\(\int \csc (e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx\) [5]

Optimal result

Integrand size = 30, antiderivative size = 63 \[ \int \csc (e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=\frac {1}{2} a^2 c x-\frac {a^2 c \text {arctanh}(\cos (e+f x))}{f}+\frac {a^2 c \cos (e+f x)}{f}+\frac {a^2 c \cos (e+f x) \sin (e+f x)}{2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3045, 3855, 2718, 2715, 8} \[ \int \csc (e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=-\frac {a^2 c \text {arctanh}(\cos (e+f x))}{f}+\frac {a^2 c \cos (e+f x)}{f}+\frac {a^2 c \sin (e+f x) \cos (e+f x)}{2 f}+\frac {1}{2} a^2 c x \]

[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 8

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

Rule 2715

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] && Integ
erQ[2*n]

Rule 2718

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

Rule 3045

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 3855

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

Rubi steps \begin{align*} \text {integral}& = \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 \text {arctanh}(\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 \text {arctanh}(\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] (verified)

Time = 0.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int \csc (e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=\frac {a^2 c \left (-2 e+2 f x+4 \cos (e+f x)-4 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )+4 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )+\sin (2 (e+f x))\right )}{4 f} \]

[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)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.71

method	result	size
parallelrisch	\(\frac {a^{2} c \left (2 f x +4 \cos \left (f x +e \right )+4 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\sin \left (2 f x +2 e \right )+4\right )}{4 f}\)	\(45\)
derivativedivides	\(\frac {-a^{2} c \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+a^{2} c \cos \left (f x +e \right )+a^{2} c \left (f x +e \right )+a^{2} c \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{f}\)	\(76\)
default	\(\frac {-a^{2} c \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+a^{2} c \cos \left (f x +e \right )+a^{2} c \left (f x +e \right )+a^{2} c \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{f}\)	\(76\)
risch	\(\frac {a^{2} c x}{2}+\frac {a^{2} c \,{\mathrm e}^{i \left (f x +e \right )}}{2 f}+\frac {a^{2} c \,{\mathrm e}^{-i \left (f x +e \right )}}{2 f}-\frac {a^{2} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{f}+\frac {a^{2} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{f}+\frac {a^{2} c \sin \left (2 f x +2 e \right )}{4 f}\)	\(104\)
norman	\(\frac {\frac {a^{2} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {a^{2} c x}{2}-\frac {2 a^{2} c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {2 a^{2} c \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {4 a^{2} c \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {a^{2} c \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {3 a^{2} c x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {3 a^{2} c x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {a^{2} c x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}+\frac {a^{2} c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}\)	\(195\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.19 \[ \int \csc (e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=\frac {a^{2} c f x + a^{2} c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 2 \, a^{2} c \cos \left (f x + e\right ) - a^{2} c \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + a^{2} c \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{2 \, f} \]

[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

Sympy [F]

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

[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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.16 \[ \int \csc (e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=-\frac {{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c - 4 \, {\left (f x + e\right )} a^{2} c - 4 \, a^{2} c \cos \left (f x + e\right ) + 4 \, a^{2} c \log \left (\cot \left (f x + e\right ) + \csc \left (f x + e\right )\right )}{4 \, f} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.67 \[ \int \csc (e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=\frac {{\left (f x + e\right )} a^{2} c + 2 \, a^{2} c \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) - \frac {2 \, {\left (a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a^{2} c\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2}}}{2 \, f} \]

[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

Mupad [B] (verification not implemented)

Time = 12.67 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.40 \[ \int \csc (e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=\frac {a^2\,c\,\left (\cos \left (e+f\,x\right )+\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )+\frac {\sin \left (2\,e+2\,f\,x\right )}{4}+\mathrm {atan}\left (\frac {\sqrt {5}\,\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{5\,\cos \left (\frac {e}{2}+\mathrm {atan}\left (\frac {1}{2}\right )+\frac {f\,x}{2}\right )}\right )\right )}{f} \]

[In]

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

[Out]

(a^2*c*(cos(e + f*x) + log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)) + sin(2*e + 2*f*x)/4 + atan((5^(1/2)*(cos(e/
2 + (f*x)/2) + 2*sin(e/2 + (f*x)/2)))/(5*cos(e/2 + atan(1/2) + (f*x)/2)))))/f