Integrand size = 29, antiderivative size = 171 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {a^3 \csc (c+d x)}{d}-\frac {3 a (a+b) (5 a+3 b) \log (1-\sin (c+d x))}{16 d}+\frac {3 a^2 b \log (\sin (c+d x))}{d}+\frac {3 a (5 a-3 b) (a-b) \log (1+\sin (c+d x))}{16 d}+\frac {b \sec ^4(c+d x) \left (3 a^2+b^2+a \left (3+\frac {a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac {a b \sec ^2(c+d x) \left (12 a+\left (9+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d} \]
-a^3*csc(d*x+c)/d-3/16*a*(a+b)*(5*a+3*b)*ln(1-sin(d*x+c))/d+3*a^2*b*ln(sin (d*x+c))/d+3/16*a*(5*a-3*b)*(a-b)*ln(1+sin(d*x+c))/d+1/4*b*sec(d*x+c)^4*(3 *a^2+b^2+a*(3+a^2/b^2)*b*sin(d*x+c))/d+1/8*a*b*sec(d*x+c)^2*(12*a+(9+7*a^2 /b^2)*b*sin(d*x+c))/d
Time = 0.99 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.94 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {16 a^3 \csc (c+d x)+3 a (a+b) (5 a+3 b) \log (1-\sin (c+d x))-48 a^2 b \log (\sin (c+d x))-3 a (5 a-3 b) (a-b) \log (1+\sin (c+d x))-\frac {(a+b)^3}{(-1+\sin (c+d x))^2}+\frac {(a+b)^2 (7 a+b)}{-1+\sin (c+d x)}+\frac {(a-b)^3}{(1+\sin (c+d x))^2}+\frac {(a-b)^2 (7 a-b)}{1+\sin (c+d x)}}{16 d} \]
-1/16*(16*a^3*Csc[c + d*x] + 3*a*(a + b)*(5*a + 3*b)*Log[1 - Sin[c + d*x]] - 48*a^2*b*Log[Sin[c + d*x]] - 3*a*(5*a - 3*b)*(a - b)*Log[1 + Sin[c + d* x]] - (a + b)^3/(-1 + Sin[c + d*x])^2 + ((a + b)^2*(7*a + b))/(-1 + Sin[c + d*x]) + (a - b)^3/(1 + Sin[c + d*x])^2 + ((a - b)^2*(7*a - b))/(1 + Sin[ c + d*x]))/d
Time = 0.66 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.27, 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, 532, 25, 2336, 25, 2333, 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.
| \(\displaystyle \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{\sin (c+d x)^2 \cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\csc ^2(c+d x) (a+b \sin (c+d x))^3}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^7 \int \frac {\csc ^2(c+d x) (a+b \sin (c+d x))^3}{b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 532 |
| \(\displaystyle \frac {b^7 \left (\frac {a b \left (\frac {a^2}{b^2}+3\right ) \sin (c+d x)+3 a^2+b^2}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {\csc ^2(c+d x) \left (4 a^3+12 b \sin (c+d x) a^2+3 \left (\frac {a^2}{b^2}+3\right ) b^2 \sin ^2(c+d x) a\right )}{b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^7 \left (\frac {\int \frac {\csc ^2(c+d x) \left (4 a^3+12 b \sin (c+d x) a^2+3 \left (\frac {a^2}{b^2}+3\right ) b^2 \sin ^2(c+d x) a\right )}{b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {a b \left (\frac {a^2}{b^2}+3\right ) \sin (c+d x)+3 a^2+b^2}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle \frac {b^7 \left (\frac {\frac {a \left (b \left (\frac {7 a^2}{b^2}+9\right ) \sin (c+d x)+12 a\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int -\frac {\csc ^2(c+d x) \left (8 a^3+24 b \sin (c+d x) a^2+\left (\frac {7 a^2}{b^2}+9\right ) b^2 \sin ^2(c+d x) a\right )}{b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}}{4 b^2}+\frac {a b \left (\frac {a^2}{b^2}+3\right ) \sin (c+d x)+3 a^2+b^2}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^7 \left (\frac {\frac {\int \frac {\csc ^2(c+d x) \left (8 a^3+24 b \sin (c+d x) a^2+\left (\frac {7 a^2}{b^2}+9\right ) b^2 \sin ^2(c+d x) a\right )}{b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}+\frac {a \left (b \left (\frac {7 a^2}{b^2}+9\right ) \sin (c+d x)+12 a\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}+\frac {a b \left (\frac {a^2}{b^2}+3\right ) \sin (c+d x)+3 a^2+b^2}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle \frac {b^7 \left (\frac {\frac {\int \left (\frac {8 \csc ^2(c+d x) a^3}{b^4}+\frac {24 \csc (c+d x) a^2}{b^3}+\frac {3 (a+b) (5 a+3 b) a}{2 b^3 (b-b \sin (c+d x))}+\frac {3 (5 a-3 b) (a-b) a}{2 b^3 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 b^2}+\frac {a \left (b \left (\frac {7 a^2}{b^2}+9\right ) \sin (c+d x)+12 a\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}+\frac {a b \left (\frac {a^2}{b^2}+3\right ) \sin (c+d x)+3 a^2+b^2}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^7 \left (\frac {a b \left (\frac {a^2}{b^2}+3\right ) \sin (c+d x)+3 a^2+b^2}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}+\frac {\frac {a \left (b \left (\frac {7 a^2}{b^2}+9\right ) \sin (c+d x)+12 a\right )}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {-\frac {8 a^3 \csc (c+d x)}{b^3}+\frac {24 a^2 \log (b \sin (c+d x))}{b^2}-\frac {3 a (a+b) (5 a+3 b) \log (b-b \sin (c+d x))}{2 b^3}+\frac {3 a (5 a-3 b) (a-b) \log (b \sin (c+d x)+b)}{2 b^3}}{2 b^2}}{4 b^2}\right )}{d}\) |
