Integrand size = 18, antiderivative size = 62 \[ \int \csc (a+b x) \csc ^3(2 a+2 b x) \, dx=\frac {5 \text {arctanh}(\sin (a+b x))}{16 b}-\frac {\csc (a+b x)}{4 b}-\frac {\csc ^3(a+b x)}{24 b}+\frac {\sec (a+b x) \tan (a+b x)}{16 b} \] Output:
5/16*arctanh(sin(b*x+a))/b-1/4*csc(b*x+a)/b-1/24*csc(b*x+a)^3/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 = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.50 \[ \int \csc (a+b x) \csc ^3(2 a+2 b x) \, dx=-\frac {\csc ^3(a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2,-\frac {1}{2},\sin ^2(a+b x)\right )}{24 b} \] Input:
Integrate[Csc[a + b*x]*Csc[2*a + 2*b*x]^3,x]
Output:
-1/24*(Csc[a + b*x]^3*Hypergeometric2F1[-3/2, 2, -1/2, Sin[a + b*x]^2])/b
Time = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, 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.
\(\displaystyle \int \csc (a+b x) \csc ^3(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x) \sin (2 a+2 b x)^3}dx\) |
\(\Big \downarrow \) 4776 |
\(\displaystyle \frac {1}{8} \int \csc ^4(a+b x) \sec ^3(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{8} \int \csc (a+b x)^4 \sec (a+b x)^3dx\) |
\(\Big \downarrow \) 3101 |
\(\displaystyle -\frac {\int \frac {\csc ^6(a+b x)}{\left (1-\csc ^2(a+b x)\right )^2}d\csc (a+b x)}{8 b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \int \frac {\csc ^4(a+b x)}{1-\csc ^2(a+b x)}d\csc (a+b x)}{8 b}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle -\frac {\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \int \left (-\csc ^2(a+b x)+\frac {1}{1-\csc ^2(a+b x)}-1\right )d\csc (a+b x)}{8 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \left (\text {arctanh}(\csc (a+b x))-\frac {1}{3} \csc ^3(a+b x)-\csc (a+b x)\right )}{8 b}\) |
Input:
Int[Csc[a + b*x]*Csc[2*a + 2*b*x]^3,x]
Output:
-1/8*(Csc[a + b*x]^5/(2*(1 - Csc[a + b*x]^2)) - (5*(ArcTanh[Csc[a + b*x]] - Csc[a + b*x] - Csc[a + b*x]^3/3))/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 = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.11
method | result | size |
default | \(\frac {-\frac {1}{3 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{2}}+\frac {5}{6 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}-\frac {5}{2 \sin \left (b x +a \right )}+\frac {5 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{8 b}\) | \(69\) |
risch | \(-\frac {i \left (15 \,{\mathrm e}^{9 i \left (b x +a \right )}-20 \,{\mathrm e}^{7 i \left (b x +a \right )}-22 \,{\mathrm e}^{5 i \left (b x +a \right )}-20 \,{\mathrm e}^{3 i \left (b x +a \right )}+15 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{24 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}-\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{16 b}+\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{16 b}\) | \(126\) |
Input:
int(csc(b*x+a)*csc(2*b*x+2*a)^3,x,method=_RETURNVERBOSE)
Output:
1/8/b*(-1/3/sin(b*x+a)^3/cos(b*x+a)^2+5/6/sin(b*x+a)/cos(b*x+a)^2-5/2/sin( b*x+a)+5/2*ln(sec(b*x+a)+tan(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (54) = 108\).
Time = 0.08 (sec) , antiderivative size = 130, normalized size of antiderivative = 2.10 \[ \int \csc (a+b x) \csc ^3(2 a+2 b x) \, dx=-\frac {30 \, \cos \left (b x + a\right )^{4} - 15 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 15 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 40 \, \cos \left (b x + a\right )^{2} + 6}{96 \, {\left (b \cos \left (b x + a\right )^{4} - b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \] Input:
integrate(csc(b*x+a)*csc(2*b*x+2*a)^3,x, algorithm="fricas")
Output:
-1/96*(30*cos(b*x + a)^4 - 15*(cos(b*x + a)^4 - cos(b*x + a)^2)*log(sin(b* x + a) + 1)*sin(b*x + a) + 15*(cos(b*x + a)^4 - cos(b*x + a)^2)*log(-sin(b *x + a) + 1)*sin(b*x + a) - 40*cos(b*x + a)^2 + 6)/((b*cos(b*x + a)^4 - b* cos(b*x + a)^2)*sin(b*x + a))
\[ \int \csc (a+b x) \csc ^3(2 a+2 b x) \, dx=\int \csc {\left (a + b x \right )} \csc ^{3}{\left (2 a + 2 b x \right )}\, dx \] Input:
integrate(csc(b*x+a)*csc(2*b*x+2*a)**3,x)
Output:
Integral(csc(a + b*x)*csc(2*a + 2*b*x)**3, x)
Leaf count of result is larger than twice the leaf count of optimal. 1780 vs. \(2 (54) = 108\).
