Integrand size = 18, antiderivative size = 47 \[ \int \csc (a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {3 \text {arctanh}(\cos (a+b x))}{8 b}-\frac {\cot (a+b x) \csc (a+b x)}{8 b}+\frac {\sec (a+b x)}{4 b} \] Output:
-3/8*arctanh(cos(b*x+a))/b-1/8*cot(b*x+a)*csc(b*x+a)/b+1/4*sec(b*x+a)/b
Leaf count is larger than twice the leaf count of optimal. \(143\) vs. \(2(47)=94\).
Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 3.04 \[ \int \csc (a+b x) \csc ^2(2 a+2 b x) \, dx=\frac {\csc ^4(a+b x) \left (2-6 \cos (2 (a+b x))+2 \cos (3 (a+b x))+3 \cos (3 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-3 \cos (3 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )+\cos (a+b x) \left (-2-3 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+3 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )}{8 b \left (\csc ^2\left (\frac {1}{2} (a+b x)\right )-\sec ^2\left (\frac {1}{2} (a+b x)\right )\right )} \] Input:
Integrate[Csc[a + b*x]*Csc[2*a + 2*b*x]^2,x]
Output:
(Csc[a + b*x]^4*(2 - 6*Cos[2*(a + b*x)] + 2*Cos[3*(a + b*x)] + 3*Cos[3*(a + b*x)]*Log[Cos[(a + b*x)/2]] - 3*Cos[3*(a + b*x)]*Log[Sin[(a + b*x)/2]] + Cos[a + b*x]*(-2 - 3*Log[Cos[(a + b*x)/2]] + 3*Log[Sin[(a + b*x)/2]])))/( 8*b*(Csc[(a + b*x)/2]^2 - Sec[(a + b*x)/2]^2))
Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.15, 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, 3102, 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.
\(\displaystyle \int \csc (a+b x) \csc ^2(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x) \sin (2 a+2 b x)^2}dx\) |
\(\Big \downarrow \) 4776 |
\(\displaystyle \frac {1}{4} \int \csc ^3(a+b x) \sec ^2(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \int \csc (a+b x)^3 \sec (a+b x)^2dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int \frac {\sec ^4(a+b x)}{\left (1-\sec ^2(a+b x)\right )^2}d\sec (a+b x)}{4 b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {\sec ^3(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {3}{2} \int \frac {\sec ^2(a+b x)}{1-\sec ^2(a+b x)}d\sec (a+b x)}{4 b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {\sec ^3(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {3}{2} \left (\int \frac {1}{1-\sec ^2(a+b x)}d\sec (a+b x)-\sec (a+b x)\right )}{4 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\sec ^3(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {3}{2} (\text {arctanh}(\sec (a+b x))-\sec (a+b x))}{4 b}\) |
Input:
Int[Csc[a + b*x]*Csc[2*a + 2*b*x]^2,x]
Output:
((-3*(ArcTanh[Sec[a + b*x]] - Sec[a + b*x]))/2 + Sec[a + b*x]^3/(2*(1 - Se c[a + b*x]^2)))/(4*b)
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[((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 = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.13
method | result | size |
default | \(\frac {-\frac {1}{2 \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )}+\frac {3}{2 \cos \left (b x +a \right )}+\frac {3 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{4 b}\) | \(53\) |
risch | \(\frac {3 \,{\mathrm e}^{5 i \left (b x +a \right )}-2 \,{\mathrm e}^{3 i \left (b x +a \right )}+3 \,{\mathrm e}^{i \left (b x +a \right )}}{4 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{8 b}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{8 b}\) | \(101\) |
Input:
int(csc(b*x+a)*csc(2*b*x+2*a)^2,x,method=_RETURNVERBOSE)
Output:
1/4/b*(-1/2/sin(b*x+a)^2/cos(b*x+a)+3/2/cos(b*x+a)+3/2*ln(csc(b*x+a)-cot(b *x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (41) = 82\).
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.04 \[ \int \csc (a+b x) \csc ^2(2 a+2 b x) \, dx=\frac {6 \, \cos \left (b x + a\right )^{2} - 3 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 4}{16 \, {\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )}} \] Input:
integrate(csc(b*x+a)*csc(2*b*x+2*a)^2,x, algorithm="fricas")
Output:
1/16*(6*cos(b*x + a)^2 - 3*(cos(b*x + a)^3 - cos(b*x + a))*log(1/2*cos(b*x + a) + 1/2) + 3*(cos(b*x + a)^3 - cos(b*x + a))*log(-1/2*cos(b*x + a) + 1 /2) - 4)/(b*cos(b*x + a)^3 - b*cos(b*x + a))
\[ \int \csc (a+b x) \csc ^2(2 a+2 b x) \, dx=\int \csc {\left (a + b x \right )} \csc ^{2}{\left (2 a + 2 b x \right )}\, dx \] Input:
integrate(csc(b*x+a)*csc(2*b*x+2*a)**2,x)
Output:
Integral(csc(a + b*x)*csc(2*a + 2*b*x)**2, x)
Leaf count of result is larger than twice the leaf count of optimal. 974 vs. \(2 (41) = 82\).
