3.41 \(\int \csc (a+b x) \csc (2 a+2 b x) \, dx\)

Optimal. Leaf size=28 \[ \frac {\tanh ^{-1}(\sin (a+b x))}{2 b}-\frac {\csc (a+b x)}{2 b} \]

[Out]

1/2*arctanh(sin(b*x+a))/b-1/2*csc(b*x+a)/b

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Rubi [A]  time = 0.04, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4288, 2621, 321, 207} \[ \frac {\tanh ^{-1}(\sin (a+b x))}{2 b}-\frac {\csc (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]*Csc[2*a + 2*b*x],x]

[Out]

ArcTanh[Sin[a + b*x]]/(2*b) - Csc[a + b*x]/(2*b)

Rule 207

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

Rule 321

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 2621

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 4288

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 (2 a+2 b x) \, dx &=\frac {1}{2} \int \csc ^2(a+b x) \sec (a+b x) \, dx\\ &=-\frac {\operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{2 b}\\ &=-\frac {\csc (a+b x)}{2 b}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{2 b}\\ &=\frac {\tanh ^{-1}(\sin (a+b x))}{2 b}-\frac {\csc (a+b x)}{2 b}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 29, normalized size = 1.04 \[ -\frac {\csc (a+b x) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\sin ^2(a+b x)\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 50, normalized size = 1.79 \[ \frac {\log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 2}{4 \, b \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(log(sin(b*x + a) + 1)*sin(b*x + a) - log(-sin(b*x + a) + 1)*sin(b*x + a) - 2)/(b*sin(b*x + a))

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giac [B]  time = 1.70, size = 484, normalized size = 17.29 \[ -\frac {\frac {\tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{12} - 12 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{10} + 6 \, \tan \left (\frac {1}{2} \, a\right )^{11} - 27 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{8} - 2 \, \tan \left (\frac {1}{2} \, a\right )^{9} - 36 \, \tan \left (\frac {1}{2} \, a\right )^{7} + 27 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{4} - 36 \, \tan \left (\frac {1}{2} \, a\right )^{5} + 12 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{2} - 2 \, \tan \left (\frac {1}{2} \, a\right )^{3} - \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) + 6 \, \tan \left (\frac {1}{2} \, a\right )}{{\left (3 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{5} - \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{6} - 10 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right )^{2} \tan \left (\frac {1}{2} \, a\right )^{3} + 15 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{4} - 3 \, \tan \left (\frac {1}{2} \, a\right )^{5} + 3 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right )^{2} \tan \left (\frac {1}{2} \, a\right ) - 15 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{2} + 10 \, \tan \left (\frac {1}{2} \, a\right )^{3} + \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) - 3 \, \tan \left (\frac {1}{2} \, a\right )\right )} {\left (3 \, \tan \left (\frac {1}{2} \, a\right )^{5} - 10 \, \tan \left (\frac {1}{2} \, a\right )^{3} + 3 \, \tan \left (\frac {1}{2} \, a\right )\right )}} - 2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{3} + 3 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{2} - \tan \left (\frac {1}{2} \, a\right )^{3} - 3 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right ) + 3 \, \tan \left (\frac {1}{2} \, a\right )^{2} - \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) + 3 \, \tan \left (\frac {1}{2} \, a\right ) - 1 \right |}\right ) + 2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{3} - 3 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right )^{2} + \tan \left (\frac {1}{2} \, a\right )^{3} - 3 \, \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) \tan \left (\frac {1}{2} \, a\right ) + 3 \, \tan \left (\frac {1}{2} \, a\right )^{2} + \tan \left (\frac {1}{2} \, b x + 2 \, a\right ) - 3 \, \tan \left (\frac {1}{2} \, a\right ) - 1 \right |}\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((tan(1/2*b*x + 2*a)*tan(1/2*a)^12 - 12*tan(1/2*b*x + 2*a)*tan(1/2*a)^10 + 6*tan(1/2*a)^11 - 27*tan(1/2*b
*x + 2*a)*tan(1/2*a)^8 - 2*tan(1/2*a)^9 - 36*tan(1/2*a)^7 + 27*tan(1/2*b*x + 2*a)*tan(1/2*a)^4 - 36*tan(1/2*a)
^5 + 12*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 - 2*tan(1/2*a)^3 - tan(1/2*b*x + 2*a) + 6*tan(1/2*a))/((3*tan(1/2*b*x
+ 2*a)^2*tan(1/2*a)^5 - tan(1/2*b*x + 2*a)*tan(1/2*a)^6 - 10*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^3 + 15*tan(1/2*b*
x + 2*a)*tan(1/2*a)^4 - 3*tan(1/2*a)^5 + 3*tan(1/2*b*x + 2*a)^2*tan(1/2*a) - 15*tan(1/2*b*x + 2*a)*tan(1/2*a)^
2 + 10*tan(1/2*a)^3 + tan(1/2*b*x + 2*a) - 3*tan(1/2*a))*(3*tan(1/2*a)^5 - 10*tan(1/2*a)^3 + 3*tan(1/2*a))) -
2*log(abs(tan(1/2*b*x + 2*a)*tan(1/2*a)^3 + 3*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 - tan(1/2*a)^3 - 3*tan(1/2*b*x +
2*a)*tan(1/2*a) + 3*tan(1/2*a)^2 - tan(1/2*b*x + 2*a) + 3*tan(1/2*a) - 1)) + 2*log(abs(tan(1/2*b*x + 2*a)*tan
(1/2*a)^3 - 3*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 + tan(1/2*a)^3 - 3*tan(1/2*b*x + 2*a)*tan(1/2*a) + 3*tan(1/2*a)^
2 + tan(1/2*b*x + 2*a) - 3*tan(1/2*a) - 1)))/b

