Integrand size = 27, antiderivative size = 165 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {\left (8 a^3+9 a^2 b-b^3\right ) \log (1-\sin (c+d x))}{16 d}+\frac {a^3 \log (\sin (c+d x))}{d}-\frac {\left (8 a^3-9 a^2 b+b^3\right ) \log (1+\sin (c+d x))}{16 d}+\frac {\sec ^2(c+d x) \left (4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d} \]
-1/16*(8*a^3+9*a^2*b-b^3)*ln(1-sin(d*x+c))/d+a^3*ln(sin(d*x+c))/d-1/16*(8* a^3-9*a^2*b+b^3)*ln(1+sin(d*x+c))/d+1/8*sec(d*x+c)^2*(4*a^3+b*(9*a^2-b^2)* sin(d*x+c))/d+1/4*sec(d*x+c)^4*(a*(a^2+3*b^2)+b*(3*a^2+b^2)*sin(d*x+c))/d
Time = 0.40 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.95 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-\left (\left (8 a^3+9 a^2 b-b^3\right ) \log (1-\sin (c+d x))\right )+16 a^3 \log (\sin (c+d x))-\left (8 a^3-9 a^2 b+b^3\right ) \log (1+\sin (c+d x))+\frac {(a+b)^3}{(-1+\sin (c+d x))^2}-\frac {(5 a-b) (a+b)^2}{-1+\sin (c+d x)}+\frac {(a-b)^3}{(1+\sin (c+d x))^2}+\frac {(a-b)^2 (5 a+b)}{1+\sin (c+d x)}}{16 d} \]
(-((8*a^3 + 9*a^2*b - b^3)*Log[1 - Sin[c + d*x]]) + 16*a^3*Log[Sin[c + d*x ]] - (8*a^3 - 9*a^2*b + b^3)*Log[1 + Sin[c + d*x]] + (a + b)^3/(-1 + Sin[c + d*x])^2 - ((5*a - b)*(a + b)^2)/(-1 + Sin[c + d*x]) + (a - b)^3/(1 + Si n[c + d*x])^2 + ((a - b)^2*(5*a + b))/(1 + Sin[c + d*x]))/(16*d)
Time = 0.47 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.31, 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, 532, 25, 532, 25, 523, 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.
| \(\displaystyle \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{\sin (c+d x) \cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\csc (c+d x) (a+b \sin (c+d x))^3}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^6 \int \frac {\csc (c+d x) (a+b \sin (c+d x))^3}{b \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 532 |
| \(\displaystyle \frac {b^6 \left (\frac {b \left (3 a^2+b^2\right ) \sin (c+d x)+a \left (a^2+3 b^2\right )}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {\csc (c+d x) \left (4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^6 \left (\frac {\int \frac {\csc (c+d x) \left (4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {b \left (3 a^2+b^2\right ) \sin (c+d x)+a \left (a^2+3 b^2\right )}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 532 |
| \(\displaystyle \frac {b^6 \left (\frac {\frac {4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int -\frac {\csc (c+d x) \left (8 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{b \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}}{4 b^2}+\frac {b \left (3 a^2+b^2\right ) \sin (c+d x)+a \left (a^2+3 b^2\right )}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^6 \left (\frac {\frac {\int \frac {\csc (c+d x) \left (8 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{b \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}+\frac {4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}+\frac {b \left (3 a^2+b^2\right ) \sin (c+d x)+a \left (a^2+3 b^2\right )}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 523 |
| \(\displaystyle \frac {b^6 \left (\frac {\frac {\int \left (\frac {8 \csc (c+d x) a^3}{b^3}+\frac {8 a^3+9 b a^2-b^3}{2 b^2 (b-b \sin (c+d x))}+\frac {-8 a^3+9 b a^2-b^3}{2 b^2 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 b^2}+\frac {4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}+\frac {b \left (3 a^2+b^2\right ) \sin (c+d x)+a \left (a^2+3 b^2\right )}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^6 \left (\frac {b \left (3 a^2+b^2\right ) \sin (c+d x)+a \left (a^2+3 b^2\right )}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}+\frac {\frac {4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {\frac {8 a^3 \log (b \sin (c+d x))}{b^2}-\frac {\left (8 a^3+9 a^2 b-b^3\right ) \log (b-b \sin (c+d x))}{2 b^2}-\frac {\left (8 a^3-9 a^2 b+b^3\right ) \log (b \sin (c+d x)+b)}{2 b^2}}{2 b^2}}{4 b^2}\right )}{d}\) |
