\(\int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx\) [70]

Optimal result

Integrand size = 20, antiderivative size = 70 \[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {15 \text {arctanh}(\cos (a+b x))}{32 b}+\frac {15 \sec (a+b x)}{32 b}-\frac {5 \csc ^2(a+b x) \sec (a+b x)}{32 b}-\frac {\csc ^4(a+b x) \sec (a+b x)}{16 b} \]

[Out]

-15/32*arctanh(cos(b*x+a))/b+15/32*sec(b*x+a)/b-5/32*csc(b*x+a)^2*sec(b*x+a)/b-1/16*csc(b*x+a)^4*sec(b*x+a)/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, 2702, 294, 327, 213} \[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {15 \text {arctanh}(\cos (a+b x))}{32 b}+\frac {15 \sec (a+b x)}{32 b}-\frac {\csc ^4(a+b x) \sec (a+b x)}{16 b}-\frac {5 \csc ^2(a+b x) \sec (a+b x)}{32 b} \]

[In]

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

[Out]

(-15*ArcTanh[Cos[a + b*x]])/(32*b) + (15*Sec[a + b*x])/(32*b) - (5*Csc[a + b*x]^2*Sec[a + b*x])/(32*b) - (Csc[
a + b*x]^4*Sec[a + b*x])/(16*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 2702

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[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])

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}{4} \int \csc ^5(a+b x) \sec ^2(a+b x) \, dx \\ & = \frac {\text {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^3} \, dx,x,\sec (a+b x)\right )}{4 b} \\ & = -\frac {\csc ^4(a+b x) \sec (a+b x)}{16 b}+\frac {5 \text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (a+b x)\right )}{16 b} \\ & = -\frac {5 \csc ^2(a+b x) \sec (a+b x)}{32 b}-\frac {\csc ^4(a+b x) \sec (a+b x)}{16 b}+\frac {15 \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (a+b x)\right )}{32 b} \\ & = \frac {15 \sec (a+b x)}{32 b}-\frac {5 \csc ^2(a+b x) \sec (a+b x)}{32 b}-\frac {\csc ^4(a+b x) \sec (a+b x)}{16 b}+\frac {15 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (a+b x)\right )}{32 b} \\ & = -\frac {15 \text {arctanh}(\cos (a+b x))}{32 b}+\frac {15 \sec (a+b x)}{32 b}-\frac {5 \csc ^2(a+b x) \sec (a+b x)}{32 b}-\frac {\csc ^4(a+b x) \sec (a+b x)}{16 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.63 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.84 \[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {14 \csc ^2\left (\frac {1}{2} (a+b x)\right )+\csc ^4\left (\frac {1}{2} (a+b x)\right )+\frac {\sec ^2\left (\frac {1}{2} (a+b x)\right ) \left (78+\cos (a+b x) \left (-8 \left (8+15 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-15 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right )+\sec ^4\left (\frac {1}{2} (a+b x)\right )\right )-14 \tan ^2\left (\frac {1}{2} (a+b x)\right )\right )}{-1+\tan ^2\left (\frac {1}{2} (a+b x)\right )}}{256 b} \]

[In]

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

[Out]

-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[Co
s[(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

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {-\frac {1}{4 \cos \left (x b +a \right ) \sin \left (x b +a \right )^{4}}-\frac {5}{8 \sin \left (x b +a \right )^{2} \cos \left (x b +a \right )}+\frac {15}{8 \cos \left (x b +a \right )}+\frac {15 \ln \left (\csc \left (x b +a \right )-\cot \left (x b +a \right )\right )}{8}}{4 b}\)	\(71\)
risch	\(\frac {15 \,{\mathrm e}^{9 i \left (x b +a \right )}-40 \,{\mathrm e}^{7 i \left (x b +a \right )}+18 \,{\mathrm e}^{5 i \left (x b +a \right )}-40 \,{\mathrm e}^{3 i \left (x b +a \right )}+15 \,{\mathrm e}^{i \left (x b +a \right )}}{16 b \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}-\frac {15 \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{32 b}+\frac {15 \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{32 b}\)	\(123\)

[In]

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

[Out]

1/4/b*(-1/4/cos(b*x+a)/sin(b*x+a)^4-5/8/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))
)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (62) = 124\).

Time = 0.25 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.89 \[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=\frac {30 \, \cos \left (b x + a\right )^{4} - 50 \, \cos \left (b x + a\right )^{2} - 15 \, {\left (\cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} + \cos \left (b x + a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 16}{64 \, {\left (b \cos \left (b x + a\right )^{5} - 2 \, b \cos \left (b x + a\right )^{3} + b \cos \left (b x + a\right )\right )}} \]

[In]

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

[Out]

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*co
s(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))

Sympy [F]

\[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=\int \csc ^{3}{\left (a + b x \right )} \csc ^{2}{\left (2 a + 2 b x \right )}\, dx \]

[In]

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

[Out]

Integral(csc(a + b*x)**3*csc(2*a + 2*b*x)**2, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2237 vs. \(2 (62) = 124\).

Time = 0.25 (sec) , antiderivative size = 2237, normalized size of antiderivative = 31.96 \[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

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*c
os(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*c
os(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(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 + 2*cos(b*x)*cos(a) + cos(a)^2 + sin
(b*x)^2 - 2*sin(b*x)*sin(a) + sin(a)^2) - 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(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 - 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 + 2*sin(b*x)*sin(a) + sin(
a)^2) + 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))*sin(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))*sin(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))*
sin(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))*sin(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))*sin(6*b*x + 6*a) + 72*(2*sin(4*b*x + 4*a) - 3*sin(2
*b*x + 2*a))*sin(5*b*x + 5*a) - 40*(8*sin(3*b*x + 3*a) - 3*sin(b*x + a))*sin(4*b*x + 4*a) + 480*sin(3*b*x + 3*
a)*sin(2*b*x + 2*a) - 180*sin(2*b*x + 2*a)*sin(b*x + a) + 60*cos(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*cos(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*co
s(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))*sin(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (62) = 124\).

Time = 0.32 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.29 \[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=\frac {\frac {{\left (\frac {16 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {90 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} - \frac {16 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {128}{\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1} + 60 \, \log \left (-\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1}\right )}{256 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.94 \[ \int \csc ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=\frac {\frac {15\,{\cos \left (a+b\,x\right )}^4}{32}-\frac {25\,{\cos \left (a+b\,x\right )}^2}{32}+\frac {1}{4}}{b\,\left ({\cos \left (a+b\,x\right )}^5-2\,{\cos \left (a+b\,x\right )}^3+\cos \left (a+b\,x\right )\right )}-\frac {15\,\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{32\,b} \]

[In]

int(1/(sin(a + b*x)^3*sin(2*a + 2*b*x)^2),x)

[Out]

((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)