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

Optimal. Leaf size=76 \[ -\frac{a^3 A \cos ^3(c+d x)}{3 d}+\frac{a^3 A \cos (c+d x)}{d}+\frac{a^3 A \sin (c+d x) \cos (c+d x)}{d}-\frac{a^3 A \tanh ^{-1}(\cos (c+d x))}{d}+a^3 A x \]

[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

________________________________________________________________________________________

Rubi [A]  time = 0.104153, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2966, 3770, 2635, 8, 2633} \[ -\frac{a^3 A \cos ^3(c+d x)}{3 d}+\frac{a^3 A \cos (c+d x)}{d}+\frac{a^3 A \sin (c+d x) \cos (c+d x)}{d}-\frac{a^3 A \tanh ^{-1}(\cos (c+d x))}{d}+a^3 A x \]

Antiderivative was successfully verified.

[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 2966

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 3770

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

Rule 2635

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] && Integer
Q[2*n]

Rule 8

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

Rule 2633

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]

Rubi steps

\begin{align*} \int \csc (c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx &=\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 \tanh ^{-1}(\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 ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=a^3 A x-\frac{a^3 A \tanh ^{-1}(\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]  time = 0.153143, size = 74, normalized size = 0.97 \[ \frac{a^3 A \left (9 \cos (c+d x)-\cos (3 (c+d x))+6 \left (\sin (2 (c+d x))+2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-2 c+2 d x\right )\right )}{12 d} \]

Antiderivative was successfully verified.

[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]  time = 0.05, size = 99, normalized size = 1.3 \begin{align*}{\frac{A\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{a}^{3}}{3\,d}}+{\frac{2\,{a}^{3}A\cos \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}A\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+{a}^{3}Ax+{\frac{{a}^{3}Ac}{d}}+{\frac{{a}^{3}A\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.969253, size = 115, normalized size = 1.51 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Fricas [A]  time = 2.0146, size = 247, normalized size = 3.25 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.15356, size = 144, normalized size = 1.89 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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