Integrand size = 29, antiderivative size = 388 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {4 a^4 \left (a^2-2 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}-\frac {a^4 \left (2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}-\frac {2 a^2 \left (a^4-3 a^2 b^2+6 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 a^5 \cos (c+d x)}{2 b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))} \] Output:
4*a^4*(a^2-2*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b^2/(a^ 2-b^2)^(7/2)/d-a^4*(2*a^2+b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^( 1/2))/b^2/(a^2-b^2)^(7/2)/d-2*a^2*(a^4-3*a^2*b^2+6*b^4)*arctan((b+a*tan(1/ 2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b^2/(a^2-b^2)^(7/2)/d+1/2*cos(d*x+c)/(a+b)^ 3/d/(1-sin(d*x+c))-1/2*cos(d*x+c)/(a-b)^3/d/(1+sin(d*x+c))-1/2*a^4*cos(d*x +c)/b/(a^2-b^2)^2/d/(a+b*sin(d*x+c))^2-3/2*a^5*cos(d*x+c)/b/(a^2-b^2)^3/d/ (a+b*sin(d*x+c))+2*a^3*(a^2-2*b^2)*cos(d*x+c)/b/(a^2-b^2)^3/d/(a+b*sin(d*x +c))
Time = 4.48 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.50 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {-\frac {6 a^2 \left (a^2+4 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {2}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2}{(a-b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )+\frac {a^3 \cos (c+d x) \left (-7 a b+\left (a^2-8 b^2\right ) \sin (c+d x)\right )}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))^2}}{2 d} \] Input:
Integrate[(Sin[c + d*x]^2*Tan[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]
Output:
((-6*a^2*(a^2 + 4*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/( a^2 - b^2)^(7/2) + Sin[(c + d*x)/2]*(2/((a + b)^3*(Cos[(c + d*x)/2] - Sin[ (c + d*x)/2])) + 2/((a - b)^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) + (a ^3*Cos[c + d*x]*(-7*a*b + (a^2 - 8*b^2)*Sin[c + d*x]))/((a - b)^3*(a + b)^ 3*(a + b*Sin[c + d*x])^2))/(2*d)
Time = 0.95 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3376, 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.
\(\displaystyle \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^4}{\cos (c+d x)^2 (a+b \sin (c+d x))^3}dx\) |
\(\Big \downarrow \) 3376 |
\(\displaystyle \int \left (\frac {a^4}{b^2 \left (b^2-a^2\right ) (a+b \sin (c+d x))^3}+\frac {a^6-3 a^4 b^2+6 a^2 b^4}{b^2 \left (b^2-a^2\right )^3 (a+b \sin (c+d x))}-\frac {2 \left (2 a^3 b^2-a^5\right )}{b^2 \left (b^2-a^2\right )^2 (a+b \sin (c+d x))^2}-\frac {1}{2 (a+b)^3 (\sin (c+d x)-1)}+\frac {1}{2 (a-b)^3 (\sin (c+d x)+1)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 a^5 \cos (c+d x)}{2 b d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {a^4 \left (2 a^2+b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}+\frac {4 a^4 \left (a^2-2 b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}-\frac {2 a^2 \left (a^4-3 a^2 b^2+6 b^4\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}-\frac {a^4 \cos (c+d x)}{2 b d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^3 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 d (a-b)^3 (\sin (c+d x)+1)}\) |
Input:
Int[(Sin[c + d*x]^2*Tan[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]
Output:
(4*a^4*(a^2 - 2*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^ 2*(a^2 - b^2)^(7/2)*d) - (a^4*(2*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2] )/Sqrt[a^2 - b^2]])/(b^2*(a^2 - b^2)^(7/2)*d) - (2*a^2*(a^4 - 3*a^2*b^2 + 6*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^2*(a^2 - b^2)^ (7/2)*d) + Cos[c + d*x]/(2*(a + b)^3*d*(1 - Sin[c + d*x])) - Cos[c + d*x]/ (2*(a - b)^3*d*(1 + Sin[c + d*x])) - (a^4*Cos[c + d*x])/(2*b*(a^2 - b^2)^2 *d*(a + b*Sin[c + d*x])^2) - (3*a^5*Cos[c + d*x])/(2*b*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])) + (2*a^3*(a^2 - 2*b^2)*Cos[c + d*x])/(b*(a^2 - b^2)^3*d *(a + b*Sin[c + d*x]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(d*sin[ e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x] /; Fr eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && ( LtQ[m, -1] || (EqQ[m, -1] && GtQ[p, 0]))
Time = 2.29 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.57
method | result | size |
derivativedivides | \(\frac {-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 a^{2} \left (\frac {\frac {a \left (a^{2}+6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2}+\frac {7 b \left (a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-\frac {a \left (a^{2}-22 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {7 a^{2} b}{2}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {3 \left (a^{2}+4 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(222\) |
