Integrand size = 24, antiderivative size = 77 \[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {15 \text {arctanh}(\cos (c+d x))}{8 a d}-\frac {9 \cot (c+d x) \csc (c+d x)}{8 a d}-\frac {\cot ^3(c+d x) \csc (c+d x)}{4 a d}+\frac {\sec (c+d x)}{a d} \] Output:
-15/8*arctanh(cos(d*x+c))/a/d-9/8*cot(d*x+c)*csc(d*x+c)/a/d-1/4*cot(d*x+c) ^3*csc(d*x+c)/a/d+sec(d*x+c)/a/d
Time = 4.93 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.71 \[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {14 \csc ^2\left (\frac {1}{2} (c+d x)\right )+\csc ^4\left (\frac {1}{2} (c+d x)\right )+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (78+\cos (c+d x) \left (-8 \left (8+15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-15 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec ^4\left (\frac {1}{2} (c+d x)\right )\right )-14 \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}{64 a d} \] Input:
Integrate[Csc[c + d*x]^5/(a - a*Sin[c + d*x]^2),x]
Output:
-1/64*(14*Csc[(c + d*x)/2]^2 + Csc[(c + d*x)/2]^4 + (Sec[(c + d*x)/2]^2*(7 8 + Cos[c + d*x]*(-8*(8 + 15*Log[Cos[(c + d*x)/2]] - 15*Log[Sin[(c + d*x)/ 2]]) + Sec[(c + d*x)/2]^4) - 14*Tan[(c + d*x)/2]^2))/(-1 + Tan[(c + d*x)/2 ]^2))/(a*d)
Time = 0.32 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3654, 3042, 3102, 25, 252, 252, 262, 219}
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.
| \(\displaystyle \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^5 \left (a-a \sin (c+d x)^2\right )}dx\) |
| \(\Big \downarrow \) 3654 |
| \(\displaystyle \frac {\int \csc ^5(c+d x) \sec ^2(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc (c+d x)^5 \sec (c+d x)^2dx}{a}\) |
| \(\Big \downarrow \) 3102 |
| \(\displaystyle \frac {\int -\frac {\sec ^6(c+d x)}{\left (1-\sec ^2(c+d x)\right )^3}d\sec (c+d x)}{a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\sec ^6(c+d x)}{\left (1-\sec ^2(c+d x)\right )^3}d\sec (c+d x)}{a d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {\frac {5}{4} \int \frac {\sec ^4(c+d x)}{\left (1-\sec ^2(c+d x)\right )^2}d\sec (c+d x)-\frac {\sec ^5(c+d x)}{4 \left (1-\sec ^2(c+d x)\right )^2}}{a d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {\frac {5}{4} \left (\frac {\sec ^3(c+d x)}{2 \left (1-\sec ^2(c+d x)\right )}-\frac {3}{2} \int \frac {\sec ^2(c+d x)}{1-\sec ^2(c+d x)}d\sec (c+d x)\right )-\frac {\sec ^5(c+d x)}{4 \left (1-\sec ^2(c+d x)\right )^2}}{a d}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {\frac {5}{4} \left (\frac {\sec ^3(c+d x)}{2 \left (1-\sec ^2(c+d x)\right )}-\frac {3}{2} \left (\int \frac {1}{1-\sec ^2(c+d x)}d\sec (c+d x)-\sec (c+d x)\right )\right )-\frac {\sec ^5(c+d x)}{4 \left (1-\sec ^2(c+d x)\right )^2}}{a d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {5}{4} \left (\frac {\sec ^3(c+d x)}{2 \left (1-\sec ^2(c+d x)\right )}-\frac {3}{2} (\text {arctanh}(\sec (c+d x))-\sec (c+d x))\right )-\frac {\sec ^5(c+d x)}{4 \left (1-\sec ^2(c+d x)\right )^2}}{a d}\) |
Input:
Int[Csc[c + d*x]^5/(a - a*Sin[c + d*x]^2),x]
Output:
(-1/4*Sec[c + d*x]^5/(1 - Sec[c + d*x]^2)^2 + (5*((-3*(ArcTanh[Sec[c + d*x ]] - Sec[c + d*x]))/2 + Sec[c + d*x]^3/(2*(1 - Sec[c + d*x]^2))))/4)/(a*d)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
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[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[ a^p Int[ActivateTrig[u*cos[e + f*x]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]
Time = 0.74 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.13
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{\cos \left (d x +c \right )}-\frac {1}{16 \left (\cos \left (d x +c \right )-1\right )^{2}}+\frac {7}{16 \left (\cos \left (d x +c \right )-1\right )}+\frac {15 \ln \left (\cos \left (d x +c \right )-1\right )}{16}+\frac {1}{16 \left (\cos \left (d x +c \right )+1\right )^{2}}+\frac {7}{16 \left (\cos \left (d x +c \right )+1\right )}-\frac {15 \ln \left (\cos \left (d x +c \right )+1\right )}{16}}{d a}\) | \(87\) |
| default | \(\frac {\frac {1}{\cos \left (d x +c \right )}-\frac {1}{16 \left (\cos \left (d x +c \right )-1\right )^{2}}+\frac {7}{16 \left (\cos \left (d x +c \right )-1\right )}+\frac {15 \ln \left (\cos \left (d x +c \right )-1\right )}{16}+\frac {1}{16 \left (\cos \left (d x +c \right )+1\right )^{2}}+\frac {7}{16 \left (\cos \left (d x +c \right )+1\right )}-\frac {15 \ln \left (\cos \left (d x +c \right )+1\right )}{16}}{d a}\) | \(87\) |
| parallelrisch | \(\frac {-160+120 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+15 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{64 d a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\) | \(99\) |
