3.20 \(\int \csc ^3(2 a+2 b x) \sin ^2(a+b x) \, dx\)
Optimal. Leaf size=30 \[ \frac{\tan ^2(a+b x)}{16 b}+\frac{\log (\tan (a+b x))}{8 b} \]
[Out]
Log[Tan[a + b*x]]/(8*b) + Tan[a + b*x]^2/(16*b)
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Rubi [A] time = 0.0478498, antiderivative size = 30, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used =
{4288, 2620, 14} \[ \frac{\tan ^2(a+b x)}{16 b}+\frac{\log (\tan (a+b x))}{8 b} \]
Antiderivative was successfully verified.
[In]
Int[Csc[2*a + 2*b*x]^3*Sin[a + b*x]^2,x]
[Out]
Log[Tan[a + b*x]]/(8*b) + Tan[a + b*x]^2/(16*b)
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]
Rule 2620
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]
Rule 14
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&& !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Rubi steps
\begin{align*} \int \csc ^3(2 a+2 b x) \sin ^2(a+b x) \, dx &=\frac{1}{8} \int \csc (a+b x) \sec ^3(a+b x) \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{x} \, dx,x,\tan (a+b x)\right )}{8 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{x}+x\right ) \, dx,x,\tan (a+b x)\right )}{8 b}\\ &=\frac{\log (\tan (a+b x))}{8 b}+\frac{\tan ^2(a+b x)}{16 b}\\ \end{align*}
Mathematica [A] time = 0.0355133, size = 36, normalized size = 1.2 \[ -\frac{-\sec ^2(a+b x)-2 \log (\sin (a+b x))+2 \log (\cos (a+b x))}{16 b} \]
Antiderivative was successfully verified.
[In]
Integrate[Csc[2*a + 2*b*x]^3*Sin[a + b*x]^2,x]
[Out]
-(2*Log[Cos[a + b*x]] - 2*Log[Sin[a + b*x]] - Sec[a + b*x]^2)/(16*b)
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Maple [A] time = 0.036, size = 27, normalized size = 0.9 \begin{align*}{\frac{1}{16\,b \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}+{\frac{\ln \left ( \tan \left ( bx+a \right ) \right ) }{8\,b}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(2*b*x+2*a)^3*sin(b*x+a)^2,x)
[Out]
1/16/b/cos(b*x+a)^2+1/8*ln(tan(b*x+a))/b
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Maxima [B] time = 1.12181, size = 865, normalized size = 28.83 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(2*b*x+2*a)^3*sin(b*x+a)^2,x, algorithm="maxima")
[Out]
1/16*(4*cos(4*b*x + 4*a)*cos(2*b*x + 2*a) + 8*cos(2*b*x + 2*a)^2 - (2*(2*cos(2*b*x + 2*a) + 1)*cos(4*b*x + 4*a
) + cos(4*b*x + 4*a)^2 + 4*cos(2*b*x + 2*a)^2 + sin(4*b*x + 4*a)^2 + 4*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 4*s
in(2*b*x + 2*a)^2 + 4*cos(2*b*x + 2*a) + 1)*log(cos(2*b*x)^2 + 2*cos(2*b*x)*cos(2*a) + cos(2*a)^2 + sin(2*b*x)
^2 - 2*sin(2*b*x)*sin(2*a) + sin(2*a)^2) + (2*(2*cos(2*b*x + 2*a) + 1)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 +
4*cos(2*b*x + 2*a)^2 + sin(4*b*x + 4*a)^2 + 4*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 4*sin(2*b*x + 2*a)^2 + 4*co
s(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)
+ (2*(2*cos(2*b*x + 2*a) + 1)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + 4*cos(2*b*x + 2*a)^2 + sin(4*b*x + 4*a)
^2 + 4*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 4*sin(2*b*x + 2*a)^2 + 4*cos(2*b*x + 2*a) + 1)*log(cos(b*x)^2 - 2*c
