Integrand size = 20, antiderivative size = 68 \[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {15 \text {arctanh}(\cos (a+b x))}{32 b}-\frac {9 \cot (a+b x) \csc (a+b x)}{32 b}-\frac {\cot ^3(a+b x) \csc (a+b x)}{16 b}+\frac {\sec (a+b x)}{4 b} \] Output:
-15/32*arctanh(cos(b*x+a))/b-9/32*cot(b*x+a)*csc(b*x+a)/b-1/16*cot(b*x+a)^ 3*csc(b*x+a)/b+1/4*sec(b*x+a)/b
Time = 3.02 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.90 \[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {14 \csc ^2\left (\frac {1}{2} (a+b x)\right )+\csc ^4\left (\frac {1}{2} (a+b x)\right )+\frac {\sec ^2\left (\frac {1}{2} (a+b x)\right ) \left (78+\cos (a+b x) \left (-8 \left (8+15 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-15 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right )+\sec ^4\left (\frac {1}{2} (a+b x)\right )\right )-14 \tan ^2\left (\frac {1}{2} (a+b x)\right )\right )}{-1+\tan ^2\left (\frac {1}{2} (a+b x)\right )}}{256 b} \] Input:
Integrate[Csc[a + b*x]^3*Csc[2*a + 2*b*x]^2,x]
Output:
-1/256*(14*Csc[(a + b*x)/2]^2 + Csc[(a + b*x)/2]^4 + (Sec[(a + b*x)/2]^2*( 78 + Cos[a + b*x]*(-8*(8 + 15*Log[Cos[(a + b*x)/2]] - 15*Log[Sin[(a + b*x) /2]]) + Sec[(a + b*x)/2]^4) - 14*Tan[(a + b*x)/2]^2))/(-1 + Tan[(a + b*x)/ 2]^2))/b
Time = 0.30 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.25, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {3042, 4776, 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.
\(\displaystyle \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x)^3 \sin (2 a+2 b x)^2}dx\) |
\(\Big \downarrow \) 4776 |
\(\displaystyle \frac {1}{4} \int \csc ^5(a+b x) \sec ^2(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \int \csc (a+b x)^5 \sec (a+b x)^2dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int -\frac {\sec ^6(a+b x)}{\left (1-\sec ^2(a+b x)\right )^3}d\sec (a+b x)}{4 b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\sec ^6(a+b x)}{\left (1-\sec ^2(a+b x)\right )^3}d\sec (a+b x)}{4 b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {5}{4} \int \frac {\sec ^4(a+b x)}{\left (1-\sec ^2(a+b x)\right )^2}d\sec (a+b x)-\frac {\sec ^5(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}}{4 b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {5}{4} \left (\frac {\sec ^3(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {3}{2} \int \frac {\sec ^2(a+b x)}{1-\sec ^2(a+b x)}d\sec (a+b x)\right )-\frac {\sec ^5(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}}{4 b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {5}{4} \left (\frac {\sec ^3(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {3}{2} \left (\int \frac {1}{1-\sec ^2(a+b x)}d\sec (a+b x)-\sec (a+b x)\right )\right )-\frac {\sec ^5(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}}{4 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {5}{4} \left (\frac {\sec ^3(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {3}{2} (\text {arctanh}(\sec (a+b x))-\sec (a+b x))\right )-\frac {\sec ^5(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}}{4 b}\) |
Input:
Int[Csc[a + b*x]^3*Csc[2*a + 2*b*x]^2,x]
Output:
(-1/4*Sec[a + b*x]^5/(1 - Sec[a + b*x]^2)^2 + (5*((-3*(ArcTanh[Sec[a + b*x ]] - Sec[a + b*x]))/2 + Sec[a + b*x]^3/(2*(1 - Sec[a + b*x]^2))))/4)/(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 = 0.33 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.04
method | result | size |
default | \(\frac {-\frac {1}{4 \sin \left (b x +a \right )^{4} \cos \left (b x +a \right )}-\frac {5}{8 \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )}+\frac {15}{8 \cos \left (b x +a \right )}+\frac {15 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{8}}{4 b}\) | \(71\) |
risch | \(\frac {15 \,{\mathrm e}^{9 i \left (b x +a \right )}-40 \,{\mathrm e}^{7 i \left (b x +a \right )}+18 \,{\mathrm e}^{5 i \left (b x +a \right )}-40 \,{\mathrm e}^{3 i \left (b x +a \right )}+15 \,{\mathrm e}^{i \left (b x +a \right )}}{16 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}-\frac {15 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{32 b}+\frac {15 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{32 b}\) | \(123\) |
Input:
int(csc(b*x+a)^3*csc(2*b*x+2*a)^2,x,method=_RETURNVERBOSE)
Output:
1/4/b*(-1/4/sin(b*x+a)^4/cos(b*x+a)-5/8/sin(b*x+a)^2/cos(b*x+a)+15/8/cos(b *x+a)+15/8*ln(csc(b*x+a)-cot(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (60) = 120\).