Time = 0.27 (sec) , antiderivative size = 1780, normalized size of antiderivative = 28.71 \[ \int \csc (a+b x) \csc ^3(2 a+2 b x) \, dx=\text {Too large to display} \] Input:
integrate(csc(b*x+a)*csc(2*b*x+2*a)^3,x, algorithm="maxima")
Output:
1/96*(4*(15*sin(9*b*x + 9*a) - 20*sin(7*b*x + 7*a) - 22*sin(5*b*x + 5*a) - 20*sin(3*b*x + 3*a) + 15*sin(b*x + a))*cos(10*b*x + 10*a) + 60*(sin(8*b*x + 8*a) + 2*sin(6*b*x + 6*a) - 2*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*cos( 9*b*x + 9*a) + 4*(20*sin(7*b*x + 7*a) + 22*sin(5*b*x + 5*a) + 20*sin(3*b*x + 3*a) - 15*sin(b*x + a))*cos(8*b*x + 8*a) - 80*(2*sin(6*b*x + 6*a) - 2*s in(4*b*x + 4*a) - sin(2*b*x + 2*a))*cos(7*b*x + 7*a) + 8*(22*sin(5*b*x + 5 *a) + 20*sin(3*b*x + 3*a) - 15*sin(b*x + a))*cos(6*b*x + 6*a) + 88*(2*sin( 4*b*x + 4*a) + sin(2*b*x + 2*a))*cos(5*b*x + 5*a) - 40*(4*sin(3*b*x + 3*a) - 3*sin(b*x + a))*cos(4*b*x + 4*a) + 15*(2*(cos(8*b*x + 8*a) + 2*cos(6*b* x + 6*a) - 2*cos(4*b*x + 4*a) - cos(2*b*x + 2*a) + 1)*cos(10*b*x + 10*a) - cos(10*b*x + 10*a)^2 - 2*(2*cos(6*b*x + 6*a) - 2*cos(4*b*x + 4*a) - cos(2 *b*x + 2*a) + 1)*cos(8*b*x + 8*a) - cos(8*b*x + 8*a)^2 + 4*(2*cos(4*b*x + 4*a) + cos(2*b*x + 2*a) - 1)*cos(6*b*x + 6*a) - 4*cos(6*b*x + 6*a)^2 - 4*( cos(2*b*x + 2*a) - 1)*cos(4*b*x + 4*a) - 4*cos(4*b*x + 4*a)^2 - cos(2*b*x + 2*a)^2 + 2*(sin(8*b*x + 8*a) + 2*sin(6*b*x + 6*a) - 2*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*sin(10*b*x + 10*a) - sin(10*b*x + 10*a)^2 - 2*(2*sin(6* b*x + 6*a) - 2*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*sin(8*b*x + 8*a) - sin (8*b*x + 8*a)^2 + 4*(2*sin(4*b*x + 4*a) + sin(2*b*x + 2*a))*sin(6*b*x + 6* a) - 4*sin(6*b*x + 6*a)^2 - 4*sin(4*b*x + 4*a)^2 - 4*sin(4*b*x + 4*a)*sin( 2*b*x + 2*a) - sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) - 1)*log((cos(b*...
Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.16 \[ \int \csc (a+b x) \csc ^3(2 a+2 b x) \, dx=-\frac {\frac {6 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} + \frac {4 \, {\left (6 \, \sin \left (b x + a\right )^{2} + 1\right )}}{\sin \left (b x + a\right )^{3}} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (-\sin \left (b x + a\right ) + 1\right )}{96 \, b} \] Input:
integrate(csc(b*x+a)*csc(2*b*x+2*a)^3,x, algorithm="giac")
Output:
-1/96*(6*sin(b*x + a)/(sin(b*x + a)^2 - 1) + 4*(6*sin(b*x + a)^2 + 1)/sin( b*x + a)^3 - 15*log(sin(b*x + a) + 1) + 15*log(-sin(b*x + a) + 1))/b
Time = 0.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.98 \[ \int \csc (a+b x) \csc ^3(2 a+2 b x) \, dx=\frac {5\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{16\,b}-\frac {-\frac {5\,{\sin \left (a+b\,x\right )}^4}{16}+\frac {5\,{\sin \left (a+b\,x\right )}^2}{24}+\frac {1}{24}}{b\,\left ({\sin \left (a+b\,x\right )}^3-{\sin \left (a+b\,x\right )}^5\right )} \] Input:
int(1/(sin(a + b*x)*sin(2*a + 2*b*x)^3),x)
Output:
(5*atanh(sin(a + b*x)))/(16*b) - ((5*sin(a + b*x)^2)/24 - (5*sin(a + b*x)^ 4)/16 + 1/24)/(b*(sin(a + b*x)^3 - sin(a + b*x)^5))
\[ \int \csc (a+b x) \csc ^3(2 a+2 b x) \, dx=\int \csc \left (2 b x +2 a \right )^{3} \csc \left (b x +a \right )d x \] Input:
int(csc(b*x+a)*csc(2*b*x+2*a)^3,x)
Output:
int(csc(2*a + 2*b*x)**3*csc(a + b*x),x)