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

Optimal. Leaf size=78 \[ -\frac{a^3 A \cot ^3(c+d x)}{3 d}-\frac{a^3 A \cot (c+d x)}{d}+\frac{a^3 A \tanh ^{-1}(\cos (c+d x))}{d}-\frac{a^3 A \cot (c+d x) \csc (c+d x)}{d}-a^3 A x \]

[Out]

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

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Rubi [A]  time = 0.130712, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2966, 3770, 3768, 3767} \[ -\frac{a^3 A \cot ^3(c+d x)}{3 d}-\frac{a^3 A \cot (c+d x)}{d}+\frac{a^3 A \tanh ^{-1}(\cos (c+d x))}{d}-\frac{a^3 A \cot (c+d x) \csc (c+d x)}{d}-a^3 A x \]

Antiderivative was successfully verified.

[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 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 3768

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 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]

Rubi steps

\begin{align*} \int \csc ^4(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx &=\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 \tanh ^{-1}(\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 ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=-a^3 A x+\frac{a^3 A \tanh ^{-1}(\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]  time = 0.461471, size = 141, normalized size = 1.81 \[ -\frac{a^3 A \left (-8 \tan \left (\frac{1}{2} (c+d x)\right )+8 \cot \left (\frac{1}{2} (c+d x)\right )+6 \csc ^2\left (\frac{1}{2} (c+d x)\right )-6 \sec ^2\left (\frac{1}{2} (c+d x)\right )+24 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-24 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-8 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)+\frac{1}{2} \sin (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right )+24 c+24 d x\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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[Sin[(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]))/(24*d)

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Maple [A]  time = 0.057, size = 103, normalized size = 1.3 \begin{align*} -{a}^{3}Ax-{\frac{A{a}^{3}c}{d}}-{\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 ) \csc \left ( dx+c \right ) }{d}}-{\frac{2\,{a}^{3}A\cot \left ( dx+c \right ) }{3\,d}}-{\frac{{a}^{3}A\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.988819, size = 158, normalized size = 2.03 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [B]  time = 1.93814, size = 435, normalized size = 5.58 \begin{align*} -\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

[Out]

Timed out

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Giac [A]  time = 1.17242, size = 203, normalized size = 2.6 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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