Integrand size = 29, antiderivative size = 183 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a x}{a^2-b^2}+\frac {a^3 x}{b^2 \left (a^2-b^2\right )}-\frac {2 a^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{3/2} d}+\frac {a^2 \cos (c+d x)}{b \left (a^2-b^2\right ) d}-\frac {b \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d} \] Output:
-a*x/(a^2-b^2)+a^3*x/b^2/(a^2-b^2)-2*a^4*arctan((b+a*tan(1/2*d*x+1/2*c))/( a^2-b^2)^(1/2))/b^2/(a^2-b^2)^(3/2)/d+a^2*cos(d*x+c)/b/(a^2-b^2)/d-b*cos(d *x+c)/(a^2-b^2)/d-b*sec(d*x+c)/(a^2-b^2)/d+a*tan(d*x+c)/(a^2-b^2)/d
Time = 1.66 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.02 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {b^3-a^3 (c+d x)+a b^2 (c+d x)}{-a^2 b^2+b^4}-\frac {2 a^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{3/2}}+\frac {\cos (c+d x)}{b}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{(a-b) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}}{d} \] Input:
Integrate[(Sin[c + d*x]^2*Tan[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
Output:
((b^3 - a^3*(c + d*x) + a*b^2*(c + d*x))/(-(a^2*b^2) + b^4) - (2*a^4*ArcTa n[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^2*(a^2 - b^2)^(3/2)) + Cos [c + d*x]/b + Sin[(c + d*x)/2]/((a + b)*(Cos[(c + d*x)/2] - Sin[(c + d*x)/ 2])) + Sin[(c + d*x)/2]/((a - b)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))/d
Time = 0.91 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.88, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.586, Rules used = {3042, 3381, 3042, 3070, 244, 2009, 3225, 25, 27, 3042, 3214, 3042, 3139, 1083, 217, 3954, 24}
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.
| \(\displaystyle \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4}{\cos (c+d x)^2 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3381 |
| \(\displaystyle -\frac {a^2 \int \frac {\sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{a^2-b^2}+\frac {a \int \tan ^2(c+d x)dx}{a^2-b^2}-\frac {b \int \sin (c+d x) \tan ^2(c+d x)dx}{a^2-b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a^2 \int \frac {\sin (c+d x)^2}{a+b \sin (c+d x)}dx}{a^2-b^2}+\frac {a \int \tan (c+d x)^2dx}{a^2-b^2}-\frac {b \int \sin (c+d x) \tan (c+d x)^2dx}{a^2-b^2}\) |
| \(\Big \downarrow \) 3070 |
| \(\displaystyle -\frac {a^2 \int \frac {\sin (c+d x)^2}{a+b \sin (c+d x)}dx}{a^2-b^2}+\frac {a \int \tan (c+d x)^2dx}{a^2-b^2}+\frac {b \int \left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)d\cos (c+d x)}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle -\frac {a^2 \int \frac {\sin (c+d x)^2}{a+b \sin (c+d x)}dx}{a^2-b^2}+\frac {a \int \tan (c+d x)^2dx}{a^2-b^2}+\frac {b \int \left (\sec ^2(c+d x)-1\right )d\cos (c+d x)}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^2 \int \frac {\sin (c+d x)^2}{a+b \sin (c+d x)}dx}{a^2-b^2}+\frac {a \int \tan (c+d x)^2dx}{a^2-b^2}+\frac {b (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3225 |
| \(\displaystyle \frac {a \int \tan (c+d x)^2dx}{a^2-b^2}-\frac {a^2 \left (\frac {\int -\frac {a \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}-\frac {\cos (c+d x)}{b d}\right )}{a^2-b^2}+\frac {b (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {a \int \tan (c+d x)^2dx}{a^2-b^2}-\frac {a^2 \left (-\frac {\int \frac {a \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}-\frac {\cos (c+d x)}{b d}\right )}{a^2-b^2}+\frac {b (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \int \tan (c+d x)^2dx}{a^2-b^2}-\frac {a^2 \left (-\frac {a \int \frac {\sin (c+d x)}{a+b \sin (c+d x)}dx}{b}-\frac {\cos (c+d x)}{b d}\right )}{a^2-b^2}+\frac {b (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \int \tan (c+d x)^2dx}{a^2-b^2}-\frac {a^2 \left (-\frac {a \int \frac {\sin (c+d x)}{a+b \sin (c+d x)}dx}{b}-\frac {\cos (c+d x)}{b d}\right )}{a^2-b^2}+\frac {b (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {a \int \tan (c+d x)^2dx}{a^2-b^2}-\frac {a^2 \left (-\frac {a \left (\frac {x}{b}-\frac {a \int \frac {1}{a+b \sin (c+d x)}dx}{b}\right )}{b}-\frac {\cos (c+d x)}{b d}\right )}{a^2-b^2}+\frac {b (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \int \tan (c+d x)^2dx}{a^2-b^2}-\frac {a^2 \left (-\frac {a \left (\frac {x}{b}-\frac {a \int \frac {1}{a+b \sin (c+d x)}dx}{b}\right )}{b}-\frac {\cos (c+d x)}{b