default | \(\frac {-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 a^{2} \left (\frac {\frac {a \left (a^{2}+6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2}+\frac {7 b \left (a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-\frac {a \left (a^{2}-22 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {7 a^{2} b}{2}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {3 \left (a^{2}+4 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(222\) |
risch | \(-\frac {i \left (-30 i a^{4} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+2 i b^{7} {\mathrm e}^{i \left (d x +c \right )}+2 i b^{7} {\mathrm e}^{5 i \left (d x +c \right )}+4 i a^{2} b^{5} {\mathrm e}^{3 i \left (d x +c \right )}+9 i a^{4} b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+6 i a^{2} b^{5} {\mathrm e}^{5 i \left (d x +c \right )}-2 i a^{6} b \,{\mathrm e}^{5 i \left (d x +c \right )}+2 a^{7} {\mathrm e}^{4 i \left (d x +c \right )}-15 a^{5} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-30 a^{3} b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-2 a \,b^{6} {\mathrm e}^{4 i \left (d x +c \right )}-4 i b^{7} {\mathrm e}^{3 i \left (d x +c \right )}-31 i a^{4} b^{3} {\mathrm e}^{i \left (d x +c \right )}+2 i a^{6} b \,{\mathrm e}^{i \left (d x +c \right )}-18 i a^{2} b^{5} {\mathrm e}^{i \left (d x +c \right )}+2 a^{7} {\mathrm e}^{2 i \left (d x +c \right )}-24 a^{5} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-4 a^{3} b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-4 a \,b^{6} {\mathrm e}^{2 i \left (d x +c \right )}-a^{5} b^{2}+10 a^{3} b^{4}+6 a \,b^{6}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} b^{2} d \left (a^{2}-b^{2}\right )^{3}}-\frac {3 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 )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {6 a^{2} \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 ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {3 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 )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {6 a^{2} \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 ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\) | \(731\) |
Input:
int(sin(d*x+c)^2*tan(d*x+c)^2/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)-2*a^2/(a-b)^3/(a+b)^3*((1/2*a*(a^2+ 6*b^2)*tan(1/2*d*x+1/2*c)^3+7/2*b*(a^2+2*b^2)*tan(1/2*d*x+1/2*c)^2-1/2*a*( a^2-22*b^2)*tan(1/2*d*x+1/2*c)+7/2*a^2*b)/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan( 1/2*d*x+1/2*c)+a)^2+3/2*(a^2+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)))-1/(a-b)^3/(tan(1/2*d*x+1/2*c)+1))
Time = 0.13 (sec) , antiderivative size = 906, normalized size of antiderivative = 2.34 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="frica s")
Output:
[1/4*(4*a^6*b - 12*a^4*b^3 + 12*a^2*b^5 - 4*b^7 + 2*(11*a^6*b - 5*a^4*b^3 - 8*a^2*b^5 + 2*b^7)*cos(d*x + c)^2 + 3*((a^4*b^2 + 4*a^2*b^4)*cos(d*x + c )^3 - 2*(a^5*b + 4*a^3*b^3)*cos(d*x + c)*sin(d*x + c) - (a^6 + 5*a^4*b^2 + 4*a^2*b^4)*cos(d*x + c))*sqrt(-a^2 + b^2)*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*c os(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*(2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 + (a^7 - 11*a^5*b ^2 + 4*a^3*b^4 + 6*a*b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b^2 - 4*a^6* b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*b^ 3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) - (a^10 - 3 *a^8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10)*d*cos(d*x + c)), 1/2* (2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 + (11*a^6*b - 5*a^4*b^3 - 8*a^2*b ^5 + 2*b^7)*cos(d*x + c)^2 + 3*((a^4*b^2 + 4*a^2*b^4)*cos(d*x + c)^3 - 2*( a^5*b + 4*a^3*b^3)*cos(d*x + c)*sin(d*x + c) - (a^6 + 5*a^4*b^2 + 4*a^2*b^ 4)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - (2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 + (a^7 - 11 *a^5*b^2 + 4*a^3*b^4 + 6*a*b^6)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d*x + c)^3 - 2*(a^9*b - 4* a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) - (a^ 10 - 3*a^8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10)*d*cos(d*x + ...
