Integrand size = 27, antiderivative size = 194 \[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 b^5 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2} d}-\frac {\text {arctanh}(\cos (c+d x))}{a d}+\frac {\sec (c+d x)}{a d}+\frac {\sec ^3(c+d x)}{3 a d}+\frac {b \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a \left (a^2-b^2\right ) d}-\frac {b \sec (c+d x) \left (3 b^3+a \left (2 a^2-5 b^2\right ) \sin (c+d x)\right )}{3 a \left (a^2-b^2\right )^2 d} \] Output:
-2*b^5*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a/(a^2-b^2)^(5/2)/ d-arctanh(cos(d*x+c))/a/d+sec(d*x+c)/a/d+1/3*sec(d*x+c)^3/a/d+1/3*b*sec(d* x+c)^3*(b-a*sin(d*x+c))/a/(a^2-b^2)/d-1/3*b*sec(d*x+c)*(3*b^3+a*(2*a^2-5*b ^2)*sin(d*x+c))/a/(a^2-b^2)^2/d
Leaf count is larger than twice the leaf count of optimal. \(408\) vs. \(2(194)=388\).
Time = 7.14 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.10 \[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 b^5 \arctan \left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (b \cos \left (\frac {1}{2} (c+d x)\right )+a \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2} d}-\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a d}+\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a d}+\frac {1}{12 (a+b) d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{6 (a+b) d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{6 (a-b) d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {1}{12 (a-b) d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {-7 a \sin \left (\frac {1}{2} (c+d x)\right )+10 b \sin \left (\frac {1}{2} (c+d x)\right )}{6 (a-b)^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {7 a \sin \left (\frac {1}{2} (c+d x)\right )+10 b \sin \left (\frac {1}{2} (c+d x)\right )}{6 (a+b)^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:
Integrate[(Csc[c + d*x]*Sec[c + d*x]^4)/(a + b*Sin[c + d*x]),x]
Output:
(-2*b^5*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2]) )/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(5/2)*d) - Log[Cos[(c + d*x)/2]]/(a*d) + Log[Sin[(c + d*x)/2]]/(a*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*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (-7*a*Sin[(c + d*x)/2] + 10*b*Sin[(c + d*x)/2])/(6*(a - b)^2*d*(Cos[(c + d *x)/2] + Sin[(c + d*x)/2])) + (7*a*Sin[(c + d*x)/2] + 10*b*Sin[(c + d*x)/2 ])/(6*(a + b)^2*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))
Time = 0.66 (sec) , antiderivative size = 194, 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, 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.
| \(\displaystyle \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x) \cos (c+d x)^4 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3377 |
| \(\displaystyle \int \left (\frac {\csc (c+d x) \sec ^4(c+d x)}{a}-\frac {b \sec ^4(c+d x)}{a (a+b \sin (c+d x))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 b^5 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a d \left (a^2-b^2\right )^{5/2}}+\frac {b \sec ^3(c+d x) (b-a \sin (c+d x))}{3 a d \left (a^2-b^2\right )}-\frac {b \sec (c+d x) \left (a \left (2 a^2-5 b^2\right ) \sin (c+d x)+3 b^3\right )}{3 a d \left (a^2-b^2\right )^2}-\frac {\text {arctanh}(\cos (c+d x))}{a d}+\frac {\sec ^3(c+d x)}{3 a d}+\frac {\sec (c+d x)}{a d}\) |
Input:
Int[(Csc[c + d*x]*Sec[c + d*x]^4)/(a + b*Sin[c + d*x]),x]
Output:
(-2*b^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^( 5/2)*d) - ArcTanh[Cos[c + d*x]]/(a*d) + Sec[c + d*x]/(a*d) + Sec[c + d*x]^ 3/(3*a*d) + (b*Sec[c + d*x]^3*(b - a*Sin[c + d*x]))/(3*a*(a^2 - b^2)*d) - (b*Sec[c + d*x]*(3*b^3 + a*(2*a^2 - 5*b^2)*Sin[c + d*x]))/(3*a*(a^2 - b^2) ^2*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 = 0.88 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.14
| method | result | size |
