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

Optimal. Leaf size=79 \[ \frac{2 a^3 A \cos (c+d x)}{d}-\frac{a^3 A \cot (c+d x)}{d}+\frac{a^3 A \sin (c+d x) \cos (c+d x)}{2 d}-\frac{2 a^3 A \tanh ^{-1}(\cos (c+d x))}{d}-\frac{1}{2} a^3 A x \]

[Out]

-(a^3*A*x)/2 - (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)

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Rubi [A]  time = 0.178912, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.219, Rules used = {2950, 2709, 3770, 3767, 8, 2638, 2635} \[ \frac{2 a^3 A \cos (c+d x)}{d}-\frac{a^3 A \cot (c+d x)}{d}+\frac{a^3 A \sin (c+d x) \cos (c+d x)}{2 d}-\frac{2 a^3 A \tanh ^{-1}(\cos (c+d x))}{d}-\frac{1}{2} a^3 A x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*A*x)/2 - (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 2950

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 2709

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 3770

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

Rule 3767

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 8

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[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]

Rubi steps

\begin{align*} \int \csc ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx &=(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 \tanh ^{-1}(\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 ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}\\ &=-\frac{1}{2} a^3 A x-\frac{2 a^3 A \tanh ^{-1}(\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]  time = 0.186276, size = 77, normalized size = 0.97 \[ \frac{a^3 A \left (-8 \sin (c) \sin (d x)+\sin (2 (c+d x))+8 \cos (c) \cos (d x)-4 \cot (c+d x)+8 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-8 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-2 c-2 d x\right )}{4 d} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.046, size = 95, normalized size = 1.2 \begin{align*}{\frac{{a}^{3}A\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}-{\frac{{a}^{3}Ax}{2}}-{\frac{{a}^{3}Ac}{2\,d}}+2\,{\frac{{a}^{3}A\cos \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{3}A\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{3}A\cot \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.99272, size = 112, normalized size = 1.42 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.95138, size = 297, normalized size = 3.76 \begin{align*} -\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

[Out]

Timed out

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Giac [B]  time = 1.18799, size = 207, normalized size = 2.62 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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