Integrand size = 27, antiderivative size = 366 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {2 a^3 \left (a^2-3 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{7/2} d}+\frac {a^3 \left (2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{7/2} d}+\frac {2 a b \left (a^2+3 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}+\frac {a^3 \cos (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac {3 a^4 \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {a^2 \left (a^2-3 b^2\right ) \cos (c+d x)}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))} \] Output:
-2*a^3*(a^2-3*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b/(a^2 -b^2)^(7/2)/d+a^3*(2*a^2+b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1 /2))/b/(a^2-b^2)^(7/2)/d+2*a*b*(a^2+3*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c)) /(a^2-b^2)^(1/2))/(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^3*cos(d*x+c)/(a^2-b^2)^2/d /(a+b*sin(d*x+c))^2+3/2*a^4*cos(d*x+c)/(a^2-b^2)^3/d/(a+b*sin(d*x+c))-a^2* (a^2-3*b^2)*cos(d*x+c)/(a^2-b^2)^3/d/(a+b*sin(d*x+c))
Time = 3.90 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.56 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {6 a b \left (3 a^2+2 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {2}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {2}{(a-b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )+\frac {a^2 \cos (c+d x) \left (2 a^3+5 a b^2+b \left (a^2+6 b^2\right ) \sin (c+d x)\right )}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))^2}}{2 d} \] Input:
Integrate[(Sin[c + d*x]*Tan[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]
Output:
((6*a*b*(3*a^2 + 2*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^2*Cos[c + d*x]*(2*a^3 + 5*a*b^2 + b*(a^2 + 6*b^2)*Sin[c + d*x]))/((a - b )^3*(a + b)^3*(a + b*Sin[c + d*x])^2))/(2*d)
Time = 0.85 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
\(\displaystyle \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3}{\cos (c+d x)^2 (a+b \sin (c+d x))^3}dx\) |
\(\Big \downarrow \) 3376 |
\(\displaystyle \int \left (\frac {3 a^2 b^2-a^4}{b \left (b^2-a^2\right )^2 (a+b \sin (c+d x))^2}+\frac {a^3}{b \left (a^2-b^2\right ) (a+b \sin (c+d x))^3}-\frac {b \left (a^3+3 a b^2\right )}{\left (b^2-a^2\right )^3 (a+b \sin (c+d x))}-\frac {1}{2 (a+b)^3 (\sin (c+d x)-1)}-\frac {1}{2 (a-b)^3 (\sin (c+d x)+1)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 a b \left (a^2+3 b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}-\frac {a^2 \left (a^2-3 b^2\right ) \cos (c+d x)}{d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {3 a^4 \cos (c+d x)}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {a^3 \left (2 a^2+b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{7/2}}-\frac {2 a^3 \left (a^2-3 b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{7/2}}+\frac {a^3 \cos (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {\cos (c+d x)}{2 d (a+b)^3 (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a-b)^3 (\sin (c+d x)+1)}\) |
Input:
Int[(Sin[c + d*x]*Tan[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]
Output:
(-2*a^3*(a^2 - 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b *(a^2 - b^2)^(7/2)*d) + (a^3*(2*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2]) /Sqrt[a^2 - b^2]])/(b*(a^2 - b^2)^(7/2)*d) + (2*a*b*(a^2 + 3*b^2)*ArcTan[( b + a*Tan[(c + d*x)/2])/Sqrt[a^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^3*Cos[c + d*x])/(2*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x ])^2) + (3*a^4*Cos[c + d*x])/(2*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])) - (a ^2*(a^2 - 3*b^2)*Cos[c + d*x])/((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 = 2.24 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.66
method | result | size |
