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

Optimal. Leaf size=74 \[ \frac{1}{2} b x \left (6 a^2+b^2\right )-\frac{a^3 \tanh ^{-1}(\cos (e+f x))}{f}-\frac{5 a b^2 \cos (e+f x)}{2 f}-\frac{b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.114637, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2793, 3023, 2735, 3770} \[ \frac{1}{2} b x \left (6 a^2+b^2\right )-\frac{a^3 \tanh ^{-1}(\cos (e+f x))}{f}-\frac{5 a b^2 \cos (e+f x)}{2 f}-\frac{b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2793

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

Rule 3023

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

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

Mathematica [A]  time = 0.159826, size = 81, normalized size = 1.09 \[ -\frac{-2 b \left (6 a^2+b^2\right ) (e+f x)-4 a^3 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )+4 a^3 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )+12 a b^2 \cos (e+f x)+b^3 \sin (2 (e+f x))}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-2*b*(6*a^2 + b^2)*(e + f*x) + 12*a*b^2*Cos[e + f*x] + 4*a^3*Log[Cos[(e + f*x)/2]] - 4*a^3*Log[Sin[(e + f*x)
/2]] + b^3*Sin[2*(e + f*x)])/(4*f)

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Maple [A]  time = 0.056, size = 92, normalized size = 1.2 \begin{align*}{\frac{{a}^{3}\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{f}}+3\,{a}^{2}bx+3\,{\frac{{a}^{2}be}{f}}-3\,{\frac{a{b}^{2}\cos \left ( fx+e \right ) }{f}}-{\frac{{b}^{3}\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2\,f}}+{\frac{{b}^{3}x}{2}}+{\frac{{b}^{3}e}{2\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.72161, size = 96, normalized size = 1.3 \begin{align*} \frac{12 \,{\left (f x + e\right )} a^{2} b +{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} b^{3} - 12 \, a b^{2} \cos \left (f x + e\right ) - 4 \, a^{3} \log \left (\cot \left (f x + e\right ) + \csc \left (f x + e\right )\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.85242, size = 208, normalized size = 2.81 \begin{align*} -\frac{b^{3} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 6 \, a b^{2} \cos \left (f x + e\right ) + a^{3} \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) - a^{3} \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) -{\left (6 \, a^{2} b + b^{3}\right )} f x}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sin{\left (e + f x \right )}\right )^{3} \csc{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.25302, size = 154, normalized size = 2.08 \begin{align*} \frac{2 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) \right |}\right ) +{\left (6 \, a^{2} b + b^{3}\right )}{\left (f x + e\right )} + \frac{2 \,{\left (b^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 6 \, a b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - b^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 6 \, a b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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