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]

1/8*ln(tan(b*x+a))/b+1/16*tan(b*x+a)^2/b

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Rubi [A]  time = 0.05, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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 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]]

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 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 ^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*}

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Mathematica [A]  time = 0.04, size = 36, normalized size = 1.20 \[ -\frac {-\sec ^2(a+b x)-2 \log (\sin (a+b x))+2 \log (\cos (a+b x))}{16 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(2*Log[Cos[a + b*x]] - 2*Log[Sin[a + b*x]] - Sec[a + b*x]^2)/b

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fricas [B]  time = 0.43, size = 56, normalized size = 1.87 \[ -\frac {\cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right )^{2}\right ) - \cos \left (b x + a\right )^{2} \log \left (-\frac {1}{4} \, \cos \left (b x + a\right )^{2} + \frac {1}{4}\right ) - 1}{16 \, b \cos \left (b x + a\right )^{2}} \]

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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giac [B]  time = 3.55, size = 734, normalized size = 24.47 \[ \text {result too large to display} \]

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

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maple [A]  time = 0.87, size = 27, normalized size = 0.90 \[ \frac {1}{16 b \cos \left (b x +a \right )^{2}}+\frac {\ln \left (\tan \left (b x +a \right )\right )}{8 b} \]

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 = 0.35, size = 641, normalized size = 21.37 \[ \frac {4 \, \cos \left (4 \, b x + 4 \, a\right ) \cos \left (2 \, b x + 2 \, a\right ) + 8 \, \cos \left (2 \, b x + 2 \, a\right )^{2} - {\left (2 \, {\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \cos \left (4 \, b x + 4 \, a\right ) + \cos \left (4 \, b x + 4 \, a\right )^{2} + 4 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 4 \, \sin \left (4 \, b x + 4 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 4 \, \sin \left (2 \, b x + 2 \, a\right )^{2} + 4 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, a\right ) + \sin \left (2 \, a\right )^{2}\right ) + {\left (2 \, {\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \cos \left (4 \, b x + 4 \, a\right ) + \cos \left (4 \, b x + 4 \, a\right )^{2} + 4 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 4 \, \sin \left (4 \, b x + 4 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 4 \, \sin \left (2 \, b x + 2 \, a\right )^{2} + 4 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \log \left (\cos \left (b x\right )^{2} + 2 \, \cos \left (b x\right ) \cos \relax (a) + \cos \relax (a)^{2} + \sin \left (b x\right )^{2} - 2 \, \sin \left (b x\right ) \sin \relax (a) + \sin \relax (a)^{2}\right ) + {\left (2 \, {\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \cos \left (4 \, b x + 4 \, a\right ) + \cos \left (4 \, b x + 4 \, a\right )^{2} + 4 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 4 \, \sin \left (4 \, b x + 4 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 4 \, \sin \left (2 \, b x + 2 \, a\right )^{2} + 4 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \log \left (\cos \left (b x\right )^{2} - 2 \, \cos \left (b x\right ) \cos \relax (a) + \cos \relax (a)^{2} + \sin \left (b x\right )^{2} + 2 \, \sin \left (b x\right ) \sin \relax (a) + \sin \relax (a)^{2}\right ) + 4 \, \sin \left (4 \, b x + 4 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 8 \, \sin \left (2 \, b x + 2 \, a\right )^{2} + 4 \, \cos \left (2 \, b x + 2 \, a\right )}{16 \, {\left (b \cos \left (4 \, b x + 4 \, a\right )^{2} + 4 \, b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (4 \, b x + 4 \, a\right )^{2} + 4 \, b \sin \left (4 \, b x + 4 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 4 \, b \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, {\left (2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )} \cos \left (4 \, b x + 4 \, a\right ) + 4 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )}} \]

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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mupad [B]  time = 0.15, size = 35, normalized size = 1.17 \[ \frac {\frac {\ln \left ({\sin \left (a+b\,x\right )}^2\right )}{16}-\frac {\ln \left (\cos \left (a+b\,x\right )\right )}{8}+\frac {1}{16\,{\cos \left (a+b\,x\right )}^2}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(sin(a + b*x)^2)/16 - log(cos(a + b*x))/8 + 1/(16*cos(a + b*x)^2))/b

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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