Integrand size = 29, antiderivative size = 220 \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 b^6 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{5/2} d}+\frac {b \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cot (c+d x)}{a d}+\frac {b \left (-a^2+2 b^2\right ) \sec (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {b \sec ^3(c+d x) (-a+b \sin (c+d x))}{3 a \left (a^2-b^2\right ) d}+\frac {\left (6 a^4-10 a^2 b^2+b^4\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right )^2 d}+\frac {\tan ^3(c+d x)}{3 a d} \] Output:
2*b^6*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^2/(a^2-b^2)^(5/2) /d+b*arctanh(cos(d*x+c))/a^2/d-cot(d*x+c)/a/d+b*(-a^2+2*b^2)*sec(d*x+c)/(a ^2-b^2)^2/d+1/3*b*sec(d*x+c)^3*(-a+b*sin(d*x+c))/a/(a^2-b^2)/d+1/3*(6*a^4- 10*a^2*b^2+b^4)*tan(d*x+c)/a/(a^2-b^2)^2/d+1/3*tan(d*x+c)^3/a/d
Leaf count is larger than twice the leaf count of optimal. \(450\) vs. \(2(220)=440\).
Time = 7.32 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.05 \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 b^6 \arctan \left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (b \cos \left (\frac {1}{2} (c+d x)\right )+a \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{5/2} d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{2 a d}+\frac {b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 d}-\frac {b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 d}+\frac {1}{12 (a+b) d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{6 (a+b) d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{6 (a-b) d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {1}{12 (a-b) d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {10 a \sin \left (\frac {1}{2} (c+d x)\right )-13 b \sin \left (\frac {1}{2} (c+d x)\right )}{6 (a-b)^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {10 a \sin \left (\frac {1}{2} (c+d x)\right )+13 b \sin \left (\frac {1}{2} (c+d x)\right )}{6 (a+b)^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{2 a d} \] Input:
Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^4)/(a + b*Sin[c + d*x]),x]
Output:
(2*b^6*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2])) /Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(5/2)*d) - Cot[(c + d*x)/2]/(2*a*d) + (b*Log[Cos[(c + d*x)/2]])/(a^2*d) - (b*Log[Sin[(c + d*x)/2]])/(a^2*d) + 1/ (12*(a + b)*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + Sin[(c + d*x)/2]/ (6*(a + b)*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + Sin[(c + d*x)/2]/( 6*(a - b)*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - 1/(12*(a - b)*d*(Co s[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (10*a*Sin[(c + d*x)/2] - 13*b*Sin[ (c + d*x)/2])/(6*(a - b)^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (10* a*Sin[(c + d*x)/2] + 13*b*Sin[(c + d*x)/2])/(6*(a + b)^2*d*(Cos[(c + d*x)/ 2] - Sin[(c + d*x)/2])) + Tan[(c + d*x)/2]/(2*a*d)
Time = 0.69 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3377, 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.
| \(\displaystyle \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^2 \cos (c+d x)^4 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (\frac {b^2 \sec ^4(c+d x)}{a^2 (a+b \sin (c+d x))}-\frac {b \csc (c+d x) \sec ^4(c+d x)}{a^2}+\frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 b^6 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{5/2}}+\frac {b \text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {b^2 \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a^2 d \left (a^2-b^2\right )}+\frac {b^2 \sec (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)+3 b^3\right )}{3 a^2 d \left (a^2-b^2\right )^2}-\frac {b \sec ^3(c+d x)}{3 a^2 d}-\frac {b \sec (c+d x)}{a^2 d}+\frac {\tan ^3(c+d x)}{3 a d}+\frac {2 \tan (c+d x)}{a d}-\frac {\cot (c+d x)}{a d}\) |
Input:
Int[(Csc[c + d*x]^2*Sec[c + d*x]^4)/(a + b*Sin[c + d*x]),x]
Output:
(2*b^6*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^ (5/2)*d) + (b*ArcTanh[Cos[c + d*x]])/(a^2*d) - Cot[c + d*x]/(a*d) - (b*Sec [c + d*x])/(a^2*d) - (b*Sec[c + d*x]^3)/(3*a^2*d) - (b^2*Sec[c + d*x]^3*(b - a*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)*d) + (b^2*Sec[c + d*x]*(3*b^3 + a*( 2*a^2 - 5*b^2)*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)^2*d) + (2*Tan[c + d*x])/( a*d) + Tan[c + d*x]^3/(3*a*d)
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a _) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/ 2, 0])
Time = 1.04 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.15
