Integrand size = 29, antiderivative size = 185 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {2 a b \csc (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {\left (12 a^2+15 a b+4 b^2\right ) \log (1-\sin (c+d x))}{8 d}+\frac {\left (3 a^2+b^2\right ) \log (\sin (c+d x))}{d}-\frac {\left (12 a^2-15 a b+4 b^2\right ) \log (1+\sin (c+d x))}{8 d}+\frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}+\frac {\sec ^2(c+d x) \left (2 \left (2 a^2+b^2\right )+7 a b \sin (c+d x)\right )}{4 d} \]
-2*a*b*csc(d*x+c)/d-1/2*a^2*csc(d*x+c)^2/d-1/8*(12*a^2+15*a*b+4*b^2)*ln(1- sin(d*x+c))/d+(3*a^2+b^2)*ln(sin(d*x+c))/d-1/8*(12*a^2-15*a*b+4*b^2)*ln(1+ sin(d*x+c))/d+1/4*sec(d*x+c)^4*(a^2+b^2+2*a*b*sin(d*x+c))/d+1/4*sec(d*x+c) ^2*(4*a^2+2*b^2+7*a*b*sin(d*x+c))/d
Time = 3.69 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.98 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-32 a b \csc (c+d x)-8 a^2 \csc ^2(c+d x)-2 \left (12 a^2+15 a b+4 b^2\right ) \log (1-\sin (c+d x))+16 \left (3 a^2+b^2\right ) \log (\sin (c+d x))-2 \left (12 a^2-15 a b+4 b^2\right ) \log (1+\sin (c+d x))+\frac {(a+b)^2}{(-1+\sin (c+d x))^2}-\frac {(a+b) (9 a+5 b)}{-1+\sin (c+d x)}+\frac {(a-b)^2}{(1+\sin (c+d x))^2}+\frac {(9 a-5 b) (a-b)}{1+\sin (c+d x)}}{16 d} \]
(-32*a*b*Csc[c + d*x] - 8*a^2*Csc[c + d*x]^2 - 2*(12*a^2 + 15*a*b + 4*b^2) *Log[1 - Sin[c + d*x]] + 16*(3*a^2 + b^2)*Log[Sin[c + d*x]] - 2*(12*a^2 - 15*a*b + 4*b^2)*Log[1 + Sin[c + d*x]] + (a + b)^2/(-1 + Sin[c + d*x])^2 - ((a + b)*(9*a + 5*b))/(-1 + Sin[c + d*x]) + (a - b)^2/(1 + Sin[c + d*x])^2 + ((9*a - 5*b)*(a - b))/(1 + Sin[c + d*x]))/(16*d)
Time = 0.69 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.32, 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, 27, 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.
| \(\displaystyle \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^2}{\sin (c+d x)^3 \cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\csc ^3(c+d x) (a+b \sin (c+d x))^2}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^8 \int \frac {\csc ^3(c+d x) (a+b \sin (c+d x))^2}{b^3 \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 532 |
| \(\displaystyle \frac {b^8 \left (\frac {b^2 \left (\frac {a^2}{b^2}+1\right )+2 a b \sin (c+d x)}{4 b^4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {2 \csc ^3(c+d x) \left (3 a b \sin ^3(c+d x)+2 \left (\frac {a^2}{b^2}+1\right ) b^2 \sin ^2(c+d x)+4 a b \sin (c+d x)+2 a^2\right )}{b^3 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^8 \left (\frac {\int \frac {\csc ^3(c+d x) \left (3 a b \sin ^3(c+d x)+2 \left (\frac {a^2}{b^2}+1\right ) b^2 \sin ^2(c+d x)+4 a b \sin (c+d x)+2 a^2\right )}{b^3 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{2 b^2}+\frac {b^2 \left (\frac {a^2}{b^2}+1\right )+2 a b \sin (c+d x)}{4 b^4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle \frac {b^8 \left (\frac {\frac {2 b^2 \left (\frac {2 a^2}{b^2}+1\right )+7 a b \sin (c+d x)}{2 b^4 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int -\frac {\csc ^3(c+d x) \left (7 a b \sin ^3(c+d x)+4 \left (\frac {2 a^2}{b^2}+1\right ) b^2 \sin ^2(c+d x)+8 a b \sin (c+d x)+4 a^2\right )}{b^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}}{2 b^2}+\frac {b^2 \left (\frac {a^2}{b^2}+1\right )+2 a b \sin (c+d x)}{4 b^4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^8 \left (\frac {\frac {\int \frac {\csc ^3(c+d x) \left (7 a b \sin ^3(c+d x)+4 \left (\frac {2 a^2}{b^2}+1\right ) b^2 \sin ^2(c+d x)+8 a b \sin (c+d x)+4 a^2\right )}{b^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}+\frac {2 b^2 \left (\frac {2 a^2}{b^2}+1\right )+7 a b \sin (c+d x)}{2 b^4 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{2 b^2}+\frac {b^2 \left (\frac {a^2}{b^2}+1\right )+2 a b \sin (c+d x)}{4 b^4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle \frac {b^8 \left (\frac {\frac {\int \left (\frac {4 a^2 \csc ^3(c+d x)}{b^5}+\frac {8 a \csc ^2(c+d x)}{b^4}+\frac {4 \left (3 a^2+b^2\right ) \csc (c+d x)}{b^5}+\frac {12 a^2+15 b a+4 b^2}{2 b^4 (b-b \sin (c+d x))}+\frac {-12 a^2+15 b a-4 b^2}{2 b^4 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 b^2}+\frac {2 b^2 \left (\frac {2 a^2}{b^2}+1\right )+7 a b \sin (c+d x)}{2 b^4 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{2 b^2}+\frac {b^2 \left (\frac {a^2}{b^2}+1\right )+2 a b \sin (c+d x)}{4 b^4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^8 \left (\frac {b^2 \left (\frac {a^2}{b^2}+1\right )+2 a b \sin (c+d x)}{4 b^4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}+\frac {\frac {2 b^2 \left (\frac {2 a^2}{b^2}+1\right )+7 a b \sin (c+d x)}{2 b^4 \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {-\frac {2 a^2 \csc ^2(c+d x)}{b^4}+\frac {4 \left (3 a^2+b^2\right ) \log (b \sin (c+d x))}{b^4}-\frac {\left (12 a^2+15 a b+4 b^2\right ) \log (b-b \sin (c+d x))}{2 b^4}-\frac {\left (12 a^2-15 a b+4 b^2\right ) \log (b \sin (c+d x)+b)}{2 b^4}-\frac {8 a \csc (c+d x)}{b^3}}{2 b^2}}{2 b^2}\right )}{d}\) |
(b^8*(((1 + a^2/b^2)*b^2 + 2*a*b*Sin[c + d*x])/(4*b^4*(b^2 - b^2*Sin[c + d *x]^2)^2) + (((-8*a*Csc[c + d*x])/b^3 - (2*a^2*Csc[c + d*x]^2)/b^4 + (4*(3 *a^2 + b^2)*Log[b*Sin[c + d*x]])/b^4 - ((12*a^2 + 15*a*b + 4*b^2)*Log[b - b*Sin[c + d*x]])/(2*b^4) - ((12*a^2 - 15*a*b + 4*b^2)*Log[b + b*Sin[c + d* x]])/(2*b^4))/(2*b^2) + (2*(1 + (2*a^2)/b^2)*b^2 + 7*a*b*Sin[c + d*x])/(2* b^4*(b^2 - b^2*Sin[c + d*x]^2)))/(2*b^2)))/d
3.15.100.3.1 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 = 1.18 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.89
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )+2 a b \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(165\) |
| default | \(\frac {a^{2} \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )+2 a b \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(165\) |
| parallelrisch | \(\frac {-384 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+\frac {5}{4} a b +\frac {1}{3} b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-384 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}-\frac {5}{4} a b +\frac {1}{3} b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+384 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+\frac {b^{2}}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-27 \left (\cos \left (2 d x +2 c \right )+\frac {2 \cos \left (4 d x +4 c \right )}{9}-\frac {\cos \left (6 d x +6 c \right )}{9}+\frac {2}{27}\right ) \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-160 b \left (\cos \left (2 d x +2 c \right )+\frac {3 \cos \left (4 d x +4 c \right )}{8}+\frac {9}{40}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) a \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-32 b^{2} \left (\frac {3 \cos \left (4 d x +4 c \right )}{4}-\frac {7}{4}+\cos \left (2 d x +2 c \right )\right )}{32 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(303\) |
