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

Optimal result

Integrand size = 32, antiderivative size = 53 \[ \int \csc ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=-a^2 c x-\frac {a^2 c \text {arctanh}(\cos (e+f x))}{f}+\frac {a^2 c \cos (e+f x)}{f}-\frac {a^2 c \cot (e+f x)}{f} \]

[Out]

-a^2*c*x-a^2*c*arctanh(cos(f*x+e))/f+a^2*c*cos(f*x+e)/f-a^2*c*cot(f*x+e)/f

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3029, 2789, 2672, 327, 212, 3554, 8} \[ \int \csc ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=-\frac {a^2 c \text {arctanh}(\cos (e+f x))}{f}+\frac {a^2 c \cos (e+f x)}{f}-\frac {a^2 c \cot (e+f x)}{f}+a^2 (-c) x \]

[In]

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

Rule 8

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2672

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 2789

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2
, 0] && IGtQ[m, 0]

Rule 3029

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 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps \begin{align*} \text {integral}& = (a c) \int \cot ^2(e+f x) (a+a \sin (e+f x)) \, dx \\ & = (a c) \int \left (a \cos (e+f x) \cot (e+f x)+a \cot ^2(e+f x)\right ) \, dx \\ & = \left (a^2 c\right ) \int \cos (e+f x) \cot (e+f x) \, dx+\left (a^2 c\right ) \int \cot ^2(e+f x) \, dx \\ & = -\frac {a^2 c \cot (e+f x)}{f}-\left (a^2 c\right ) \int 1 \, dx-\frac {\left (a^2 c\right ) \text {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\cos (e+f x)\right )}{f} \\ & = -a^2 c x+\frac {a^2 c \cos (e+f x)}{f}-\frac {a^2 c \cot (e+f x)}{f}-\frac {\left (a^2 c\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\cos (e+f x)\right )}{f} \\ & = -a^2 c x-\frac {a^2 c \text {arctanh}(\cos (e+f x))}{f}+\frac {a^2 c \cos (e+f x)}{f}-\frac {a^2 c \cot (e+f x)}{f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.83 \[ \int \csc ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=-a^2 c x+\frac {a^2 c \cos (e) \cos (f x)}{f}-\frac {a^2 c \cot (e+f x)}{f}-\frac {a^2 c \log \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f}+\frac {a^2 c \log \left (\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f}-\frac {a^2 c \sin (e) \sin (f x)}{f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {a^{2} c \cos \left (f x +e \right )-a^{2} c \left (f x +e \right )+a^{2} c \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )-a^{2} c \cot \left (f x +e \right )}{f}\)	\(61\)
default	\(\frac {a^{2} c \cos \left (f x +e \right )-a^{2} c \left (f x +e \right )+a^{2} c \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )-a^{2} c \cot \left (f x +e \right )}{f}\)	\(61\)
parallelrisch	\(\frac {a^{2} c \left (-2 f x -2+2 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 \cos \left (f x +e \right )+2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sec \left (\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}\)	\(67\)
risch	\(-a^{2} c x +\frac {a^{2} c \,{\mathrm e}^{i \left (f x +e \right )}}{2 f}+\frac {a^{2} c \,{\mathrm e}^{-i \left (f x +e \right )}}{2 f}-\frac {2 i a^{2} c}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}-\frac {a^{2} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{f}+\frac {a^{2} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{f}\)	\(109\)
norman	\(\frac {\frac {a^{2} c \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {a^{2} c}{2 f}-\frac {2 a^{2} c \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {2 a^{2} c \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {4 a^{2} c \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {a^{2} c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {a^{2} c \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}-a^{2} c x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-3 a^{2} c x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-3 a^{2} c x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-a^{2} c x \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}+\frac {a^{2} c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}\)	\(246\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.87 \[ \int \csc ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=-\frac {a^{2} c \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) \sin \left (f x + e\right ) - a^{2} c \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) \sin \left (f x + e\right ) + 2 \, a^{2} c \cos \left (f x + e\right ) + 2 \, {\left (a^{2} c f x - a^{2} c \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \, f \sin \left (f x + e\right )} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \csc ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=- a^{2} c \left (\int \left (- \sin {\left (e + f x \right )} \csc ^{2}{\left (e + f x \right )}\right )\, dx + \int \sin ^{2}{\left (e + f x \right )} \csc ^{2}{\left (e + f x \right )}\, dx + \int \sin ^{3}{\left (e + f x \right )} \csc ^{2}{\left (e + f x \right )}\, dx + \int \left (- \csc ^{2}{\left (e + f x \right )}\right )\, dx\right ) \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.30 \[ \int \csc ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=-\frac {2 \, {\left (f x + e\right )} a^{2} c + a^{2} c {\left (\log \left (\cos \left (f x + e\right ) + 1\right ) - \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - 2 \, a^{2} c \cos \left (f x + e\right ) + \frac {2 \, a^{2} c}{\tan \left (f x + e\right )}}{2 \, f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (53) = 106\).

Time = 0.32 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.43 \[ \int \csc ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=-\frac {6 \, {\left (f x + e\right )} a^{2} c - 6 \, a^{2} c \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) - 3 \, a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + \frac {2 \, a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 \, a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 10 \, a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, a^{2} c}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}}{6 \, f} \]

[In]

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

[Out]

-1/6*(6*(f*x + e)*a^2*c - 6*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) + (2*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 - 10*a^2*c*tan(1/2*f*x + 1/2*e) + 3*a^2*c)/(tan(1/2*f*x +
1/2*e)^3 + tan(1/2*f*x + 1/2*e)))/f

Mupad [B] (verification not implemented)

Time = 12.17 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.08 \[ \int \csc ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=\frac {a^2\,c\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{2\,\cos \left (\frac {e}{2}-\frac {\pi }{4}+\frac {f\,x}{2}\right )}\right )+\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )\right )}{f}-\frac {a^2\,c\,\left (\cos \left (e+f\,x\right )-\frac {\sin \left (2\,e+2\,f\,x\right )}{2}\right )}{f\,\sin \left (e+f\,x\right )} \]

[In]

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

[Out]

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