3.1.43 \(\int \csc (a+b x) \csc ^3(2 a+2 b x) \, dx\) [43]

Optimal. Leaf size=66 \[ \frac {5 \tanh ^{-1}(\sin (a+b x))}{16 b}-\frac {5 \csc (a+b x)}{16 b}-\frac {5 \csc ^3(a+b x)}{48 b}+\frac {\csc ^3(a+b x) \sec ^2(a+b x)}{16 b} \]

[Out]

5/16*arctanh(sin(b*x+a))/b-5/16*csc(b*x+a)/b-5/48*csc(b*x+a)^3/b+1/16*csc(b*x+a)^3*sec(b*x+a)^2/b

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Rubi [A]
time = 0.05, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {4373, 2701, 294, 308, 213} \begin {gather*} -\frac {5 \csc ^3(a+b x)}{48 b}-\frac {5 \csc (a+b x)}{16 b}+\frac {5 \tanh ^{-1}(\sin (a+b x))}{16 b}+\frac {\csc ^3(a+b x) \sec ^2(a+b x)}{16 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[Sin[a + b*x]])/(16*b) - (5*Csc[a + b*x])/(16*b) - (5*Csc[a + b*x]^3)/(48*b) + (Csc[a + b*x]^3*Sec[a
+ b*x]^2)/(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 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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*} \int \csc (a+b x) \csc ^3(2 a+2 b x) \, dx &=\frac {1}{8} \int \csc ^4(a+b x) \sec ^3(a+b x) \, dx\\ &=-\frac {\text {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^2} \, dx,x,\csc (a+b x)\right )}{8 b}\\ &=\frac {\csc ^3(a+b x) \sec ^2(a+b x)}{16 b}-\frac {5 \text {Subst}\left (\int \frac {x^4}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{16 b}\\ &=\frac {\csc ^3(a+b x) \sec ^2(a+b x)}{16 b}-\frac {5 \text {Subst}\left (\int \left (1+x^2+\frac {1}{-1+x^2}\right ) \, dx,x,\csc (a+b x)\right )}{16 b}\\ &=-\frac {5 \csc (a+b x)}{16 b}-\frac {5 \csc ^3(a+b x)}{48 b}+\frac {\csc ^3(a+b x) \sec ^2(a+b x)}{16 b}-\frac {5 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{16 b}\\ &=\frac {5 \tanh ^{-1}(\sin (a+b x))}{16 b}-\frac {5 \csc (a+b x)}{16 b}-\frac {5 \csc ^3(a+b x)}{48 b}+\frac {\csc ^3(a+b x) \sec ^2(a+b x)}{16 b}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 31, normalized size = 0.47 \begin {gather*} -\frac {\csc ^3(a+b x) \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};\sin ^2(a+b x)\right )}{24 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(Csc[a + b*x]^3*Hypergeometric2F1[-3/2, 2, -1/2, Sin[a + b*x]^2])/b

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Maple [A]
time = 0.09, size = 69, normalized size = 1.05