d}\right )}{a^2-b^2}+\frac {b (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {a \int \tan (c+d x)^2dx}{a^2-b^2}-\frac {a^2 \left (-\frac {a \left (\frac {x}{b}-\frac {2 a \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}\right )}{b}-\frac {\cos (c+d x)}{b d}\right )}{a^2-b^2}+\frac {b (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {a \int \tan (c+d x)^2dx}{a^2-b^2}-\frac {a^2 \left (-\frac {a \left (\frac {4 a \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}+\frac {x}{b}\right )}{b}-\frac {\cos (c+d x)}{b d}\right )}{a^2-b^2}+\frac {b (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {a \int \tan (c+d x)^2dx}{a^2-b^2}-\frac {a^2 \left (-\frac {a \left (\frac {x}{b}-\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}\right )}{b}-\frac {\cos (c+d x)}{b d}\right )}{a^2-b^2}+\frac {b (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle \frac {a \left (\frac {\tan (c+d x)}{d}-\int 1dx\right )}{a^2-b^2}-\frac {a^2 \left (-\frac {a \left (\frac {x}{b}-\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}\right )}{b}-\frac {\cos (c+d x)}{b d}\right )}{a^2-b^2}+\frac {b (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle -\frac {a^2 \left (-\frac {a \left (\frac {x}{b}-\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}\right )}{b}-\frac {\cos (c+d x)}{b d}\right )}{a^2-b^2}+\frac {a \left (\frac {\tan (c+d x)}{d}-x\right )}{a^2-b^2}+\frac {b (-\cos (c+d x)-\sec (c+d x))}{d \left (a^2-b^2\right )}\) |
Input:
Int[(Sin[c + d*x]^2*Tan[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
Output:
-((a^2*(-((a*(x/b - (2*a*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d)))/b) - Cos[c + d*x]/(b*d)))/(a^2 - b^2)) + (b*(-Cos[c + d*x] - Sec[c + d*x]))/((a^2 - b^2)*d) + (a*(-x + Tan[c + d*x] /d))/(a^2 - b^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((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[((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_)])^2/((c_.) + (d_.)*sin[(e_.) + (f _.)*(x_)]), x_Symbol] :> Simp[(-b^2)*(Cos[e + f*x]/(d*f)), x] + Simp[1/d Int[Simp[a^2*d - b*(b*c - 2*a*d)*Sin[e + f*x], x]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 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[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Time = 0.72 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {2 b}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}-\frac {32}{\left (32 a +32 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {32}{\left (32 a -32 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) b^{2} \sqrt {a^{2}-b^{2}}}}{d}\) | \(150\) |
| default | \(\frac {\frac {\frac {2 b}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}-\frac {32}{\left (32 a +32 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {32}{\left (32 a -32 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) b^{2} \sqrt {a^{2}-b^{2}}}}{d}\) | \(150\) |
| risch | \(\frac {a x}{b^{2}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b d}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\frac {-2 i a +2 b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left (-a^{2}+b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}\) | \(254\) |
Input:
int(sin(d*x+c)^2*tan(d*x+c)^2/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(2/b^2*(b/(1+tan(1/2*d*x+1/2*c)^2)+a*arctan(tan(1/2*d*x+1/2*c)))-32/(3 2*a+32*b)/(tan(1/2*d*x+1/2*c)-1)-32/(32*a-32*b)/(tan(1/2*d*x+1/2*c)+1)-2/( a-b)/(a+b)*a^4/b^2/(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)))
Time = 0.11 (sec) , antiderivative size = 431, normalized size of antiderivative = 2.36 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {\sqrt {-a^{2} + b^{2}} a^{4} \cos \left (d x + c\right ) \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, a^{2} b^{3} + 2 \, b^{5} + 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )}, \frac {\sqrt {a^{2} - b^{2}} a^{4} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - a^{2} b^{3} + b^{5} + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )}\right ] \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
