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

Optimal. Leaf size=59 \[ -\frac{\left (a^2+2 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{a^2 \cot (e+f x) \csc (e+f x)}{2 f}-\frac{2 a b \cot (e+f x)}{f} \]

[Out]

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

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Rubi [A]  time = 0.0756538, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2789, 3767, 8, 3012, 3770} \[ -\frac{\left (a^2+2 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{a^2 \cot (e+f x) \csc (e+f x)}{2 f}-\frac{2 a b \cot (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3012

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

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

Mathematica [B]  time = 0.458003, size = 133, normalized size = 2.25 \[ \frac{a^2 \left (-\csc ^2\left (\frac{1}{2} (e+f x)\right )\right )+a^2 \sec ^2\left (\frac{1}{2} (e+f x)\right )+4 a^2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )-4 a^2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )+8 a b \tan \left (\frac{1}{2} (e+f x)\right )-8 a b \cot \left (\frac{1}{2} (e+f x)\right )+8 b^2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )-8 b^2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )}{8 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.052, size = 82, normalized size = 1.4 \begin{align*} -{\frac{{a}^{2}\cot \left ( fx+e \right ) \csc \left ( fx+e \right ) }{2\,f}}+{\frac{{a}^{2}\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{2\,f}}-2\,{\frac{ab\cot \left ( fx+e \right ) }{f}}+{\frac{{b}^{2}\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{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))^2,x)

[Out]

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

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Maxima [A]  time = 1.973, size = 120, normalized size = 2.03 \begin{align*} \frac{a^{2}{\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 )} - 2 \, 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{8 \, a 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))^2,x, algorithm="maxima")

[Out]

1/4*(a^2*(2*cos(f*x + e)/(cos(f*x + e)^2 - 1) - log(cos(f*x + e) + 1) + log(cos(f*x + e) - 1)) - 2*b^2*(log(co
s(f*x + e) + 1) - log(cos(f*x + e) - 1)) - 8*a*b/tan(f*x + e))/f

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Fricas [B]  time = 1.92625, size = 316, normalized size = 5.36 \begin{align*} \frac{8 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 2 \, a^{2} \cos \left (f x + e\right ) -{\left ({\left (a^{2} + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) +{\left ({\left (a^{2} + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, b^{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))^2,x, algorithm="fricas")

[Out]

1/4*(8*a*b*cos(f*x + e)*sin(f*x + e) + 2*a^2*cos(f*x + e) - ((a^2 + 2*b^2)*cos(f*x + e)^2 - a^2 - 2*b^2)*log(1
/2*cos(f*x + e) + 1/2) + ((a^2 + 2*b^2)*cos(f*x + e)^2 - a^2 - 2*b^2)*log(-1/2*cos(f*x + e) + 1/2))/(f*cos(f*x
+ e)^2 - f)

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

[Out]

Timed out

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Giac [B]  time = 1.66352, size = 169, normalized size = 2.86 \begin{align*} \frac{a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 8 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 4 \,{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) \right |}\right ) - \frac{6 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 12 \, b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 8 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a^{2}}{\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))^2,x, algorithm="giac")

[Out]

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