method	result	size
default	\(\frac {-\frac {1}{3 \sin \left (x b +a \right )^{3} \cos \left (x b +a \right )^{2}}+\frac {5}{6 \sin \left (x b +a \right ) \cos \left (x b +a \right )^{2}}-\frac {5}{2 \sin \left (x b +a \right )}+\frac {5 \ln \left (\sec \left (x b +a \right )+\tan \left (x b +a \right )\right )}{2}}{8 b}\)	\(69\)
risch	\(-\frac {i \left (15 \,{\mathrm e}^{9 i \left (x b +a \right )}-20 \,{\mathrm e}^{7 i \left (x b +a \right )}-22 \,{\mathrm e}^{5 i \left (x b +a \right )}-20 \,{\mathrm e}^{3 i \left (x b +a \right )}+15 \,{\mathrm e}^{i \left (x b +a \right )}\right )}{24 b \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )^{2}}-\frac {5 \ln \left ({\mathrm e}^{i \left (x b +a \right )}-i\right )}{16 b}+\frac {5 \ln \left (i+{\mathrm e}^{i \left (x b +a \right )}\right )}{16 b}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/b*(-1/3/sin(b*x+a)^3/cos(b*x+a)^2+5/6/sin(b*x+a)/cos(b*x+a)^2-5/2/sin(b*x+a)+5/2*ln(sec(b*x+a)+tan(b*x+a))
)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 1780 vs. \(2 (58) = 116\).
time = 0.55, size = 1780, normalized size = 26.97 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(4*(15*sin(9*b*x + 9*a) - 20*sin(7*b*x + 7*a) - 22*sin(5*b*x + 5*a) - 20*sin(3*b*x + 3*a) + 15*sin(b*x +
a))*cos(10*b*x + 10*a) + 60*(sin(8*b*x + 8*a) + 2*sin(6*b*x + 6*a) - 2*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*co
s(9*b*x + 9*a) + 4*(20*sin(7*b*x + 7*a) + 22*sin(5*b*x + 5*a) + 20*sin(3*b*x + 3*a) - 15*sin(b*x + a))*cos(8*b
*x + 8*a) - 80*(2*sin(6*b*x + 6*a) - 2*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*cos(7*b*x + 7*a) + 8*(22*sin(5*b*x
+ 5*a) + 20*sin(3*b*x + 3*a) - 15*sin(b*x + a))*cos(6*b*x + 6*a) + 88*(2*sin(4*b*x + 4*a) + sin(2*b*x + 2*a))
*cos(5*b*x + 5*a) - 40*(4*sin(3*b*x + 3*a) - 3*sin(b*x + a))*cos(4*b*x + 4*a) + 15*(2*(cos(8*b*x + 8*a) + 2*co
s(6*b*x + 6*a) - 2*cos(4*b*x + 4*a) - cos(2*b*x + 2*a) + 1)*cos(10*b*x + 10*a) - cos(10*b*x + 10*a)^2 - 2*(2*c
os(6*b*x + 6*a) - 2*cos(4*b*x + 4*a) - cos(2*b*x + 2*a) + 1)*cos(8*b*x + 8*a) - cos(8*b*x + 8*a)^2 + 4*(2*cos(
4*b*x + 4*a) + cos(2*b*x + 2*a) - 1)*cos(6*b*x + 6*a) - 4*cos(6*b*x + 6*a)^2 - 4*(cos(2*b*x + 2*a) - 1)*cos(4*
b*x + 4*a) - 4*cos(4*b*x + 4*a)^2 - cos(2*b*x + 2*a)^2 + 2*(sin(8*b*x + 8*a) + 2*sin(6*b*x + 6*a) - 2*sin(4*b*
x + 4*a) - sin(2*b*x + 2*a))*sin(10*b*x + 10*a) - sin(10*b*x + 10*a)^2 - 2*(2*sin(6*b*x + 6*a) - 2*sin(4*b*x +
4*a) - sin(2*b*x + 2*a))*sin(8*b*x + 8*a) - sin(8*b*x + 8*a)^2 + 4*(2*sin(4*b*x + 4*a) + sin(2*b*x + 2*a))*si
n(6*b*x + 6*a) - 4*sin(6*b*x + 6*a)^2 - 4*sin(4*b*x + 4*a)^2 - 4*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*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) - 20*cos(7*b*x + 7*a) - 22*cos(5*b*x + 5
*a) - 20*cos(3*b*x + 3*a) + 15*cos(b*x + a))*sin(10*b*x + 10*a) - 60*(cos(8*b*x + 8*a) + 2*cos(6*b*x + 6*a) -
2*cos(4*b*x + 4*a) - cos(2*b*x + 2*a) + 1)*sin(9*b*x + 9*a) - 4*(20*cos(7*b*x + 7*a) + 22*cos(5*b*x + 5*a) + 2
0*cos(3*b*x + 3*a) - 15*cos(b*x + a))*sin(8*b*x + 8*a) + 80*(2*cos(6*b*x + 6*a) - 2*cos(4*b*x + 4*a) - cos(2*b
*x + 2*a) + 1)*sin(7*b*x + 7*a) - 8*(22*cos(5*b*x + 5*a) + 20*cos(3*b*x + 3*a) - 15*cos(b*x + a))*sin(6*b*x +
