Integrand size = 29, antiderivative size = 268 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {3 b x}{2 \left (a^2-b^2\right )}-\frac {a^2 \left (2 a^2+b^2\right ) x}{2 b^3 \left (a^2-b^2\right )}+\frac {2 a^5 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{3/2} d}+\frac {a \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac {a^3 \cos (c+d x)}{b^2 \left (a^2-b^2\right ) d}+\frac {a \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d}-\frac {3 b \tan (c+d x)}{2 \left (a^2-b^2\right ) d}+\frac {b \sin ^2(c+d x) \tan (c+d x)}{2 \left (a^2-b^2\right ) d} \]
3/2*b*x/(a^2-b^2)-1/2*a^2*(2*a^2+b^2)*x/b^3/(a^2-b^2)+2*a^5*arctan((b+a*ta n(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b^3/(a^2-b^2)^(3/2)/d+a*cos(d*x+c)/(a^2 -b^2)/d-a^3*cos(d*x+c)/b^2/(a^2-b^2)/d+a*sec(d*x+c)/(a^2-b^2)/d+1/2*a^2*co s(d*x+c)*sin(d*x+c)/b/(a^2-b^2)/d-3/2*b*tan(d*x+c)/(a^2-b^2)/d+1/2*b*sin(d *x+c)^2*tan(d*x+c)/(a^2-b^2)/d
Time = 1.35 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.82 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {-4 a b^3+4 a^4 (c+d x)+2 a^2 b^2 (c+d x)-6 b^4 (c+d x)}{-a^2 b^3+b^5}+\frac {8 a^5 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{3/2}}-\frac {4 a \cos (c+d x)}{b^2}+\frac {4 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {4 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\sin (2 (c+d x))}{b}}{4 d} \]
((-4*a*b^3 + 4*a^4*(c + d*x) + 2*a^2*b^2*(c + d*x) - 6*b^4*(c + d*x))/(-(a ^2*b^3) + b^5) + (8*a^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/ (b^3*(a^2 - b^2)^(3/2)) - (4*a*Cos[c + d*x])/b^2 + (4*Sin[(c + d*x)/2])/(( a + b)*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - (4*Sin[(c + d*x)/2])/((a - b)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + Sin[2*(c + d*x)]/b)/(4*d)
Time = 1.15 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.88, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.655, Rules used = {3042, 3381, 3042, 3070, 244, 2009, 3071, 252, 262, 216, 3272, 3042, 3502, 3042, 3214, 3042, 3139, 1083, 217}
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.
| \(\displaystyle \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^5}{\cos (c+d x)^2 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3381 |
| \(\displaystyle -\frac {a^2 \int \frac {\sin ^3(c+d x)}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {b \int \sin ^2(c+d x) \tan ^2(c+d x)dx}{a^2-b^2}+\frac {a \int \sin (c+d x) \tan ^2(c+d x)dx}{a^2-b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a^2 \int \frac {\sin (c+d x)^3}{a+b \sin (c+d x)}dx}{a^2-b^2}+\frac {a \int \sin (c+d x) \tan (c+d x)^2dx}{a^2-b^2}-\frac {b \int \sin (c+d x)^2 \tan (c+d x)^2dx}{a^2-b^2}\) |
| \(\Big \downarrow \) 3070 |
| \(\displaystyle -\frac {a^2 \int \frac {\sin (c+d x)^3}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {b \int \sin (c+d x)^2 \tan (c+d x)^2dx}{a^2-b^2}-\frac {a \int \left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)d\cos (c+d x)}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle -\frac {a^2 \int \frac {\sin (c+d x)^3}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {b \int \sin (c+d x)^2 \tan (c+d x)^2dx}{a^2-b^2}-\frac {a \int \left (\sec ^2(c+d x)-1\right )d\cos (c+d x)}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^2 \int \frac {\sin (c+d x)^3}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {b \int \sin (c+d x)^2 \tan (c+d x)^2dx}{a^2-b^2}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3071 |
