Integrand size = 20, antiderivative size = 77 \[ \int \csc ^3(a+b x) \csc ^3(2 a+2 b x) \, dx=\frac {7 \text {arctanh}(\sin (a+b x))}{16 b}-\frac {3 \csc (a+b x)}{8 b}-\frac {\csc ^3(a+b x)}{12 b}-\frac {\csc ^5(a+b x)}{40 b}+\frac {\sec (a+b x) \tan (a+b x)}{16 b} \] Output:
7/16*arctanh(sin(b*x+a))/b-3/8*csc(b*x+a)/b-1/12*csc(b*x+a)^3/b-1/40*csc(b *x+a)^5/b+1/16*sec(b*x+a)*tan(b*x+a)/b
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.40 \[ \int \csc ^3(a+b x) \csc ^3(2 a+2 b x) \, dx=-\frac {\csc ^5(a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},2,-\frac {3}{2},\sin ^2(a+b x)\right )}{40 b} \] Input:
Integrate[Csc[a + b*x]^3*Csc[2*a + 2*b*x]^3,x]
Output:
-1/40*(Csc[a + b*x]^5*Hypergeometric2F1[-5/2, 2, -3/2, Sin[a + b*x]^2])/b
Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 4776, 3042, 3101, 252, 254, 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.
\(\displaystyle \int \csc ^3(a+b x) \csc ^3(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x)^3 \sin (2 a+2 b x)^3}dx\) |
\(\Big \downarrow \) 4776 |
\(\displaystyle \frac {1}{8} \int \csc ^6(a+b x) \sec ^3(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{8} \int \csc (a+b x)^6 \sec (a+b x)^3dx\) |
\(\Big \downarrow \) 3101 |
\(\displaystyle -\frac {\int \frac {\csc ^8(a+b x)}{\left (1-\csc ^2(a+b x)\right )^2}d\csc (a+b x)}{8 b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {\csc ^7(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {7}{2} \int \frac {\csc ^6(a+b x)}{1-\csc ^2(a+b x)}d\csc (a+b x)}{8 b}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle -\frac {\frac {\csc ^7(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {7}{2} \int \left (-\csc ^4(a+b x)-\csc ^2(a+b x)+\frac {1}{1-\csc ^2(a+b x)}-1\right )d\csc (a+b x)}{8 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {\csc ^7(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {7}{2} \left (\text {arctanh}(\csc (a+b x))-\frac {1}{5} \csc ^5(a+b x)-\frac {1}{3} \csc ^3(a+b x)-\csc (a+b x)\right )}{8 b}\) |
Input:
Int[Csc[a + b*x]^3*Csc[2*a + 2*b*x]^3,x]
Output:
-1/8*(Csc[a + b*x]^7/(2*(1 - Csc[a + b*x]^2)) - (7*(ArcTanh[Csc[a + b*x]] - Csc[a + b*x] - Csc[a + b*x]^3/3 - Csc[a + b*x]^5/5))/2)/b
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[(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_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_S ymbol] :> Simp[-(f*a^n)^(-1) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_ Symbol] :> Simp[2^p/f^p Int[Cos[a + b*x]^p*(f*Sin[a + b*x])^(n + p), x], x] /; FreeQ[{a, b, c, d, f, n}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && I ntegerQ[p]
Time = 0.42 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.13
method | result | size |
default | \(\frac {-\frac {1}{5 \sin \left (b x +a \right )^{5} \cos \left (b x +a \right )^{2}}-\frac {7}{15 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{2}}+\frac {7}{6 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}-\frac {7}{2 \sin \left (b x +a \right )}+\frac {7 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{8 b}\) | \(87\) |
risch | \(-\frac {i \left (105 \,{\mathrm e}^{13 i \left (b x +a \right )}-350 \,{\mathrm e}^{11 i \left (b x +a \right )}+231 \,{\mathrm e}^{9 i \left (b x +a \right )}+412 \,{\mathrm e}^{7 i \left (b x +a \right )}+231 \,{\mathrm e}^{5 i \left (b x +a \right )}-350 \,{\mathrm e}^{3 i \left (b x +a \right )}+105 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{120 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{5} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}-\frac {7 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{16 b}+\frac {7 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{16 b}\) | \(148\) |
Input:
int(csc(b*x+a)^3*csc(2*b*x+2*a)^3,x,method=_RETURNVERBOSE)
Output:
1/8/b*(-1/5/sin(b*x+a)^5/cos(b*x+a)^2-7/15/sin(b*x+a)^3/cos(b*x+a)^2+7/6/s in(b*x+a)/cos(b*x+a)^2-7/2/sin(b*x+a)+7/2*ln(sec(b*x+a)+tan(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (67) = 134\).