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maple [A]  time = 0.67, size = 34, normalized size = 1.21 \[ -\frac {1}{2 b \sin \left (b x +a \right )}+\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)*csc(2*b*x+2*a),x)

[Out]

-1/2/b/sin(b*x+a)+1/2/b*ln(sec(b*x+a)+tan(b*x+a))

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maxima [B]  time = 0.48, size = 233, normalized size = 8.32 \[ -\frac {{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \log \left (\frac {\cos \left (b x + 2 \, a\right )^{2} + \cos \relax (a)^{2} - 2 \, \cos \relax (a) \sin \left (b x + 2 \, a\right ) + \sin \left (b x + 2 \, a\right )^{2} + 2 \, \cos \left (b x + 2 \, a\right ) \sin \relax (a) + \sin \relax (a)^{2}}{\cos \left (b x + 2 \, a\right )^{2} + \cos \relax (a)^{2} + 2 \, \cos \relax (a) \sin \left (b x + 2 \, a\right ) + \sin \left (b x + 2 \, a\right )^{2} - 2 \, \cos \left (b x + 2 \, a\right ) \sin \relax (a) + \sin \relax (a)^{2}}\right ) + 4 \, \cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right ) - 4 \, \cos \left (2 \, b x + 2 \, a\right ) \sin \left (b x + a\right ) + 4 \, \sin \left (b x + a\right )}{4 \, {\left (b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((cos(2*b*x + 2*a)^2 + 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*cos(b*x + a)*sin(2*b*x +
2*a) - 4*cos(2*b*x + 2*a)*sin(b*x + a) + 4*sin(b*x + a))/(b*cos(2*b*x + 2*a)^2 + b*sin(2*b*x + 2*a)^2 - 2*b*c
os(2*b*x + 2*a) + b)

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mupad [B]  time = 0.11, size = 26, normalized size = 0.93 \[ \frac {\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{2\,b}-\frac {1}{2\,b\,\sin \left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atanh(sin(a + b*x))/(2*b) - 1/(2*b*sin(a + b*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \csc {\left (a + b x \right )} \csc {\left (2 a + 2 b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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