3.30 \(\int \csc ^3(2 a+2 b x) \sin ^3(a+b x) \, dx\)

Optimal. Leaf size=34 \[ \frac {\tanh ^{-1}(\sin (a+b x))}{16 b}+\frac {\tan (a+b x) \sec (a+b x)}{16 b} \]

[Out]

1/16*arctanh(sin(b*x+a))/b+1/16*sec(b*x+a)*tan(b*x+a)/b

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Rubi [A]  time = 0.04, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4288, 3768, 3770} \[ \frac {\tanh ^{-1}(\sin (a+b x))}{16 b}+\frac {\tan (a+b x) \sec (a+b x)}{16 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[a + b*x]]/(16*b) + (Sec[a + b*x]*Tan[a + b*x])/(16*b)

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 ^3(a+b x) \, dx &=\frac {1}{8} \int \sec ^3(a+b x) \, dx\\ &=\frac {\sec (a+b x) \tan (a+b x)}{16 b}+\frac {1}{16} \int \sec (a+b x) \, dx\\ &=\frac {\tanh ^{-1}(\sin (a+b x))}{16 b}+\frac {\sec (a+b x) \tan (a+b x)}{16 b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 38, normalized size = 1.12 \[ \frac {1}{8} \left (\frac {\tanh ^{-1}(\sin (a+b x))}{2 b}+\frac {\tan (a+b x) \sec (a+b x)}{2 b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTanh[Sin[a + b*x]]/(2*b) + (Sec[a + b*x]*Tan[a + b*x])/(2*b))/8

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fricas [B]  time = 0.45, size = 61, normalized size = 1.79 \[ \frac {\cos \left (b x + a\right )^{2} \log \left (\sin \left (b x + a\right ) + 1\right ) - \cos \left (b x + a\right )^{2} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, \sin \left (b x + a\right )}{32 \, b \cos \left (b x + a\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(cos(b*x + a)^2*log(sin(b*x + a) + 1) - cos(b*x + a)^2*log(-sin(b*x + a) + 1) + 2*sin(b*x + a))/(b*cos(b*
x + a)^2)