| risch | \(-\frac {i \left (12 i a^{2} {\mathrm e}^{10 i \left (d x +c \right )}+4 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+15 a b \,{\mathrm e}^{11 i \left (d x +c \right )}+24 i a^{2} {\mathrm e}^{8 i \left (d x +c \right )}+4 i b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+25 a b \,{\mathrm e}^{9 i \left (d x +c \right )}+8 i b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+24 i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-22 a b \,{\mathrm e}^{7 i \left (d x +c \right )}+8 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-8 i a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+22 a b \,{\mathrm e}^{5 i \left (d x +c \right )}+12 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-24 i b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-25 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-15 a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{4 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{4 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{d}\) | \(426\) |
1/d*(a^2*(1/4/sin(d*x+c)^2/cos(d*x+c)^4+3/4/sin(d*x+c)^2/cos(d*x+c)^2-3/2/ sin(d*x+c)^2+3*ln(tan(d*x+c)))+2*a*b*(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)))+b^2*(1 /4/cos(d*x+c)^4+1/2/cos(d*x+c)^2+ln(tan(d*x+c))))
Time = 0.28 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.54 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {4 \, {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 2 \, b^{2} + 8 \, {\left ({\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{6} - {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left ({\left (12 \, a^{2} - 15 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - {\left (12 \, a^{2} - 15 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (12 \, a^{2} + 15 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - {\left (12 \, a^{2} + 15 \, a b + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, a b \cos \left (d x + c\right )^{4} - 5 \, a b \cos \left (d x + c\right )^{2} - 2 \, a b\right )} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right )^{6} - d \cos \left (d x + c\right )^{4}\right )}} \]
1/8*(4*(3*a^2 + b^2)*cos(d*x + c)^4 - 2*(3*a^2 + b^2)*cos(d*x + c)^2 - 2*a ^2 - 2*b^2 + 8*((3*a^2 + b^2)*cos(d*x + c)^6 - (3*a^2 + b^2)*cos(d*x + c)^ 4)*log(1/2*sin(d*x + c)) - ((12*a^2 - 15*a*b + 4*b^2)*cos(d*x + c)^6 - (12 *a^2 - 15*a*b + 4*b^2)*cos(d*x + c)^4)*log(sin(d*x + c) + 1) - ((12*a^2 + 15*a*b + 4*b^2)*cos(d*x + c)^6 - (12*a^2 + 15*a*b + 4*b^2)*cos(d*x + c)^4) *log(-sin(d*x + c) + 1) + 2*(15*a*b*cos(d*x + c)^4 - 5*a*b*cos(d*x + c)^2 - 2*a*b)*sin(d*x + c))/(d*cos(d*x + c)^6 - d*cos(d*x + c)^4)