Time = 0.08 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.94 \[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=\frac {30 \, \cos \left (b x + a\right )^{4} - 50 \, \cos \left (b x + a\right )^{2} - 15 \, {\left (\cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 16}{64 \, {\left (b \cos \left (b x + a\right )^{5} - 2 \, b \cos \left (b x + a\right )^{3} + b \cos \left (b x + a\right )\right )}} \] Input:
integrate(csc(b*x+a)^3*csc(2*b*x+2*a)^2,x, algorithm="fricas")
Output:
1/64*(30*cos(b*x + a)^4 - 50*cos(b*x + a)^2 - 15*(cos(b*x + a)^5 - 2*cos(b *x + a)^3 + cos(b*x + a))*log(1/2*cos(b*x + a) + 1/2) + 15*(cos(b*x + a)^5 - 2*cos(b*x + a)^3 + cos(b*x + a))*log(-1/2*cos(b*x + a) + 1/2) + 16)/(b* cos(b*x + a)^5 - 2*b*cos(b*x + a)^3 + b*cos(b*x + a))
\[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=\int \csc ^{3}{\left (a + b x \right )} \csc ^{2}{\left (2 a + 2 b x \right )}\, dx \] Input:
integrate(csc(b*x+a)**3*csc(2*b*x+2*a)**2,x)
Output:
Integral(csc(a + b*x)**3*csc(2*a + 2*b*x)**2, x)
Leaf count of result is larger than twice the leaf count of optimal. 2237 vs. \(2 (60) = 120\).
Time = 0.25 (sec) , antiderivative size = 2237, normalized size of antiderivative = 32.90 \[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=\text {Too large to display} \] Input:
integrate(csc(b*x+a)^3*csc(2*b*x+2*a)^2,x, algorithm="maxima")
Output:
1/64*(4*(15*cos(9*b*x + 9*a) - 40*cos(7*b*x + 7*a) + 18*cos(5*b*x + 5*a) - 40*cos(3*b*x + 3*a) + 15*cos(b*x + a))*cos(10*b*x + 10*a) - 60*(3*cos(8*b *x + 8*a) - 2*cos(6*b*x + 6*a) - 2*cos(4*b*x + 4*a) + 3*cos(2*b*x + 2*a) - 1)*cos(9*b*x + 9*a) + 12*(40*cos(7*b*x + 7*a) - 18*cos(5*b*x + 5*a) + 40* cos(3*b*x + 3*a) - 15*cos(b*x + a))*cos(8*b*x + 8*a) - 160*(2*cos(6*b*x + 6*a) + 2*cos(4*b*x + 4*a) - 3*cos(2*b*x + 2*a) + 1)*cos(7*b*x + 7*a) + 8*( 18*cos(5*b*x + 5*a) - 40*cos(3*b*x + 3*a) + 15*cos(b*x + a))*cos(6*b*x + 6 *a) + 72*(2*cos(4*b*x + 4*a) - 3*cos(2*b*x + 2*a) + 1)*cos(5*b*x + 5*a) - 40*(8*cos(3*b*x + 3*a) - 3*cos(b*x + a))*cos(4*b*x + 4*a) + 160*(3*cos(2*b *x + 2*a) - 1)*cos(3*b*x + 3*a) - 180*cos(2*b*x + 2*a)*cos(b*x + a) + 15*( 2*(3*cos(8*b*x + 8*a) - 2*cos(6*b*x + 6*a) - 2*cos(4*b*x + 4*a) + 3*cos(2* b*x + 2*a) - 1)*cos(10*b*x + 10*a) - cos(10*b*x + 10*a)^2 + 6*(2*cos(6*b*x + 6*a) + 2*cos(4*b*x + 4*a) - 3*cos(2*b*x + 