Integrand size = 29, antiderivative size = 182 \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b (3 a+b) \log (1-\sin (c+d x))}{16 (a+b)^3 d}-\frac {(3 a-b) b \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {a^3 b^2 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a^3-b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d} \]
1/16*b*(3*a+b)*ln(1-sin(d*x+c))/(a+b)^3/d-1/16*(3*a-b)*b*ln(1+sin(d*x+c))/ (a-b)^3/d+a^3*b^2*ln(a+b*sin(d*x+c))/(a^2-b^2)^3/d+1/4*sec(d*x+c)^4*(a-b*s in(d*x+c))/(a^2-b^2)/d-1/8*sec(d*x+c)^2*(4*a^3-b*(5*a^2-b^2)*sin(d*x+c))/( a^2-b^2)^2/d
Time = 0.96 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {b (3 a+b) \log (1-\sin (c+d x))}{(a+b)^3}-\frac {(3 a-b) b \log (1+\sin (c+d x))}{(a-b)^3}+\frac {16 a^3 b^2 \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}+\frac {1}{(a+b) (-1+\sin (c+d x))^2}+\frac {3 a+b}{(a+b)^2 (-1+\sin (c+d x))}+\frac {1}{(a-b) (1+\sin (c+d x))^2}+\frac {-3 a+b}{(a-b)^2 (1+\sin (c+d x))}}{16 d} \]
((b*(3*a + b)*Log[1 - Sin[c + d*x]])/(a + b)^3 - ((3*a - b)*b*Log[1 + Sin[ c + d*x]])/(a - b)^3 + (16*a^3*b^2*Log[a + b*Sin[c + d*x]])/((a - b)^3*(a + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x])^2) + (3*a + b)/((a + b)^2*(-1 + S in[c + d*x])) + 1/((a - b)*(1 + Sin[c + d*x])^2) + (-3*a + b)/((a - b)^2*( 1 + Sin[c + d*x])))/(16*d)
Time = 0.57 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.36, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 3316, 27, 601, 25, 27, 663, 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.
| \(\displaystyle \int \frac {\tan ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3}{\cos (c+d x)^5 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\sin ^3(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^2 \int \frac {b^3 \sin ^3(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 601 |
| \(\displaystyle \frac {b^2 \left (\frac {b^2 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {b^2 \left (\frac {a b^2}{a^2-b^2}-\frac {b \left (4 a^2-b^2\right ) \sin (c+d x)}{a^2-b^2}\right )}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^2 \left (\frac {\int \frac {b^2 \left (a b^2-b \left (4 a^2-b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right ) (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {b^2 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^2 \left (\frac {\int \frac {a b^2-b \left (4 a^2-b^2\right ) \sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 \left (a^2-b^2\right )}+\frac {b^2 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 663 |
| \(\displaystyle \frac {b^2 \left (\frac {\int \left (\frac {4 a^3}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac {(a-b) (3 a+b)}{4 b (a+b)^2 (b-b \sin (c+d x))}-\frac {(3 a-b) (a+b)}{4 (a-b)^2 b (\sin (c+d x) b+b)}-\frac {4 a^2-b a-b^2}{4 b (a+b) (b-b \sin (c+d x))^2}+\frac {4 a^2+b a-b^2}{4 (a-b) b (\sin (c+d x) b+b)^2}\right )d(b \sin (c+d x))}{4 \left (a^2-b^2\right )}+\frac {b^2 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^2 \left (\frac {b^2 (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}+\frac {-\frac {4 a^2-a b-b^2}{4 b (a+b) (b-b \sin (c+d x))}-\frac {4 a^2+a b-b^2}{4 b (a-b) (b \sin (c+d x)+b)}+\frac {4 a^3 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac {(a-b) (3 a+b) \log (b-b \sin (c+d x))}{4 b (a+b)^2}-\frac {(3 a-b) (a+b) \log (b \sin (c+d x)+b)}{4 b (a-b)^2}}{4 \left (a^2-b^2\right )}\right )}{d}\) |
(b^2*((b^2*(a - b*Sin[c + d*x]))/(4*(a^2 - b^2)*(b^2 - b^2*Sin[c + d*x]^2) ^2) + (((a - b)*(3*a + b)*Log[b - b*Sin[c + d*x]])/(4*b*(a + b)^2) + (4*a^ 3*Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^2 - ((3*a - b)*(a + b)*Log[b + b*Si n[c + d*x]])/(4*(a - b)^2*b) - (4*a^2 - a*b - b^2)/(4*b*(a + b)*(b - b*Sin [c + d*x])) - (4*a^2 + a*b - b^2)/(4*(a - b)*b*(b + b*Sin[c + d*x])))/(4*( a^2 - b^2))))/d
3.14.64.3.1 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_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* (2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ )^2)^(p_.), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Simp[1/c^p Int[ExpandI ntegrand[(d + e*x)^m*(f + g*x)^n*(-q + c*x)^p*(q + c*x)^p, x], x], x] /; ! FractionalPowerFactorQ[q]] /; FreeQ[{a, c, d, e, f, g}, x] && ILtQ[p, -1] & & IntegersQ[m, n] && NiceSqrtQ[(-a)*c]
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 = 1.02 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.97