| derivativedivides | \(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 b^{5} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} a \sqrt {a^{2}-b^{2}}}-\frac {1}{2 \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{3 \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-3 a +4 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{3 \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3 a +4 b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\) | \(221\) |
| default | \(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 b^{5} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} a \sqrt {a^{2}-b^{2}}}-\frac {1}{2 \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{3 \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-3 a +4 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{3 \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3 a +4 b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\) | \(221\) |
| risch | \(\frac {2 i \left (-3 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}+6 i a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-10 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+16 i a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}+6 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-6 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2} b +12 \,{\mathrm e}^{2 i \left (d x +c \right )} b^{3}-2 a^{2} b +5 b^{3}\right )}{3 d \left (-a^{2}+b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {i b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d a}-\frac {i b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d a}\) | \(389\) |
Input:
int(csc(d*x+c)*sec(d*x+c)^4/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/a*ln(tan(1/2*d*x+1/2*c))-2*b^5/(a-b)^2/(a+b)^2/a/(a^2-b^2)^(1/2)*ar ctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-1/2/(a-b)/(tan(1/2* d*x+1/2*c)+1)^2+1/3/(a-b)/(tan(1/2*d*x+1/2*c)+1)^3-1/2/(a-b)^2*(-3*a+4*b)/ (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*(3*a+4*b)/(a+b)^2/(tan(1/2*d*x+1/2*c)-1))
Time = 0.46 (sec) , antiderivative size = 680, normalized size of antiderivative = 3.51 \[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
[-1/6*(3*sqrt(-a^2 + b^2)*b^5*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)) - 2*a^6 + 4*a^4*b^2 - 2*a^2*b^4 + 3*(a^6 - 3*a^4*b^2 + 3*a^ 2*b^4 - b^6)*cos(d*x + c)^3*log(1/2*cos(d*x + c) + 1/2) - 3*(a^6 - 3*a^4*b ^2 + 3*a^2*b^4 - b^6)*cos(d*x + c)^3*log(-1/2*cos(d*x + c) + 1/2) - 6*(a^6 - 3*a^4*b^2 + 2*a^2*b^4)*cos(d*x + c)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5 + (2*a^5*b - 7*a^3*b^3 + 5*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^7 - 3*a^ 5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^3), 1/6*(6*sqrt(a^2 - b^2)*b^5*a rctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*cos(d*x + c)^3 + 2*a^6 - 4*a^4*b^2 + 2*a^2*b^4 - 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*c os(d*x + c)^3*log(1/2*cos(d*x + c) + 1/2) + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos(d*x + c)^3*log(-1/2*cos(d*x + c) + 1/2) + 6*(a^6 - 3*a^4*b^2 + 2*a^2*b^4)*cos(d*x + c)^2 - 2*(a^5*b - 2*a^3*b^3 + a*b^5 + (2*a^5*b - 7*a ^3*b^3 + 5*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^7 - 3*a^5*b^2 + 3*a^3* b^4 - a*b^6)*d*cos(d*x + c)^3)]
\[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\csc {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(csc(d*x+c)*sec(d*x+c)**4/(a+b*sin(d*x+c)),x)
Output:
Integral(csc(c + d*x)*sec(c + d*x)**4/(a + b*sin(c + d*x)), x)
Exception generated. \[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(csc(d*x+c)*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 = 0.23 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.59 \[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {6 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{5}}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {2 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 9 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{3} + 7 \, a b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
