Integrand size = 27, antiderivative size = 115 \[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{a d}+\frac {\sec (c+d x)}{a d}+\frac {\sec ^3(c+d x)}{3 a d}+\frac {\sec ^5(c+d x)}{5 a d}-\frac {\tan (c+d x)}{a d}-\frac {2 \tan ^3(c+d x)}{3 a d}-\frac {\tan ^5(c+d x)}{5 a d} \] Output:
-arctanh(cos(d*x+c))/a/d+sec(d*x+c)/a/d+1/3*sec(d*x+c)^3/a/d+1/5*sec(d*x+c )^5/a/d-tan(d*x+c)/a/d-2/3*tan(d*x+c)^3/a/d-1/5*tan(d*x+c)^5/a/d
Leaf count is larger than twice the leaf count of optimal. \(267\) vs. \(2(115)=230\).
Time = 1.76 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.32 \[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sec ^3(c+d x) \left (-100-76 \cos (2 (c+d x))+\frac {149}{4} \cos (3 (c+d x))-8 \cos (4 (c+d x))+30 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (\frac {447}{4}+90 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-90 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-30 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-22 \sin (c+d x)+\frac {149}{4} \sin (2 (c+d x))+30 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (2 (c+d x))-30 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (2 (c+d x))-14 \sin (3 (c+d x))+\frac {149}{8} \sin (4 (c+d x))+15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 (c+d x))-15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 (c+d x))\right )}{120 a d (1+\sin (c+d x))} \] Input:
Integrate[(Csc[c + d*x]*Sec[c + d*x]^4)/(a + a*Sin[c + d*x]),x]
Output:
-1/120*(Sec[c + d*x]^3*(-100 - 76*Cos[2*(c + d*x)] + (149*Cos[3*(c + d*x)] )/4 - 8*Cos[4*(c + d*x)] + 30*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] + Cos [c + d*x]*(447/4 + 90*Log[Cos[(c + d*x)/2]] - 90*Log[Sin[(c + d*x)/2]]) - 30*Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 22*Sin[c + d*x] + (149*Sin[2*( c + d*x)])/4 + 30*Log[Cos[(c + d*x)/2]]*Sin[2*(c + d*x)] - 30*Log[Sin[(c + d*x)/2]]*Sin[2*(c + d*x)] - 14*Sin[3*(c + d*x)] + (149*Sin[4*(c + d*x)])/ 8 + 15*Log[Cos[(c + d*x)/2]]*Sin[4*(c + d*x)] - 15*Log[Sin[(c + d*x)/2]]*S in[4*(c + d*x)]))/(a*d*(1 + Sin[c + d*x]))
Time = 0.43 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.77, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3318, 3042, 3102, 25, 254, 2009, 4254, 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.
| \(\displaystyle \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x) \cos (c+d x)^4 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3318 |
| \(\displaystyle \frac {\int \csc (c+d x) \sec ^6(c+d x)dx}{a}-\frac {\int \sec ^6(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc (c+d x) \sec (c+d x)^6dx}{a}-\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^6dx}{a}\) |
| \(\Big \downarrow \) 3102 |
| \(\displaystyle \frac {\int -\frac {\sec ^6(c+d x)}{1-\sec ^2(c+d x)}d\sec (c+d x)}{a d}-\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^6dx}{a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^6dx}{a}-\frac {\int \frac {\sec ^6(c+d x)}{1-\sec ^2(c+d x)}d\sec (c+d x)}{a d}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle -\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^6dx}{a}-\frac {\int \left (-\sec ^4(c+d x)-\sec ^2(c+d x)+\frac {1}{1-\sec ^2(c+d x)}-1\right )d\sec (c+d x)}{a d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\text {arctanh}(\sec (c+d x))+\frac {1}{5} \sec ^5(c+d x)+\frac {1}{3} \sec ^3(c+d x)+\sec (c+d x)}{a d}-\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^6dx}{a}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\int \left (\tan ^4(c+d x)+2 \tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{a d}+\frac {-\text {arctanh}(\sec (c+d x))+\frac {1}{5} \sec ^5(c+d x)+\frac {1}{3} \sec ^3(c+d x)+\sec (c+d x)}{a d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\text {arctanh}(\sec (c+d x))+\frac {1}{5} \sec ^5(c+d x)+\frac {1}{3} \sec ^3(c+d x)+\sec (c+d x)}{a d}+\frac {-\frac {1}{5} \tan ^5(c+d x)-\frac {2}{3} \tan ^3(c+d x)-\tan (c+d x)}{a d}\) |
