\(\int \csc ^4(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx\) [231]

Optimal result

Integrand size = 32, antiderivative size = 78 \[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-a^3 A x+\frac {a^3 A \text {arctanh}(\cos (c+d x))}{d}-\frac {a^3 A \cot (c+d x)}{d}-\frac {a^3 A \cot ^3(c+d x)}{3 d}-\frac {a^3 A \cot (c+d x) \csc (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3045, 3855, 3853, 3852} \[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^3 A \text {arctanh}(\cos (c+d x))}{d}-\frac {a^3 A \cot ^3(c+d x)}{3 d}-\frac {a^3 A \cot (c+d x)}{d}-\frac {a^3 A \cot (c+d x) \csc (c+d x)}{d}-a^3 A x \]

[In]

Int[Csc[c + d*x]^4*(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*Cot[c + d*x])/d - (a^3*A*Cot[c + d*x]^3)/(3*d) - (a^3*A*
Cot[c + d*x]*Csc[c + d*x])/d

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 3852

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]

Rule 3853

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 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^3 A-2 a^3 A \csc (c+d x)+2 a^3 A \csc ^3(c+d x)+a^3 A \csc ^4(c+d x)\right ) \, dx \\ & = -a^3 A x+\left (a^3 A\right ) \int \csc ^4(c+d x) \, dx-\left (2 a^3 A\right ) \int \csc (c+d x) \, dx+\left (2 a^3 A\right ) \int \csc ^3(c+d x) \, dx \\ & = -a^3 A x+\frac {2 a^3 A \text {arctanh}(\cos (c+d x))}{d}-\frac {a^3 A \cot (c+d x) \csc (c+d x)}{d}+\left (a^3 A\right ) \int \csc (c+d x) \, dx-\frac {\left (a^3 A\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d} \\ & = -a^3 A x+\frac {a^3 A \text {arctanh}(\cos (c+d x))}{d}-\frac {a^3 A \cot (c+d x)}{d}-\frac {a^3 A \cot ^3(c+d x)}{3 d}-\frac {a^3 A \cot (c+d x) \csc (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.81 \[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {a^3 A \left (24 c+24 d x+8 \cot \left (\frac {1}{2} (c+d x)\right )+6 \csc ^2\left (\frac {1}{2} (c+d x)\right )-24 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+24 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 \sec ^2\left (\frac {1}{2} (c+d x)\right )-8 \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )+\frac {1}{2} \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)-8 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{24 d} \]

[In]

Integrate[Csc[c + d*x]^4*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

-1/24*(a^3*A*(24*c + 24*d*x + 8*Cot[(c + d*x)/2] + 6*Csc[(c + d*x)/2]^2 - 24*Log[Cos[(c + d*x)/2]] + 24*Log[Si
n[(c + d*x)/2]] - 6*Sec[(c + d*x)/2]^2 - 8*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + (Csc[(c + d*x)/2]^4*Sin[c + d*x
])/2 - 8*Tan[(c + d*x)/2]))/d

Maple [A] (verified)

Time = 1.19 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.27

method	result	size
parallelrisch	\(\frac {A \,a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )-\left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 d x +9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-9 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{24 d}\)	\(99\)
derivativedivides	\(\frac {-A \,a^{3} \left (d x +c \right )-2 A \,a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+2 A \,a^{3} \left (-\frac {\csc \left (d x +c \right ) \cot \left (d x +c \right )}{2}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{3}\right ) \cot \left (d x +c \right )}{d}\)	\(101\)
default	\(\frac {-A \,a^{3} \left (d x +c \right )-2 A \,a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+2 A \,a^{3} \left (-\frac {\csc \left (d x +c \right ) \cot \left (d x +c \right )}{2}+\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{3}\right ) \cot \left (d x +c \right )}{d}\)	\(101\)
risch	\(-a^{3} A x +\frac {2 A \,a^{3} \left (3 \,{\mathrm e}^{5 i \left (d x +c \right )}+6 i {\mathrm e}^{2 i \left (d x +c \right )}-2 i-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}-\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\)	\(109\)
norman	\(\frac {\frac {2 A \,a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {A \,a^{3}}{24 d}+\frac {6 A \,a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {11 A \,a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {21 A \,a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {A \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {13 A \,a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {11 A \,a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {7 A \,a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {7 A \,a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {11 A \,a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {13 A \,a^{3} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {A \,a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {A \,a^{3} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-a^{3} A x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 a^{3} A x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 a^{3} A x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 a^{3} A x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a^{3} A x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(406\)