-1/3*(6*(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^5/((a^5 - 2*a^3*b^2 + a*b^4)*sqrt(a^2 - b ^2)) - 3*log(abs(tan(1/2*d*x + 1/2*c)))/a - 2*(3*a^2*b*tan(1/2*d*x + 1/2*c )^5 - 6*b^3*tan(1/2*d*x + 1/2*c)^5 - 6*a^3*tan(1/2*d*x + 1/2*c)^4 + 9*a*b^ 2*tan(1/2*d*x + 1/2*c)^4 - 2*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 8*b^3*tan(1/2* d*x + 1/2*c)^3 + 6*a^3*tan(1/2*d*x + 1/2*c)^2 - 12*a*b^2*tan(1/2*d*x + 1/2 *c)^2 + 3*a^2*b*tan(1/2*d*x + 1/2*c) - 6*b^3*tan(1/2*d*x + 1/2*c) - 4*a^3 + 7*a*b^2)/((a^4 - 2*a^2*b^2 + b^4)*(tan(1/2*d*x + 1/2*c)^2 - 1)^3))/d
Time = 22.73 (sec) , antiderivative size = 2162, normalized size of antiderivative = 11.14 \[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
int(1/(cos(c + d*x)^4*sin(c + d*x)*(a + b*sin(c + d*x))),x)
Output:
-(a^5*((15*b^5*sin(c + d*x))/4 + (7*b^5*sin(3*c + 3*d*x))/4) - a^7*((9*b^3 *sin(c + d*x))/4 + (11*b^3*sin(3*c + 3*d*x))/12) - a^3*((11*b^7*sin(c + d* x))/4 + (17*b^7*sin(3*c + 3*d*x))/12) + a^9*((b*sin(c + d*x))/2 + (b*sin(3 *c + 3*d*x))/6) - a^10*(cos(c + d*x) + cos(2*c + 2*d*x)/2 + cos(3*c + 3*d* x)/3 + (3*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (lo g(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4 + 5/6) + a*(( 3*b^9*sin(c + d*x))/4 + (5*b^9*sin(3*c + 3*d*x))/12) - a^2*((7*b^8*cos(c + d*x))/4 + (4*b^8)/3 + b^8*cos(2*c + 2*d*x) + (7*b^8*cos(3*c + 3*d*x))/12 + (15*b^8*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (5* b^8*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4) - a^6* ((33*b^4*cos(c + d*x))/4 + (13*b^4)/2 + (9*b^4*cos(2*c + 2*d*x))/2 + (11*b ^4*cos(3*c + 3*d*x))/4 + (15*b^4*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c /2 + (d*x)/2)))/2 + (5*b^4*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos( 3*c + 3*d*x))/2) + a^8*((19*b^2*cos(c + d*x))/4 + (23*b^2)/6 + (5*b^2*cos( 2*c + 2*d*x))/2 + (19*b^2*cos(3*c + 3*d*x))/12 + (15*b^2*cos(c + d*x)*log( sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (5*b^2*log(sin(c/2 + (d*x)/2)/ cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/4) + a^4*((25*b^6*cos(c + d*x))/4 + (29*b^6)/6 + (7*b^6*cos(2*c + 2*d*x))/2 + (25*b^6*cos(3*c + 3*d*x))/12 + ( 15*b^6*cos(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + (5*b^6 *log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2) + (3*b...
Time = 0.19 (sec) , antiderivative size = 635, normalized size of antiderivative = 3.27 \[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(csc(d*x+c)*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))*c os(c + d*x)*sin(c + d*x)**2*b**5 + 6*sqrt(a**2 - b**2)*atan((tan((c + d*x) /2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*b**5 + 3*cos(c + d*x)*log(tan(( c + d*x)/2))*sin(c + d*x)**2*a**6 - 9*cos(c + d*x)*log(tan((c + d*x)/2))*s in(c + d*x)**2*a**4*b**2 + 9*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d* x)**2*a**2*b**4 - 3*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*b** 6 - 3*cos(c + d*x)*log(tan((c + d*x)/2))*a**6 + 9*cos(c + d*x)*log(tan((c + d*x)/2))*a**4*b**2 - 9*cos(c + d*x)*log(tan((c + d*x)/2))*a**2*b**4 + 3* cos(c + d*x)*log(tan((c + d*x)/2))*b**6 - cos(c + d*x)*sin(c + d*x)**2*a** 4*b**2 + cos(c + d*x)*sin(c + d*x)**2*a**2*b**4 + cos(c + d*x)*a**4*b**2 - cos(c + d*x)*a**2*b**4 - 2*sin(c + d*x)**3*a**5*b + 7*sin(c + d*x)**3*a** 3*b**3 - 5*sin(c + d*x)**3*a*b**5 + 3*sin(c + d*x)**2*a**6 - 9*sin(c + d*x )**2*a**4*b**2 + 6*sin(c + d*x)**2*a**2*b**4 + 3*sin(c + d*x)*a**5*b - 9*s in(c + d*x)*a**3*b**3 + 6*sin(c + d*x)*a*b**5 - 4*a**6 + 11*a**4*b**2 - 7* a**2*b**4)/(3*cos(c + d*x)*a*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 + b**6))