Integrand size = 20, antiderivative size = 68 \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {15 \text {arctanh}(\sin (a+b x))}{256 b}-\frac {\csc (a+b x)}{32 b}+\frac {9 \sec (a+b x) \tan (a+b x)}{256 b}+\frac {\sec (a+b x) \tan ^3(a+b x)}{128 b} \] Output:
15/256*arctanh(sin(b*x+a))/b-1/32*csc(b*x+a)/b+9/256*sec(b*x+a)*tan(b*x+a) /b+1/128*sec(b*x+a)*tan(b*x+a)^3/b
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.43 \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=-\frac {\csc (a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3,\frac {1}{2},\sin ^2(a+b x)\right )}{32 b} \] Input:
Integrate[Csc[2*a + 2*b*x]^5*Sin[a + b*x]^3,x]
Output:
-1/32*(Csc[a + b*x]*Hypergeometric2F1[-1/2, 3, 1/2, Sin[a + b*x]^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, 3101, 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.
\(\displaystyle \int \sin ^3(a+b x) \csc ^5(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (a+b x)^3}{\sin (2 a+2 b x)^5}dx\) |
\(\Big \downarrow \) 4776 |
\(\displaystyle \frac {1}{32} \int \csc ^2(a+b x) \sec ^5(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{32} \int \csc (a+b x)^2 \sec (a+b x)^5dx\) |
\(\Big \downarrow \) 3101 |
\(\displaystyle -\frac {\int -\frac {\csc ^6(a+b x)}{\left (1-\csc ^2(a+b x)\right )^3}d\csc (a+b x)}{32 b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\csc ^6(a+b x)}{\left (1-\csc ^2(a+b x)\right )^3}d\csc (a+b x)}{32 b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {5}{4} \int \frac {\csc ^4(a+b x)}{\left (1-\csc ^2(a+b x)\right )^2}d\csc (a+b x)-\frac {\csc ^5(a+b x)}{4 \left (1-\csc ^2(a+b x)\right )^2}}{32 b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {5}{4} \left (\frac {\csc ^3(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {3}{2} \int \frac {\csc ^2(a+b x)}{1-\csc ^2(a+b x)}d\csc (a+b x)\right )-\frac {\csc ^5(a+b x)}{4 \left (1-\csc ^2(a+b x)\right )^2}}{32 b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {\frac {5}{4} \left (\frac {\csc ^3(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {3}{2} \left (\int \frac {1}{1-\csc ^2(a+b x)}d\csc (a+b x)-\csc (a+b x)\right )\right )-\frac {\csc ^5(a+b x)}{4 \left (1-\csc ^2(a+b x)\right )^2}}{32 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {5}{4} \left (\frac {\csc ^3(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {3}{2} (\text {arctanh}(\csc (a+b x))-\csc (a+b x))\right )-\frac {\csc ^5(a+b x)}{4 \left (1-\csc ^2(a+b x)\right )^2}}{32 b}\) |
Input:
Int[Csc[2*a + 2*b*x]^5*Sin[a + b*x]^3,x]
Output:
-1/32*(-1/4*Csc[a + b*x]^5/(1 - Csc[a + b*x]^2)^2 + (5*((-3*(ArcTanh[Csc[a + b*x]] - Csc[a + b*x]))/2 + Csc[a + b*x]^3/(2*(1 - Csc[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_)]*(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 = 8.44 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01
method | result | size |
default | \(\frac {\frac {1}{4 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{4}}+\frac {5}{8 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}-\frac {15}{8 \sin \left (b x +a \right )}+\frac {15 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{32 b}\) | \(69\) |
risch | \(-\frac {i \left (15 \,{\mathrm e}^{9 i \left (b x +a \right )}+40 \,{\mathrm e}^{7 i \left (b x +a \right )}+18 \,{\mathrm e}^{5 i \left (b x +a \right )}+40 \,{\mathrm e}^{3 i \left (b x +a \right )}+15 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{128 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{4} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\frac {15 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{256 b}-\frac {15 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{256 b}\) | \(126\) |
Input:
int(csc(2*b*x+2*a)^5*sin(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/32/b*(1/4/sin(b*x+a)/cos(b*x+a)^4+5/8/sin(b*x+a)/cos(b*x+a)^2-15/8/sin(b *x+a)+15/8*ln(sec(b*x+a)+tan(b*x+a)))
Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.40 \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {15 \, \cos \left (b x + a\right )^{4} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 15 \, \cos \left (b x + a\right )^{4} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 30 \, \cos \left (b x + a\right )^{4} + 10 \, \cos \left (b x + a\right )^{2} + 4}{512 \, b \cos \left (b x + a\right )^{4} \sin \left (b x + a\right )} \] Input:
integrate(csc(2*b*x+2*a)^5*sin(b*x+a)^3,x, algorithm="fricas")
Output:
1/512*(15*cos(b*x + a)^4*log(sin(b*x + a) + 1)*sin(b*x + a) - 15*cos(b*x + a)^4*log(-sin(b*x + a) + 1)*sin(b*x + a) - 30*cos(b*x + a)^4 + 10*cos(b*x + a)^2 + 4)/(b*cos(b*x + a)^4*sin(b*x + a))
Timed out. \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=\text {Timed out} \] Input:
integrate(csc(2*b*x+2*a)**5*sin(b*x+a)**3,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 1805 vs. \(2 (60) = 120\).