Input:
Int[(Csc[c + d*x]*Sec[c + d*x]^4)/(a + a*Sin[c + d*x]),x]
Output:
(-ArcTanh[Sec[c + d*x]] + Sec[c + d*x] + Sec[c + d*x]^3/3 + Sec[c + d*x]^5 /5)/(a*d) + (-Tan[c + d*x] - (2*Tan[c + d*x]^3)/3 - Tan[c + d*x]^5/5)/(a*d )
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S ymbol] :> Simp[1/(f*a^n) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/ 2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1 )/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d) Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Time = 0.78 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.21
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {7}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {23}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(139\) |
| default | \(\frac {-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {7}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {23}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\) | \(139\) |
| risch | \(\frac {4 i {\mathrm e}^{6 i \left (d x +c \right )}+2 \,{\mathrm e}^{7 i \left (d x +c \right )}+\frac {40 i {\mathrm e}^{4 i \left (d x +c \right )}}{3}+\frac {14 \,{\mathrm e}^{5 i \left (d x +c \right )}}{3}+\frac {92 i {\mathrm e}^{2 i \left (d x +c \right )}}{15}+\frac {26 \,{\mathrm e}^{3 i \left (d x +c \right )}}{15}+\frac {16 i}{15}+\frac {2 \,{\mathrm e}^{i \left (d x +c \right )}}{15}}{\left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d a}\) | \(158\) |
| parallelrisch | \(\frac {15 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+30 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-30 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-130 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-50 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+146 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+62 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-62 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-46}{15 d a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(164\) |
| norman | \(\frac {-\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d a}-\frac {46}{15 a d}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d a}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d a}-\frac {26 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d a}-\frac {62 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 d a}+\frac {62 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{15 d a}+\frac {146 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{15 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}\) | \(186\) |
Input:
int(csc(d*x+c)*sec(d*x+c)^4/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/6/(tan(1/2*d*x+1/2*c)-1)^3-1/4/(tan(1/2*d*x+1/2*c)-1)^2-7/8/(tan (1/2*d*x+1/2*c)-1)+2/5/(tan(1/2*d*x+1/2*c)+1)^5-1/(tan(1/2*d*x+1/2*c)+1)^4 +2/(tan(1/2*d*x+1/2*c)+1)^3-2/(tan(1/2*d*x+1/2*c)+1)^2+23/8/(tan(1/2*d*x+1 /2*c)+1)+ln(tan(1/2*d*x+1/2*c)))
Time = 0.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.30 \[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {16 \, \cos \left (d x + c\right )^{4} + 22 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (7 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) + 8}{30 \, {\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/30*(16*cos(d*x + c)^4 + 22*cos(d*x + c)^2 - 15*(cos(d*x + c)^3*sin(d*x + c) + cos(d*x + c)^3)*log(1/2*cos(d*x + c) + 1/2) + 15*(cos(d*x + c)^3*sin (d*x + c) + cos(d*x + c)^3)*log(-1/2*cos(d*x + c) + 1/2) + 2*(7*cos(d*x + c)^2 + 1)*sin(d*x + c) + 8)/(a*d*cos(d*x + c)^3*sin(d*x + c) + a*d*cos(d*x + c)^3)
\[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\csc {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)**4/(a+a*sin(d*x+c)),x)
Output:
Integral(csc(c + d*x)*sec(c + d*x)**4/(sin(c + d*x) + 1), x)/a
Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (107) = 214\).