Timed out. \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.99 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {{\left (12 \, a^{2} - 15 \, a b + 4 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (12 \, a^{2} + 15 \, a b + 4 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - 8 \, {\left (3 \, a^{2} + b^{2}\right )} \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (15 \, a b \sin \left (d x + c\right )^{5} - 25 \, a b \sin \left (d x + c\right )^{3} + 2 \, {\left (3 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{4} + 8 \, a b \sin \left (d x + c\right ) - 3 \, {\left (3 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{2} + 2 \, a^{2}\right )}}{\sin \left (d x + c\right )^{6} - 2 \, \sin \left (d x + c\right )^{4} + \sin \left (d x + c\right )^{2}}}{8 \, d} \]
-1/8*((12*a^2 - 15*a*b + 4*b^2)*log(sin(d*x + c) + 1) + (12*a^2 + 15*a*b + 4*b^2)*log(sin(d*x + c) - 1) - 8*(3*a^2 + b^2)*log(sin(d*x + c)) + 2*(15* a*b*sin(d*x + c)^5 - 25*a*b*sin(d*x + c)^3 + 2*(3*a^2 + b^2)*sin(d*x + c)^ 4 + 8*a*b*sin(d*x + c) - 3*(3*a^2 + b^2)*sin(d*x + c)^2 + 2*a^2)/(sin(d*x + c)^6 - 2*sin(d*x + c)^4 + sin(d*x + c)^2))/d
Time = 0.40 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.03 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {{\left (12 \, a^{2} - 15 \, a b + 4 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + {\left (12 \, a^{2} + 15 \, a b + 4 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 8 \, {\left (3 \, a^{2} + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac {2 \, {\left (15 \, a b \sin \left (d x + c\right )^{5} + 6 \, a^{2} \sin \left (d x + c\right )^{4} + 2 \, b^{2} \sin \left (d x + c\right )^{4} - 25 \, a b \sin \left (d x + c\right )^{3} - 9 \, a^{2} \sin \left (d x + c\right )^{2} - 3 \, b^{2} \sin \left (d x + c\right )^{2} + 8 \, a b \sin \left (d x + c\right ) + 2 \, a^{2}\right )}}{{\left (\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )\right )}^{2}}}{8 \, d} \]
-1/8*((12*a^2 - 15*a*b + 4*b^2)*log(abs(sin(d*x + c) + 1)) + (12*a^2 + 15* a*b + 4*b^2)*log(abs(sin(d*x + c) - 1)) - 8*(3*a^2 + b^2)*log(abs(sin(d*x + c))) + 2*(15*a*b*sin(d*x + c)^5 + 6*a^2*sin(d*x + c)^4 + 2*b^2*sin(d*x + c)^4 - 25*a*b*sin(d*x + c)^3 - 9*a^2*sin(d*x + c)^2 - 3*b^2*sin(d*x + c)^ 2 + 8*a*b*sin(d*x + c) + 2*a^2)/(sin(d*x + c)^3 - sin(d*x + c))^2)/d
Time = 12.45 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.05 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )\right )\,\left (3\,a^2+b^2\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {3\,a^2}{2}-\frac {15\,a\,b}{8}+\frac {b^2}{2}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {3\,a^2}{2}+\frac {15\,a\,b}{8}+\frac {b^2}{2}\right )}{d}-\frac {\frac {a^2}{2}+{\sin \left (c+d\,x\right )}^4\,\left (\frac {3\,a^2}{2}+\frac {b^2}{2}\right )-{\sin \left (c+d\,x\right )}^2\,\left (\frac {9\,a^2}{4}+\frac {3\,b^2}{4}\right )+2\,a\,b\,\sin \left (c+d\,x\right )-\frac {25\,a\,b\,{\sin \left (c+d\,x\right )}^3}{4}+\frac {15\,a\,b\,{\sin \left (c+d\,x\right )}^5}{4}}{d\,\left ({\sin \left (c+d\,x\right )}^6-2\,{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^2\right )} \]
(log(sin(c + d*x))*(3*a^2 + b^2))/d - (log(sin(c + d*x) + 1)*((3*a^2)/2 - (15*a*b)/8 + b^2/2))/d - (log(sin(c + d*x) - 1)*((15*a*b)/8 + (3*a^2)/2 + b^2/2))/d - (a^2/2 + sin(c + d*x)^4*((3*a^2)/2 + b^2/2) - sin(c + d*x)^2*( (9*a^2)/4 + (3*b^2)/4) + 2*a*b*sin(c + d*x) - (25*a*b*sin(c + d*x)^3)/4 + (15*a*b*sin(c + d*x)^5)/4)/(d*(sin(c + d*x)^2 - 2*sin(c + d*x)^4 + sin(c + d*x)^6))