Time = 0.08 (sec) , antiderivative size = 166, normalized size of antiderivative = 2.16 \[ \int \csc ^3(a+b x) \csc ^3(2 a+2 b x) \, dx=-\frac {210 \, \cos \left (b x + a\right )^{6} - 490 \, \cos \left (b x + a\right )^{4} - 105 \, {\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 105 \, {\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 322 \, \cos \left (b x + a\right )^{2} - 30}{480 \, {\left (b \cos \left (b x + a\right )^{6} - 2 \, b \cos \left (b x + a\right )^{4} + b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \] Input:
integrate(csc(b*x+a)^3*csc(2*b*x+2*a)^3,x, algorithm="fricas")
Output:
-1/480*(210*cos(b*x + a)^6 - 490*cos(b*x + a)^4 - 105*(cos(b*x + a)^6 - 2* cos(b*x + a)^4 + cos(b*x + a)^2)*log(sin(b*x + a) + 1)*sin(b*x + a) + 105* (cos(b*x + a)^6 - 2*cos(b*x + a)^4 + cos(b*x + a)^2)*log(-sin(b*x + a) + 1 )*sin(b*x + a) + 322*cos(b*x + a)^2 - 30)/((b*cos(b*x + a)^6 - 2*b*cos(b*x + a)^4 + b*cos(b*x + a)^2)*sin(b*x + a))
\[ \int \csc ^3(a+b x) \csc ^3(2 a+2 b x) \, dx=\int \csc ^{3}{\left (a + b x \right )} \csc ^{3}{\left (2 a + 2 b x \right )}\, dx \] Input:
integrate(csc(b*x+a)**3*csc(2*b*x+2*a)**3,x)
Output:
Integral(csc(a + b*x)**3*csc(2*a + 2*b*x)**3, x)
Leaf count of result is larger than twice the leaf count of optimal. 3095 vs. \(2 (67) = 134\).
Time = 0.49 (sec) , antiderivative size = 3095, normalized size of antiderivative = 40.19 \[ \int \csc ^3(a+b x) \csc ^3(2 a+2 b x) \, dx=\text {Too large to display} \] Input:
integrate(csc(b*x+a)^3*csc(2*b*x+2*a)^3,x, algorithm="maxima")
Output:
1/480*(4*(105*sin(13*b*x + 13*a) - 350*sin(11*b*x + 11*a) + 231*sin(9*b*x + 9*a) + 412*sin(7*b*x + 7*a) + 231*sin(5*b*x + 5*a) - 350*sin(3*b*x + 3*a ) + 105*sin(b*x + a))*cos(14*b*x + 14*a) + 420*(3*sin(12*b*x + 12*a) - sin (10*b*x + 10*a) - 5*sin(8*b*x + 8*a) + 5*sin(6*b*x + 6*a) + sin(4*b*x + 4* a) - 3*sin(2*b*x + 2*a))*cos(13*b*x + 13*a) + 12*(350*sin(11*b*x + 11*a) - 231*sin(9*b*x + 9*a) - 412*sin(7*b*x + 7*a) - 231*sin(5*b*x + 5*a) + 350* sin(3*b*x + 3*a) - 105*sin(b*x + a))*cos(12*b*x + 12*a) + 1400*(sin(10*b*x + 10*a) + 5*sin(8*b*x + 8*a) - 5*sin(6*b*x + 6*a) - sin(4*b*x + 4*a) + 3* sin(2*b*x + 2*a))*cos(11*b*x + 11*a) + 4*(231*sin(9*b*x + 9*a) + 412*sin(7 *b*x + 7*a) + 231*sin(5*b*x + 5*a) - 350*sin(3*b*x + 3*a) + 105*sin(b*x + a))*cos(10*b*x + 10*a) - 924*(5*sin(8*b*x + 8*a) - 5*sin(6*b*x + 6*a) - si n(4*b*x + 4*a) + 3*sin(2*b*x + 2*a))*cos(9*b*x + 9*a) + 20*(412*sin(7*b*x + 7*a) + 231*sin(5*b*x + 5*a) - 350*sin(3*b*x + 3*a) + 105*sin(b*x + a))*c os(8*b*x + 8*a) + 1648*(5*sin(6*b*x + 6*a) + sin(4*b*x + 4*a) - 3*sin(2*b* x + 2*a))*cos(7*b*x + 7*a) - 140*(33*sin(5*b*x + 5*a) - 50*sin(3*b*x + 3*a ) + 15*sin(b*x + a))*cos(6*b*x + 6*a) + 924*(sin(4*b*x + 4*a) - 3*sin(2*b* x + 2*a))*cos(5*b*x + 5*a) + 140*(10*sin(3*b*x + 3*a) - 3*sin(b*x + a))*co s(4*b*x + 4*a) + 105*(2*(3*cos(12*b*x + 12*a) - cos(10*b*x + 10*a) - 5*cos (8*b*x + 8*a) + 5*cos(6*b*x + 6*a) + cos(4*b*x + 4*a) - 3*cos(2*b*x + 2*a) + 1)*cos(14*b*x + 14*a) - cos(14*b*x + 14*a)^2 + 6*(cos(10*b*x + 10*a)...
Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.06 \[ \int \csc ^3(a+b x) \csc ^3(2 a+2 b x) \, dx=-\frac {\frac {30 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} + \frac {4 \, {\left (45 \, \sin \left (b x + a\right )^{4} + 10 \, \sin \left (b x + a\right )^{2} + 3\right )}}{\sin \left (b x + a\right )^{5}} - 105 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 105 \, \log \left (-\sin \left (b x + a\right ) + 1\right )}{480 \, b} \] Input:
integrate(csc(b*x+a)^3*csc(2*b*x+2*a)^3,x, algorithm="giac")
Output:
-1/480*(30*sin(b*x + a)/(sin(b*x + a)^2 - 1) + 4*(45*sin(b*x + a)^4 + 10*s in(b*x + a)^2 + 3)/sin(b*x + a)^5 - 105*log(sin(b*x + a) + 1) + 105*log(-s in(b*x + a) + 1))/b
Time = 19.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92 \[ \int \csc ^3(a+b x) \csc ^3(2 a+2 b x) \, dx=\frac {7\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{16\,b}-\frac {-\frac {7\,{\sin \left (a+b\,x\right )}^6}{16}+\frac {7\,{\sin \left (a+b\,x\right )}^4}{24}+\frac {7\,{\sin \left (a+b\,x\right )}^2}{120}+\frac {1}{40}}{b\,\left ({\sin \left (a+b\,x\right )}^5-{\sin \left (a+b\,x\right )}^7\right )} \] Input:
int(1/(sin(a + b*x)^3*sin(2*a + 2*b*x)^3),x)
Output:
(7*atanh(sin(a + b*x)))/(16*b) - ((7*sin(a + b*x)^2)/120 + (7*sin(a + b*x) ^4)/24 - (7*sin(a + b*x)^6)/16 + 1/40)/(b*(sin(a + b*x)^5 - sin(a + b*x)^7 ))
\[ \int \csc ^3(a+b x) \csc ^3(2 a+2 b x) \, dx=\int \csc \left (2 b x +2 a \right )^{3} \csc \left (b x +a \right )^{3}d x \] Input:
int(csc(b*x+a)^3*csc(2*b*x+2*a)^3,x)
Output:
int(csc(2*a + 2*b*x)**3*csc(a + b*x)**3,x)