3.172 \(\int \csc ^3(e+f x) (a+b \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=79 \[ -\frac{a \left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{5 a^2 b \cot (e+f x)}{2 f}-\frac{a^2 \cot (e+f x) \csc (e+f x) (a+b \sin (e+f x))}{2 f}+b^3 x \]

[Out]

b^3*x - (a*(a^2 + 6*b^2)*ArcTanh[Cos[e + f*x]])/(2*f) - (5*a^2*b*Cot[e + f*x])/(2*f) - (a^2*Cot[e + f*x]*Csc[e
+ f*x]*(a + b*Sin[e + f*x]))/(2*f)

________________________________________________________________________________________

Rubi [A]  time = 0.133165, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2792, 3021, 2735, 3770} \[ -\frac{a \left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{5 a^2 b \cot (e+f x)}{2 f}-\frac{a^2 \cot (e+f x) \csc (e+f x) (a+b \sin (e+f x))}{2 f}+b^3 x \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^3*(a + b*Sin[e + f*x])^3,x]

[Out]

b^3*x - (a*(a^2 + 6*b^2)*ArcTanh[Cos[e + f*x]])/(2*f) - (5*a^2*b*Cot[e + f*x])/(2*f) - (a^2*Cot[e + f*x]*Csc[e
+ f*x]*(a + b*Sin[e + f*x]))/(2*f)

Rule 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 3770

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

Rubi steps

\begin{align*} \int \csc ^3(e+f x) (a+b \sin (e+f x))^3 \, dx &=-\frac{a^2 \cot (e+f x) \csc (e+f x) (a+b \sin (e+f x))}{2 f}+\frac{1}{2} \int \csc ^2(e+f x) \left (5 a^2 b+a \left (a^2+6 b^2\right ) \sin (e+f x)+2 b^3 \sin ^2(e+f x)\right ) \, dx\\ &=-\frac{5 a^2 b \cot (e+f x)}{2 f}-\frac{a^2 \cot (e+f x) \csc (e+f x) (a+b \sin (e+f x))}{2 f}+\frac{1}{2} \int \csc (e+f x) \left (a \left (a^2+6 b^2\right )+2 b^3 \sin (e+f x)\right ) \, dx\\ &=b^3 x-\frac{5 a^2 b \cot (e+f x)}{2 f}-\frac{a^2 \cot (e+f x) \csc (e+f x) (a+b \sin (e+f x))}{2 f}+\frac{1}{2} \left (a \left (a^2+6 b^2\right )\right ) \int \csc (e+f x) \, dx\\ &=b^3 x-\frac{a \left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{5 a^2 b \cot (e+f x)}{2 f}-\frac{a^2 \cot (e+f x) \csc (e+f x) (a+b \sin (e+f x))}{2 f}\\ \end{align*}

Mathematica [A]  time = 0.64197, size = 152, normalized size = 1.92 \[ \frac{12 a^2 b \tan \left (\frac{1}{2} (e+f x)\right )-12 a^2 b \cot \left (\frac{1}{2} (e+f x)\right )+a^3 \left (-\csc ^2\left (\frac{1}{2} (e+f x)\right )\right )+a^3 \sec ^2\left (\frac{1}{2} (e+f x)\right )+4 a^3 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )-4 a^3 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )+24 a b^2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )-24 a b^2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )+8 b^3 e+8 b^3 f x}{8 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^3*(a + b*Sin[e + f*x])^3,x]

[Out]

(8*b^3*e + 8*b^3*f*x - 12*a^2*b*Cot[(e + f*x)/2] - a^3*Csc[(e + f*x)/2]^2 - 4*a^3*Log[Cos[(e + f*x)/2]] - 24*a
*b^2*Log[Cos[(e + f*x)/2]] + 4*a^3*Log[Sin[(e + f*x)/2]] + 24*a*b^2*Log[Sin[(e + f*x)/2]] + a^3*Sec[(e + f*x)/
2]^2 + 12*a^2*b*Tan[(e + f*x)/2])/(8*f)