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 a -b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {\left (3 a -b \right ) b \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {a^{3} b^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-3 a -b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (3 a +b \right ) b \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}}{d}\) | \(176\) |
| default | \(\frac {\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3 a -b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {\left (3 a -b \right ) b \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {a^{3} b^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-3 a -b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (3 a +b \right ) b \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}}{d}\) | \(176\) |
| parallelrisch | \(\frac {8 b^{2} \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a^{3} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+3 b \left (a +\frac {b}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-2 \left (\frac {3 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -\frac {b}{3}\right ) \left (a +b \right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\left (a -b \right ) \left (\left (a^{3}-a \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\frac {\left (-a^{3}-a \,b^{2}\right ) \cos \left (4 d x +4 c \right )}{4}+\frac {\left (-5 a^{2} b +b^{3}\right ) \sin \left (3 d x +3 c \right )}{4}+\frac {\left (3 a^{2} b -7 b^{3}\right ) \sin \left (d x +c \right )}{4}-\frac {3 a^{3}}{4}+\frac {5 a \,b^{2}}{4}\right )\right ) \left (a +b \right )}{2 \left (a -b \right )^{3} \left (a +b \right )^{3} d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(304\) |
| norman | \(\frac {-\frac {2 a \,b^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 a \,b^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {4 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {b \left (3 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}+\frac {b \left (3 a^{2}+b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {\left (11 a^{2}-7 b^{2}\right ) b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {\left (11 a^{2}-7 b^{2}\right ) b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {a^{3} b^{2} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {\left (3 a -b \right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}+\frac {\left (3 a +b \right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}\) | \(460\) |
| risch | \(\frac {3 i a b c}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {2 i a^{3} b^{2} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {i b^{2} c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {i b^{2} x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {3 i a b x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {i b^{2} c}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {3 i a b c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {i b^{2} x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 i a b x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {2 i a^{3} b^{2} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}+\frac {i \left (8 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-5 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+16 i b^{2} a \,{\mathrm e}^{4 i \left (d x +c \right )}+3 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-7 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+8 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-3 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+7 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+5 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}+\frac {a^{3} b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\) | \(752\) |
1/d*(1/2/(8*a-8*b)/(1+sin(d*x+c))^2-1/16*(3*a-b)/(a-b)^2/(1+sin(d*x+c))-1/ 16*(3*a-b)/(a-b)^3*b*ln(1+sin(d*x+c))+a^3/(a+b)^3*b^2/(a-b)^3*ln(a+b*sin(d *x+c))+1/2/(8*a+8*b)/(sin(d*x+c)-1)^2-1/16*(-3*a-b)/(a+b)^2/(sin(d*x+c)-1) +1/16*(3*a+b)/(a+b)^3*b*ln(sin(d*x+c)-1))
Time = 0.48 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.43 \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {16 \, a^{3} b^{2} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (3 \, a^{4} b + 8 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, a^{4} b - 8 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{5} - 8 \, a^{3} b^{2} + 4 \, a b^{4} - 8 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - {\left (5 \, a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \]