6*a) - 88*(2*cos(4*b*x + 4*a) + cos(2*b*x + 2*a) - 1)*sin(5*b*x + 5*a) + 40*(4*cos(3*b*x + 3*a) - 3*cos(b*x +
a))*sin(4*b*x + 4*a) - 80*(cos(2*b*x + 2*a) - 1)*sin(3*b*x + 3*a) + 80*cos(3*b*x + 3*a)*sin(2*b*x + 2*a) - 60*
cos(b*x + a)*sin(2*b*x + 2*a) + 60*cos(2*b*x + 2*a)*sin(b*x + a) - 60*sin(b*x + a))/(b*cos(10*b*x + 10*a)^2 +
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 + b*cos(2*b*x + 2*a)^2 + b*sin(10*b*x +
10*a)^2 + 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 + 4*b*sin(4*b*x + 4*a)*sin(2
*b*x + 2*a) + b*sin(2*b*x + 2*a)^2 - 2*(b*cos(8*b*x + 8*a) + 2*b*cos(6*b*x + 6*a) - 2*b*cos(4*b*x + 4*a) - b*c
os(2*b*x + 2*a) + b)*cos(10*b*x + 10*a) + 2*(2*b*cos(6*b*x + 6*a) - 2*b*cos(4*b*x + 4*a) - b*cos(2*b*x + 2*a)
+ b)*cos(8*b*x + 8*a) - 4*(2*b*cos(4*b*x + 4*a) + b*cos(2*b*x + 2*a) - b)*cos(6*b*x + 6*a) + 4*(b*cos(2*b*x +
2*a) - b)*cos(4*b*x + 4*a) - 2*b*cos(2*b*x + 2*a) - 2*(b*sin(8*b*x + 8*a) + 2*b*sin(6*b*x + 6*a) - 2*b*sin(4*b
*x + 4*a) - b*sin(2*b*x + 2*a))*sin(10*b*x + 10*a) + 2*(2*b*sin(6*b*x + 6*a) - 2*b*sin(4*b*x + 4*a) - b*sin(2*
b*x + 2*a))*sin(8*b*x + 8*a) - 4*(2*b*sin(4*b*x + 4*a) + b*sin(2*b*x + 2*a))*sin(6*b*x + 6*a) + b)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (58) = 116\).
time = 3.45, size = 130, normalized size = 1.97 \begin {gather*} -\frac {30 \, \cos \left (b x + a\right )^{4} - 15 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 15 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 40 \, \cos \left (b x + a\right )^{2} + 6}{96 \, {\left (b \cos \left (b x + a\right )^{4} - b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(30*cos(b*x + a)^4 - 15*(cos(b*x + a)^4 - cos(b*x + a)^2)*log(sin(b*x + a) + 1)*sin(b*x + a) + 15*(cos(b
*x + a)^4 - cos(b*x + a)^2)*log(-sin(b*x + a) + 1)*sin(b*x + a) - 40*cos(b*x + a)^2 + 6)/((b*cos(b*x + a)^4 -
b*cos(b*x + a)^2)*sin(b*x + a))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \csc {\left (a + b x \right )} \csc ^{3}{\left (2 a + 2 b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.50, size = 72, normalized size = 1.09 \begin {gather*} -\frac {\frac {6 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} + \frac {4 \, {\left (6 \, \sin \left (b x + a\right )^{2} + 1\right )}}{\sin \left (b x + a\right )^{3}} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (-\sin \left (b x + a\right ) + 1\right )}{96 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(6*sin(b*x + a)/(sin(b*x + a)^2 - 1) + 4*(6*sin(b*x + a)^2 + 1)/sin(b*x + a)^3 - 15*log(sin(b*x + a) + 1
) + 15*log(-sin(b*x + a) + 1))/b

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Mupad [B]
time = 0.19, size = 61, normalized size = 0.92 \begin {gather*} \frac {5\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{16\,b}-\frac {-\frac {5\,{\sin \left (a+b\,x\right )}^4}{16}+\frac {5\,{\sin \left (a+b\,x\right )}^2}{24}+\frac {1}{24}}{b\,\left ({\sin \left (a+b\,x\right )}^3-{\sin \left (a+b\,x\right )}^5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*atanh(sin(a + b*x)))/(16*b) - ((5*sin(a + b*x)^2)/24 - (5*sin(a + b*x)^4)/16 + 1/24)/(b*(sin(a + b*x)^3 - s
in(a + b*x)^5))

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