Time = 0.04 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.78 \[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (\frac {31 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {31 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {73 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {25 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {65 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 23\right )}}{a + \frac {2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {6 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {2 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {2 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac {15 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{15 \, d} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
1/15*(2*(31*sin(d*x + c)/(cos(d*x + c) + 1) - 31*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 73*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 25*sin(d*x + c)^4/(c os(d*x + c) + 1)^4 + 65*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 23)/ (a + 2*a*sin(d*x + c)/(cos(d*x + c) + 1) - 2*a*sin(d*x + c)^2/(cos(d*x + c ) + 1)^2 - 6*a*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 6*a*sin(d*x + c)^5/(c os(d*x + c) + 1)^5 + 2*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 2*a*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) + 15 *log(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d
Time = 0.17 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.18 \[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {5 \, {\left (21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 19\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} + \frac {3 \, {\left (115 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 380 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 530 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 340 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 91\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/120*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a - 5*(21*tan(1/2*d*x + 1/2*c)^2 - 36*tan(1/2*d*x + 1/2*c) + 19)/(a*(tan(1/2*d*x + 1/2*c) - 1)^3) + 3*(115 *tan(1/2*d*x + 1/2*c)^4 + 380*tan(1/2*d*x + 1/2*c)^3 + 530*tan(1/2*d*x + 1 /2*c)^2 + 340*tan(1/2*d*x + 1/2*c) + 91)/(a*(tan(1/2*d*x + 1/2*c) + 1)^5)) /d
Time = 35.41 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.24 \[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {26\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+\frac {10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-\frac {146\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{15}-\frac {62\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {62\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+\frac {46}{15}}{a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \] Input:
int(1/(cos(c + d*x)^4*sin(c + d*x)*(a + a*sin(c + d*x))),x)
Output:
log(tan(c/2 + (d*x)/2))/(a*d) - ((62*tan(c/2 + (d*x)/2))/15 - (62*tan(c/2 + (d*x)/2)^2)/15 - (146*tan(c/2 + (d*x)/2)^3)/15 + (10*tan(c/2 + (d*x)/2)^ 4)/3 + (26*tan(c/2 + (d*x)/2)^5)/3 + 2*tan(c/2 + (d*x)/2)^6 - 2*tan(c/2 + (d*x)/2)^7 + 46/15)/(a*d*(tan(c/2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2) + 1)^5)
Time = 0.18 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.02 \[ \int \frac {\csc (c+d x) \sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}+15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )-15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )-8 \cos \left (d x +c \right )-8 \sin \left (d x +c \right )^{4}+7 \sin \left (d x +c \right )^{3}+27 \sin \left (d x +c \right )^{2}-8 \sin \left (d x +c \right )-23}{15 \cos \left (d x +c \right ) a d \left (\sin \left (d x +c \right )^{3}+\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )-1\right )} \] Input:
int(csc(d*x+c)*sec(d*x+c)^4/(a+a*sin(d*x+c)),x)
Output:
(15*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**3 + 15*cos(c + d*x)*l og(tan((c + d*x)/2))*sin(c + d*x)**2 - 15*cos(c + d*x)*log(tan((c + d*x)/2 ))*sin(c + d*x) - 15*cos(c + d*x)*log(tan((c + d*x)/2)) + 8*cos(c + d*x)*s in(c + d*x)**3 + 8*cos(c + d*x)*sin(c + d*x)**2 - 8*cos(c + d*x)*sin(c + d *x) - 8*cos(c + d*x) - 8*sin(c + d*x)**4 + 7*sin(c + d*x)**3 + 27*sin(c + d*x)**2 - 8*sin(c + d*x) - 23)/(15*cos(c + d*x)*a*d*(sin(c + d*x)**3 + sin (c + d*x)**2 - sin(c + d*x) - 1))