________________________________________________________________________________________

Maple [A]  time = 0.063, size = 99, normalized size = 1.3 \begin{align*} -{\frac{{a}^{3}\csc \left ( fx+e \right ) \cot \left ( fx+e \right ) }{2\,f}}+{\frac{{a}^{3}\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{2\,f}}-3\,{\frac{{a}^{2}b\cot \left ( fx+e \right ) }{f}}+3\,{\frac{a{b}^{2}\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{f}}+{b}^{3}x+{\frac{{b}^{3}e}{f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*a^3*csc(f*x+e)*cot(f*x+e)+1/2/f*a^3*ln(csc(f*x+e)-cot(f*x+e))-3*a^2*b*cot(f*x+e)/f+3/f*a*b^2*ln(csc(f*x
+e)-cot(f*x+e))+b^3*x+1/f*b^3*e

________________________________________________________________________________________

Maxima [A]  time = 2.37047, size = 138, normalized size = 1.75 \begin{align*} \frac{4 \,{\left (f x + e\right )} b^{3} + a^{3}{\left (\frac{2 \, \cos \left (f x + e\right )}{\cos \left (f x + e\right )^{2} - 1} - \log \left (\cos \left (f x + e\right ) + 1\right ) + \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - 6 \, a b^{2}{\left (\log \left (\cos \left (f x + e\right ) + 1\right ) - \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - \frac{12 \, a^{2} b}{\tan \left (f x + e\right )}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(f*x + e)*b^3 + a^3*(2*cos(f*x + e)/(cos(f*x + e)^2 - 1) - log(cos(f*x + e) + 1) + log(cos(f*x + e) - 1
)) - 6*a*b^2*(log(cos(f*x + e) + 1) - log(cos(f*x + e) - 1)) - 12*a^2*b/tan(f*x + e))/f

________________________________________________________________________________________

Fricas [B]  time = 1.72809, size = 383, normalized size = 4.85 \begin{align*} \frac{4 \, b^{3} f x \cos \left (f x + e\right )^{2} - 4 \, b^{3} f x + 12 \, a^{2} b \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 2 \, a^{3} \cos \left (f x + e\right ) +{\left (a^{3} + 6 \, a b^{2} -{\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) -{\left (a^{3} + 6 \, a b^{2} -{\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right )}{4 \,{\left (f \cos \left (f x + e\right )^{2} - f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*b^3*f*x*cos(f*x + e)^2 - 4*b^3*f*x + 12*a^2*b*cos(f*x + e)*sin(f*x + e) + 2*a^3*cos(f*x + e) + (a^3 + 6
*a*b^2 - (a^3 + 6*a*b^2)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) - (a^3 + 6*a*b^2 - (a^3 + 6*a*b^2)*cos(f*
x + e)^2)*log(-1/2*cos(f*x + e) + 1/2))/(f*cos(f*x + e)^2 - f)

________________________________________________________________________________________

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(f*x+e)**3*(a+b*sin(f*x+e))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 2.01914, size = 192, normalized size = 2.43 \begin{align*} \frac{a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 8 \,{\left (f x + e\right )} b^{3} + 12 \, a^{2} b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 4 \,{\left (a^{3} + 6 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) \right |}\right ) - \frac{6 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 36 \, a b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 12 \, a^{2} b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a^{3}}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^3*tan(1/2*f*x + 1/2*e)^2 + 8*(f*x + e)*b^3 + 12*a^2*b*tan(1/2*f*x + 1/2*e) + 4*(a^3 + 6*a*b^2)*log(abs(
tan(1/2*f*x + 1/2*e))) - (6*a^3*tan(1/2*f*x + 1/2*e)^2 + 36*a*b^2*tan(1/2*f*x + 1/2*e)^2 + 12*a^2*b*tan(1/2*f*
x + 1/2*e) + a^3)/tan(1/2*f*x + 1/2*e)^2)/f