Integrand size = 27, antiderivative size = 126 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {a (4 a+3 b) \log (1-\sin (c+d x))}{8 d}+\frac {a^2 \log (\sin (c+d x))}{d}-\frac {a (4 a-3 b) \log (1+\sin (c+d x))}{8 d}+\frac {a \sec ^2(c+d x) (2 a+3 b \sin (c+d x))}{4 d}+\frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d} \]
-1/8*a*(4*a+3*b)*ln(1-sin(d*x+c))/d+a^2*ln(sin(d*x+c))/d-1/8*a*(4*a-3*b)*l n(1+sin(d*x+c))/d+1/4*a*sec(d*x+c)^2*(2*a+3*b*sin(d*x+c))/d+1/4*sec(d*x+c) ^4*(a^2+b^2+2*a*b*sin(d*x+c))/d
Time = 0.62 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.09 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-2 a (4 a+3 b) \log (1-\sin (c+d x))+16 a^2 \log (\sin (c+d x))-2 a (4 a-3 b) \log (1+\sin (c+d x))+\frac {(a+b)^2}{(-1+\sin (c+d x))^2}-\frac {(a+b) (5 a+b)}{-1+\sin (c+d x)}+\frac {(a-b)^2}{(1+\sin (c+d x))^2}+\frac {(a-b) (5 a-b)}{1+\sin (c+d x)}}{16 d} \]
(-2*a*(4*a + 3*b)*Log[1 - Sin[c + d*x]] + 16*a^2*Log[Sin[c + d*x]] - 2*a*( 4*a - 3*b)*Log[1 + Sin[c + d*x]] + (a + b)^2/(-1 + Sin[c + d*x])^2 - ((a + b)*(5*a + b))/(-1 + Sin[c + d*x]) + (a - b)^2/(1 + Sin[c + d*x])^2 + ((a - b)*(5*a - b))/(1 + Sin[c + d*x]))/(16*d)
Time = 0.42 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.37, 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, 27, 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.
| \(\displaystyle \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^2}{\sin (c+d x) \cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\csc (c+d x) (a+b \sin (c+d x))^2}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^6 \int \frac {\csc (c+d x) (a+b \sin (c+d x))^2}{b \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 532 |
| \(\displaystyle \frac {b^6 \left (\frac {a^2+2 a b \sin (c+d x)+b^2}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {2 a \csc (c+d x) (2 a+3 b \sin (c+d x))}{b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^6 \left (\frac {a \int \frac {\csc (c+d x) (2 a+3 b \sin (c+d x))}{b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{2 b^2}+\frac {a^2+2 a b \sin (c+d x)+b^2}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 532 |
| \(\displaystyle \frac {b^6 \left (\frac {a \left (\frac {2 a+3 b \sin (c+d x)}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int -\frac {\csc (c+d x) (4 a+3 b \sin (c+d x))}{b \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}\right )}{2 b^2}+\frac {a^2+2 a b \sin (c+d x)+b^2}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^6 \left (\frac {a \left (\frac {\int \frac {\csc (c+d x) (4 a+3 b \sin (c+d x))}{b \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}+\frac {2 a+3 b \sin (c+d x)}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{2 b^2}+\frac {a^2+2 a b \sin (c+d x)+b^2}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 523 |
| \(\displaystyle \frac {b^6 \left (\frac {a \left (\frac {\int \left (\frac {3 b-4 a}{2 b^2 (\sin (c+d x) b+b)}+\frac {4 a \csc (c+d x)}{b^3}+\frac {4 a+3 b}{2 b^2 (b-b \sin (c+d x))}\right )d(b \sin (c+d x))}{2 b^2}+\frac {2 a+3 b \sin (c+d x)}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}\right )}{2 b^2}+\frac {a^2+2 a b \sin (c+d x)+b^2}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^6 \left (\frac {a^2+2 a b \sin (c+d x)+b^2}{4 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}+\frac {a \left (\frac {2 a+3 b \sin (c+d x)}{2 b^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {\frac {4 a \log (b \sin (c+d x))}{b^2}-\frac {(4 a+3 b) \log (b-b \sin (c+d x))}{2 b^2}-\frac {(4 a-3 b) \log (b \sin (c+d x)+b)}{2 b^2}}{2 b^2}\right )}{2 b^2}\right )}{d}\) |
(b^6*((a^2 + b^2 + 2*a*b*Sin[c + d*x])/(4*b^2*(b^2 - b^2*Sin[c + d*x]^2)^2 ) + (a*(((4*a*Log[b*Sin[c + d*x]])/b^2 - ((4*a + 3*b)*Log[b - b*Sin[c + d* x]])/(2*b^2) - ((4*a - 3*b)*Log[b + b*Sin[c + d*x]])/(2*b^2))/(2*b^2) + (2 *a + 3*b*Sin[c + d*x])/(2*b^2*(b^2 - b^2*Sin[c + d*x]^2))))/(2*b^2)))/d
3.15.98.3.1 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 = 0.92 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.79
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+2 a b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b^{2}}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(99\) |
| default | \(\frac {a^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+2 a b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b^{2}}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(99\) |