| norman | \(\frac {\frac {1}{64 a d}+\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{64 a d}+\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{64 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{64 a d}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 a d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}\) | \(132\) |
| risch | \(\frac {15 \,{\mathrm e}^{9 i \left (d x +c \right )}-40 \,{\mathrm e}^{7 i \left (d x +c \right )}+18 \,{\mathrm e}^{5 i \left (d x +c \right )}-40 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}}{4 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 a d}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 a d}\) | \(132\) |
Input:
int(csc(d*x+c)^5/(a-a*sin(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/cos(d*x+c)-1/16/(cos(d*x+c)-1)^2+7/16/(cos(d*x+c)-1)+15/16*ln(cos (d*x+c)-1)+1/16/(cos(d*x+c)+1)^2+7/16/(cos(d*x+c)+1)-15/16*ln(cos(d*x+c)+1 ))
Time = 0.09 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.75 \[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {30 \, \cos \left (d x + c\right )^{4} - 50 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (d x + c\right )^{5} - 2 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 16}{16 \, {\left (a d \cos \left (d x + c\right )^{5} - 2 \, a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )\right )}} \] Input:
integrate(csc(d*x+c)^5/(a-a*sin(d*x+c)^2),x, algorithm="fricas")
Output:
1/16*(30*cos(d*x + c)^4 - 50*cos(d*x + c)^2 - 15*(cos(d*x + c)^5 - 2*cos(d *x + c)^3 + cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 15*(cos(d*x + c)^5 - 2*cos(d*x + c)^3 + cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 16)/(a* d*cos(d*x + c)^5 - 2*a*d*cos(d*x + c)^3 + a*d*cos(d*x + c))
\[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=- \frac {\int \frac {\csc ^{5}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} - 1}\, dx}{a} \] Input:
integrate(csc(d*x+c)**5/(a-a*sin(d*x+c)**2),x)
Output:
-Integral(csc(c + d*x)**5/(sin(c + d*x)**2 - 1), x)/a
Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.17 \[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 25 \, \cos \left (d x + c\right )^{2} + 8\right )}}{a \cos \left (d x + c\right )^{5} - 2 \, a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )} - \frac {15 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a}}{16 \, d} \] Input:
integrate(csc(d*x+c)^5/(a-a*sin(d*x+c)^2),x, algorithm="maxima")
Output:
1/16*(2*(15*cos(d*x + c)^4 - 25*cos(d*x + c)^2 + 8)/(a*cos(d*x + c)^5 - 2* a*cos(d*x + c)^3 + a*cos(d*x + c)) - 15*log(cos(d*x + c) + 1)/a + 15*log(c os(d*x + c) - 1)/a)/d
Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (71) = 142\).
Time = 0.47 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.35 \[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {{\left (\frac {16 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {90 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}} + \frac {60 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a} - \frac {\frac {16 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2}} + \frac {128}{a {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}}}{64 \, d} \] Input:
integrate(csc(d*x+c)^5/(a-a*sin(d*x+c)^2),x, algorithm="giac")
Output:
1/64*((16*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 90*(cos(d*x + c) - 1)^2/ (cos(d*x + c) + 1)^2 - 1)*(cos(d*x + c) + 1)^2/(a*(cos(d*x + c) - 1)^2) + 60*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a - (16*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/ a^2 + 128/(a*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)))/d
Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96 \[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {15\,{\cos \left (c+d\,x\right )}^4}{8}-\frac {25\,{\cos \left (c+d\,x\right )}^2}{8}+1}{d\,\left (a\,{\cos \left (c+d\,x\right )}^5-2\,a\,{\cos \left (c+d\,x\right )}^3+a\,\cos \left (c+d\,x\right )\right )}-\frac {15\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{8\,a\,d} \] Input:
int(1/(sin(c + d*x)^5*(a - a*sin(c + d*x)^2)),x)
Output:
((15*cos(c + d*x)^4)/8 - (25*cos(c + d*x)^2)/8 + 1)/(d*(a*cos(c + d*x) - 2 *a*cos(c + d*x)^3 + a*cos(c + d*x)^5)) - (15*atanh(cos(c + d*x)))/(8*a*d)
Time = 0.18 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.14 \[ \int \frac {\csc ^5(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4}-10 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+15 \sin \left (d x +c \right )^{4}-5 \sin \left (d x +c \right )^{2}-2}{8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a d} \] Input:
int(csc(d*x+c)^5/(a-a*sin(d*x+c)^2),x)
Output:
(15*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**4 - 10*cos(c + d*x)*s in(c + d*x)**4 + 15*sin(c + d*x)**4 - 5*sin(c + d*x)**2 - 2)/(8*cos(c + d* x)*sin(c + d*x)**4*a*d)