| \(\displaystyle -\frac {a^2 \int \frac {\sin (c+d x)^3}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {b \int \frac {\tan ^4(c+d x)}{\left (\tan ^2(c+d x)+1\right )^2}d\tan (c+d x)}{d \left (a^2-b^2\right )}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {a^2 \int \frac {\sin (c+d x)^3}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {b \left (\frac {3}{2} \int \frac {\tan ^2(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d \left (a^2-b^2\right )}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle -\frac {a^2 \int \frac {\sin (c+d x)^3}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {b \left (\frac {3}{2} \left (\tan (c+d x)-\int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)\right )-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d \left (a^2-b^2\right )}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {a^2 \int \frac {\sin (c+d x)^3}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {b \left (\frac {3}{2} (\tan (c+d x)-\arctan (\tan (c+d x)))-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d \left (a^2-b^2\right )}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3272 |
| \(\displaystyle -\frac {a^2 \left (\frac {\int \frac {-2 a \sin ^2(c+d x)+b \sin (c+d x)+a}{a+b \sin (c+d x)}dx}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\right )}{a^2-b^2}-\frac {b \left (\frac {3}{2} (\tan (c+d x)-\arctan (\tan (c+d x)))-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d \left (a^2-b^2\right )}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a^2 \left (\frac {\int \frac {-2 a \sin (c+d x)^2+b \sin (c+d x)+a}{a+b \sin (c+d x)}dx}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\right )}{a^2-b^2}-\frac {b \left (\frac {3}{2} (\tan (c+d x)-\arctan (\tan (c+d x)))-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d \left (a^2-b^2\right )}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle -\frac {a^2 \left (\frac {\frac {\int \frac {a b+\left (2 a^2+b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}+\frac {2 a \cos (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\right )}{a^2-b^2}-\frac {b \left (\frac {3}{2} (\tan (c+d x)-\arctan (\tan (c+d x)))-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d \left (a^2-b^2\right )}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a^2 \left (\frac {\frac {\int \frac {a b+\left (2 a^2+b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}+\frac {2 a \cos (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\right )}{a^2-b^2}-\frac {b \left (\frac {3}{2} (\tan (c+d x)-\arctan (\tan (c+d x)))-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d \left (a^2-b^2\right )}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle -\frac {a^2 \left (\frac {\frac {\frac {x \left (2 a^2+b^2\right )}{b}-\frac {2 a^3 \int \frac {1}{a+b \sin (c+d x)}dx}{b}}{b}+\frac {2 a \cos (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\right )}{a^2-b^2}-\frac {b \left (\frac {3}{2} (\tan (c+d x)-\arctan (\tan (c+d x)))-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d \left (a^2-b^2\right )}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a^2 \left (\frac {\frac {\frac {x \left (2 a^2+b^2\right )}{b}-\frac {2 a^3 \int \frac {1}{a+b \sin (c+d x)}dx}{b}}{b}+\frac {2 a \cos (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\right )}{a^2-b^2}-\frac {b \left (\frac {3}{2} (\tan (c+d x)-\arctan (\tan (c+d x)))-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d \left (a^2-b^2\right )}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle -\frac {a^2 \left (\frac {\frac {\frac {x \left (2 a^2+b^2\right )}{b}-\frac {4 a^3 \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b}+\frac {2 a \cos (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\right )}{a^2-b^2}-\frac {b \left (\frac {3}{2} (\tan (c+d x)-\arctan (\tan (c+d x)))-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d \left (a^2-b^2\right )}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {a^2 \left (\frac {\frac {\frac {8 a^3 \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}+\frac {x \left (2 a^2+b^2\right )}{b}}{b}+\frac {2 a \cos (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\right )}{a^2-b^2}-\frac {b \left (\frac {3}{2} (\tan (c+d x)-\arctan (\tan (c+d x)))-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d \left (a^2-b^2\right )}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {b \left (\frac {3}{2} (\tan (c+d x)-\arctan (\tan (c+d x)))-\frac {\tan ^3(c+d x)}{2 \left (\tan ^2(c+d x)+1\right )}\right )}{d \left (a^2-b^2\right )}-\frac {a (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}-\frac {a^2 \left (\frac {\frac {\frac {x \left (2 a^2+b^2\right )}{b}-\frac {4 a^3 \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}}{b}+\frac {2 a \cos (c+d x)}{b d}}{2 b}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\right )}{a^2-b^2}\) |
