\(\int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx\) [229]

Optimal result

Integrand size = 32, antiderivative size = 79 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {1}{2} a^3 A x-\frac {2 a^3 A \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^3 A \cos (c+d x)}{d}-\frac {a^3 A \cot (c+d x)}{d}+\frac {a^3 A \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3029, 2788, 3855, 3852, 8, 2718, 2715} \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {2 a^3 A \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^3 A \cos (c+d x)}{d}-\frac {a^3 A \cot (c+d x)}{d}+\frac {a^3 A \sin (c+d x) \cos (c+d x)}{2 d}-\frac {1}{2} a^3 A x \]

[In]

Int[Csc[c + d*x]^2*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

-1/2*(a^3*A*x) - (2*a^3*A*ArcTanh[Cos[c + d*x]])/d + (2*a^3*A*Cos[c + d*x])/d - (a^3*A*Cot[c + d*x])/d + (a^3*
A*Cos[c + d*x]*Sin[c + d*x])/(2*d)

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 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 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 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 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}& = (a A) \int \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx \\ & = \frac {A \int \left (2 a^4 \csc (c+d x)+a^4 \csc ^2(c+d x)-2 a^4 \sin (c+d x)-a^4 \sin ^2(c+d x)\right ) \, dx}{a} \\ & = \left (a^3 A\right ) \int \csc ^2(c+d x) \, dx-\left (a^3 A\right ) \int \sin ^2(c+d x) \, dx+\left (2 a^3 A\right ) \int \csc (c+d x) \, dx-\left (2 a^3 A\right ) \int \sin (c+d x) \, dx \\ & = -\frac {2 a^3 A \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^3 A \cos (c+d x)}{d}+\frac {a^3 A \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{2} \left (a^3 A\right ) \int 1 \, dx-\frac {\left (a^3 A\right ) \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d} \\ & = -\frac {1}{2} a^3 A x-\frac {2 a^3 A \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^3 A \cos (c+d x)}{d}-\frac {a^3 A \cot (c+d x)}{d}+\frac {a^3 A \cos (c+d x) \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.97 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^3 A \left (-2 c-2 d x+8 \cos (c) \cos (d x)-4 \cot (c+d x)-8 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+8 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-8 \sin (c) \sin (d x)+\sin (2 (c+d x))\right )}{4 d} \]

[In]

Integrate[Csc[c + d*x]^2*(a + a*Sin[c + d*x])^3*(A - A*Sin[c + d*x]),x]

[Out]

(a^3*A*(-2*c - 2*d*x + 8*Cos[c]*Cos[d*x] - 4*Cot[c + d*x] - 8*Log[Cos[(c + d*x)/2]] + 8*Log[Sin[(c + d*x)/2]]
- 8*Sin[c]*Sin[d*x] + Sin[2*(c + d*x)]))/(4*d)

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.01

method	result	size
derivativedivides	\(\frac {-A \,a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \,a^{3} \cos \left (d x +c \right )+2 A \,a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-A \,a^{3} \cot \left (d x +c \right )}{d}\)	\(80\)
default	\(\frac {-A \,a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \,a^{3} \cos \left (d x +c \right )+2 A \,a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-A \,a^{3} \cot \left (d x +c \right )}{d}\)	\(80\)
parallelrisch	\(-\frac {a^{3} \left (-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\cos \left (d x +c \right )-\frac {\cos \left (2 d x +2 c \right )}{2}+4 \sin \left (d x +c \right )-\frac {5}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+d x \right ) A}{2 d}\)	\(84\)
risch	\(-\frac {a^{3} A x}{2}-\frac {i A \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {A \,a^{3} {\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {A \,a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i A \,a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 i A \,a^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {2 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {2 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\)	\(146\)
norman	\(\frac {\frac {4 A \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {A \,a^{3}}{2 d}+\frac {4 A \,a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {12 A \,a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {12 A \,a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {A \,a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {A \,a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {A \,a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-2 a^{3} A x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{3} A x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 a^{3} A x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {a^{3} A x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {2 A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(284\)

[In]

int(csc(d*x+c)^2*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.41 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {A a^{3} \cos \left (d x + c\right )^{3} + 2 \, A a^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 2 \, A a^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + A a^{3} \cos \left (d x + c\right ) + {\left (A a^{3} d x - 4 \, A a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \sin \left (d x + c\right )} \]

[In]

integrate(csc(d*x+c)^2*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.05 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 4 \, A a^{3} {\left (\log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 8 \, A a^{3} \cos \left (d x + c\right ) + \frac {4 \, A a^{3}}{\tan \left (d x + c\right )}}{4 \, d} \]

[In]

integrate(csc(d*x+c)^2*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/4*((2*d*x + 2*c - sin(2*d*x + 2*c))*A*a^3 + 4*A*a^3*(log(cos(d*x + c) + 1) - log(cos(d*x + c) - 1)) - 8*A*a
^3*cos(d*x + c) + 4*A*a^3/tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (75) = 150\).

Time = 0.30 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.94 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=-\frac {{\left (d x + c\right )} A a^{3} - 4 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {4 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {2 \, {\left (A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, A a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

[In]

integrate(csc(d*x+c)^2*(a+a*sin(d*x+c))^3*(A-A*sin(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.31 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.86 \[ \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {A\,a^3\,\mathrm {atan}\left (\frac {A^2\,a^6}{4\,A^2\,a^6+A^2\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {4\,A^2\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,A^2\,a^6+A^2\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {3\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,A\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-8\,A\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+A\,a^3}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {2\,A\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {A\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

[In]

int(((A - A*sin(c + d*x))*(a + a*sin(c + d*x))^3)/sin(c + d*x)^2,x)

[Out]

(A*a^3*atan((A^2*a^6)/(4*A^2*a^6 + 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 + A
^2*a^6*tan(c/2 + (d*x)/2))))/d - (A*a^3 - 8*A*a^3*tan(c/2 + (d*x)/2) - 8*A*a^3*tan(c/2 + (d*x)/2)^3 + 3*A*a^3*
tan(c/2 + (d*x)/2)^4)/(d*(2*tan(c/2 + (d*x)/2) + 4*tan(c/2 + (d*x)/2)^3 + 2*tan(c/2 + (d*x)/2)^5)) + (2*A*a^3*
log(tan(c/2 + (d*x)/2)))/d + (A*a^3*tan(c/2 + (d*x)/2))/(2*d)