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {4 a +5 b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{3 \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {4 a -5 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{3 \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {2 b^{6} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{2} \left (a +b \right )^{2} \left (a -b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\) | \(253\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {4 a +5 b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{3 \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {4 a -5 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{3 \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {2 b^{6} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{2} \left (a +b \right )^{2} \left (a -b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}}{d}\) | \(253\) |
| risch | \(-\frac {2 \left (-14 i a^{2} b^{2}+3 a^{3} b \,{\mathrm e}^{7 i \left (d x +c \right )}-6 b^{3} a \,{\mathrm e}^{7 i \left (d x +c \right )}+8 i a^{4}+3 i b^{4} {\mathrm e}^{6 i \left (d x +c \right )}+7 a^{3} {\mathrm e}^{5 i \left (d x +c \right )} b -10 b^{3} a \,{\mathrm e}^{5 i \left (d x +c \right )}-6 i a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 i b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+3 i b^{4}-7 b \,a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+10 b^{3} a \,{\mathrm e}^{3 i \left (d x +c \right )}-28 i a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+16 i a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+9 i b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-3 b \,a^{3} {\mathrm e}^{i \left (d x +c \right )}+6 b^{3} a \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left (a^{2}-b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{2} d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{2} d}+\frac {b^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{2}}-\frac {b^{6} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{2}}\) | \(493\) |
Input:
int(csc(d*x+c)^2*sec(d*x+c)^4/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/2*tan(1/2*d*x+1/2*c)/a-1/2*(4*a+5*b)/(a+b)^2/(tan(1/2*d*x+1/2*c)-1) -1/3/(a+b)/(tan(1/2*d*x+1/2*c)-1)^3-1/2/(a+b)/(tan(1/2*d*x+1/2*c)-1)^2-1/2 *(4*a-5*b)/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)-1/3/(a-b)/(tan(1/2*d*x+1/2*c)+1) ^3+1/2/(a-b)/(tan(1/2*d*x+1/2*c)+1)^2+2/a^2/(a+b)^2/(a-b)^2*b^6/(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))-1/2/a/tan(1 /2*d*x+1/2*c)-1/a^2*b*ln(tan(1/2*d*x+1/2*c)))
Time = 0.46 (sec) , antiderivative size = 831, normalized size of antiderivative = 3.78 \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
[-1/6*(3*sqrt(-a^2 + b^2)*b^6*cos(d*x + c)^3*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))*sin(d*x + c) - 2*a^7 + 4*a^5*b^2 - 2*a^3*b^4 - 3*(a^6*b - 3* a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d*x + c)^3*log(1/2*cos(d*x + c) + 1/2)*sin( d*x + c) + 3*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d*x + c)^3*log(-1/2 *cos(d*x + c) + 1/2)*sin(d*x + c) + 2*(8*a^7 - 22*a^5*b^2 + 17*a^3*b^4 - 3 *a*b^6)*cos(d*x + c)^4 - 2*(4*a^7 - 11*a^5*b^2 + 7*a^3*b^4)*cos(d*x + c)^2 + 2*(a^6*b - 2*a^4*b^3 + a^2*b^5 + 3*(a^6*b - 3*a^4*b^3 + 2*a^2*b^5)*cos( d*x + c)^2)*sin(d*x + c))/((a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*cos(d *x + c)^3*sin(d*x + c)), -1/6*(6*sqrt(a^2 - b^2)*b^6*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*cos(d*x + c)^3*sin(d*x + c) - 2*a^ 7 + 4*a^5*b^2 - 2*a^3*b^4 - 3*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d* x + c)^3*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 3*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d*x + c)^3*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 2*(8*a^7 - 22*a^5*b^2 + 17*a^3*b^4 - 3*a*b^6)*cos(d*x + c)^4 - 2*(4*a^7 - 11*a^5*b^2 + 7*a^3*b^4)*cos(d*x + c)^2 + 2*(a^6*b - 2*a^4*b^3 + a^2*b^5 + 3*(a^6*b - 3*a^4*b^3 + 2*a^2*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*cos(d*x + c)^3*sin(d*x + c))]
Timed out. \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(csc(d*x+c)**2*sec(d*x+c)**4/(a+b*sin(d*x+c)),x)
Output:
Timed out
