Integrand size = 21, antiderivative size = 190 \[ \int \frac {\cos ^7(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {3 \left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))}{b^7 d}+\frac {a \left (10 a^2-9 b^2\right ) \sin (c+d x)}{b^6 d}-\frac {3 \left (2 a^2-b^2\right ) \sin ^2(c+d x)}{2 b^5 d}+\frac {a \sin ^3(c+d x)}{b^4 d}-\frac {\sin ^4(c+d x)}{4 b^3 d}+\frac {\left (a^2-b^2\right )^3}{2 b^7 d (a+b \sin (c+d x))^2}-\frac {6 a \left (a^2-b^2\right )^2}{b^7 d (a+b \sin (c+d x))} \]
-3*(5*a^4-6*a^2*b^2+b^4)*ln(a+b*sin(d*x+c))/b^7/d+a*(10*a^2-9*b^2)*sin(d*x +c)/b^6/d-3/2*(2*a^2-b^2)*sin(d*x+c)^2/b^5/d+a*sin(d*x+c)^3/b^4/d-1/4*sin( d*x+c)^4/b^3/d+1/2*(a^2-b^2)^3/b^7/d/(a+b*sin(d*x+c))^2-6*a*(a^2-b^2)^2/b^ 7/d/(a+b*sin(d*x+c))
Time = 0.44 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.48 \[ \int \frac {\cos ^7(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {b^6 \cos ^6(c+d x)+b^4 \cos ^4(c+d x) \left (-a^2+3 b^2+2 a b \sin (c+d x)\right )-2 \left (\left (a^2-b^2\right ) \left (19 a^4-16 a^2 b^2-3 b^4+6 a^2 \left (5 a^2-b^2\right ) \log (a+b \sin (c+d x))\right )+2 a b \left (4 a^4-17 a^2 b^2+11 b^4+6 \left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))\right ) \sin (c+d x)+2 b^2 \left (-13 a^4+10 a^2 b^2+3 \left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))\right ) \sin ^2(c+d x)-10 a b^3 \left (a^2-b^2\right ) \sin ^3(c+d x)+2 a^2 b^4 \sin ^4(c+d x)\right )}{4 b^7 d (a+b \sin (c+d x))^2} \]
(b^6*Cos[c + d*x]^6 + b^4*Cos[c + d*x]^4*(-a^2 + 3*b^2 + 2*a*b*Sin[c + d*x ]) - 2*((a^2 - b^2)*(19*a^4 - 16*a^2*b^2 - 3*b^4 + 6*a^2*(5*a^2 - b^2)*Log [a + b*Sin[c + d*x]]) + 2*a*b*(4*a^4 - 17*a^2*b^2 + 11*b^4 + 6*(5*a^4 - 6* a^2*b^2 + b^4)*Log[a + b*Sin[c + d*x]])*Sin[c + d*x] + 2*b^2*(-13*a^4 + 10 *a^2*b^2 + 3*(5*a^4 - 6*a^2*b^2 + b^4)*Log[a + b*Sin[c + d*x]])*Sin[c + d* x]^2 - 10*a*b^3*(a^2 - b^2)*Sin[c + d*x]^3 + 2*a^2*b^4*Sin[c + d*x]^4))/(4 *b^7*d*(a + b*Sin[c + d*x])^2)
Time = 0.34 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3147, 476, 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.
