\(\int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx\) [32]

Optimal result

Integrand size = 20, antiderivative size = 70 \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {15 \text {arctanh}(\sin (a+b x))}{256 b}-\frac {15 \csc (a+b x)}{256 b}+\frac {5 \csc (a+b x) \sec ^2(a+b x)}{256 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{128 b} \]

[Out]

15/256*arctanh(sin(b*x+a))/b-15/256*csc(b*x+a)/b+5/256*csc(b*x+a)*sec(b*x+a)^2/b+1/128*csc(b*x+a)*sec(b*x+a)^4
/b

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4373, 2701, 294, 327, 213} \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {15 \text {arctanh}(\sin (a+b x))}{256 b}-\frac {15 \csc (a+b x)}{256 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{128 b}+\frac {5 \csc (a+b x) \sec ^2(a+b x)}{256 b} \]

[In]

Int[Csc[2*a + 2*b*x]^5*Sin[a + b*x]^3,x]

[Out]

(15*ArcTanh[Sin[a + b*x]])/(256*b) - (15*Csc[a + b*x])/(256*b) + (5*Csc[a + b*x]*Sec[a + b*x]^2)/(256*b) + (Cs
c[a + b*x]*Sec[a + b*x]^4)/(128*b)

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2701

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(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] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 4373

Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[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]
&& IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{32} \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx \\ & = -\frac {\text {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^3} \, dx,x,\csc (a+b x)\right )}{32 b} \\ & = \frac {\csc (a+b x) \sec ^4(a+b x)}{128 b}-\frac {5 \text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\csc (a+b x)\right )}{128 b} \\ & = \frac {5 \csc (a+b x) \sec ^2(a+b x)}{256 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{128 b}-\frac {15 \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{256 b} \\ & = -\frac {15 \csc (a+b x)}{256 b}+\frac {5 \csc (a+b x) \sec ^2(a+b x)}{256 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{128 b}-\frac {15 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{256 b} \\ & = \frac {15 \text {arctanh}(\sin (a+b x))}{256 b}-\frac {15 \csc (a+b x)}{256 b}+\frac {5 \csc (a+b x) \sec ^2(a+b x)}{256 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{128 b} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.41 \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=-\frac {\csc (a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3,\frac {1}{2},\sin ^2(a+b x)\right )}{32 b} \]

[In]

Integrate[Csc[2*a + 2*b*x]^5*Sin[a + b*x]^3,x]

[Out]

-1/32*(Csc[a + b*x]*Hypergeometric2F1[-1/2, 3, 1/2, Sin[a + b*x]^2])/b

Maple [A] (verified)

Time = 13.98 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.99

method	result	size
default	\(\frac {\frac {1}{4 \sin \left (x b +a \right ) \cos \left (x b +a \right )^{4}}+\frac {5}{8 \sin \left (x b +a \right ) \cos \left (x b +a \right )^{2}}-\frac {15}{8 \sin \left (x b +a \right )}+\frac {15 \ln \left (\sec \left (x b +a \right )+\tan \left (x b +a \right )\right )}{8}}{32 b}\)	\(69\)
risch	\(-\frac {i \left (15 \,{\mathrm e}^{9 i \left (x b +a \right )}+40 \,{\mathrm e}^{7 i \left (x b +a \right )}+18 \,{\mathrm e}^{5 i \left (x b +a \right )}+40 \,{\mathrm e}^{3 i \left (x b +a \right )}+15 \,{\mathrm e}^{i \left (x b +a \right )}\right )}{128 b \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )^{4} \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )}-\frac {15 \ln \left ({\mathrm e}^{i \left (x b +a \right )}-i\right )}{256 b}+\frac {15 \ln \left (i+{\mathrm e}^{i \left (x b +a \right )}\right )}{256 b}\)	\(126\)

[In]

int(csc(2*b*x+2*a)^5*sin(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

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))
)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.36 \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {15 \, \cos \left (b x + a\right )^{4} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 15 \, \cos \left (b x + a\right )^{4} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 30 \, \cos \left (b x + a\right )^{4} + 10 \, \cos \left (b x + a\right )^{2} + 4}{512 \, b \cos \left (b x + a\right )^{4} \sin \left (b x + a\right )} \]

[In]

integrate(csc(2*b*x+2*a)^5*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

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))

Sympy [F(-1)]

Timed out. \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=\text {Timed out} \]

[In]

integrate(csc(2*b*x+2*a)**5*sin(b*x+a)**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1805 vs. \(2 (62) = 124\).

