2 + 1))/d
Time = 0.37 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.32 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=-\frac {16 \, b^{3} \sin \left (d x + c\right ) - 3 \, {\left (3 \, a^{2} b - 8 \, a b^{2} + 5 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \, {\left (3 \, a^{2} b + 8 \, a b^{2} + 5 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (18 \, a b^{2} \sin \left (d x + c\right )^{4} + 15 \, a^{2} b \sin \left (d x + c\right )^{3} + 9 \, b^{3} \sin \left (d x + c\right )^{3} + 4 \, a^{3} \sin \left (d x + c\right )^{2} - 12 \, a b^{2} \sin \left (d x + c\right )^{2} - 9 \, a^{2} b \sin \left (d x + c\right ) - 7 \, b^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
-1/16*(16*b^3*sin(d*x + c) - 3*(3*a^2*b - 8*a*b^2 + 5*b^3)*log(abs(sin(d*x + c) + 1)) + 3*(3*a^2*b + 8*a*b^2 + 5*b^3)*log(abs(sin(d*x + c) - 1)) - 2 *(18*a*b^2*sin(d*x + c)^4 + 15*a^2*b*sin(d*x + c)^3 + 9*b^3*sin(d*x + c)^3 + 4*a^3*sin(d*x + c)^2 - 12*a*b^2*sin(d*x + c)^2 - 9*a^2*b*sin(d*x + c) - 7*b^3*sin(d*x + c) - 2*a^3)/(sin(d*x + c)^2 - 1)^2)/d
Time = 11.94 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.51 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,a^3+18\,a\,b^2\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {9\,a^2\,b}{4}+\frac {15\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (4\,a^3+18\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (6\,a^2\,b+10\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (6\,a^2\,b+10\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {9\,a^2\,b}{4}+\frac {15\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {33\,a^2\,b}{2}-\frac {9\,b^3}{2}\right )-6\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-6\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,a\,b^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {3\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (a+b\right )\,\left (3\,a+5\,b\right )}{8\,d}+\frac {3\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a-b\right )\,\left (3\,a-5\,b\right )}{8\,d} \]
(tan(c/2 + (d*x)/2)^4*(18*a*b^2 + 4*a^3) - tan(c/2 + (d*x)/2)*((9*a^2*b)/4 + (15*b^3)/4) + tan(c/2 + (d*x)/2)^6*(18*a*b^2 + 4*a^3) + tan(c/2 + (d*x) /2)^3*(6*a^2*b + 10*b^3) + tan(c/2 + (d*x)/2)^7*(6*a^2*b + 10*b^3) - tan(c /2 + (d*x)/2)^9*((9*a^2*b)/4 + (15*b^3)/4) + tan(c/2 + (d*x)/2)^5*((33*a^2 *b)/2 - (9*b^3)/2) - 6*a*b^2*tan(c/2 + (d*x)/2)^2 - 6*a*b^2*tan(c/2 + (d*x )/2)^8)/(d*(2*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) + (3*a*b ^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (3*b*log(tan(c/2 + (d*x)/2) - 1)*(a + b)*(3*a + 5*b))/(8*d) + (3*b*log(tan(c/2 + (d*x)/2) + 1)*(a - b)*(3*a - 5*b))/(8*d)