\(\int \csc (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx\) [228]

Optimal result

Integrand size = 30, antiderivative size = 76 \[ \int \csc (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=a^3 A x-\frac {a^3 A \text {arctanh}(\cos (c+d x))}{d}+\frac {a^3 A \cos (c+d x)}{d}-\frac {a^3 A \cos ^3(c+d x)}{3 d}+\frac {a^3 A \cos (c+d x) \sin (c+d x)}{d} \]

[Out]

a^3*A*x-a^3*A*arctanh(cos(d*x+c))/d+a^3*A*cos(d*x+c)/d-1/3*a^3*A*cos(d*x+c)^3/d+a^3*A*cos(d*x+c)*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3045, 3855, 2715, 8, 2713} \[ \int \csc (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {a^3 A \text {arctanh}(\cos (c+d x))}{d}-\frac {a^3 A \cos ^3(c+d x)}{3 d}+\frac {a^3 A \cos (c+d x)}{d}+\frac {a^3 A \sin (c+d x) \cos (c+d x)}{d}+a^3 A x \]

[In]

Int[Csc[c + d*x]*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

a^3*A*x - (a^3*A*ArcTanh[Cos[c + d*x]])/d + (a^3*A*Cos[c + d*x])/d - (a^3*A*Cos[c + d*x]^3)/(3*d) + (a^3*A*Cos
[c + d*x]*Sin[c + d*x])/d

Rule 8

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

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 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 (2 a^3 A+a^3 A \csc (c+d x)-2 a^3 A \sin ^2(c+d x)-a^3 A \sin ^3(c+d x)\right ) \, dx \\ & = 2 a^3 A x+\left (a^3 A\right ) \int \csc (c+d x) \, dx-\left (a^3 A\right ) \int \sin ^3(c+d x) \, dx-\left (2 a^3 A\right ) \int \sin ^2(c+d x) \, dx \\ & = 2 a^3 A x-\frac {a^3 A \text {arctanh}(\cos (c+d x))}{d}+\frac {a^3 A \cos (c+d x) \sin (c+d x)}{d}-\left (a^3 A\right ) \int 1 \, dx+\frac {\left (a^3 A\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = a^3 A x-\frac {a^3 A \text {arctanh}(\cos (c+d x))}{d}+\frac {a^3 A \cos (c+d x)}{d}-\frac {a^3 A \cos ^3(c+d x)}{3 d}+\frac {a^3 A \cos (c+d x) \sin (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.97 \[ \int \csc (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^3 A \left (9 \cos (c+d x)-\cos (3 (c+d x))+6 \left (-2 c+2 d x-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\sin (2 (c+d x))\right )\right )}{12 d} \]

[In]

Integrate[Csc[c + d*x]*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

(a^3*A*(9*Cos[c + d*x] - Cos[3*(c + d*x)] + 6*(-2*c + 2*d*x - 2*Log[Cos[(c + d*x)/2]] + 2*Log[Sin[(c + d*x)/2]
] + Sin[2*(c + d*x)])))/(12*d)

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.76

method	result	size
parallelrisch	\(\frac {A \,a^{3} \left (12 d x +9 \cos \left (d x +c \right )-\cos \left (3 d x +3 c \right )+6 \sin \left (2 d x +2 c \right )+12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8\right )}{12 d}\)	\(58\)
derivativedivides	\(\frac {\frac {A \,a^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}-2 A \,a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \,a^{3} \left (d x +c \right )+A \,a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}\)	\(88\)
default	\(\frac {\frac {A \,a^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}-2 A \,a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \,a^{3} \left (d x +c \right )+A \,a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}\)	\(88\)
risch	\(a^{3} A x +\frac {3 A \,a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {3 A \,a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {A \,a^{3} \cos \left (3 d x +3 c \right )}{12 d}+\frac {A \,a^{3} \sin \left (2 d x +2 c \right )}{2 d}\)	\(121\)
norman	\(\frac {a^{3} A x +a^{3} A x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4 A \,a^{3}}{3 d}+\frac {4 A \,a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {16 A \,a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 A \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A \,a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 A \,a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 A \,a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+4 a^{3} A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 a^{3} A x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 a^{3} A x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(241\)