(b^6*((a*(a^2 + 3*b^2) + b*(3*a^2 + b^2)*Sin[c + d*x])/(4*b^2*(b^2 - b^2*S in[c + d*x]^2)^2) + (((8*a^3*Log[b*Sin[c + d*x]])/b^2 - ((8*a^3 + 9*a^2*b - b^3)*Log[b - b*Sin[c + d*x]])/(2*b^2) - ((8*a^3 - 9*a^2*b + b^3)*Log[b + b*Sin[c + d*x]])/(2*b^2))/(2*b^2) + (4*a^3 + b*(9*a^2 - b^2)*Sin[c + d*x] )/(2*b^2*(b^2 - b^2*Sin[c + d*x]^2)))/(4*b^2)))/d
3.16.6.3.1 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_)))/((a_) + (b_.)*(x_)^2), x_Symbol] :> In t[ExpandIntegrand[x^m*((c + d*x)/(a + b*x^2)), x], x] /; FreeQ[{a, b, c, d} , x] && IntegerQ[m]
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[x^m *(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*p]
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 = 1.10 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.01
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a^{2} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a \,b^{2}}{4 \cos \left (d x +c \right )^{4}}+b^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(167\) |
| default | \(\frac {a^{3} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a^{2} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a \,b^{2}}{4 \cos \left (d x +c \right )^{4}}+b^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(167\) |
| parallelrisch | \(\frac {-32 \left (a^{2}+\frac {1}{8} a b -\frac {1}{8} b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-32 \left (a^{2}-\frac {1}{8} a b -\frac {1}{8} b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+32 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-8 a^{3}-24 a \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (-6 a^{3}-6 a \,b^{2}\right ) \cos \left (4 d x +4 c \right )+\left (18 a^{2} b -2 b^{3}\right ) \sin \left (3 d x +3 c \right )+\left (66 a^{2} b +14 b^{3}\right ) \sin \left (d x +c \right )+14 a^{3}+30 a \,b^{2}}{8 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(270\) |
| risch | \(\frac {i \left (-8 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-9 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-32 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-48 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-33 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-7 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-8 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+33 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+7 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+9 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{3}}{d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{8 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{3}}{8 d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(345\) |
| norman | \(\frac {\frac {\left (4 a^{3}+6 a \,b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 a^{3}+6 a \,b^{2}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 a^{3}+24 a \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 a^{3}+24 a \,b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (8 a^{3}+18 a \,b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (8 a^{3}+18 a \,b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b \left (21 a^{2}+11 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b \left (15 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {b \left (15 a^{2}+b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {b \left (27 a^{2}+5 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {b \left (27 a^{2}+5 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {b \left (81 a^{2}+31 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {b \left (81 a^{2}+31 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (8 a^{3}-9 a^{2} b +b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (8 a^{3}+9 a^{2} b -b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\) | \(468\) |