os(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 + 2*sin(b*x)*sin(a) + sin(a)^2) + 4*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) +
8*sin(2*b*x + 2*a)^2 + 4*cos(2*b*x + 2*a))/(b*cos(4*b*x + 4*a)^2 + 4*b*cos(2*b*x + 2*a)^2 + b*sin(4*b*x + 4*a
)^2 + 4*b*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 4*b*sin(2*b*x + 2*a)^2 + 2*(2*b*cos(2*b*x + 2*a) + b)*cos(4*b*x
+ 4*a) + 4*b*cos(2*b*x + 2*a) + b)
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Fricas [B] time = 0.498528, size = 155, normalized size = 5.17 \begin{align*} -\frac{\cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right )^{2}\right ) - \cos \left (b x + a\right )^{2} \log \left (-\frac{1}{4} \, \cos \left (b x + a\right )^{2} + \frac{1}{4}\right ) - 1}{16 \, b \cos \left (b x + a\right )^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(2*b*x+2*a)^3*sin(b*x+a)^2,x, algorithm="fricas")
[Out]
-1/16*(cos(b*x + a)^2*log(cos(b*x + a)^2) - cos(b*x + a)^2*log(-1/4*cos(b*x + a)^2 + 1/4) - 1)/(b*cos(b*x + a)
^2)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(2*b*x+2*a)**3*sin(b*x+a)**2,x)
[Out]
Timed out
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Giac [B] time = 1.63354, size = 991, normalized size = 33.03 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(2*b*x+2*a)^3*sin(b*x+a)^2,x, algorithm="giac")
[Out]
1/64*((12*tan(b*x + 4*a)*tan(1/2*a)^23 - tan(1/2*a)^24 + 3888*tan(b*x + 4*a)^2*tan(1/2*a)^20 - 1444*tan(b*x +
4*a)*tan(1/2*a)^21 + 168*tan(1/2*a)^22 - 51840*tan(b*x + 4*a)^2*tan(1/2*a)^18 + 32052*tan(b*x + 4*a)*tan(1/2*a
)^19 - 4074*tan(1/2*a)^20 + 274752*tan(b*x + 4*a)^2*tan(1/2*a)^16 - 260028*tan(b*x + 4*a)*tan(1/2*a)^17 + 4948
0*tan(1/2*a)^18 - 731520*tan(b*x + 4*a)^2*tan(1/2*a)^14 + 979064*tan(b*x + 4*a)*tan(1/2*a)^15 - 276687*tan(1/2
*a)^16 + 1021728*tan(b*x + 4*a)^2*tan(1/2*a)^12 - 1873128*tan(b*x + 4*a)*tan(1/2*a)^13 + 737808*tan(1/2*a)^14
- 731520*tan(b*x + 4*a)^2*tan(1/2*a)^10 + 1873128*tan(b*x + 4*a)*tan(1/2*a)^11 - 1009292*tan(1/2*a)^12 + 27475
2*tan(b*x + 4*a)^2*tan(1/2*a)^8 - 979064*tan(b*x + 4*a)*tan(1/2*a)^9 + 737808*tan(1/2*a)^10 - 51840*tan(b*x +
4*a)^2*tan(1/2*a)^6 + 260028*tan(b*x + 4*a)*tan(1/2*a)^7 - 276687*tan(1/2*a)^8 + 3888*tan(b*x + 4*a)^2*tan(1/2
*a)^4 - 32052*tan(b*x + 4*a)*tan(1/2*a)^5 + 49480*tan(1/2*a)^6 + 1444*tan(b*x + 4*a)*tan(1/2*a)^3 - 4074*tan(1
/2*a)^4 - 12*tan(b*x + 4*a)*tan(1/2*a) + 168*tan(1/2*a)^2 - 1)/((9*tan(1/2*a)^10 - 60*tan(1/2*a)^8 + 118*tan(1
/2*a)^6 - 60*tan(1/2*a)^4 + 9*tan(1/2*a)^2)*(6*tan(b*x + 4*a)*tan(1/2*a)^5 - tan(1/2*a)^6 - 20*tan(b*x + 4*a)*
tan(1/2*a)^3 + 15*tan(1/2*a)^4 + 6*tan(b*x + 4*a)*tan(1/2*a) - 15*tan(1/2*a)^2 + 1)^2) + 8*log(abs(tan(b*x + 4
*a)*tan(1/2*a)^6 - 15*tan(b*x + 4*a)*tan(1/2*a)^4 + 6*tan(1/2*a)^5 + 15*tan(b*x + 4*a)*tan(1/2*a)^2 - 20*tan(1
/2*a)^3 - tan(b*x + 4*a) + 6*tan(1/2*a))) - 8*log(abs(6*tan(b*x + 4*a)*tan(1/2*a)^5 - tan(1/2*a)^6 - 20*tan(b*
x + 4*a)*tan(1/2*a)^3 + 15*tan(1/2*a)^4 + 6*tan(b*x + 4*a)*tan(1/2*a) - 15*tan(1/2*a)^2 + 1)))/b