derivativedivides | \(\frac {-\frac {2 a \left (\frac {-\frac {a b \left (3 a^{2}+4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2}+\left (-a^{4}-5 b^{4}-\frac {9}{2} a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {b a \left (5 a^{2}+16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-a^{4}-\frac {5 a^{2} b^{2}}{2}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}-\frac {3 b \left (3 a^{2}+2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(243\) |
default | \(\frac {-\frac {2 a \left (\frac {-\frac {a b \left (3 a^{2}+4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2}+\left (-a^{4}-5 b^{4}-\frac {9}{2} a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {b a \left (5 a^{2}+16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-a^{4}-\frac {5 a^{2} b^{2}}{2}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}-\frac {3 b \left (3 a^{2}+2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(243\) |
risch | \(-\frac {i \left (-9 i a^{3} b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-6 i a \,b^{5} {\mathrm e}^{5 i \left (d x +c \right )}+39 i a^{3} b^{3} {\mathrm e}^{i \left (d x +c \right )}+4 i a \,b^{5} {\mathrm e}^{3 i \left (d x +c \right )}+12 i a^{5} b \,{\mathrm e}^{3 i \left (d x +c \right )}+2 a^{6} {\mathrm e}^{4 i \left (d x +c \right )}+21 a^{4} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+24 a^{2} b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-2 b^{6} {\mathrm e}^{4 i \left (d x +c \right )}+2 i a \,b^{5} {\mathrm e}^{i \left (d x +c \right )}+14 i a^{3} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+4 i a^{5} b \,{\mathrm e}^{i \left (d x +c \right )}+2 a^{6} {\mathrm e}^{2 i \left (d x +c \right )}+28 a^{4} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-4 a^{2} b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+4 b^{6} {\mathrm e}^{2 i \left (d x +c \right )}-a^{4} b^{2}-12 a^{2} b^{4}-2 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 d \left (a^{2}-b^{2}\right )^{3}}+\frac {9 i b \,a^{3} \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 )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {3 i b^{3} a \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 )^{3} \left (a -b \right )^{3} d}-\frac {9 i b \,a^{3} \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 )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {3 i b^{3} a \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 )^{3} \left (a -b \right )^{3} d}\) | \(679\) |
Input:
int(sin(d*x+c)*tan(d*x+c)^2/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(-2*a/(a-b)^3/(a+b)^3*((-1/2*a*b*(3*a^2+4*b^2)*tan(1/2*d*x+1/2*c)^3+(- a^4-5*b^4-9/2*a^2*b^2)*tan(1/2*d*x+1/2*c)^2-1/2*b*a*(5*a^2+16*b^2)*tan(1/2 *d*x+1/2*c)-a^4-5/2*a^2*b^2)/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c )+a)^2-3/2*b*(3*a^2+2*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)+1/(a-b)^3/(tan (1/2*d*x+1/2*c)+1))
Time = 0.13 (sec) , antiderivative size = 917, normalized size of antiderivative = 2.51 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(sin(d*x+c)*tan(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="fricas" )
Output:
[-1/4*(4*a^7 - 12*a^5*b^2 + 12*a^3*b^4 - 4*a*b^6 + 2*(2*a^7 + 13*a^5*b^2 - 17*a^3*b^4 + 2*a*b^6)*cos(d*x + c)^2 - 3*((3*a^3*b^3 + 2*a*b^5)*cos(d*x + c)^3 - 2*(3*a^4*b^2 + 2*a^2*b^4)*cos(d*x + c)*sin(d*x + c) - (3*a^5*b + 5 *a^3*b^3 + 2*a*b^5)*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*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*(2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 - (a^6*b + 11*a^4*b^3 - 10*a^2*b^5 - 2*b^7)*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^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 + (2*a^7 + 13*a^5*b^2 - 17*a^3*b^4 + 2*a*b^6)*cos(d*x + c)^2 + 3*((3*a^3*b^3 + 2*a*b^5)*cos(d*x + c)^3 - 2*(3*a^4*b^2 + 2*a^2*b^4)*cos(d*x + c)*sin(d*x + c) - (3*a^5*b + 5*a^3*b^3 + 2*a*b^5)*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^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 - (a^6*b + 11*a^4*b^3 - 10*a^2*b^5 - 2*b^7)*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 +...