1/d*(a^3*(1/4/cos(d*x+c)^4+1/2/cos(d*x+c)^2+ln(tan(d*x+c)))+3*a^2*b*(-(-1/ 4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+3 /4*a*b^2/cos(d*x+c)^4+b^3*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/ cos(d*x+c)^2+1/8*sin(d*x+c)-1/8*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.29 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.05 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {16 \, a^{3} \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (8 \, a^{3} - 9 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, a^{3} + 9 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, a^{3} \cos \left (d x + c\right )^{2} + 4 \, a^{3} + 12 \, a b^{2} + 2 \, {\left (6 \, a^{2} b + 2 \, b^{3} + {\left (9 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
1/16*(16*a^3*cos(d*x + c)^4*log(1/2*sin(d*x + c)) - (8*a^3 - 9*a^2*b + b^3 )*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (8*a^3 + 9*a^2*b - b^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 8*a^3*cos(d*x + c)^2 + 4*a^3 + 12*a*b^2 + 2 *(6*a^2*b + 2*b^3 + (9*a^2*b - b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d *x + c)^4)
Timed out. \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.97 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {16 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) - {\left (8 \, a^{3} - 9 \, a^{2} b + b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, a^{3} + 9 \, a^{2} b - b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (4 \, a^{3} \sin \left (d x + c\right )^{2} + {\left (9 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{3} - 6 \, a^{3} - 6 \, a b^{2} - {\left (15 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
1/16*(16*a^3*log(sin(d*x + c)) - (8*a^3 - 9*a^2*b + b^3)*log(sin(d*x + c) + 1) - (8*a^3 + 9*a^2*b - b^3)*log(sin(d*x + c) - 1) - 2*(4*a^3*sin(d*x + c)^2 + (9*a^2*b - b^3)*sin(d*x + c)^3 - 6*a^3 - 6*a*b^2 - (15*a^2*b + b^3) *sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.45 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.06 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {16 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - {\left (8 \, a^{3} - 9 \, a^{2} b + b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (8 \, a^{3} + 9 \, a^{2} b - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (6 \, a^{3} \sin \left (d x + c\right )^{4} - 9 \, a^{2} b \sin \left (d x + c\right )^{3} + b^{3} \sin \left (d x + c\right )^{3} - 16 \, a^{3} \sin \left (d x + c\right )^{2} + 15 \, a^{2} b \sin \left (d x + c\right ) + b^{3} \sin \left (d x + c\right ) + 12 \, a^{3} + 6 \, a b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
1/16*(16*a^3*log(abs(sin(d*x + c))) - (8*a^3 - 9*a^2*b + b^3)*log(abs(sin( d*x + c) + 1)) - (8*a^3 + 9*a^2*b - b^3)*log(abs(sin(d*x + c) - 1)) + 2*(6 *a^3*sin(d*x + c)^4 - 9*a^2*b*sin(d*x + c)^3 + b^3*sin(d*x + c)^3 - 16*a^3 *sin(d*x + c)^2 + 15*a^2*b*sin(d*x + c) + b^3*sin(d*x + c) + 12*a^3 + 6*a* b^2)/(sin(d*x + c)^2 - 1)^2)/d
Time = 11.83 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.02 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a^3\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {a^3}{2}-\frac {9\,a^2\,b}{16}+\frac {b^3}{16}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {a^3}{2}+\frac {9\,a^2\,b}{16}-\frac {b^3}{16}\right )}{d}+\frac {\frac {3\,a\,b^2}{4}-{\sin \left (c+d\,x\right )}^3\,\left (\frac {9\,a^2\,b}{8}-\frac {b^3}{8}\right )+\frac {3\,a^3}{4}+\sin \left (c+d\,x\right )\,\left (\frac {15\,a^2\,b}{8}+\frac {b^3}{8}\right )-\frac {a^3\,{\sin \left (c+d\,x\right )}^2}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )} \]
(a^3*log(sin(c + d*x)))/d - (log(sin(c + d*x) + 1)*(a^3/2 - (9*a^2*b)/16 + b^3/16))/d - (log(sin(c + d*x) - 1)*((9*a^2*b)/16 + a^3/2 - b^3/16))/d + ((3*a*b^2)/4 - sin(c + d*x)^3*((9*a^2*b)/8 - b^3/8) + (3*a^3)/4 + sin(c + d*x)*((15*a^2*b)/8 + b^3/8) - (a^3*sin(c + d*x)^2)/2)/(d*(sin(c + d*x)^4 - 2*sin(c + d*x)^2 + 1))