[In]

int(csc(d*x+c)^4*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/24*A*a^3*(tan(1/2*d*x+1/2*c)^3-cot(1/2*d*x+1/2*c)^3+6*tan(1/2*d*x+1/2*c)^2-6*cot(1/2*d*x+1/2*c)^2-24*d*x+9*t
an(1/2*d*x+1/2*c)-9*cot(1/2*d*x+1/2*c)-24*ln(tan(1/2*d*x+1/2*c)))/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (76) = 152\).

Time = 0.29 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.24 \[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {4 \, A a^{3} \cos \left (d x + c\right )^{3} - 6 \, A a^{3} \cos \left (d x + c\right ) - 3 \, {\left (A a^{3} \cos \left (d x + c\right )^{2} - A a^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (A a^{3} \cos \left (d x + c\right )^{2} - A a^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 6 \, {\left (A a^{3} d x \cos \left (d x + c\right )^{2} - A a^{3} d x - A a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(csc(d*x+c)^4*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/6*(4*A*a^3*cos(d*x + c)^3 - 6*A*a^3*cos(d*x + c) - 3*(A*a^3*cos(d*x + c)^2 - A*a^3)*log(1/2*cos(d*x + c) +
1/2)*sin(d*x + c) + 3*(A*a^3*cos(d*x + c)^2 - A*a^3)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 6*(A*a^3*d*x*
cos(d*x + c)^2 - A*a^3*d*x - A*a^3*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^2 - d)*sin(d*x + c))

Sympy [F]

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

[In]

integrate(csc(d*x+c)**4*(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)**4, x) + Integral(2*sin(c + d*x)**3*csc(c + d*x)**4, x) + Integ
ral(sin(c + d*x)**4*csc(c + d*x)**4, x) + Integral(-csc(c + d*x)**4, x))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.50 \[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {6 \, {\left (d x + c\right )} A a^{3} - 3 \, A a^{3} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 6 \, A a^{3} {\left (\log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {2 \, {\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} A a^{3}}{\tan \left (d x + c\right )^{3}}}{6 \, d} \]

[In]

integrate(csc(d*x+c)^4*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.92 \[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, {\left (d x + c\right )} A a^{3} - 24 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 9 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {44 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

[In]

integrate(csc(d*x+c)^4*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="giac")

[Out]

1/24*(A*a^3*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 24*(d*x + c)*A*a^3 - 24*A*a^3*log(abs(ta
n(1/2*d*x + 1/2*c))) + 9*A*a^3*tan(1/2*d*x + 1/2*c) + (44*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 9*A*a^3*tan(1/2*d*x +
1/2*c)^2 - 6*A*a^3*tan(1/2*d*x + 1/2*c) - A*a^3)/tan(1/2*d*x + 1/2*c)^3)/d

Mupad [B] (verification not implemented)

Time = 13.97 (sec) , antiderivative size = 245, normalized size of antiderivative = 3.14 \[ \int \csc ^4(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}-\frac {A\,a^3\,\cos \left (3\,c+3\,d\,x\right )}{6}+\frac {A\,a^3\,\cos \left (c+d\,x\right )}{2}-\frac {A\,a^3\,\sin \left (3\,c+3\,d\,x\right )\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,\cos \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d\,x}{2}\right )}\right )}{2}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,\cos \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d\,x}{2}\right )}\right )}{2}-\frac {A\,a^3\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sin \left (3\,c+3\,d\,x\right )}{4}}{\frac {3\,d\,\sin \left (c+d\,x\right )}{4}-\frac {d\,\sin \left (3\,c+3\,d\,x\right )}{4}} \]

[In]

int(((A - A*sin(c + d*x))*(a + a*sin(c + d*x))^3)/sin(c + d*x)^4,x)

[Out]

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