[In]

int(csc(d*x+c)*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/12*A*a^3*(12*d*x+9*cos(d*x+c)-cos(3*d*x+3*c)+6*sin(2*d*x+2*c)+12*ln(tan(1/2*d*x+1/2*c))+8)/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.21 \[ \int \csc (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {2 \, A a^{3} \cos \left (d x + c\right )^{3} - 6 \, A a^{3} d x - 6 \, A a^{3} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 6 \, A a^{3} \cos \left (d x + c\right ) + 3 \, A a^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, A a^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{6 \, d} \]

[In]

integrate(csc(d*x+c)*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/6*(2*A*a^3*cos(d*x + c)^3 - 6*A*a^3*d*x - 6*A*a^3*cos(d*x + c)*sin(d*x + c) - 6*A*a^3*cos(d*x + c) + 3*A*a^
3*log(1/2*cos(d*x + c) + 1/2) - 3*A*a^3*log(-1/2*cos(d*x + c) + 1/2))/d

Sympy [F]

\[ \int \csc (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=- A a^{3} \left (\int \left (- 2 \sin {\left (c + d x \right )} \csc {\left (c + d x \right )}\right )\, dx + \int 2 \sin ^{3}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \left (- \csc {\left (c + d x \right )}\right )\, dx\right ) \]

[In]

integrate(csc(d*x+c)*(a+a*sin(d*x+c))**3*(A-A*sin(d*x+c)),x)

[Out]

-A*a**3*(Integral(-2*sin(c + d*x)*csc(c + d*x), x) + Integral(2*sin(c + d*x)**3*csc(c + d*x), x) + Integral(si
n(c + d*x)**4*csc(c + d*x), x) + Integral(-csc(c + d*x), x))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.12 \[ \int \csc (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {2 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} A a^{3} + 3 \, {\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 12 \, {\left (d x + c\right )} A a^{3} + 6 \, A a^{3} \log \left (\cot \left (d x + c\right ) + \csc \left (d x + c\right )\right )}{6 \, d} \]

[In]

integrate(csc(d*x+c)*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/6*(2*(cos(d*x + c)^3 - 3*cos(d*x + c))*A*a^3 + 3*(2*d*x + 2*c - sin(2*d*x + 2*c))*A*a^3 - 12*(d*x + c)*A*a^
3 + 6*A*a^3*log(cot(d*x + c) + csc(d*x + c)))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.41 \[ \int \csc (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {3 \, {\left (d x + c\right )} A a^{3} + 3 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \]

[In]

integrate(csc(d*x+c)*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="giac")

[Out]

1/3*(3*(d*x + c)*A*a^3 + 3*A*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(3*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^3*
tan(1/2*d*x + 1/2*c)^2 - 3*A*a^3*tan(1/2*d*x + 1/2*c) - 2*A*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

Mupad [B] (verification not implemented)

Time = 12.64 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.79 \[ \int \csc (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {-2\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,A\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {4\,A\,a^3}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2\,A\,a^3\,\mathrm {atan}\left (\frac {4\,A^2\,a^6}{4\,A^2\,a^6-4\,A^2\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {4\,A^2\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,A^2\,a^6-4\,A^2\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

[In]

int(((A - A*sin(c + d*x))*(a + a*sin(c + d*x))^3)/sin(c + d*x),x)

[Out]

((4*A*a^3)/3 + 2*A*a^3*tan(c/2 + (d*x)/2) + 4*A*a^3*tan(c/2 + (d*x)/2)^2 - 2*A*a^3*tan(c/2 + (d*x)/2)^5)/(d*(3
*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) + (2*A*a^3*atan((4*A^2*a^6)/(4*A^2
*a^6 - 4*A^2*a^6*tan(c/2 + (d*x)/2)) + (4*A^2*a^6*tan(c/2 + (d*x)/2))/(4*A^2*a^6 - 4*A^2*a^6*tan(c/2 + (d*x)/2
))))/d + (A*a^3*log(tan(c/2 + (d*x)/2)))/d