-((a*(-Cos[c + d*x] - Sec[c + d*x]))/((a^2 - b^2)*d)) - (a^2*(((((2*a^2 + b^2)*x)/b - (4*a^3*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2]) ])/(b*Sqrt[a^2 - b^2]*d))/b + (2*a*Cos[c + d*x])/(b*d))/(2*b) - (Cos[c + d *x]*Sin[c + d*x])/(2*b*d)))/(a^2 - b^2) - (b*((3*(-ArcTan[Tan[c + d*x]] + Tan[c + d*x]))/2 - Tan[c + d*x]^3/(2*(1 + Tan[c + d*x]^2))))/((a^2 - b^2)* d)
3.14.38.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
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[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Simp[-f^(-1) Subst[Int[(1 - x^2)^((m + n - 1)/2)/x^n, x], x, Cos[e + f *x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[b*(ff/f) Subst[I nt[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d *(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m ] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a*(d^2/(a^2 - b^2)) Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-Simp[ b*(d/(a^2 - b^2)) Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Simp[a^2*(d^2/(g^2*(a^2 - b^2))) Int[(g*Cos[e + f*x])^(p + 2)*((d*Sin[ e + f*x])^(n - 2)/(a + b*Sin[e + f*x])), x], x]) /; FreeQ[{a, b, d, e, f, g }, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1 ]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 0.90 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.78
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{2}+\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a b -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}}{2}+a b}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (2 a^{2}+3 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{3}}-\frac {64}{\left (64 a +64 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 a^{5} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) b^{3} \sqrt {a^{2}-b^{2}}}+\frac {64}{\left (64 a -64 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(208\) |
| default | \(\frac {-\frac {2 \left (\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{2}+\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a b -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}}{2}+a b}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (2 a^{2}+3 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{3}}-\frac {64}{\left (64 a +64 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 a^{5} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) b^{3} \sqrt {a^{2}-b^{2}}}+\frac {64}{\left (64 a -64 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(208\) |
| risch | \(-\frac {x \,a^{2}}{b^{3}}-\frac {3 x}{2 b}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 b d}-\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d \,b^{2}}-\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 b d}+\frac {2 i \left (i a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{d \left (-a^{2}+b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {i a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {i a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}\) | \(304\) |
1/d*(-2/b^3*((1/2*tan(1/2*d*x+1/2*c)^3*b^2+tan(1/2*d*x+1/2*c)^2*a*b-1/2*ta n(1/2*d*x+1/2*c)*b^2+a*b)/(1+tan(1/2*d*x+1/2*c)^2)^2+1/2*(2*a^2+3*b^2)*arc tan(tan(1/2*d*x+1/2*c)))-64/(64*a+64*b)/(tan(1/2*d*x+1/2*c)-1)+2/(a-b)/(a+ b)*a^5/b^3/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^ 2)^(1/2))+64/(64*a-64*b)/(tan(1/2*d*x+1/2*c)+1))