(b^7*((3*a^2 + b^2 + a*(3 + a^2/b^2)*b*Sin[c + d*x])/(4*b^2*(b^2 - b^2*Sin [c + d*x]^2)^2) + (((-8*a^3*Csc[c + d*x])/b^3 + (24*a^2*Log[b*Sin[c + d*x] ])/b^2 - (3*a*(a + b)*(5*a + 3*b)*Log[b - b*Sin[c + d*x]])/(2*b^3) + (3*a* (5*a - 3*b)*(a - b)*Log[b + b*Sin[c + d*x]])/(2*b^3))/(2*b^2) + (a*(12*a + (9 + (7*a^2)/b^2)*b*Sin[c + d*x]))/(2*b^2*(b^2 - b^2*Sin[c + d*x]^2)))/(4 *b^2)))/d
3.16.7.3.1 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[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[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) ^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 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 = 1.16 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.99
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{2} b \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b^{3}}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(170\) |
| default | \(\frac {a^{3} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{2} b \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b^{3}}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(170\) |
| parallelrisch | \(\frac {-120 \left (a +\frac {3 b}{5}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+120 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -\frac {3 b}{5}\right ) \left (a -b \right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+192 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 \left (\cos \left (2 d x +2 c \right )+\frac {3 \cos \left (4 d x +4 c \right )}{8}+\frac {9}{40}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-48 a^{2} b -16 b^{3}\right ) \cos \left (2 d x +2 c \right )+\left (-36 a^{2} b -4 b^{3}\right ) \cos \left (4 d x +4 c \right )+132 \sin \left (d x +c \right ) a \,b^{2}+36 \sin \left (3 d x +3 c \right ) a \,b^{2}+84 a^{2} b +20 b^{3}}{16 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(288\) |
| risch | \(\frac {-9 i a \,b^{2} {\mathrm e}^{9 i \left (d x +c \right )}-24 i a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-40 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-15 i a^{3} {\mathrm e}^{9 i \left (d x +c \right )}+24 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+66 i a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-15 i a^{3} {\mathrm e}^{i \left (d x +c \right )}+72 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+16 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-18 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-9 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-72 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-16 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-24 i a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-40 i a^{3} {\mathrm e}^{7 i \left (d x +c \right )}-24 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b}{d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{2}}{8 d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{3}}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{8 d}+\frac {3 a^{2} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(428\) |
1/d*(a^3*(1/4/sin(d*x+c)/cos(d*x+c)^4+5/8/sin(d*x+c)/cos(d*x+c)^2-15/8/sin (d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c)))+3*a^2*b*(1/4/cos(d*x+c)^4+1/2/cos( d*x+c)^2+ln(tan(d*x+c)))+3*a*b^2*(-(-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)))+1/4*b^3/cos(d*x+c)^4)
Time = 0.30 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.32 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {48 \, a^{2} b \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 3 \, {\left (5 \, a^{3} - 8 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 3 \, {\left (5 \, a^{3} + 8 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 6 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 4 \, a^{3} + 12 \, a b^{2} + 2 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (6 \, a^{2} b \cos \left (d x + c\right )^{2} + 3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )} \]