| \(\displaystyle \int \frac {\cos ^7(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^7}{(a+b \sin (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {\int \frac {\left (b^2-b^2 \sin ^2(c+d x)\right )^3}{(a+b \sin (c+d x))^3}d(b \sin (c+d x))}{b^7 d}\) |
| \(\Big \downarrow \) 476 |
| \(\displaystyle \frac {\int \left (10 \left (1-\frac {9 b^2}{10 a^2}\right ) a^3+3 b^2 \sin ^2(c+d x) a+\frac {6 \left (a^2-b^2\right )^2 a}{(a+b \sin (c+d x))^2}-b^3 \sin ^3(c+d x)-3 b \left (2 a^2-b^2\right ) \sin (c+d x)-\frac {3 \left (5 a^4-6 b^2 a^2+b^4\right )}{a+b \sin (c+d x)}-\frac {\left (a^2-b^2\right )^3}{(a+b \sin (c+d x))^3}\right )d(b \sin (c+d x))}{b^7 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {3}{2} b^2 \left (2 a^2-b^2\right ) \sin ^2(c+d x)+a b \left (10 a^2-9 b^2\right ) \sin (c+d x)-\frac {6 a \left (a^2-b^2\right )^2}{a+b \sin (c+d x)}+\frac {\left (a^2-b^2\right )^3}{2 (a+b \sin (c+d x))^2}-3 \left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))+a b^3 \sin ^3(c+d x)-\frac {1}{4} b^4 \sin ^4(c+d x)}{b^7 d}\) |
(-3*(5*a^4 - 6*a^2*b^2 + b^4)*Log[a + b*Sin[c + d*x]] + a*b*(10*a^2 - 9*b^ 2)*Sin[c + d*x] - (3*b^2*(2*a^2 - b^2)*Sin[c + d*x]^2)/2 + a*b^3*Sin[c + d *x]^3 - (b^4*Sin[c + d*x]^4)/4 + (a^2 - b^2)^3/(2*(a + b*Sin[c + d*x])^2) - (6*a*(a^2 - b^2)^2)/(a + b*Sin[c + d*x]))/(b^7*d)
3.5.49.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 1.92 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.04
| method | result | size |
| derivativedivides | \(-\frac {-\frac {-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) b^{3}}{4}+a \left (\sin ^{3}\left (d x +c \right )\right ) b^{2}-3 \left (\sin ^{2}\left (d x +c \right )\right ) a^{2} b +\frac {3 \left (\sin ^{2}\left (d x +c \right )\right ) b^{3}}{2}+10 a^{3} \sin \left (d x +c \right )-9 a \,b^{2} \sin \left (d x +c \right )}{b^{6}}+\frac {\left (15 a^{4}-18 a^{2} b^{2}+3 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{7}}-\frac {a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}{2 b^{7} \left (a +b \sin \left (d x +c \right )\right )^{2}}+\frac {6 a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{b^{7} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(198\) |
| default | \(-\frac {-\frac {-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) b^{3}}{4}+a \left (\sin ^{3}\left (d x +c \right )\right ) b^{2}-3 \left (\sin ^{2}\left (d x +c \right )\right ) a^{2} b +\frac {3 \left (\sin ^{2}\left (d x +c \right )\right ) b^{3}}{2}+10 a^{3} \sin \left (d x +c \right )-9 a \,b^{2} \sin \left (d x +c \right )}{b^{6}}+\frac {\left (15 a^{4}-18 a^{2} b^{2}+3 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{7}}-\frac {a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}{2 b^{7} \left (a +b \sin \left (d x +c \right )\right )^{2}}+\frac {6 a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{b^{7} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(198\) |
| parallelrisch | \(\frac {-1920 \left (a +b \right ) \left (a^{2}-\frac {b^{2}}{5}\right ) \left (\frac {b^{2}}{2}-\frac {b^{2} \cos \left (2 d x +2 c \right )}{2}+2 \sin \left (d x +c \right ) a b +a^{2}\right ) \left (a -b \right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+1920 \left (a +b \right ) \left (a^{2}-\frac {b^{2}}{5}\right ) \left (\frac {b^{2}}{2}-\frac {b^{2} \cos \left (2 d x +2 c \right )}{2}+2 \sin \left (d x +c \right ) a b +a^{2}\right ) \left (a -b \right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-960 a^{4} b^{2}+1232 a^{2} b^{4}-273 b^{6}\right ) \cos \left (2 d x +2 c \right )+\left (-20 a^{2} b^{4}+18 b^{6}\right ) \cos \left (4 d x +4 c \right )+\left (-160 a^{3} b^{3}+172 a \,b^{5}\right ) \sin \left (3 d x +3 c \right )+b^{6} \cos \left (6 d x +6 c \right )+4 a \,b^{5} \sin \left (5 d x +5 c \right )+\left (480 a^{3} b^{3}-536 a \,b^{5}\right ) \sin \left (d x +c \right )-960 a^{6}+2112 a^{4} b^{2}-1404 a^{2} b^{4}+190 b^{6}}{64 b^{7} d \left (4 \sin \left (d x +c \right ) a b -b^{2} \cos \left (2 d x +2 c \right )+2 a^{2}+b^{2}\right )}\) | \(345\) |