Time = 0.39 (sec) , antiderivative size = 1805, normalized size of antiderivative = 25.79 \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=\text {Too large to display} \]

[In]

integrate(csc(2*b*x+2*a)^5*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

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))*c
os(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) + 72*(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(10*b*x + 10*a) + cos(10*b*x + 1
0*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*si
n(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(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*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 + 6*cos(2*b*x + 2*a) + 1)*log((cos(b*x + 2*a)^2 + cos(a)^2 - 2
*cos(a)*sin(b*x + 2*a) + sin(b*x + 2*a)^2 + 2*cos(b*x + 2*a)*sin(a) + sin(a)^2)/(cos(b*x + 2*a)^2 + cos(a)^2 +
2*cos(a)*sin(b*x + 2*a) + sin(b*x + 2*a)^2 - 2*cos(b*x + 2*a)*sin(a) + sin(a)^2)) - 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))*sin(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)*sin(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))*sin(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)*sin(7*b*x + 7*a) - 8*(18*cos(5*b*x + 5*a) + 40*co
s(3*b*x + 3*a) + 15*cos(b*x + a))*sin(6*b*x + 6*a) - 72*(2*cos(4*b*x + 4*a) + 3*cos(2*b*x + 2*a) + 1)*sin(5*b*
x + 5*a) + 40*(8*cos(3*b*x + 3*a) + 3*cos(b*x + a))*sin(4*b*x + 4*a) - 160*(3*cos(2*b*x + 2*a) + 1)*sin(3*b*x
+ 3*a) + 480*cos(3*b*x + 3*a)*sin(2*b*x + 2*a) + 180*cos(b*x + a)*sin(2*b*x + 2*a) - 180*cos(2*b*x + 2*a)*sin(
b*x + a) - 60*sin(b*x + a))/(b*cos(10*b*x + 10*a)^2 + 9*b*cos(8*b*x + 8*a)^2 + 4*b*cos(6*b*x + 6*a)^2 + 4*b*co
s(4*b*x + 4*a)^2 + 9*b*cos(2*b*x + 2*a)^2 + b*sin(10*b*x + 10*a)^2 + 9*b*sin(8*b*x + 8*a)^2 + 4*b*sin(6*b*x +
6*a)^2 + 4*b*sin(4*b*x + 4*a)^2 + 12*b*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 9*b*sin(2*b*x + 2*a)^2 + 2*(3*b*cos
(8*b*x + 8*a) + 2*b*cos(6*b*x + 6*a) - 2*b*cos(4*b*x + 4*a) - 3*b*cos(2*b*x + 2*a) - b)*cos(10*b*x + 10*a) + 6
*(2*b*cos(6*b*x + 6*a) - 2*b*cos(4*b*x + 4*a) - 3*b*cos(2*b*x + 2*a) - b)*cos(8*b*x + 8*a) - 4*(2*b*cos(4*b*x
+ 4*a) + 3*b*cos(2*b*x + 2*a) + b)*cos(6*b*x + 6*a) + 4*(3*b*cos(2*b*x + 2*a) + b)*cos(4*b*x + 4*a) + 6*b*cos(
2*b*x + 2*a) + 2*(3*b*sin(8*b*x + 8*a) + 2*b*sin(6*b*x + 6*a) - 2*b*sin(4*b*x + 4*a) - 3*b*sin(2*b*x + 2*a))*s
in(10*b*x + 10*a) + 6*(2*b*sin(6*b*x + 6*a) - 2*b*sin(4*b*x + 4*a) - 3*b*sin(2*b*x + 2*a))*sin(8*b*x + 8*a) -
4*(2*b*sin(4*b*x + 4*a) + 3*b*sin(2*b*x + 2*a))*sin(6*b*x + 6*a) + b)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.04 \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=-\frac {\frac {2 \, {\left (7 \, \sin \left (b x + a\right )^{3} - 9 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{2}} + \frac {16}{\sin \left (b x + a\right )} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (-\sin \left (b x + a\right ) + 1\right )}{512 \, b} \]

[In]

integrate(csc(2*b*x+2*a)^5*sin(b*x+a)^3,x, algorithm="giac")

[Out]

-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

Mupad [B] (verification not implemented)

Time = 19.51 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.96 \[ \int \csc ^5(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {15\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{256\,b}-\frac {\frac {15\,{\sin \left (a+b\,x\right )}^4}{256}-\frac {25\,{\sin \left (a+b\,x\right )}^2}{256}+\frac {1}{32}}{b\,\left ({\sin \left (a+b\,x\right )}^5-2\,{\sin \left (a+b\,x\right )}^3+\sin \left (a+b\,x\right )\right )} \]

[In]

int(sin(a + b*x)^3/sin(2*a + 2*b*x)^5,x)

[Out]

(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))