1/16*(48*a^2*b*cos(d*x + c)^4*log(1/2*sin(d*x + c))*sin(d*x + c) + 3*(5*a^ 3 - 8*a^2*b + 3*a*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1)*sin(d*x + c) - 3*(5*a^3 + 8*a^2*b + 3*a*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1)*sin(d *x + c) - 6*(5*a^3 + 3*a*b^2)*cos(d*x + c)^4 + 4*a^3 + 12*a*b^2 + 2*(5*a^3 + 3*a*b^2)*cos(d*x + c)^2 + 4*(6*a^2*b*cos(d*x + c)^2 + 3*a^2*b + b^3)*si n(d*x + c))/(d*cos(d*x + c)^4*sin(d*x + c))
Timed out. \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.10 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {48 \, a^{2} b \log \left (\sin \left (d x + c\right )\right ) + 3 \, {\left (5 \, a^{3} - 8 \, a^{2} b + 3 \, a b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, a^{3} + 8 \, a^{2} b + 3 \, a b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (12 \, a^{2} b \sin \left (d x + c\right )^{3} + 3 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )^{4} + 8 \, a^{3} - 5 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} - 2 \, {\left (9 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{5} - 2 \, \sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )}}{16 \, d} \]
1/16*(48*a^2*b*log(sin(d*x + c)) + 3*(5*a^3 - 8*a^2*b + 3*a*b^2)*log(sin(d *x + c) + 1) - 3*(5*a^3 + 8*a^2*b + 3*a*b^2)*log(sin(d*x + c) - 1) - 2*(12 *a^2*b*sin(d*x + c)^3 + 3*(5*a^3 + 3*a*b^2)*sin(d*x + c)^4 + 8*a^3 - 5*(5* a^3 + 3*a*b^2)*sin(d*x + c)^2 - 2*(9*a^2*b + b^3)*sin(d*x + c))/(sin(d*x + c)^5 - 2*sin(d*x + c)^3 + sin(d*x + c)))/d
Time = 0.43 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.23 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {48 \, a^{2} b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 3 \, {\left (5 \, a^{3} - 8 \, a^{2} b + 3 \, a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (5 \, a^{3} + 8 \, a^{2} b + 3 \, a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {16 \, {\left (3 \, a^{2} b \sin \left (d x + c\right ) + a^{3}\right )}}{\sin \left (d x + c\right )} + \frac {2 \, {\left (18 \, a^{2} b \sin \left (d x + c\right )^{4} - 7 \, a^{3} \sin \left (d x + c\right )^{3} - 9 \, a b^{2} \sin \left (d x + c\right )^{3} - 48 \, a^{2} b \sin \left (d x + c\right )^{2} + 9 \, a^{3} \sin \left (d x + c\right ) + 15 \, a b^{2} \sin \left (d x + c\right ) + 36 \, a^{2} b + 2 \, b^{3}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
1/16*(48*a^2*b*log(abs(sin(d*x + c))) + 3*(5*a^3 - 8*a^2*b + 3*a*b^2)*log( abs(sin(d*x + c) + 1)) - 3*(5*a^3 + 8*a^2*b + 3*a*b^2)*log(abs(sin(d*x + c ) - 1)) - 16*(3*a^2*b*sin(d*x + c) + a^3)/sin(d*x + c) + 2*(18*a^2*b*sin(d *x + c)^4 - 7*a^3*sin(d*x + c)^3 - 9*a*b^2*sin(d*x + c)^3 - 48*a^2*b*sin(d *x + c)^2 + 9*a^3*sin(d*x + c) + 15*a*b^2*sin(d*x + c) + 36*a^2*b + 2*b^3) /(sin(d*x + c)^2 - 1)^2)/d
Time = 0.15 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.06 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3\,a^2\,b\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^4\,\left (\frac {15\,a^3}{8}+\frac {9\,a\,b^2}{8}\right )-{\sin \left (c+d\,x\right )}^2\,\left (\frac {25\,a^3}{8}+\frac {15\,a\,b^2}{8}\right )+a^3-\sin \left (c+d\,x\right )\,\left (\frac {9\,a^2\,b}{4}+\frac {b^3}{4}\right )+\frac {3\,a^2\,b\,{\sin \left (c+d\,x\right )}^3}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^5-2\,{\sin \left (c+d\,x\right )}^3+\sin \left (c+d\,x\right )\right )}+\frac {3\,a\,\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (a-b\right )\,\left (5\,a-3\,b\right )}{16\,d}-\frac {3\,a\,\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (a+b\right )\,\left (5\,a+3\,b\right )}{16\,d} \]
(3*a^2*b*log(sin(c + d*x)))/d - (sin(c + d*x)^4*((9*a*b^2)/8 + (15*a^3)/8) - sin(c + d*x)^2*((15*a*b^2)/8 + (25*a^3)/8) + a^3 - sin(c + d*x)*((9*a^2 *b)/4 + b^3/4) + (3*a^2*b*sin(c + d*x)^3)/2)/(d*(sin(c + d*x) - 2*sin(c + d*x)^3 + sin(c + d*x)^5)) + (3*a*log(sin(c + d*x) + 1)*(a - b)*(5*a - 3*b) )/(16*d) - (3*a*log(sin(c + d*x) - 1)*(a + b)*(5*a + 3*b))/(16*d)