Exception generated. \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^4/(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 = 0.18 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.62 \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {12 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{6}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {6 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} + \frac {3 \, {\left (2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {4 \, {\left (6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 9 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 14 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{2} b + 7 \, b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
1/6*(12*(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)))*b^6/((a^6 - 2*a^4*b^2 + a^2*b^4)*sqrt(a^2 - b^2)) - 6*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 3*tan(1/2*d*x + 1/2*c)/a + 3*(2*b*tan(1/2*d*x + 1/2*c) - a)/(a^2*tan(1/2*d*x + 1/2*c)) - 4*(6*a^3* tan(1/2*d*x + 1/2*c)^5 - 9*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*a^2*b*tan(1/2* d*x + 1/2*c)^4 + 9*b^3*tan(1/2*d*x + 1/2*c)^4 - 8*a^3*tan(1/2*d*x + 1/2*c) ^3 + 14*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 12 *b^3*tan(1/2*d*x + 1/2*c)^2 + 6*a^3*tan(1/2*d*x + 1/2*c) - 9*a*b^2*tan(1/2 *d*x + 1/2*c) - 4*a^2*b + 7*b^3)/((a^4 - 2*a^2*b^2 + b^4)*(tan(1/2*d*x + 1 /2*c)^2 - 1)^3))/d
Time = 23.88 (sec) , antiderivative size = 2317, normalized size of antiderivative = 10.53 \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
int(1/(cos(c + d*x)^4*sin(c + d*x)^2*(a + b*sin(c + d*x))),x)
Output:
(a*((3*b^10)/8 + (b^10*cos(2*c + 2*d*x))/2 + (b^10*cos(4*c + 4*d*x))/8) - a^10*((7*b*sin(c + d*x))/12 + (b*sin(2*c + 2*d*x))/3 + (b*sin(3*c + 3*d*x) )/4 + (b*sin(4*c + 4*d*x))/6 + (b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2 ))*sin(2*c + 2*d*x))/4 + (b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin (4*c + 4*d*x))/8) - a^6*((17*b^5*sin(c + d*x))/4 + (11*b^5*sin(2*c + 2*d*x ))/4 + (9*b^5*sin(3*c + 3*d*x))/4 + (11*b^5*sin(4*c + 4*d*x))/8 + (5*b^5*l og(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(2*c + 2*d*x))/2 + (5*b^5*log (sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(4*c + 4*d*x))/4) - a^2*((5*b^9 *sin(c + d*x))/6 + (7*b^9*sin(2*c + 2*d*x))/12 + (b^9*sin(3*c + 3*d*x))/2 + (7*b^9*sin(4*c + 4*d*x))/24 + (5*b^9*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d *x)/2))*sin(2*c + 2*d*x))/4 + (5*b^9*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x )/2))*sin(4*c + 4*d*x))/8) + a^8*((31*b^3*sin(c + d*x))/12 + (19*b^3*sin(2 *c + 2*d*x))/12 + (5*b^3*sin(3*c + 3*d*x))/4 + (19*b^3*sin(4*c + 4*d*x))/2 4 + (5*b^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(2*c + 2*d*x))/4 + (5*b^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(4*c + 4*d*x))/8) + a^4*((37*b^7*sin(c + d*x))/12 + (25*b^7*sin(2*c + 2*d*x))/12 + (7*b^7*sin (3*c + 3*d*x))/4 + (25*b^7*sin(4*c + 4*d*x))/24 + (5*b^7*log(sin(c/2 + (d* x)/2)/cos(c/2 + (d*x)/2))*sin(2*c + 2*d*x))/2 + (5*b^7*log(sin(c/2 + (d*x) /2)/cos(c/2 + (d*x)/2))*sin(4*c + 4*d*x))/4) - a^7*((9*b^4)/8 + 6*b^4*cos( 2*c + 2*d*x) + (23*b^4*cos(4*c + 4*d*x))/8) + a^9*(b^2/4 + (19*b^2*cos(...
Time = 0.20 (sec) , antiderivative size = 770, normalized size of antiderivative = 3.50 \[ \int \frac {\csc ^2(c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(csc(d*x+c)^2*sec(d*x+c)^4/(a+b*sin(d*x+c)),x)
Output:
(6*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)**3*b**6 - 6*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)*b**6 - 3*cos(c + d*x) *log(tan((c + d*x)/2))*sin(c + d*x)**3*a**6*b + 9*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**3*a**4*b**3 - 9*cos(c + d*x)*log(tan((c + d*x)/2) )*sin(c + d*x)**3*a**2*b**5 + 3*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**3*b**7 + 3*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)*a**6*b - 9*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)*a**4*b**3 + 9*cos(c + d *x)*log(tan((c + d*x)/2))*sin(c + d*x)*a**2*b**5 - 3*cos(c + d*x)*log(tan( (c + d*x)/2))*sin(c + d*x)*b**7 + cos(c + d*x)*sin(c + d*x)**3*a**4*b**3 - cos(c + d*x)*sin(c + d*x)**3*a**2*b**5 - cos(c + d*x)*sin(c + d*x)*a**4*b **3 + cos(c + d*x)*sin(c + d*x)*a**2*b**5 + 8*sin(c + d*x)**4*a**7 - 22*si n(c + d*x)**4*a**5*b**2 + 17*sin(c + d*x)**4*a**3*b**4 - 3*sin(c + d*x)**4 *a*b**6 - 3*sin(c + d*x)**3*a**6*b + 9*sin(c + d*x)**3*a**4*b**3 - 6*sin(c + d*x)**3*a**2*b**5 - 12*sin(c + d*x)**2*a**7 + 33*sin(c + d*x)**2*a**5*b **2 - 27*sin(c + d*x)**2*a**3*b**4 + 6*sin(c + d*x)**2*a*b**6 + 4*sin(c + d*x)*a**6*b - 11*sin(c + d*x)*a**4*b**3 + 7*sin(c + d*x)*a**2*b**5 + 3*a** 7 - 9*a**5*b**2 + 9*a**3*b**4 - 3*a*b**6)/(3*cos(c + d*x)*sin(c + d*x)*a** 2*d*(sin(c + d*x)**2*a**6 - 3*sin(c + d*x)**2*a**4*b**2 + 3*sin(c + d*x)** 2*a**2*b**4 - sin(c + d*x)**2*b**6 - a**6 + 3*a**4*b**2 - 3*a**2*b**4 +...