3.7 \(\int \csc ^3(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx\)

Optimal. Leaf size=64 \[ -\frac{a^2 c \cot (e+f x)}{f}+\frac{a^2 c \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{a^2 c \cot (e+f x) \csc (e+f x)}{2 f}+a^2 (-c) x \]

[Out]

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

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Rubi [A]  time = 0.111457, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {2966, 3770, 3767, 8, 3768} \[ -\frac{a^2 c \cot (e+f x)}{f}+\frac{a^2 c \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{a^2 c \cot (e+f x) \csc (e+f x)}{2 f}+a^2 (-c) x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*c*x) + (a^2*c*ArcTanh[Cos[e + f*x]])/(2*f) - (a^2*c*Cot[e + f*x])/f - (a^2*c*Cot[e + f*x]*Csc[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 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 3768

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

Rubi steps

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

Mathematica [A]  time = 0.711202, size = 95, normalized size = 1.48 \[ -\frac{a^2 c \left (-4 \tan \left (\frac{1}{2} (e+f x)\right )+4 \cot \left (\frac{1}{2} (e+f x)\right )+\csc ^2\left (\frac{1}{2} (e+f x)\right )-\sec ^2\left (\frac{1}{2} (e+f x)\right )+4 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )-4 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )+8 e+8 f x\right )}{8 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.049, size = 80, normalized size = 1.3 \begin{align*} -{a}^{2}cx-{\frac{{a}^{2}ce}{f}}-{\frac{{a}^{2}c\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{2\,f}}-{\frac{{a}^{2}c\cot \left ( fx+e \right ) }{f}}-{\frac{{a}^{2}c\cot \left ( fx+e \right ) \csc \left ( fx+e \right ) }{2\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.969109, size = 142, normalized size = 2.22 \begin{align*} -\frac{4 \,{\left (f x + e\right )} a^{2} c - a^{2} c{\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 \, a^{2} c{\left (\log \left (\cos \left (f x + e\right ) + 1\right ) - \log \left (\cos \left (f x + e\right ) - 1\right )\right )} + \frac{4 \, a^{2} c}{\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+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.06303, size = 343, normalized size = 5.36 \begin{align*} -\frac{4 \, a^{2} c f x \cos \left (f x + e\right )^{2} - 4 \, a^{2} c f x - 4 \, 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 ) -{\left (a^{2} c \cos \left (f x + e\right )^{2} - a^{2} c\right )} \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) +{\left (a^{2} c \cos \left (f x + e\right )^{2} - a^{2} c\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+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.3244, size = 166, normalized size = 2.59 \begin{align*} \frac{a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 8 \,{\left (f x + e\right )} a^{2} c - 4 \, a^{2} c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) \right |}\right ) + 4 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \frac{6 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 4 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - a^{2} c}{\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+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x, algorithm="giac")

[Out]

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