Integrand size = 27, antiderivative size = 115 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {15 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {15 a \csc (c+d x)}{8 d}+\frac {b \log (\tan (c+d x))}{d}+\frac {5 a \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {a \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {b \tan ^2(c+d x)}{d}+\frac {b \tan ^4(c+d x)}{4 d} \]
15/8*a*arctanh(sin(d*x+c))/d-15/8*a*csc(d*x+c)/d+b*ln(tan(d*x+c))/d+5/8*a* csc(d*x+c)*sec(d*x+c)^2/d+1/4*a*csc(d*x+c)*sec(d*x+c)^4/d+b*tan(d*x+c)^2/d +1/4*b*tan(d*x+c)^4/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.75 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \csc (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3,\frac {1}{2},\sin ^2(c+d x)\right )}{d}-\frac {b \log (\cos (c+d x))}{d}+\frac {b \log (\sin (c+d x))}{d}+\frac {b \sec ^2(c+d x)}{2 d}+\frac {b \sec ^4(c+d x)}{4 d} \]
-((a*Csc[c + d*x]*Hypergeometric2F1[-1/2, 3, 1/2, Sin[c + d*x]^2])/d) - (b *Log[Cos[c + d*x]])/d + (b*Log[Sin[c + d*x]])/d + (b*Sec[c + d*x]^2)/(2*d) + (b*Sec[c + d*x]^4)/(4*d)
Time = 0.44 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.09, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.481, Rules used = {3042, 3313, 3042, 3100, 243, 49, 2009, 3101, 25, 252, 252, 262, 219}
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.
| \(\displaystyle \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a+b \sin (c+d x)}{\sin (c+d x)^2 \cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3313 |
| \(\displaystyle a \int \csc ^2(c+d x) \sec ^5(c+d x)dx+b \int \csc (c+d x) \sec ^5(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \csc (c+d x)^2 \sec (c+d x)^5dx+b \int \csc (c+d x) \sec (c+d x)^5dx\) |
| \(\Big \downarrow \) 3100 |
| \(\displaystyle a \int \csc (c+d x)^2 \sec (c+d x)^5dx+\frac {b \int \cot (c+d x) \left (\tan ^2(c+d x)+1\right )^2d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle a \int \csc (c+d x)^2 \sec (c+d x)^5dx+\frac {b \int \cot (c+d x) \left (\tan ^2(c+d x)+1\right )^2d\tan ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle a \int \csc (c+d x)^2 \sec (c+d x)^5dx+\frac {b \int \left (\tan ^2(c+d x)+\cot (c+d x)+2\right )d\tan ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a \int \csc (c+d x)^2 \sec (c+d x)^5dx+\frac {b \left (\frac {1}{2} \tan ^4(c+d x)+2 \tan ^2(c+d x)+\log \left (\tan ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 3101 |
| \(\displaystyle \frac {b \left (\frac {1}{2} \tan ^4(c+d x)+2 \tan ^2(c+d x)+\log \left (\tan ^2(c+d x)\right )\right )}{2 d}-\frac {a \int -\frac {\csc ^6(c+d x)}{\left (1-\csc ^2(c+d x)\right )^3}d\csc (c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {a \int \frac {\csc ^6(c+d x)}{\left (1-\csc ^2(c+d x)\right )^3}d\csc (c+d x)}{d}+\frac {b \left (\frac {1}{2} \tan ^4(c+d x)+2 \tan ^2(c+d x)+\log \left (\tan ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {b \left (\frac {1}{2} \tan ^4(c+d x)+2 \tan ^2(c+d x)+\log \left (\tan ^2(c+d x)\right )\right )}{2 d}-\frac {a \left (\frac {5}{4} \int \frac {\csc ^4(c+d x)}{\left (1-\csc ^2(c+d x)\right )^2}d\csc (c+d x)-\frac {\csc ^5(c+d x)}{4 \left (1-\csc ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {b \left (\frac {1}{2} \tan ^4(c+d x)+2 \tan ^2(c+d x)+\log \left (\tan ^2(c+d x)\right )\right )}{2 d}-\frac {a \left (\frac {5}{4} \left (\frac {\csc ^3(c+d x)}{2 \left (1-\csc ^2(c+d x)\right )}-\frac {3}{2} \int \frac {\csc ^2(c+d x)}{1-\csc ^2(c+d x)}d\csc (c+d x)\right )-\frac {\csc ^5(c+d x)}{4 \left (1-\csc ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {b \left (\frac {1}{2} \tan ^4(c+d x)+2 \tan ^2(c+d x)+\log \left (\tan ^2(c+d x)\right )\right )}{2 d}-\frac {a \left (\frac {5}{4} \left (\frac {\csc ^3(c+d x)}{2 \left (1-\csc ^2(c+d x)\right )}-\frac {3}{2} \left (\int \frac {1}{1-\csc ^2(c+d x)}d\csc (c+d x)-\csc (c+d x)\right )\right )-\frac {\csc ^5(c+d x)}{4 \left (1-\csc ^2(c+d x)\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {b \left (\frac {1}{2} \tan ^4(c+d x)+2 \tan ^2(c+d x)+\log \left (\tan ^2(c+d x)\right )\right )}{2 d}-\frac {a \left (\frac {5}{4} \left (\frac {\csc ^3(c+d x)}{2 \left (1-\csc ^2(c+d x)\right )}-\frac {3}{2} (\text {arctanh}(\csc (c+d x))-\csc (c+d x))\right )-\frac {\csc ^5(c+d x)}{4 \left (1-\csc ^2(c+d x)\right )^2}\right )}{d}\) |
-((a*(-1/4*Csc[c + d*x]^5/(1 - Csc[c + d*x]^2)^2 + (5*((-3*(ArcTanh[Csc[c + d*x]] - Csc[c + d*x]))/2 + Csc[c + d*x]^3/(2*(1 - Csc[c + d*x]^2))))/4)) /d) + (b*(Log[Tan[c + d*x]^2] + 2*Tan[c + d*x]^2 + Tan[c + d*x]^4/2))/(2*d )
3.15.89.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Simp[1/f Subst[Int[(1 + x^2)^((m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]] , x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_S ymbol] :> Simp[-(f*a^n)^(-1) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_ ) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[Cos[e + f*x]^ p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[Cos[e + f*x]^p*(d*Sin[e + f*x ])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2 ] && IntegerQ[n] && ((LtQ[p, 0] && NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] | | LtQ[p + 1, -n, 2*p + 1])
Time = 0.94 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(101\) |
| default | \(\frac {a \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(101\) |
| parallelrisch | \(\frac {-15 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +\frac {8 b}{15}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -\frac {8 b}{15}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \left (\cos \left (2 d x +2 c \right )+\frac {3 \cos \left (4 d x +4 c \right )}{8}+\frac {9}{40}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) a \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b \left (\frac {3 \cos \left (4 d x +4 c \right )}{4}-\frac {7}{4}+\cos \left (2 d x +2 c \right )\right )}{2 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(216\) |
| risch | \(-\frac {i \left (15 a \,{\mathrm e}^{9 i \left (d x +c \right )}+40 a \,{\mathrm e}^{7 i \left (d x +c \right )}+8 i b \,{\mathrm e}^{8 i \left (d x +c \right )}+18 a \,{\mathrm e}^{5 i \left (d x +c \right )}+24 i b \,{\mathrm e}^{6 i \left (d x +c \right )}+40 a \,{\mathrm e}^{3 i \left (d x +c \right )}-24 i b \,{\mathrm e}^{4 i \left (d x +c \right )}+15 a \,{\mathrm e}^{i \left (d x +c \right )}-8 i b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {15 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{d}+\frac {15 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{d}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(240\) |
| norman | \(\frac {\frac {4 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a}{2 d}+\frac {13 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {5 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {13 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {a \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (15 a -8 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (15 a +8 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\) | \(250\) |