| risch | \(-\frac {i a \,{\mathrm e}^{-3 i \left (d x +c \right )}}{8 b^{4} d}+\frac {3 i x}{b^{3}}-\frac {33 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 b^{4} d}+\frac {6 i c}{b^{3} d}+\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{4 b^{5} d}-\frac {5 \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 b^{3} d}+\frac {5 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{b^{6} d}-\frac {36 i a^{2} c}{b^{5} d}-\frac {5 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{b^{6} d}+\frac {33 i a \,{\mathrm e}^{i \left (d x +c \right )}}{8 b^{4} d}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{4 b^{5} d}-\frac {5 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 b^{3} d}+\frac {i a \,{\mathrm e}^{3 i \left (d x +c \right )}}{8 b^{4} d}+\frac {15 i x \,a^{4}}{b^{7}}+\frac {30 i a^{4} c}{b^{7} d}-\frac {18 i x \,a^{2}}{b^{5}}-\frac {2 \left (-6 i a^{5} b \,{\mathrm e}^{3 i \left (d x +c \right )}+12 i a^{3} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-6 i b^{5} a \,{\mathrm e}^{3 i \left (d x +c \right )}+6 i a^{5} b \,{\mathrm e}^{i \left (d x +c \right )}-12 i a^{3} b^{3} {\mathrm e}^{i \left (d x +c \right )}+6 i b^{5} a \,{\mathrm e}^{i \left (d x +c \right )}+11 a^{6} {\mathrm e}^{2 i \left (d x +c \right )}-21 a^{4} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+9 a^{2} b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+b^{6} {\mathrm e}^{2 i \left (d x +c \right )}\right )}{\left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+i b \right )^{2} d \,b^{7}}-\frac {15 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) a^{4}}{b^{7} d}+\frac {18 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) a^{2}}{b^{5} d}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{b^{3} d}-\frac {\cos \left (4 d x +4 c \right )}{32 b^{3} d}\) | \(590\) |
-1/d*(-1/b^6*(-1/4*sin(d*x+c)^4*b^3+a*sin(d*x+c)^3*b^2-3*sin(d*x+c)^2*a^2* b+3/2*sin(d*x+c)^2*b^3+10*a^3*sin(d*x+c)-9*a*b^2*sin(d*x+c))+(15*a^4-18*a^ 2*b^2+3*b^4)/b^7*ln(a+b*sin(d*x+c))-1/2/b^7*(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/ (a+b*sin(d*x+c))^2+6*a/b^7*(a^4-2*a^2*b^2+b^4)/(a+b*sin(d*x+c)))
Time = 0.31 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.60 \[ \int \frac {\cos ^7(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {8 \, b^{6} \cos \left (d x + c\right )^{6} - 176 \, a^{6} + 928 \, a^{4} b^{2} - 685 \, a^{2} b^{4} + 3 \, b^{6} - 8 \, {\left (5 \, a^{2} b^{4} - 3 \, b^{6}\right )} \cos \left (d x + c\right )^{4} - {\left (544 \, a^{4} b^{2} - 560 \, a^{2} b^{4} + 51 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - 96 \, {\left (5 \, a^{6} - a^{4} b^{2} - 5 \, a^{2} b^{4} + b^{6} - {\left (5 \, a^{4} b^{2} - 6 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, a^{5} b - 6 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 2 \, {\left (8 \, a b^{5} \cos \left (d x + c\right )^{4} + 64 \, a^{5} b + 176 \, a^{3} b^{3} - 205 \, a b^{5} - 80 \, {\left (a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{32 \, {\left (b^{9} d \cos \left (d x + c\right )^{2} - 2 \, a b^{8} d \sin \left (d x + c\right ) - {\left (a^{2} b^{7} + b^{9}\right )} d\right )}} \]
-1/32*(8*b^6*cos(d*x + c)^6 - 176*a^6 + 928*a^4*b^2 - 685*a^2*b^4 + 3*b^6 - 8*(5*a^2*b^4 - 3*b^6)*cos(d*x + c)^4 - (544*a^4*b^2 - 560*a^2*b^4 + 51*b ^6)*cos(d*x + c)^2 - 96*(5*a^6 - a^4*b^2 - 5*a^2*b^4 + b^6 - (5*a^4*b^2 - 6*a^2*b^4 + b^6)*cos(d*x + c)^2 + 2*(5*a^5*b - 6*a^3*b^3 + a*b^5)*sin(d*x + c))*log(b*sin(d*x + c) + a) + 2*(8*a*b^5*cos(d*x + c)^4 + 64*a^5*b + 176 *a^3*b^3 - 205*a*b^5 - 80*(a^3*b^3 - a*b^5)*cos(d*x + c)^2)*sin(d*x + c))/ (b^9*d*cos(d*x + c)^2 - 2*a*b^8*d*sin(d*x + c) - (a^2*b^7 + b^9)*d)