2*a) + 1)*cos(8*b*x + 8*a) - 9*cos(8*b*x + 8*a)^2 - 4*(2*cos(4*b*x + 4*a) - 3*cos(2*b*x + 2*a) + 1)*cos (6*b*x + 6*a) - 4*cos(6*b*x + 6*a)^2 + 4*(3*cos(2*b*x + 2*a) - 1)*cos(4*b* x + 4*a) - 4*cos(4*b*x + 4*a)^2 - 9*cos(2*b*x + 2*a)^2 + 2*(3*sin(8*b*x + 8*a) - 2*sin(6*b*x + 6*a) - 2*sin(4*b*x + 4*a) + 3*sin(2*b*x + 2*a))*sin(1 0*b*x + 10*a) - sin(10*b*x + 10*a)^2 + 6*(2*sin(6*b*x + 6*a) + 2*sin(4*b*x + 4*a) - 3*sin(2*b*x + 2*a))*sin(8*b*x + 8*a) - 9*sin(8*b*x + 8*a)^2 - 4* (2*sin(4*b*x + 4*a) - 3*sin(2*b*x + 2*a))*sin(6*b*x + 6*a) - 4*sin(6*b*...
Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (60) = 120\).
Time = 0.17 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.35 \[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=\frac {\frac {{\left (\frac {16 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {90 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} - \frac {16 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {128}{\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1} + 60 \, \log \left (-\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1}\right )}{256 \, b} \] Input:
integrate(csc(b*x+a)^3*csc(2*b*x+2*a)^2,x, algorithm="giac")
Output:
1/256*((16*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 90*(cos(b*x + a) - 1)^2 /(cos(b*x + a) + 1)^2 - 1)*(cos(b*x + a) + 1)^2/(cos(b*x + a) - 1)^2 - 16* (cos(b*x + a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 128/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1) + 60*log(-(cos(b *x + a) - 1)/(cos(b*x + a) + 1)))/b
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=\frac {\frac {15\,{\cos \left (a+b\,x\right )}^4}{32}-\frac {25\,{\cos \left (a+b\,x\right )}^2}{32}+\frac {1}{4}}{b\,\left ({\cos \left (a+b\,x\right )}^5-2\,{\cos \left (a+b\,x\right )}^3+\cos \left (a+b\,x\right )\right )}-\frac {15\,\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{32\,b} \] Input:
int(1/(sin(a + b*x)^3*sin(2*a + 2*b*x)^2),x)
Output:
((15*cos(a + b*x)^4)/32 - (25*cos(a + b*x)^2)/32 + 1/4)/(b*(cos(a + b*x) - 2*cos(a + b*x)^3 + cos(a + b*x)^5)) - (15*atanh(cos(a + b*x)))/(32*b)
\[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=\int \csc \left (2 b x +2 a \right )^{2} \csc \left (b x +a \right )^{3}d x \] Input:
int(csc(b*x+a)^3*csc(2*b*x+2*a)^2,x)
Output:
int(csc(2*a + 2*b*x)**2*csc(a + b*x)**3,x)