3.10 \(\int \csc ^6(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx\)

Optimal. Leaf size=105 \[ -\frac{a^2 c \cot ^5(e+f x)}{5 f}-\frac{a^2 c \cot ^3(e+f x)}{3 f}+\frac{a^2 c \tanh ^{-1}(\cos (e+f x))}{8 f}-\frac{a^2 c \cot (e+f x) \csc ^3(e+f x)}{4 f}+\frac{a^2 c \cot (e+f x) \csc (e+f x)}{8 f} \]

[Out]

(a^2*c*ArcTanh[Cos[e + f*x]])/(8*f) - (a^2*c*Cot[e + f*x]^3)/(3*f) - (a^2*c*Cot[e + f*x]^5)/(5*f) + (a^2*c*Cot
[e + f*x]*Csc[e + f*x])/(8*f) - (a^2*c*Cot[e + f*x]*Csc[e + f*x]^3)/(4*f)

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Rubi [A]  time = 0.161548, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2966, 3768, 3770, 3767} \[ -\frac{a^2 c \cot ^5(e+f x)}{5 f}-\frac{a^2 c \cot ^3(e+f x)}{3 f}+\frac{a^2 c \tanh ^{-1}(\cos (e+f x))}{8 f}-\frac{a^2 c \cot (e+f x) \csc ^3(e+f x)}{4 f}+\frac{a^2 c \cot (e+f x) \csc (e+f x)}{8 f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x]),x]

[Out]

(a^2*c*ArcTanh[Cos[e + f*x]])/(8*f) - (a^2*c*Cot[e + f*x]^3)/(3*f) - (a^2*c*Cot[e + f*x]^5)/(5*f) + (a^2*c*Cot
[e + f*x]*Csc[e + f*x])/(8*f) - (a^2*c*Cot[e + f*x]*Csc[e + f*x]^3)/(4*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 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]

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]

Rubi steps

\begin{align*} \int \csc ^6(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx &=\int \left (-a^2 c \csc ^3(e+f x)-a^2 c \csc ^4(e+f x)+a^2 c \csc ^5(e+f x)+a^2 c \csc ^6(e+f x)\right ) \, dx\\ &=-\left (\left (a^2 c\right ) \int \csc ^3(e+f x) \, dx\right )-\left (a^2 c\right ) \int \csc ^4(e+f x) \, dx+\left (a^2 c\right ) \int \csc ^5(e+f x) \, dx+\left (a^2 c\right ) \int \csc ^6(e+f x) \, dx\\ &=\frac{a^2 c \cot (e+f x) \csc (e+f x)}{2 f}-\frac{a^2 c \cot (e+f x) \csc ^3(e+f x)}{4 f}-\frac{1}{2} \left (a^2 c\right ) \int \csc (e+f x) \, dx+\frac{1}{4} \left (3 a^2 c\right ) \int \csc ^3(e+f x) \, dx+\frac{\left (a^2 c\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (e+f x)\right )}{f}-\frac{\left (a^2 c\right ) \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, 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 ^5(e+f x)}{5 f}+\frac{a^2 c \cot (e+f x) \csc (e+f x)}{8 f}-\frac{a^2 c \cot (e+f x) \csc ^3(e+f x)}{4 f}+\frac{1}{8} \left (3 a^2 c\right ) \int \csc (e+f x) \, dx\\ &=\frac{a^2 c \tanh ^{-1}(\cos (e+f x))}{8 f}-\frac{a^2 c \cot ^3(e+f x)}{3 f}-\frac{a^2 c \cot ^5(e+f x)}{5 f}+\frac{a^2 c \cot (e+f x) \csc (e+f x)}{8 f}-\frac{a^2 c \cot (e+f x) \csc ^3(e+f x)}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.0563219, size = 204, normalized size = 1.94 \[ \frac{2 a^2 c \cot (e+f x)}{15 f}-\frac{a^2 c \csc ^4\left (\frac{1}{2} (e+f x)\right )}{64 f}+\frac{a^2 c \csc ^2\left (\frac{1}{2} (e+f x)\right )}{32 f}+\frac{a^2 c \sec ^4\left (\frac{1}{2} (e+f x)\right )}{64 f}-\frac{a^2 c \sec ^2\left (\frac{1}{2} (e+f x)\right )}{32 f}-\frac{a^2 c \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )}{8 f}+\frac{a^2 c \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )}{8 f}-\frac{a^2 c \cot (e+f x) \csc ^4(e+f x)}{5 f}+\frac{a^2 c \cot (e+f x) \csc ^2(e+f x)}{15 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^6*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x]),x]

[Out]

(2*a^2*c*Cot[e + f*x])/(15*f) + (a^2*c*Csc[(e + f*x)/2]^2)/(32*f) - (a^2*c*Csc[(e + f*x)/2]^4)/(64*f) + (a^2*c
*Cot[e + f*x]*Csc[e + f*x]^2)/(15*f) - (a^2*c*Cot[e + f*x]*Csc[e + f*x]^4)/(5*f) + (a^2*c*Log[Cos[(e + f*x)/2]
])/(8*f) - (a^2*c*Log[Sin[(e + f*x)/2]])/(8*f) - (a^2*c*Sec[(e + f*x)/2]^2)/(32*f) + (a^2*c*Sec[(e + f*x)/2]^4
)/(64*f)