Timed out. \[ \int \frac {\cos ^7(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^7(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {2 \, {\left (11 \, a^{6} - 21 \, a^{4} b^{2} + 9 \, a^{2} b^{4} + b^{6} + 12 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )\right )}}{b^{9} \sin \left (d x + c\right )^{2} + 2 \, a b^{8} \sin \left (d x + c\right ) + a^{2} b^{7}} + \frac {b^{3} \sin \left (d x + c\right )^{4} - 4 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \, {\left (2 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{2} - 4 \, {\left (10 \, a^{3} - 9 \, a b^{2}\right )} \sin \left (d x + c\right )}{b^{6}} + \frac {12 \, {\left (5 \, a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{7}}}{4 \, d} \]
-1/4*(2*(11*a^6 - 21*a^4*b^2 + 9*a^2*b^4 + b^6 + 12*(a^5*b - 2*a^3*b^3 + a *b^5)*sin(d*x + c))/(b^9*sin(d*x + c)^2 + 2*a*b^8*sin(d*x + c) + a^2*b^7) + (b^3*sin(d*x + c)^4 - 4*a*b^2*sin(d*x + c)^3 + 6*(2*a^2*b - b^3)*sin(d*x + c)^2 - 4*(10*a^3 - 9*a*b^2)*sin(d*x + c))/b^6 + 12*(5*a^4 - 6*a^2*b^2 + b^4)*log(b*sin(d*x + c) + a)/b^7)/d
Time = 0.37 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.29 \[ \int \frac {\cos ^7(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {12 \, {\left (5 \, a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{7}} - \frac {2 \, {\left (45 \, a^{4} b^{2} \sin \left (d x + c\right )^{2} - 54 \, a^{2} b^{4} \sin \left (d x + c\right )^{2} + 9 \, b^{6} \sin \left (d x + c\right )^{2} + 78 \, a^{5} b \sin \left (d x + c\right ) - 84 \, a^{3} b^{3} \sin \left (d x + c\right ) + 6 \, a b^{5} \sin \left (d x + c\right ) + 34 \, a^{6} - 33 \, a^{4} b^{2} - b^{6}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2} b^{7}} + \frac {b^{9} \sin \left (d x + c\right )^{4} - 4 \, a b^{8} \sin \left (d x + c\right )^{3} + 12 \, a^{2} b^{7} \sin \left (d x + c\right )^{2} - 6 \, b^{9} \sin \left (d x + c\right )^{2} - 40 \, a^{3} b^{6} \sin \left (d x + c\right ) + 36 \, a b^{8} \sin \left (d x + c\right )}{b^{12}}}{4 \, d} \]
-1/4*(12*(5*a^4 - 6*a^2*b^2 + b^4)*log(abs(b*sin(d*x + c) + a))/b^7 - 2*(4 5*a^4*b^2*sin(d*x + c)^2 - 54*a^2*b^4*sin(d*x + c)^2 + 9*b^6*sin(d*x + c)^ 2 + 78*a^5*b*sin(d*x + c) - 84*a^3*b^3*sin(d*x + c) + 6*a*b^5*sin(d*x + c) + 34*a^6 - 33*a^4*b^2 - b^6)/((b*sin(d*x + c) + a)^2*b^7) + (b^9*sin(d*x + c)^4 - 4*a*b^8*sin(d*x + c)^3 + 12*a^2*b^7*sin(d*x + c)^2 - 6*b^9*sin(d* x + c)^2 - 40*a^3*b^6*sin(d*x + c) + 36*a*b^8*sin(d*x + c))/b^12)/d
Time = 0.13 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.23 \[ \int \frac {\cos ^7(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {3}{2\,b^3}-\frac {3\,a^2}{b^5}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^4}{4\,b^3\,d}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {8\,a^3}{b^6}+\frac {3\,a\,\left (\frac {3}{b^3}-\frac {6\,a^2}{b^5}\right )}{b}\right )}{d}-\frac {\frac {11\,a^6-21\,a^4\,b^2+9\,a^2\,b^4+b^6}{2\,b}+\sin \left (c+d\,x\right )\,\left (6\,a^5-12\,a^3\,b^2+6\,a\,b^4\right )}{d\,\left (a^2\,b^6+2\,a\,b^7\,\sin \left (c+d\,x\right )+b^8\,{\sin \left (c+d\,x\right )}^2\right )}+\frac {a\,{\sin \left (c+d\,x\right )}^3}{b^4\,d}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (15\,a^4-18\,a^2\,b^2+3\,b^4\right )}{b^7\,d} \]
(sin(c + d*x)^2*(3/(2*b^3) - (3*a^2)/b^5))/d - sin(c + d*x)^4/(4*b^3*d) - (sin(c + d*x)*((8*a^3)/b^6 + (3*a*(3/b^3 - (6*a^2)/b^5))/b))/d - ((11*a^6 + b^6 + 9*a^2*b^4 - 21*a^4*b^2)/(2*b) + sin(c + d*x)*(6*a*b^4 + 6*a^5 - 12 *a^3*b^2))/(d*(a^2*b^6 + b^8*sin(c + d*x)^2 + 2*a*b^7*sin(c + d*x))) + (a* sin(c + d*x)^3)/(b^4*d) - (log(a + b*sin(c + d*x))*(15*a^4 + 3*b^4 - 18*a^ 2*b^2))/(b^7*d)