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giac [B]  time = 2.95, size = 1111, normalized size = 32.68 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2*(tan(1/2*b*x + 2*a)^3*tan(1/2*a)^24 + 30*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^22 - 6*tan(1/2*b*x + 2*a)^2*
tan(1/2*a)^23 + tan(1/2*b*x + 2*a)*tan(1/2*a)^24 - 756*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^20 + 614*tan(1/2*b*x +
2*a)^2*tan(1/2*a)^21 - 114*tan(1/2*b*x + 2*a)*tan(1/2*a)^22 + 6*tan(1/2*a)^23 + 2058*tan(1/2*b*x + 2*a)^3*tan(
1/2*a)^18 - 4578*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^19 + 1932*tan(1/2*b*x + 2*a)*tan(1/2*a)^20 - 182*tan(1/2*a)^2
1 - 27*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^16 + 6210*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^17 - 7462*tan(1/2*b*x + 2*a)*
tan(1/2*a)^18 + 1554*tan(1/2*a)^19 - 9396*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^14 + 15588*tan(1/2*b*x + 2*a)^2*tan(
1/2*a)^15 - 2331*tan(1/2*b*x + 2*a)*tan(1/2*a)^16 - 2178*tan(1/2*a)^17 - 21924*tan(1/2*b*x + 2*a)^2*tan(1/2*a)
^13 + 26028*tan(1/2*b*x + 2*a)*tan(1/2*a)^14 - 5668*tan(1/2*a)^15 + 9396*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^10 -
21924*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^11 + 6468*tan(1/2*a)^13 + 27*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^8 + 15588*t
an(1/2*b*x + 2*a)^2*tan(1/2*a)^9 - 26028*tan(1/2*b*x + 2*a)*tan(1/2*a)^10 + 6468*tan(1/2*a)^11 - 2058*tan(1/2*
b*x + 2*a)^3*tan(1/2*a)^6 + 6210*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^7 + 2331*tan(1/2*b*x + 2*a)*tan(1/2*a)^8 - 56
68*tan(1/2*a)^9 + 756*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^4 - 4578*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^5 + 7462*tan(1/
2*b*x + 2*a)*tan(1/2*a)^6 - 2178*tan(1/2*a)^7 - 30*tan(1/2*b*x + 2*a)^3*tan(1/2*a)^2 + 614*tan(1/2*b*x + 2*a)^
2*tan(1/2*a)^3 - 1932*tan(1/2*b*x + 2*a)*tan(1/2*a)^4 + 1554*tan(1/2*a)^5 - tan(1/2*b*x + 2*a)^3 - 6*tan(1/2*b
*x + 2*a)^2*tan(1/2*a) + 114*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 - 182*tan(1/2*a)^3 - tan(1/2*b*x + 2*a) + 6*tan(1
/2*a))/((tan(1/2*a)^12 - 30*tan(1/2*a)^10 + 255*tan(1/2*a)^8 - 452*tan(1/2*a)^6 + 255*tan(1/2*a)^4 - 30*tan(1/
2*a)^2 + 1)*(tan(1/2*b*x + 2*a)^2*tan(1/2*a)^6 - 15*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^4 + 12*tan(1/2*b*x + 2*a)*
tan(1/2*a)^5 - tan(1/2*a)^6 + 15*tan(1/2*b*x + 2*a)^2*tan(1/2*a)^2 - 40*tan(1/2*b*x + 2*a)*tan(1/2*a)^3 + 15*t
an(1/2*a)^4 - tan(1/2*b*x + 2*a)^2 + 12*tan(1/2*b*x + 2*a)*tan(1/2*a) - 15*tan(1/2*a)^2 + 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)) + log(abs(tan(1/2*b*x + 2*a)*tan(1/2*a)^3 - 3*ta
n(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 = 10.77, size = 38, normalized size = 1.12 \[ \frac {\sec \left (b x +a \right ) \tan \left (b x +a \right )}{16 b}+\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{16 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*sec(b*x+a)*tan(b*x+a)/b+1/16/b*ln(sec(b*x+a)+tan(b*x+a))

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maxima [B]  time = 0.47, size = 480, normalized size = 14.12 \[ \frac {4 \, {\left (\sin \left (3 \, b x + 3 \, a\right ) - \sin \left (b x + a\right )\right )} \cos \left (4 \, b x + 4 \, a\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 (\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 \, {\left (\cos \left (3 \, b x + 3 \, a\right ) - \cos \left (b x + a\right )\right )} \sin \left (4 \, b x + 4 \, a\right ) + 4 \, {\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \sin \left (3 \, b x + 3 \, a\right ) - 8 \, \cos \left (3 \, b x + 3 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 8 \, \cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right ) - 8 \, \cos \left (2 \, b x + 2 \, a\right ) \sin \left (b x + a\right ) - 4 \, \sin \left (b x + a\right )}{32 \, {\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)^3,x, algorithm="maxima")

[Out]

1/32*(4*(sin(3*b*x + 3*a) - sin(b*x + a))*cos(4*b*x + 4*a) - (2*(2*cos(2*b*x + 2*a) + 1)*cos(4*b*x + 4*a) + co
s(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*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(3*b*x + 3*a) - cos(b*x + a))*sin(4*b*x + 4*a) + 4*(2*c
os(2*b*x + 2*a) + 1)*sin(3*b*x + 3*a) - 8*cos(3*b*x + 3*a)*sin(2*b*x + 2*a) + 8*cos(b*x + a)*sin(2*b*x + 2*a)
- 8*cos(2*b*x + 2*a)*sin(b*x + a) - 4*sin(b*x + 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)*co
s(4*b*x + 4*a) + 4*b*cos(2*b*x + 2*a) + b)

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mupad [B]  time = 0.17, size = 36, normalized size = 1.06 \[ \frac {\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{16\,b}-\frac {\sin \left (a+b\,x\right )}{16\,b\,\left ({\sin \left (a+b\,x\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atanh(sin(a + b*x))/(16*b) - sin(a + b*x)/(16*b*(sin(a + b*x)^2 - 1))

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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)**3,x)

[Out]

Timed out

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