Time = 0.07 (sec) , antiderivative size = 974, normalized size of antiderivative = 20.72 \[ \int \csc (a+b x) \csc ^2(2 a+2 b x) \, dx=\text {Too large to display} \] Input:
integrate(csc(b*x+a)*csc(2*b*x+2*a)^2,x, algorithm="maxima")
Output:
1/16*(4*(3*cos(5*b*x + 5*a) - 2*cos(3*b*x + 3*a) + 3*cos(b*x + a))*cos(6*b *x + 6*a) - 12*(cos(4*b*x + 4*a) + cos(2*b*x + 2*a) - 1)*cos(5*b*x + 5*a) + 4*(2*cos(3*b*x + 3*a) - 3*cos(b*x + a))*cos(4*b*x + 4*a) + 8*(cos(2*b*x + 2*a) - 1)*cos(3*b*x + 3*a) - 12*cos(2*b*x + 2*a)*cos(b*x + a) + 3*(2*(co s(4*b*x + 4*a) + cos(2*b*x + 2*a) - 1)*cos(6*b*x + 6*a) - cos(6*b*x + 6*a) ^2 - 2*(cos(2*b*x + 2*a) - 1)*cos(4*b*x + 4*a) - cos(4*b*x + 4*a)^2 - cos( 2*b*x + 2*a)^2 + 2*(sin(4*b*x + 4*a) + sin(2*b*x + 2*a))*sin(6*b*x + 6*a) - sin(6*b*x + 6*a)^2 - sin(4*b*x + 4*a)^2 - 2*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*x)^2 + 2*co s(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 - 2*sin(b*x)*sin(a) + sin(a)^2) - 3* (2*(cos(4*b*x + 4*a) + cos(2*b*x + 2*a) - 1)*cos(6*b*x + 6*a) - cos(6*b*x + 6*a)^2 - 2*(cos(2*b*x + 2*a) - 1)*cos(4*b*x + 4*a) - cos(4*b*x + 4*a)^2 - cos(2*b*x + 2*a)^2 + 2*(sin(4*b*x + 4*a) + sin(2*b*x + 2*a))*sin(6*b*x + 6*a) - sin(6*b*x + 6*a)^2 - sin(4*b*x + 4*a)^2 - 2*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*x)^2 - 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 + 2*sin(b*x)*sin(a) + sin(a)^2 ) + 4*(3*sin(5*b*x + 5*a) - 2*sin(3*b*x + 3*a) + 3*sin(b*x + a))*sin(6*b*x + 6*a) - 12*(sin(4*b*x + 4*a) + sin(2*b*x + 2*a))*sin(5*b*x + 5*a) + 4*(2 *sin(3*b*x + 3*a) - 3*sin(b*x + a))*sin(4*b*x + 4*a) + 8*sin(3*b*x + 3*a)* sin(2*b*x + 2*a) - 12*sin(2*b*x + 2*a)*sin(b*x + a) + 12*cos(b*x + a))/...
Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (41) = 82\).
Time = 0.15 (sec) , antiderivative size = 137, normalized size of antiderivative = 2.91 \[ \int \csc (a+b x) \csc ^2(2 a+2 b x) \, dx=\frac {\frac {\frac {14 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {3 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1}{\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}}} - \frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 6 \, \log \left (-\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1}\right )}{32 \, b} \] Input:
integrate(csc(b*x+a)*csc(2*b*x+2*a)^2,x, algorithm="giac")
Output:
1/32*((14*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 3*(cos(b*x + a) - 1)^2/( cos(b*x + a) + 1)^2 + 1)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2) - (cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 6*log(-(cos(b*x + a) - 1)/(cos(b*x + a) + 1)))/b
Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.04 \[ \int \csc (a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {3\,\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{8\,b}-\frac {\frac {3\,{\cos \left (a+b\,x\right )}^2}{8}-\frac {1}{4}}{b\,\left (\cos \left (a+b\,x\right )-{\cos \left (a+b\,x\right )}^3\right )} \] Input:
int(1/(sin(a + b*x)*sin(2*a + 2*b*x)^2),x)
Output:
- (3*atanh(cos(a + b*x)))/(8*b) - ((3*cos(a + b*x)^2)/8 - 1/4)/(b*(cos(a + b*x) - cos(a + b*x)^3))
\[ \int \csc (a+b x) \csc ^2(2 a+2 b x) \, dx=\int \csc \left (2 b x +2 a \right )^{2} \csc \left (b x +a \right )d x \] Input:
int(csc(b*x+a)*csc(2*b*x+2*a)^2,x)
Output:
int(csc(2*a + 2*b*x)**2*csc(a + b*x),x)