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

Optimal. Leaf size=61 \[ -\frac{a A \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac{a A \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{a A \cot (c+d x) \csc (c+d x)}{8 d} \]

[Out]

-(a*A*ArcTanh[Cos[c + d*x]])/(8*d) - (a*A*Cot[c + d*x]*Csc[c + d*x])/(8*d) + (a*A*Cot[c + d*x]*Csc[c + d*x]^3)
/(4*d)

________________________________________________________________________________________

Rubi [A]  time = 0.0924962, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3962, 2611, 3768, 3770} \[ -\frac{a A \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac{a A \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{a A \cot (c+d x) \csc (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*A*ArcTanh[Cos[c + d*x]])/(8*d) - (a*A*Cot[c + d*x]*Csc[c + d*x])/(8*d) + (a*A*Cot[c + d*x]*Csc[c + d*x]^3)
/(4*d)

Rule 3962

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)
]*(d_.) + (c_))^(n_.), x_Symbol] :> Dist[(-(a*c))^m, Int[ExpandTrig[(g*csc[e + f*x])^p*cot[e + f*x]^(2*m), (c
+ d*csc[e + f*x])^(n - m), x], x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2
- b^2, 0] && IntegersQ[m, n] && GeQ[n - m, 0] && GtQ[m*n, 0]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

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 3770

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

Rubi steps

\begin{align*} \int \csc ^3(c+d x) (a-a \csc (c+d x)) (A+A \csc (c+d x)) \, dx &=-\left ((a A) \int \cot ^2(c+d x) \csc ^3(c+d x) \, dx\right )\\ &=\frac{a A \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{1}{4} (a A) \int \csc ^3(c+d x) \, dx\\ &=-\frac{a A \cot (c+d x) \csc (c+d x)}{8 d}+\frac{a A \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{1}{8} (a A) \int \csc (c+d x) \, dx\\ &=-\frac{a A \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a A \cot (c+d x) \csc (c+d x)}{8 d}+\frac{a A \cot (c+d x) \csc ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.044857, size = 117, normalized size = 1.92 \[ -a A \left (-\frac{\csc ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}+\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right )}{32 d}+\frac{\sec ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}-\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right )}{32 d}-\frac{\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}+\frac{\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*A*(Csc[(c + d*x)/2]^2/(32*d) - Csc[(c + d*x)/2]^4/(64*d) + Log[Cos[(c + d*x)/2]]/(8*d) - Log[Sin[(c + d*x)
/2]]/(8*d) - Sec[(c + d*x)/2]^2/(32*d) + Sec[(c + d*x)/2]^4/(64*d)))

________________________________________________________________________________________

Maple [A]  time = 0.026, size = 65, normalized size = 1.1 \begin{align*} -{\frac{Aa\cot \left ( dx+c \right ) \csc \left ( dx+c \right ) }{8\,d}}+{\frac{Aa\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{Aa\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{3}}{4\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a*A*cot(d*x+c)*csc(d*x+c)/d+1/8/d*A*a*ln(csc(d*x+c)-cot(d*x+c))+1/4*a*A*cot(d*x+c)*csc(d*x+c)^3/d

________________________________________________________________________________________

Maxima [B]  time = 1.01409, size = 162, normalized size = 2.66 \begin{align*} -\frac{A a{\left (\frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 4 \, A a{\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 )}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(A*a*(2*(3*cos(d*x + c)^3 - 5*cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - 3*log(cos(d*x + c)
+ 1) + 3*log(cos(d*x + c) - 1)) - 4*A*a*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) - log(cos(d*x + c) + 1) + log(co
s(d*x + c) - 1)))/d

________________________________________________________________________________________

Fricas [B]  time = 0.496599, size = 344, normalized size = 5.64 \begin{align*} \frac{2 \, A a \cos \left (d x + c\right )^{3} + 2 \, A a \cos \left (d x + c\right ) -{\left (A a \cos \left (d x + c\right )^{4} - 2 \, A a \cos \left (d x + c\right )^{2} + A a\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (A a \cos \left (d x + c\right )^{4} - 2 \, A a \cos \left (d x + c\right )^{2} + A a\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{16 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*A*a*cos(d*x + c)^3 + 2*A*a*cos(d*x + c) - (A*a*cos(d*x + c)^4 - 2*A*a*cos(d*x + c)^2 + A*a)*log(1/2*co
s(d*x + c) + 1/2) + (A*a*cos(d*x + c)^4 - 2*A*a*cos(d*x + c)^2 + A*a)*log(-1/2*cos(d*x + c) + 1/2))/(d*cos(d*x
+ c)^4 - 2*d*cos(d*x + c)^2 + d)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - A a \left (\int - \csc ^{3}{\left (c + d x \right )}\, dx + \int \csc ^{5}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*a*(Integral(-csc(c + d*x)**3, x) + Integral(csc(c + d*x)**5, x))

________________________________________________________________________________________

Giac [A]  time = 1.38111, size = 144, normalized size = 2.36 \begin{align*} \frac{4 \, A a \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - \frac{A a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{{\left (A a - \frac{2 \, A a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(4*A*a*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - A*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2
+ (A*a - 2*A*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)^2/(cos(d*x + c) - 1)^2)/d