3.171 \(\int \csc ^2(e+f x) (a+b \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=68 \[ \frac{b \left (a^2-b^2\right ) \cos (e+f x)}{f}-\frac{3 a^2 b \tanh ^{-1}(\cos (e+f x))}{f}-\frac{a^2 \cot (e+f x) (a+b \sin (e+f x))}{f}+3 a b^2 x \]

[Out]

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

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Rubi [A]  time = 0.119885, antiderivative size = 68, 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, 3023, 2735, 3770} \[ \frac{b \left (a^2-b^2\right ) \cos (e+f x)}{f}-\frac{3 a^2 b \tanh ^{-1}(\cos (e+f x))}{f}-\frac{a^2 \cot (e+f x) (a+b \sin (e+f x))}{f}+3 a b^2 x \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^2*(a + b*Sin[e + f*x])^3,x]

[Out]

3*a*b^2*x - (3*a^2*b*ArcTanh[Cos[e + f*x]])/f + (b*(a^2 - b^2)*Cos[e + f*x])/f - (a^2*Cot[e + f*x]*(a + b*Sin[
e + f*x]))/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 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 ^2(e+f x) (a+b \sin (e+f x))^3 \, dx &=-\frac{a^2 \cot (e+f x) (a+b \sin (e+f x))}{f}+\int \csc (e+f x) \left (3 a^2 b+3 a b^2 \sin (e+f x)-b \left (a^2-b^2\right ) \sin ^2(e+f x)\right ) \, dx\\ &=\frac{b \left (a^2-b^2\right ) \cos (e+f x)}{f}-\frac{a^2 \cot (e+f x) (a+b \sin (e+f x))}{f}+\int \csc (e+f x) \left (3 a^2 b+3 a b^2 \sin (e+f x)\right ) \, dx\\ &=3 a b^2 x+\frac{b \left (a^2-b^2\right ) \cos (e+f x)}{f}-\frac{a^2 \cot (e+f x) (a+b \sin (e+f x))}{f}+\left (3 a^2 b\right ) \int \csc (e+f x) \, dx\\ &=3 a b^2 x-\frac{3 a^2 b \tanh ^{-1}(\cos (e+f x))}{f}+\frac{b \left (a^2-b^2\right ) \cos (e+f x)}{f}-\frac{a^2 \cot (e+f x) (a+b \sin (e+f x))}{f}\\ \end{align*}

Mathematica [A]  time = 0.517057, size = 87, normalized size = 1.28 \[ \frac{a^3 \tan \left (\frac{1}{2} (e+f x)\right )+a^3 \left (-\cot \left (\frac{1}{2} (e+f x)\right )\right )+6 a b \left (a \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )-a \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )+b (e+f x)\right )-2 b^3 \cos (e+f x)}{2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^2*(a + b*Sin[e + f*x])^3,x]

[Out]

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

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Maple [A]  time = 0.052, size = 72, normalized size = 1.1 \begin{align*} 3\,a{b}^{2}x-{\frac{{a}^{3}\cot \left ( fx+e \right ) }{f}}-{\frac{{b}^{3}\cos \left ( fx+e \right ) }{f}}+3\,{\frac{{a}^{2}b\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{f}}+3\,{\frac{a{b}^{2}e}{f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.65011, size = 92, normalized size = 1.35 \begin{align*} \frac{6 \,{\left (f x + e\right )} a b^{2} - 3 \, a^{2} b{\left (\log \left (\cos \left (f x + e\right ) + 1\right ) - \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - 2 \, b^{3} \cos \left (f x + e\right ) - \frac{2 \, a^{3}}{\tan \left (f x + e\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.73625, size = 266, normalized size = 3.91 \begin{align*} -\frac{3 \, a^{2} b \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) \sin \left (f x + e\right ) - 3 \, a^{2} b \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) \sin \left (f x + e\right ) + 2 \, a^{3} \cos \left (f x + e\right ) - 2 \,{\left (3 \, a b^{2} f x - b^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \, f \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(f*x+e)**2*(a+b*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [B]  time = 2.33008, size = 197, normalized size = 2.9 \begin{align*} \frac{6 \,{\left (f x + e\right )} a b^{2} + 6 \, a^{2} b \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) \right |}\right ) + a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \frac{2 \, a^{2} b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, a^{2} b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 4 \, b^{3} \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 )^{3} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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