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Maple [A]  time = 0.149, size = 132, normalized size = 1.3 \begin{align*}{\frac{{a}^{2}c\cot \left ( fx+e \right ) \csc \left ( fx+e \right ) }{8\,f}}-{\frac{{a}^{2}c\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{8\,f}}+{\frac{2\,{a}^{2}c\cot \left ( fx+e \right ) }{15\,f}}+{\frac{{a}^{2}c\cot \left ( fx+e \right ) \left ( \csc \left ( fx+e \right ) \right ) ^{2}}{15\,f}}-{\frac{{a}^{2}c\cot \left ( fx+e \right ) \left ( \csc \left ( fx+e \right ) \right ) ^{3}}{4\,f}}-{\frac{{a}^{2}c\cot \left ( fx+e \right ) \left ( \csc \left ( fx+e \right ) \right ) ^{4}}{5\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6*(a+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x)

[Out]

1/8*a^2*c*cot(f*x+e)*csc(f*x+e)/f-1/8/f*a^2*c*ln(csc(f*x+e)-cot(f*x+e))+2/15*a^2*c*cot(f*x+e)/f+1/15/f*a^2*c*c
ot(f*x+e)*csc(f*x+e)^2-1/4*a^2*c*cot(f*x+e)*csc(f*x+e)^3/f-1/5/f*a^2*c*cot(f*x+e)*csc(f*x+e)^4

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Maxima [A]  time = 0.966041, size = 252, normalized size = 2.4 \begin{align*} \frac{15 \, a^{2} c{\left (\frac{2 \,{\left (3 \, \cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\cos \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - 60 \, 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 )} + \frac{80 \,{\left (3 \, \tan \left (f x + e\right )^{2} + 1\right )} a^{2} c}{\tan \left (f x + e\right )^{3}} - \frac{16 \,{\left (15 \, \tan \left (f x + e\right )^{4} + 10 \, \tan \left (f x + e\right )^{2} + 3\right )} a^{2} c}{\tan \left (f x + e\right )^{5}}}{240 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6*(a+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x, algorithm="maxima")

[Out]

1/240*(15*a^2*c*(2*(3*cos(f*x + e)^3 - 5*cos(f*x + e))/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1) - 3*log(cos(f*x
+ e) + 1) + 3*log(cos(f*x + e) - 1)) - 60*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)) + 80*(3*tan(f*x + e)^2 + 1)*a^2*c/tan(f*x + e)^3 - 16*(15*tan(f*x + e)^4 + 10*tan(f*x
+ e)^2 + 3)*a^2*c/tan(f*x + e)^5)/f

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Fricas [B]  time = 2.08506, size = 520, normalized size = 4.95 \begin{align*} \frac{32 \, a^{2} c \cos \left (f x + e\right )^{5} - 80 \, a^{2} c \cos \left (f x + e\right )^{3} + 15 \,{\left (a^{2} c \cos \left (f x + e\right )^{4} - 2 \, 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 ) - 15 \,{\left (a^{2} c \cos \left (f x + e\right )^{4} - 2 \, 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 ) - 30 \,{\left (a^{2} c \cos \left (f x + e\right )^{3} + a^{2} c \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{240 \,{\left (f \cos \left (f x + e\right )^{4} - 2 \, 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)^6*(a+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x, algorithm="fricas")

[Out]

1/240*(32*a^2*c*cos(f*x + e)^5 - 80*a^2*c*cos(f*x + e)^3 + 15*(a^2*c*cos(f*x + e)^4 - 2*a^2*c*cos(f*x + e)^2 +
a^2*c)*log(1/2*cos(f*x + e) + 1/2)*sin(f*x + e) - 15*(a^2*c*cos(f*x + e)^4 - 2*a^2*c*cos(f*x + e)^2 + a^2*c)*
log(-1/2*cos(f*x + e) + 1/2)*sin(f*x + e) - 30*(a^2*c*cos(f*x + e)^3 + a^2*c*cos(f*x + e))*sin(f*x + e))/((f*c
os(f*x + e)^4 - 2*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)**6*(a+a*sin(f*x+e))**2*(c-c*sin(f*x+e)),x)

[Out]

Timed out

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Giac [A]  time = 1.25409, size = 248, normalized size = 2.36 \begin{align*} \frac{6 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 15 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 10 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 120 \, a^{2} c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) \right |}\right ) - 60 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \frac{274 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 60 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 10 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 15 \, a^{2} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 6 \, a^{2} c}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5}}}{960 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6*(a+a*sin(f*x+e))^2*(c-c*sin(f*x+e)),x, algorithm="giac")

[Out]

1/960*(6*a^2*c*tan(1/2*f*x + 1/2*e)^5 + 15*a^2*c*tan(1/2*f*x + 1/2*e)^4 + 10*a^2*c*tan(1/2*f*x + 1/2*e)^3 - 12
0*a^2*c*log(abs(tan(1/2*f*x + 1/2*e))) - 60*a^2*c*tan(1/2*f*x + 1/2*e) + (274*a^2*c*tan(1/2*f*x + 1/2*e)^5 + 6
0*a^2*c*tan(1/2*f*x + 1/2*e)^4 - 10*a^2*c*tan(1/2*f*x + 1/2*e)^2 - 15*a^2*c*tan(1/2*f*x + 1/2*e) - 6*a^2*c)/ta
n(1/2*f*x + 1/2*e)^5)/f