3.5 \(\int \csc (e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx\)

Optimal. Leaf size=63 \[ \frac{a^2 c \cos (e+f x)}{f}+\frac{a^2 c \sin (e+f x) \cos (e+f x)}{2 f}-\frac{a^2 c \tanh ^{-1}(\cos (e+f x))}{f}+\frac{1}{2} a^2 c x \]

[Out]

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

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Rubi [A]  time = 0.0898987, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2966, 3770, 2638, 2635, 8} \[ \frac{a^2 c \cos (e+f x)}{f}+\frac{a^2 c \sin (e+f x) \cos (e+f x)}{2 f}-\frac{a^2 c \tanh ^{-1}(\cos (e+f x))}{f}+\frac{1}{2} a^2 c x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2966

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

Rule 3770

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

Rule 2638

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

Rule 2635

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

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.0841551, size = 61, normalized size = 0.97 \[ \frac{a^2 c \left (\sin (2 (e+f x))+4 \cos (e+f x)+4 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )-4 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )-2 e+2 f x\right )}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c*(-2*e + 2*f*x + 4*Cos[e + f*x] - 4*Log[Cos[(e + f*x)/2]] + 4*Log[Sin[(e + f*x)/2]] + Sin[2*(e + f*x)]))
/(4*f)

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Maple [A]  time = 0.039, size = 78, normalized size = 1.2 \begin{align*}{\frac{{a}^{2}c\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) }{2\,f}}+{\frac{{a}^{2}cx}{2}}+{\frac{{a}^{2}ce}{2\,f}}+{\frac{{a}^{2}c\cos \left ( fx+e \right ) }{f}}+{\frac{{a}^{2}c\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)*(a+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x)

[Out]

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

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Maxima [A]  time = 0.973882, size = 99, normalized size = 1.57 \begin{align*} -\frac{{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c - 4 \,{\left (f x + e\right )} a^{2} c - 4 \, a^{2} c \cos \left (f x + e\right ) + 4 \, a^{2} c \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+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.1364, size = 201, normalized size = 3.19 \begin{align*} \frac{a^{2} c f x + a^{2} c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 2 \, a^{2} c \cos \left (f x + e\right ) - a^{2} c \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) + a^{2} c \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right )}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - a^{2} c \left (\int - \sin{\left (e + f x \right )} \csc{\left (e + f x \right )}\, dx + \int \sin ^{2}{\left (e + f x \right )} \csc{\left (e + f x \right )}\, dx + \int \sin ^{3}{\left (e + f x \right )} \csc{\left (e + f x \right )}\, dx + \int - \csc{\left (e + f x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(a+a*sin(f*x+e))**2*(c-c*sin(f*x+e)),x)

[Out]

-a**2*c*(Integral(-sin(e + f*x)*csc(e + f*x), x) + Integral(sin(e + f*x)**2*csc(e + f*x), x) + Integral(sin(e
+ f*x)**3*csc(e + f*x), x) + Integral(-csc(e + f*x), x))

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Giac [A]  time = 1.16443, size = 150, normalized size = 2.38 \begin{align*} \frac{{\left (f x + e\right )} a^{2} c + 2 \, a^{2} c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) \right |}\right ) - \frac{2 \,{\left (a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, a^{2} c\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+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x, algorithm="giac")

[Out]

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