Time = 0.46 (sec) , antiderivative size = 521, normalized size of antiderivative = 1.94 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {\sqrt {-a^{2} + b^{2}} a^{5} \cos \left (d x + c\right ) \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{3} b^{3} - 2 \, a b^{5} - {\left (2 \, a^{6} - a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} d x \cos \left (d x + c\right ) - 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{2} - {\left (2 \, a^{2} b^{4} - 2 \, b^{6} - {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d \cos \left (d x + c\right )}, -\frac {2 \, \sqrt {a^{2} - b^{2}} a^{5} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - 2 \, a^{3} b^{3} + 2 \, a b^{5} + {\left (2 \, a^{6} - a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, a^{2} b^{4} - 2 \, b^{6} - {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d \cos \left (d x + c\right )}\right ] \]
[1/2*(sqrt(-a^2 + b^2)*a^5*cos(d*x + c)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos (d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^ 2 - b^2)) + 2*a^3*b^3 - 2*a*b^5 - (2*a^6 - a^4*b^2 - 4*a^2*b^4 + 3*b^6)*d* x*cos(d*x + c) - 2*(a^5*b - 2*a^3*b^3 + a*b^5)*cos(d*x + c)^2 - (2*a^2*b^4 - 2*b^6 - (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^4 *b^3 - 2*a^2*b^5 + b^7)*d*cos(d*x + c)), -1/2*(2*sqrt(a^2 - b^2)*a^5*arcta n(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*cos(d*x + c) - 2*a ^3*b^3 + 2*a*b^5 + (2*a^6 - a^4*b^2 - 4*a^2*b^4 + 3*b^6)*d*x*cos(d*x + c) + 2*(a^5*b - 2*a^3*b^3 + a*b^5)*cos(d*x + c)^2 + (2*a^2*b^4 - 2*b^6 - (a^4 *b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^4*b^3 - 2*a^2*b^ 5 + b^7)*d*cos(d*x + c))]
Timed out. \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.38 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.78 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {4 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{5}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt {a^{2} - b^{2}}} + \frac {4 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}} - \frac {{\left (2 \, a^{2} + 3 \, b^{2}\right )} {\left (d x + c\right )}}{b^{3}} - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \]
1/2*(4*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))*a^5/((a^2*b^3 - b^5)*sqrt(a^2 - b^2)) + 4*(b *tan(1/2*d*x + 1/2*c) - a)/((a^2 - b^2)*(tan(1/2*d*x + 1/2*c)^2 - 1)) - (2 *a^2 + 3*b^2)*(d*x + c)/b^3 - 2*(b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d* x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c) + 2*a)/((tan(1/2*d*x + 1/2*c)^2 + 1) ^2*b^2))/d
Time = 17.09 (sec) , antiderivative size = 2098, normalized size of antiderivative = 7.83 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]
(4*a^5*cos(c + d*x) + (5*a^5)/2 + (3*a^5*cos(2*c + 2*d*x))/2)/(d*cos(c + d *x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (2*a^8*atan(sin(c/2 + (d*x)/2)/ cos(c/2 + (d*x)/2)))/(b^3*d*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b*((11 *a^4*sin(c + d*x))/8 + (3*a^4*sin(3*c + 3*d*x))/8 - 3*a^4*cos(c + d*x)*ata n(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))))/(d*cos(c + d*x)*(a^2 - b^2)*(a^ 4 + b^4 - 2*a^2*b^2)) + (b^4*((3*a)/2 + 2*a*cos(c + d*x) + (a*cos(2*c + 2* d*x))/2))/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b^5*((9* sin(c + d*x))/8 + sin(3*c + 3*d*x)/8 - 3*cos(c + d*x)*atan(sin(c/2 + (d*x) /2)/cos(c/2 + (d*x)/2))))/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b ^2)) - (a^7*cos(c + d*x) + a^7/2 + (a^7*cos(2*c + 2*d*x))/2)/(b^2*d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b^2*(5*a^3*cos(c + d*x) + ( 7*a^3)/2 + (3*a^3*cos(2*c + 2*d*x))/2))/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + ((a^6*sin(c + d*x))/8 + (a^6*sin(3*c + 3*d*x))/8 + 3* a^6*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b*d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + (b^3*((19*a^2*sin(c + d*x))/8 + (3*a^2*sin(3*c + 3*d*x))/8 - 7*a^2*cos(c + d*x)*atan(sin(c/2 + (d*x)/2) /cos(c/2 + (d*x)/2))))/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2) ) + (a^5*atan((a^12*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2) ^(3/2)*8i + a^18*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1 /2)*8i - b^18*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1...