3.44 \(\int \csc ^3(c+d x) (a+a \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=88 \[ -\frac{2 a^4}{d (a-a \cos (c+d x))}+\frac{a^3 \sec ^2(c+d x)}{2 d}+\frac{3 a^3 \sec (c+d x)}{d}+\frac{5 a^3 \log (1-\cos (c+d x))}{d}-\frac{5 a^3 \log (\cos (c+d x))}{d} \]

[Out]

(-2*a^4)/(d*(a - a*Cos[c + d*x])) + (5*a^3*Log[1 - Cos[c + d*x]])/d - (5*a^3*Log[Cos[c + d*x]])/d + (3*a^3*Sec
[c + d*x])/d + (a^3*Sec[c + d*x]^2)/(2*d)

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Rubi [A]  time = 0.156347, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3872, 2836, 12, 77} \[ -\frac{2 a^4}{d (a-a \cos (c+d x))}+\frac{a^3 \sec ^2(c+d x)}{2 d}+\frac{3 a^3 \sec (c+d x)}{d}+\frac{5 a^3 \log (1-\cos (c+d x))}{d}-\frac{5 a^3 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^4)/(d*(a - a*Cos[c + d*x])) + (5*a^3*Log[1 - Cos[c + d*x]])/d - (5*a^3*Log[Cos[c + d*x]])/d + (3*a^3*Sec
[c + d*x])/d + (a^3*Sec[c + d*x]^2)/(2*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \csc ^3(c+d x) (a+a \sec (c+d x))^3 \, dx &=-\int (-a-a \cos (c+d x))^3 \csc ^3(c+d x) \sec ^3(c+d x) \, dx\\ &=\frac{a^3 \operatorname{Subst}\left (\int \frac{a^3 (-a+x)}{(-a-x)^2 x^3} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac{a^6 \operatorname{Subst}\left (\int \frac{-a+x}{(-a-x)^2 x^3} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac{a^6 \operatorname{Subst}\left (\int \left (-\frac{1}{a x^3}+\frac{3}{a^2 x^2}-\frac{5}{a^3 x}+\frac{2}{a^2 (a+x)^2}+\frac{5}{a^3 (a+x)}\right ) \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=-\frac{2 a^4}{d (a-a \cos (c+d x))}+\frac{5 a^3 \log (1-\cos (c+d x))}{d}-\frac{5 a^3 \log (\cos (c+d x))}{d}+\frac{3 a^3 \sec (c+d x)}{d}+\frac{a^3 \sec ^2(c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.861069, size = 88, normalized size = 1. \[ -\frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \left (2 \csc ^2\left (\frac{1}{2} (c+d x)\right )-\sec ^2(c+d x)-6 \sec (c+d x)+10 \left (\log (\cos (c+d x))-2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(2*Csc[(c + d*x)/2]^2 + 10*(Log[Cos[c + d*x]] - 2*Log[Sin[(c + d
*x)/2]]) - 6*Sec[c + d*x] - Sec[c + d*x]^2))/(16*d)

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Maple [A]  time = 0.074, size = 67, normalized size = 0.8 \begin{align*}{\frac{{a}^{3} \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+3\,{\frac{{a}^{3}\sec \left ( dx+c \right ) }{d}}-2\,{\frac{{a}^{3}}{d \left ( -1+\sec \left ( dx+c \right ) \right ) }}+5\,{\frac{{a}^{3}\ln \left ( -1+\sec \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^3*sec(d*x+c)^2/d+3*a^3*sec(d*x+c)/d-2/d*a^3/(-1+sec(d*x+c))+5/d*a^3*ln(-1+sec(d*x+c))

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Maxima [A]  time = 1.00001, size = 113, normalized size = 1.28 \begin{align*} \frac{10 \, a^{3} \log \left (\cos \left (d x + c\right ) - 1\right ) - 10 \, a^{3} \log \left (\cos \left (d x + c\right )\right ) + \frac{10 \, a^{3} \cos \left (d x + c\right )^{2} - 5 \, a^{3} \cos \left (d x + c\right ) - a^{3}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(10*a^3*log(cos(d*x + c) - 1) - 10*a^3*log(cos(d*x + c)) + (10*a^3*cos(d*x + c)^2 - 5*a^3*cos(d*x + c) - a
^3)/(cos(d*x + c)^3 - cos(d*x + c)^2))/d

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Fricas [A]  time = 1.74009, size = 319, normalized size = 3.62 \begin{align*} \frac{10 \, a^{3} \cos \left (d x + c\right )^{2} - 5 \, a^{3} \cos \left (d x + c\right ) - a^{3} - 10 \,{\left (a^{3} \cos \left (d x + c\right )^{3} - a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (-\cos \left (d x + c\right )\right ) + 10 \,{\left (a^{3} \cos \left (d x + c\right )^{3} - a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{2 \,{\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(10*a^3*cos(d*x + c)^2 - 5*a^3*cos(d*x + c) - a^3 - 10*(a^3*cos(d*x + c)^3 - a^3*cos(d*x + c)^2)*log(-cos(
d*x + c)) + 10*(a^3*cos(d*x + c)^3 - a^3*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2))/(d*cos(d*x + c)^3 - d*c
os(d*x + c)^2)

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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)**3*(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.37479, size = 255, normalized size = 2.9 \begin{align*} \frac{10 \, a^{3} \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 10 \, a^{3} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac{2 \,{\left (a^{3} - \frac{5 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}{\cos \left (d x + c\right ) - 1} + \frac{27 \, a^{3} + \frac{38 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{15 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(10*a^3*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 10*a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) - 1)) + 2*(a^3 - 5*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/(cos(d*x + c) - 1) +
(27*a^3 + 38*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 15*a^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/((c
os(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2)/d