Time = 0.44 (sec) , antiderivative size = 1805, normalized size of antiderivative = 26.54 \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=\text {Too large to display} \] Input:
integrate(csc(2*b*x+2*a)^5*sin(b*x+a)^3,x, algorithm="maxima")
Output:
1/512*(4*(15*sin(9*b*x + 9*a) + 40*sin(7*b*x + 7*a) + 18*sin(5*b*x + 5*a) + 40*sin(3*b*x + 3*a) + 15*sin(b*x + a))*cos(10*b*x + 10*a) - 60*(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)) *cos(9*b*x + 9*a) + 12*(40*sin(7*b*x + 7*a) + 18*sin(5*b*x + 5*a) + 40*sin (3*b*x + 3*a) + 15*sin(b*x + a))*cos(8*b*x + 8*a) - 160*(2*sin(6*b*x + 6*a ) - 2*sin(4*b*x + 4*a) - 3*sin(2*b*x + 2*a))*cos(7*b*x + 7*a) + 8*(18*sin( 5*b*x + 5*a) + 40*sin(3*b*x + 3*a) + 15*sin(b*x + a))*cos(6*b*x + 6*a) + 7 2*(2*sin(4*b*x + 4*a) + 3*sin(2*b*x + 2*a))*cos(5*b*x + 5*a) - 40*(8*sin(3 *b*x + 3*a) + 3*sin(b*x + a))*cos(4*b*x + 4*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(1 0*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(10*b*x + 10*a) + sin(1 0*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*x + 6*a)^2 + 4*sin(4*b*x + 4*a)^2 + 12*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 9*sin(2*b*x + 2*a)^2...
Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07 \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=-\frac {\frac {2 \, {\left (7 \, \sin \left (b x + a\right )^{3} - 9 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{2}} + \frac {16}{\sin \left (b x + a\right )} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (-\sin \left (b x + a\right ) + 1\right )}{512 \, b} \] Input:
integrate(csc(2*b*x+2*a)^5*sin(b*x+a)^3,x, algorithm="giac")
Output:
-1/512*(2*(7*sin(b*x + a)^3 - 9*sin(b*x + a))/(sin(b*x + a)^2 - 1)^2 + 16/ sin(b*x + a) - 15*log(sin(b*x + a) + 1) + 15*log(-sin(b*x + a) + 1))/b
Time = 18.45 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99 \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {15\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{256\,b}-\frac {\frac {15\,{\sin \left (a+b\,x\right )}^4}{256}-\frac {25\,{\sin \left (a+b\,x\right )}^2}{256}+\frac {1}{32}}{b\,\left ({\sin \left (a+b\,x\right )}^5-2\,{\sin \left (a+b\,x\right )}^3+\sin \left (a+b\,x\right )\right )} \] Input:
int(sin(a + b*x)^3/sin(2*a + 2*b*x)^5,x)
Output:
(15*atanh(sin(a + b*x)))/(256*b) - ((15*sin(a + b*x)^4)/256 - (25*sin(a + b*x)^2)/256 + 1/32)/(b*(sin(a + b*x) - 2*sin(a + b*x)^3 + sin(a + b*x)^5))
\[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=\int \csc \left (2 b x +2 a \right )^{5} \sin \left (b x +a \right )^{3}d x \] Input:
int(csc(2*b*x+2*a)^5*sin(b*x+a)^3,x)
Output:
int(csc(2*a + 2*b*x)**5*sin(a + b*x)**3,x)