\[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int \frac {\sin {\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate(sin(d*x+c)*tan(d*x+c)**2/(a+b*sin(d*x+c))**3,x)
Output:
Integral(sin(c + d*x)*tan(c + d*x)**2/(a + b*sin(c + d*x))**3, x)
Exception generated. \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sin(d*x+c)*tan(d*x+c)^2/(a+b*sin(d*x+c))^3,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 = 0.56 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.03 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {3 \, {\left (3 \, a^{3} b + 2 \, a b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3} - 3 \, a b^{2}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}} + \frac {3 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 10 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 16 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a^{5} + 5 \, a^{3} b^{2}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}}}{d} \] Input:
integrate(sin(d*x+c)*tan(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="giac")
Output:
(3*(3*a^3*b + 2*a*b^3)*(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*(3*a^2*b*tan(1/2*d*x + 1/2*c) + b^3*tan(1/2* d*x + 1/2*c) - a^3 - 3*a*b^2)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(tan(1/ 2*d*x + 1/2*c)^2 - 1)) + (3*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 4*a^2*b^3*tan(1 /2*d*x + 1/2*c)^3 + 2*a^5*tan(1/2*d*x + 1/2*c)^2 + 9*a^3*b^2*tan(1/2*d*x + 1/2*c)^2 + 10*a*b^4*tan(1/2*d*x + 1/2*c)^2 + 5*a^4*b*tan(1/2*d*x + 1/2*c) + 16*a^2*b^3*tan(1/2*d*x + 1/2*c) + 2*a^5 + 5*a^3*b^2)/((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 = 21.88 (sec) , antiderivative size = 583, normalized size of antiderivative = 1.59 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {3\,a\,b\,\mathrm {atan}\left (\frac {\frac {3\,a\,b\,\left (3\,a^2+2\,b^2\right )\,\left (2\,a^6\,b-6\,a^4\,b^3+6\,a^2\,b^5-2\,b^7\right )}{2\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}+\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2+2\,b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}}{9\,a^3\,b+6\,a\,b^3}\right )\,\left (3\,a^2+2\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {4\,a^5+11\,a^3\,b^2}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a^3\,b^2+2\,a\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^4\,b+8\,a^2\,b^3+4\,b^5\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^5+13\,a\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (7\,a^4+38\,a^2\,b^2\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {3\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (3\,a^2+2\,b^2\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-a^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2+4\,b^2\right )+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \] Input:
int((sin(c + d*x)*tan(c + d*x)^2)/(a + b*sin(c + d*x))^3,x)
Output:
(3*a*b*atan(((3*a*b*(3*a^2 + 2*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^2*b*tan(c/2 + (d*x)/2)*(3*a^2 + 2*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/((a + b)^(7/2)*(a - b)^(7/2) ))/(6*a*b^3 + 9*a^3*b))*(3*a^2 + 2*b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2)) - ((4*a^5 + 11*a^3*b^2)/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (9*tan(c/2 + (d*x)/2)^4*(2*a*b^4 + 3*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)^3*(3*a^4*b + 4*b^5 + 8*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)^2*(13*a*b^4 + 2*a^5))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) + (b*tan(c/2 + (d*x)/2)*(7*a^4 + 38*a^2*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (3*a^2*b*tan(c/2 + (d*x)/2)^5*(3*a^2 + 2* 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)))
Time = 0.19 (sec) , antiderivative size = 862, normalized size of antiderivative = 2.36 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(sin(d*x+c)*tan(d*x+c)^2/(a+b*sin(d*x+c))^3,x)
Output:
(36*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**3*b**3 + 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)**2*a*b**5 + 72 *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**4*b**2 + 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)*a**2*b**4 + 36*sqrt( a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x) *a**5*b + 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b **2))*cos(c + d*x)*a**3*b**3 - cos(c + d*x)*sin(c + d*x)**2*a**5*b**2 + 17 *cos(c + d*x)*sin(c + d*x)**2*a**3*b**4 - 16*cos(c + d*x)*sin(c + d*x)**2* a*b**6 - 2*cos(c + d*x)*sin(c + d*x)*a**6*b + 34*cos(c + d*x)*sin(c + d*x) *a**4*b**3 - 32*cos(c + d*x)*sin(c + d*x)*a**2*b**5 - cos(c + d*x)*a**7 + 17*cos(c + d*x)*a**5*b**2 - 16*cos(c + d*x)*a**3*b**4 - 2*sin(c + d*x)**3* a**6*b - 22*sin(c + d*x)**3*a**4*b**3 + 20*sin(c + d*x)**3*a**2*b**5 + 4*s in(c + d*x)**3*b**7 - 4*sin(c + d*x)**2*a**7 - 26*sin(c + d*x)**2*a**5*b** 2 + 34*sin(c + d*x)**2*a**3*b**4 - 4*sin(c + d*x)**2*a*b**6 - 2*sin(c + d* x)*a**6*b + 34*sin(c + d*x)*a**4*b**3 - 32*sin(c + d*x)*a**2*b**5 + 8*a**7 + 14*a**5*b**2 - 22*a**3*b**4)/(4*cos(c + d*x)*d*(sin(c + d*x)**2*a**8*b* *2 - 4*sin(c + d*x)**2*a**6*b**4 + 6*sin(c + d*x)**2*a**4*b**6 - 4*sin(c + d*x)**2*a**2*b**8 + sin(c + d*x)**2*b**10 + 2*sin(c + d*x)*a**9*b - 8*...