1/d*(a*(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*(1/4/cos(d*x+c)^4+1/2/cos(d*x+c)^2 +ln(tan(d*x+c))))
Time = 0.30 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.38 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {16 \, b \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + {\left (15 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - {\left (15 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 30 \, a \cos \left (d x + c\right )^{4} + 10 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, b \cos \left (d x + c\right )^{2} + b\right )} \sin \left (d x + c\right ) + 4 \, a}{16 \, d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )} \]
1/16*(16*b*cos(d*x + c)^4*log(1/2*sin(d*x + c))*sin(d*x + c) + (15*a - 8*b )*cos(d*x + c)^4*log(sin(d*x + c) + 1)*sin(d*x + c) - (15*a + 8*b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1)*sin(d*x + c) - 30*a*cos(d*x + c)^4 + 10*a*c os(d*x + c)^2 + 4*(2*b*cos(d*x + c)^2 + b)*sin(d*x + c) + 4*a)/(d*cos(d*x + c)^4*sin(d*x + c))
Timed out. \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.10 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {{\left (15 \, a - 8 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (15 \, a + 8 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, b \log \left (\sin \left (d x + c\right )\right ) - \frac {2 \, {\left (15 \, a \sin \left (d x + c\right )^{4} + 4 \, b \sin \left (d x + c\right )^{3} - 25 \, a \sin \left (d x + c\right )^{2} - 6 \, b \sin \left (d x + c\right ) + 8 \, a\right )}}{\sin \left (d x + c\right )^{5} - 2 \, \sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )}}{16 \, d} \]
1/16*((15*a - 8*b)*log(sin(d*x + c) + 1) - (15*a + 8*b)*log(sin(d*x + c) - 1) + 16*b*log(sin(d*x + c)) - 2*(15*a*sin(d*x + c)^4 + 4*b*sin(d*x + c)^3 - 25*a*sin(d*x + c)^2 - 6*b*sin(d*x + c) + 8*a)/(sin(d*x + c)^5 - 2*sin(d *x + c)^3 + sin(d*x + c)))/d
Time = 0.37 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.17 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {{\left (15 \, a - 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (15 \, a + 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 16 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {16 \, {\left (b \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )} + \frac {2 \, {\left (6 \, b \sin \left (d x + c\right )^{4} - 7 \, a \sin \left (d x + c\right )^{3} - 16 \, b \sin \left (d x + c\right )^{2} + 9 \, a \sin \left (d x + c\right ) + 12 \, b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
1/16*((15*a - 8*b)*log(abs(sin(d*x + c) + 1)) - (15*a + 8*b)*log(abs(sin(d *x + c) - 1)) + 16*b*log(abs(sin(d*x + c))) - 16*(b*sin(d*x + c) + a)/sin( d*x + c) + 2*(6*b*sin(d*x + c)^4 - 7*a*sin(d*x + c)^3 - 16*b*sin(d*x + c)^ 2 + 9*a*sin(d*x + c) + 12*b)/(sin(d*x + c)^2 - 1)^2)/d
Time = 0.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.13 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {15\,a}{16}-\frac {b}{2}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {15\,a}{16}+\frac {b}{2}\right )}{d}+\frac {b\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\frac {15\,a\,{\sin \left (c+d\,x\right )}^4}{8}+\frac {b\,{\sin \left (c+d\,x\right )}^3}{2}-\frac {25\,a\,{\sin \left (c+d\,x\right )}^2}{8}-\frac {3\,b\,\sin \left (c+d\,x\right )}{4}+a}{d\,\left ({\sin \left (c+d\,x\right )}^5-2\,{\sin \left (c+d\,x\right )}^3+\sin \left (c+d\,x\right )\right )} \]