| parallelrisch | \(\frac {-16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +\frac {3 b}{4}\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-16 \left (a -\frac {3 b}{4}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-4 a^{2}-4 b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (-3 a^{2}-b^{2}\right ) \cos \left (4 d x +4 c \right )+22 a b \sin \left (d x +c \right )+6 a b \sin \left (3 d x +3 c \right )+7 a^{2}+5 b^{2}}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(227\) |
| risch | \(-\frac {i \left (4 i a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+3 a b \,{\mathrm e}^{7 i \left (d x +c \right )}+16 i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+8 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+11 a b \,{\mathrm e}^{5 i \left (d x +c \right )}+4 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-11 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{4 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(235\) |
| norman | \(\frac {\frac {\left (4 a^{2}+4 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 a^{2}+4 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 b^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (2 a^{2}+b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (2 a^{2}+b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {13 a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {7 a b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {7 a b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {13 a b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 a b \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \left (4 a -3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}-\frac {a \left (4 a +3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}\) | \(328\) |
1/d*(a^2*(1/4/cos(d*x+c)^4+1/2/cos(d*x+c)^2+ln(tan(d*x+c)))+2*a*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)))+1/4 *b^2/cos(d*x+c)^4)
Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.14 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {8 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (4 \, a^{2} - 3 \, a b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (4 \, a^{2} + 3 \, a b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, b^{2} + 2 \, {\left (3 \, a b \cos \left (d x + c\right )^{2} + 2 \, a b\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{4}} \]
1/8*(8*a^2*cos(d*x + c)^4*log(1/2*sin(d*x + c)) - (4*a^2 - 3*a*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (4*a^2 + 3*a*b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*a^2*cos(d*x + c)^2 + 2*a^2 + 2*b^2 + 2*(3*a*b*cos(d*x + c)^ 2 + 2*a*b)*sin(d*x + c))/(d*cos(d*x + c)^4)
Timed out. \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.03 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {8 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - {\left (4 \, a^{2} - 3 \, a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (4 \, a^{2} + 3 \, a b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, a b \sin \left (d x + c\right )^{3} + 2 \, a^{2} \sin \left (d x + c\right )^{2} - 5 \, a b \sin \left (d x + c\right ) - 3 \, a^{2} - b^{2}\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{8 \, d} \]
1/8*(8*a^2*log(sin(d*x + c)) - (4*a^2 - 3*a*b)*log(sin(d*x + c) + 1) - (4* a^2 + 3*a*b)*log(sin(d*x + c) - 1) - 2*(3*a*b*sin(d*x + c)^3 + 2*a^2*sin(d *x + c)^2 - 5*a*b*sin(d*x + c) - 3*a^2 - b^2)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.35 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {8 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - {\left (4 \, a^{2} - 3 \, a b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (4 \, a^{2} + 3 \, a b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (3 \, a^{2} \sin \left (d x + c\right )^{4} - 3 \, a b \sin \left (d x + c\right )^{3} - 8 \, a^{2} \sin \left (d x + c\right )^{2} + 5 \, a b \sin \left (d x + c\right ) + 6 \, a^{2} + b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{8 \, d} \]
1/8*(8*a^2*log(abs(sin(d*x + c))) - (4*a^2 - 3*a*b)*log(abs(sin(d*x + c) + 1)) - (4*a^2 + 3*a*b)*log(abs(sin(d*x + c) - 1)) + 2*(3*a^2*sin(d*x + c)^ 4 - 3*a*b*sin(d*x + c)^3 - 8*a^2*sin(d*x + c)^2 + 5*a*b*sin(d*x + c) + 6*a ^2 + b^2)/(sin(d*x + c)^2 - 1)^2)/d
Time = 0.15 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.04 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}+\frac {-\frac {a^2\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {3\,a^2}{4}-\frac {3\,a\,b\,{\sin \left (c+d\,x\right )}^3}{4}+\frac {5\,a\,b\,\sin \left (c+d\,x\right )}{4}+\frac {b^2}{4}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}-\frac {a\,\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (4\,a+3\,b\right )}{8\,d}-\frac {a\,\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (4\,a-3\,b\right )}{8\,d} \]