[1/2*(sqrt(-a^2 + b^2)*a^4*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^2*b^3 + 2*b^5 + 2*(a^5 - 2*a^3*b^2 + a*b^4)*d*x*cos(d*x + c ) + 2*(a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c)^2 + 2*(a^3*b^2 - a*b^4)*sin(d *x + c))/((a^4*b^2 - 2*a^2*b^4 + b^6)*d*cos(d*x + c)), (sqrt(a^2 - b^2)*a^ 4*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*cos(d*x + c ) - a^2*b^3 + b^5 + (a^5 - 2*a^3*b^2 + a*b^4)*d*x*cos(d*x + c) + (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c)^2 + (a^3*b^2 - a*b^4)*sin(d*x + c))/((a^4*b^ 2 - 2*a^2*b^4 + b^6)*d*cos(d*x + c))]
\[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(sin(d*x+c)**2*tan(d*x+c)**2/(a+b*sin(d*x+c)),x)
Output:
Integral(sin(c + d*x)**2*tan(c + d*x)**2/(a + b*sin(c + d*x)), x)
Exception generated. \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima" )
Output:
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 = 0.30 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{4}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {{\left (d x + c\right )} a}{b^{2}} + \frac {2 \, {\left (a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2} - 2 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )} {\left (a^{2} b - b^{3}\right )}}}{d} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
-(2*(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^4/((a^2*b^2 - b^4)*sqrt(a^2 - b^2)) - (d*x + c)*a/b^2 + 2*(a*b*tan(1/2*d*x + 1/2*c)^3 - a^2*tan(1/2*d*x + 1/2*c)^2 + a* b*tan(1/2*d*x + 1/2*c) + a^2 - 2*b^2)/((tan(1/2*d*x + 1/2*c)^4 - 1)*(a^2*b - b^3)))/d
Time = 21.77 (sec) , antiderivative size = 1656, normalized size of antiderivative = 9.05 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
int((sin(c + d*x)^2*tan(c + d*x)^2)/(a + b*sin(c + d*x)),x)
Output:
(a^5*sin(c + d*x) - 6*a^5*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)) + (2*a^7*at an(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^2*d*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + (a^6*cos(c + d*x) + a^6/2 + (a^6*cos(2*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*(5*a^2*cos(c + d* x) + (7*a^2)/2 + (3*a^2*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^2*(2*a^3*sin(c + d*x) - 6*a^3*cos(c + d*x)*a tan(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^5*(2*cos(c + d*x) + cos(2*c + 2*d*x)/2 + 3/2) )/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + (b^4*(a*sin(c + d *x) - 2*a*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))))/(d*co s(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b*(4*a^4*cos(c + d*x) + (5*a^4)/2 + (3*a^4*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^4*atan(((2*b^14*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 2*a^14*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^ 2*b^4 + 3*a^4*b^2)^(1/2) - 2*a^8*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) - 6*a^3*b^11*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 15*a^5*b^9*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 20*a^7*b^7*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 15*a^9*b^5*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*...
Time = 0.19 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.55 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) a^{4}+\cos \left (d x +c \right ) a^{5} c +\cos \left (d x +c \right ) a^{5} d x -\cos \left (d x +c \right ) a^{4} b -2 \cos \left (d x +c \right ) a^{3} b^{2} c -2 \cos \left (d x +c \right ) a^{3} b^{2} d x +3 \cos \left (d x +c \right ) a^{2} b^{3}+\cos \left (d x +c \right ) a \,b^{4} c +\cos \left (d x +c \right ) a \,b^{4} d x -2 \cos \left (d x +c \right ) b^{5}-\sin \left (d x +c \right )^{2} a^{4} b +2 \sin \left (d x +c \right )^{2} a^{2} b^{3}-\sin \left (d x +c \right )^{2} b^{5}+\sin \left (d x +c \right ) a^{3} b^{2}-\sin \left (d x +c \right ) a \,b^{4}+a^{4} b -3 a^{2} b^{3}+2 b^{5}}{\cos \left (d x +c \right ) b^{2} d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )} \] Input:
int(sin(d*x+c)^2*tan(d*x+c)^2/(a+b*sin(d*x+c)),x)
Output:
( - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*c os(c + d*x)*a**4 + cos(c + d*x)*a**5*c + cos(c + d*x)*a**5*d*x - cos(c + d *x)*a**4*b - 2*cos(c + d*x)*a**3*b**2*c - 2*cos(c + d*x)*a**3*b**2*d*x + 3 *cos(c + d*x)*a**2*b**3 + cos(c + d*x)*a*b**4*c + cos(c + d*x)*a*b**4*d*x - 2*cos(c + d*x)*b**5 - sin(c + d*x)**2*a**4*b + 2*sin(c + d*x)**2*a**2*b* *3 - sin(c + d*x)**2*b**5 + sin(c + d*x)*a**3*b**2 - sin(c + d*x)*a*b**4 + a**4*b - 3*a**2*b**3 + 2*b**5)/(cos(c + d*x)*b**2*d*(a**4 - 2*a**2*b**2 + b**4))