\[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate(sin(d*x+c)**2*tan(d*x+c)**2/(a+b*sin(d*x+c))**3,x)
Output:
Integral(sin(c + d*x)**2*tan(c + d*x)**2/(a + b*sin(c + d*x))**3, x)
Exception generated. \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="maxim a")
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 = 0.56 (sec) , antiderivative size = 351, normalized size of antiderivative = 0.90 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {3 \, {\left (a^{4} + 4 \, a^{2} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2} b - b^{3}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}} + \frac {a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 14 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 22 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, a^{4} b}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}}}{d} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="giac" )
Output:
-(3*(a^4 + 4*a^2*b^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^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + 2*(a^3*tan(1/2*d*x + 1/2*c) + 3*a*b^2*tan(1/2*d *x + 1/2*c) - 3*a^2*b - b^3)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(tan(1/2 *d*x + 1/2*c)^2 - 1)) + (a^5*tan(1/2*d*x + 1/2*c)^3 + 6*a^3*b^2*tan(1/2*d* x + 1/2*c)^3 + 7*a^4*b*tan(1/2*d*x + 1/2*c)^2 + 14*a^2*b^3*tan(1/2*d*x + 1 /2*c)^2 - a^5*tan(1/2*d*x + 1/2*c) + 22*a^3*b^2*tan(1/2*d*x + 1/2*c) + 7*a ^4*b)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b *tan(1/2*d*x + 1/2*c) + a)^2))/d
Time = 22.27 (sec) , antiderivative size = 585, normalized size of antiderivative = 1.51 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {13\,a^4\,b+2\,a^2\,b^3}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^5+6\,a^3\,b^2+8\,a\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^4\,b+9\,a^2\,b^3+4\,b^5\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-3\,a^4+40\,a^2\,b^2+8\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (a^2+4\,b^2\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {9\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2+4\,b^2\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-a^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2+4\,b^2\right )+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {3\,a^2\,\mathrm {atan}\left (\frac {\frac {3\,a^2\,\left (a^2+4\,b^2\right )\,\left (2\,a^6\,b-6\,a^4\,b^3+6\,a^2\,b^5-2\,b^7\right )}{2\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}+\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+4\,b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}}{3\,a^4+12\,a^2\,b^2}\right )\,\left (a^2+4\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \] Input:
int((sin(c + d*x)^2*tan(c + d*x)^2)/(a + b*sin(c + d*x))^3,x)
Output:
((13*a^4*b + 2*a^2*b^3)/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (2*tan(c/2 + (d*x)/2)^3*(8*a*b^4 + a^5 + 6*a^3*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^ 2) + (2*tan(c/2 + (d*x)/2)^2*(2*a^4*b + 4*b^5 + 9*a^2*b^3))/(a^6 - b^6 + 3 *a^2*b^4 - 3*a^4*b^2) + (a*tan(c/2 + (d*x)/2)*(8*b^4 - 3*a^4 + 40*a^2*b^2) )/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (3*a^3*tan(c/2 + (d*x)/2)^5*(a^2 + 4*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (9*a^2*b*tan(c/2 + (d*x)/2) ^4*(a^2 + 4*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/(d*(a^2*tan(c/2 + ( d*x)/2)^6 - a^2 - tan(c/2 + (d*x)/2)^2*(a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^ 4*(a^2 + 4*b^2) + 4*a*b*tan(c/2 + (d*x)/2)^5 - 4*a*b*tan(c/2 + (d*x)/2))) - (3*a^2*atan(((3*a^2*(a^2 + 4*b^2)*(2*a^6*b - 2*b^7 + 6*a^2*b^5 - 6*a^4*b ^3))/(2*(a + b)^(7/2)*(a - b)^(7/2)) + (3*a^3*tan(c/2 + (d*x)/2)*(a^2 + 4* b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/((a + b)^(7/2)*(a - b)^(7/2)))/( 3*a^4 + 12*a^2*b^2))*(a^2 + 4*b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2))
Time = 0.22 (sec) , antiderivative size = 938, normalized size of antiderivative = 2.42 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(sin(d*x+c)^2*tan(d*x+c)^2/(a+b*sin(d*x+c))^3,x)
Output:
( - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))* cos(c + d*x)*sin(c + d*x)**2*a**4*b**3 - 48*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**2*b** 5 - 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))* cos(c + d*x)*sin(c + d*x)*a**5*b**2 - 96*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)*a**3*b**4 - 12 *sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**6*b - 48*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a* *2 - b**2))*cos(c + d*x)*a**4*b**3 + 3*cos(c + d*x)*sin(c + d*x)**2*a**6*b **2 - 17*cos(c + d*x)*sin(c + d*x)**2*a**4*b**4 + 10*cos(c + d*x)*sin(c + d*x)**2*a**2*b**6 + 4*cos(c + d*x)*sin(c + d*x)**2*b**8 + 6*cos(c + d*x)*s in(c + d*x)*a**7*b - 34*cos(c + d*x)*sin(c + d*x)*a**5*b**3 + 20*cos(c + d *x)*sin(c + d*x)*a**3*b**5 + 8*cos(c + d*x)*sin(c + d*x)*a*b**7 + 3*cos(c + d*x)*a**8 - 17*cos(c + d*x)*a**6*b**2 + 10*cos(c + d*x)*a**4*b**4 + 4*co s(c + d*x)*a**2*b**6 - 2*sin(c + d*x)**3*a**7*b + 22*sin(c + d*x)**3*a**5* b**3 - 8*sin(c + d*x)**3*a**3*b**5 - 12*sin(c + d*x)**3*a*b**7 + 22*sin(c + d*x)**2*a**6*b**2 - 10*sin(c + d*x)**2*a**4*b**4 - 16*sin(c + d*x)**2*a* *2*b**6 + 4*sin(c + d*x)**2*b**8 + 6*sin(c + d*x)*a**7*b - 34*sin(c + d*x) *a**5*b**3 + 20*sin(c + d*x)*a**3*b**5 + 8*sin(c + d*x)*a*b**7 - 26*a**6*b **2 + 22*a**4*b**4 + 4*a**2*b**6)/(4*cos(c + d*x)*b*d*(sin(c + d*x)**2*...