1/16*(16*a^3*b^2*cos(d*x + c)^4*log(b*sin(d*x + c) + a) - (3*a^4*b + 8*a^3 *b^2 + 6*a^2*b^3 - b^5)*cos(d*x + c)^4*log(sin(d*x + c) + 1) + (3*a^4*b - 8*a^3*b^2 + 6*a^2*b^3 - b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*a^5 - 8*a^3*b^2 + 4*a*b^4 - 8*(a^5 - a^3*b^2)*cos(d*x + c)^2 - 2*(2*a^4*b - 4 *a^2*b^3 + 2*b^5 - (5*a^4*b - 6*a^2*b^3 + b^5)*cos(d*x + c)^2)*sin(d*x + c ))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x + c)^4)
Timed out. \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.47 \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, a^{3} b^{2} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (3 \, a b - b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (4 \, a^{3} \sin \left (d x + c\right )^{2} - {\left (5 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{3} - 2 \, a^{3} - 2 \, a b^{2} + {\left (3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \]
1/16*(16*a^3*b^2*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^ 6) - (3*a*b - b^2)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (3*a*b + b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 2*( 4*a^3*sin(d*x + c)^2 - (5*a^2*b - b^3)*sin(d*x + c)^3 - 2*a^3 - 2*a*b^2 + (3*a^2*b + b^3)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^4 + a^ 4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^2))/d
Time = 0.50 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.79 \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, a^{3} b^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {{\left (3 \, a b - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (6 \, a^{3} b^{2} \sin \left (d x + c\right )^{4} - 5 \, a^{4} b \sin \left (d x + c\right )^{3} + 6 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} - b^{5} \sin \left (d x + c\right )^{3} + 4 \, a^{5} \sin \left (d x + c\right )^{2} - 16 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} + 3 \, a^{4} b \sin \left (d x + c\right ) - 2 \, a^{2} b^{3} \sin \left (d x + c\right ) - b^{5} \sin \left (d x + c\right ) - 2 \, a^{5} + 6 \, a^{3} b^{2} + 2 \, a b^{4}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
1/16*(16*a^3*b^3*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b ^5 - b^7) - (3*a*b - b^2)*log(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a* b^2 - b^3) + (3*a*b + b^2)*log(abs(sin(d*x + c) - 1))/(a^3 + 3*a^2*b + 3*a *b^2 + b^3) + 2*(6*a^3*b^2*sin(d*x + c)^4 - 5*a^4*b*sin(d*x + c)^3 + 6*a^2 *b^3*sin(d*x + c)^3 - b^5*sin(d*x + c)^3 + 4*a^5*sin(d*x + c)^2 - 16*a^3*b ^2*sin(d*x + c)^2 + 3*a^4*b*sin(d*x + c) - 2*a^2*b^3*sin(d*x + c) - b^5*si n(d*x + c) - 2*a^5 + 6*a^3*b^2 + 2*a*b^4)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(sin(d*x + c)^2 - 1)^2))/d
Time = 12.48 (sec) , antiderivative size = 471, normalized size of antiderivative = 2.59 \[ \int \frac {\sec ^2(c+d x) \tan ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (3\,a+b\right )}{8\,d\,{\left (a+b\right )}^3}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (11\,a^2\,b-7\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4-2\,a^2\,b^2+b^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (3\,a^2\,b+b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (11\,a^2\,b-7\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a^4-2\,a^2\,b^2+b^4}-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2+b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}+\frac {3\,b}{8\,{\left (a-b\right )}^2}\right )}{d}+\frac {a^3\,b^2\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \]
(b*log(tan(c/2 + (d*x)/2) - 1)*(3*a + b))/(8*d*(a + b)^3) - ((tan(c/2 + (d *x)/2)^3*(11*a^2*b - 7*b^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) - (4*a^3*tan(c/2 + (d*x)/2)^4)/(a^4 + b^4 - 2*a^2*b^2) - (tan(c/2 + (d*x)/2)^7*(3*a^2*b + b ^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^5*(11*a^2*b - 7*b^3 ))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (2*a*b^2*tan(c/2 + (d*x)/2)^2)/(a^4 + b^4 - 2*a^2*b^2) + (2*a*b^2*tan(c/2 + (d*x)/2)^6)/(a^4 + b^4 - 2*a^2*b^2) - ( b*tan(c/2 + (d*x)/2)*(3*a^2 + b^2))/(4*(a^4 + b^4 - 2*a^2*b^2)))/(d*(6*tan (c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan( c/2 + (d*x)/2)^8 + 1)) - (log(tan(c/2 + (d*x)/2) + 1)*(b^2/(4*(a - b)^3) + (3*b)/(8*(a - b)^2)))/d + (a^3*b^2*log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan (c/2 + (d*x)/2)^2))/(d*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))