3.8 \(\int \csc ^4(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx\)

Optimal. Leaf size=61 \[ -\frac{a^2 c \cot ^3(e+f x)}{3 f}+\frac{a^2 c \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{a^2 c \cot (e+f x) \csc (e+f x)}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.162082, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {2950, 2706, 2607, 30, 2611, 3770} \[ -\frac{a^2 c \cot ^3(e+f x)}{3 f}+\frac{a^2 c \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{a^2 c \cot (e+f x) \csc (e+f x)}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x]),x]

[Out]

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

Rule 2950

Int[sin[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[a^n*c^n, Int[Tan[e + f*x]^p*(a + b*Sin[e + f*x])^(m - n), x], x] /; FreeQ[{a,
b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[p + 2*n, 0] && IntegerQ[n]

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

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 ^4(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx &=\left (a^2 c^2\right ) \int \frac{\cot ^4(e+f x)}{c-c \sin (e+f x)} \, dx\\ &=\left (a^2 c\right ) \int \cot ^2(e+f x) \csc (e+f x) \, dx+\left (a^2 c\right ) \int \cot ^2(e+f x) \csc ^2(e+f x) \, dx\\ &=-\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}\left (\int x^2 \, dx,x,-\cot (e+f x)\right )}{f}\\ &=\frac{a^2 c \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{a^2 c \cot ^3(e+f x)}{3 f}-\frac{a^2 c \cot (e+f x) \csc (e+f x)}{2 f}\\ \end{align*}

Mathematica [B]  time = 0.0741181, size = 172, normalized size = 2.82 \[ a^2 c \left (-\frac{\tan \left (\frac{1}{2} (e+f x)\right )}{6 f}+\frac{\cot \left (\frac{1}{2} (e+f x)\right )}{6 f}-\frac{\csc ^2\left (\frac{1}{2} (e+f x)\right )}{8 f}+\frac{\sec ^2\left (\frac{1}{2} (e+f x)\right )}{8 f}-\frac{\log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )}{2 f}+\frac{\log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )}{2 f}-\frac{\cot \left (\frac{1}{2} (e+f x)\right ) \csc ^2\left (\frac{1}{2} (e+f x)\right )}{24 f}+\frac{\tan \left (\frac{1}{2} (e+f x)\right ) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{24 f}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x]),x]

[Out]

a^2*c*(Cot[(e + f*x)/2]/(6*f) - Csc[(e + f*x)/2]^2/(8*f) - (Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^2)/(24*f) + Log[
Cos[(e + f*x)/2]]/(2*f) - Log[Sin[(e + f*x)/2]]/(2*f) + Sec[(e + f*x)/2]^2/(8*f) - Tan[(e + f*x)/2]/(6*f) + (S
ec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/(24*f))

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Maple [A]  time = 0.048, size = 86, normalized size = 1.4 \begin{align*} -{\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 ) }{3\,f}}-{\frac{{a}^{2}c\cot \left ( fx+e \right ) \csc \left ( fx+e \right ) }{2\,f}}-{\frac{{a}^{2}c\cot \left ( fx+e \right ) \left ( \csc \left ( fx+e \right ) \right ) ^{2}}{3\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^4*(a+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x)

[Out]

-1/2/f*a^2*c*ln(csc(f*x+e)-cot(f*x+e))+1/3*a^2*c*cot(f*x+e)/f-1/2*a^2*c*cot(f*x+e)*csc(f*x+e)/f-1/3/f*a^2*c*co
t(f*x+e)*csc(f*x+e)^2

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Maxima [B]  time = 0.973217, size = 162, normalized size = 2.66 \begin{align*} \frac{3 \, 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 )} + 6 \, 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{12 \, a^{2} c}{\tan \left (f x + e\right )} - \frac{4 \,{\left (3 \, \tan \left (f x + e\right )^{2} + 1\right )} a^{2} c}{\tan \left (f x + e\right )^{3}}}{12 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x, algorithm="maxima")

[Out]

1/12*(3*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)) + 6*a^2*c*
(log(cos(f*x + e) + 1) - log(cos(f*x + e) - 1)) + 12*a^2*c/tan(f*x + e) - 4*(3*tan(f*x + e)^2 + 1)*a^2*c/tan(f
*x + e)^3)/f

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Fricas [B]  time = 1.94318, size = 348, normalized size = 5.7 \begin{align*} \frac{4 \, a^{2} c \cos \left (f x + e\right )^{3} + 6 \, a^{2} c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 3 \,{\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 ) \sin \left (f x + e\right ) - 3 \,{\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 ) \sin \left (f x + e\right )}{12 \,{\left (f \cos \left (f x + e\right )^{2} - f\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x, algorithm="fricas")

[Out]

1/12*(4*a^2*c*cos(f*x + e)^3 + 6*a^2*c*cos(f*x + e)*sin(f*x + e) + 3*(a^2*c*cos(f*x + e)^2 - a^2*c)*log(1/2*co
s(f*x + e) + 1/2)*sin(f*x + e) - 3*(a^2*c*cos(f*x + e)^2 - a^2*c)*log(-1/2*cos(f*x + e) + 1/2)*sin(f*x + e))/(
(f*cos(f*x + e)^2 - 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)**4*(a+a*sin(f*x+e))**2*(c-c*sin(f*x+e)),x)

[Out]

Timed out

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Giac [B]  time = 1.31485, size = 198, normalized size = 3.25 \begin{align*} \frac{a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 3 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 12 \, a^{2} c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) \right |}\right ) - 3 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \frac{22 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 3 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 3 \, 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 )^{3}}}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x, algorithm="giac")

[Out]

1/24*(a^2*c*tan(1/2*f*x + 1/2*e)^3 + 3*a^2*c*tan(1/2*f*x + 1/2*e)^2 - 12*a^2*c*log(abs(tan(1/2*f*x + 1/2*e)))
- 3*a^2*c*tan(1/2*f*x + 1/2*e) + (22*a^2*c*tan(1/2*f*x + 1/2*e)^3 + 3*a^2*c*tan(1/2*f*x + 1/2*e)^2 - 3*a^2*c*t
an(1/2*f*x + 1/2*e) - a^2*c)/tan(1/2*f*x + 1/2*e)^3)/f