3.43 \(\int \cot ^3(c+d x) (a+a \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=40 \[ -\frac {2 a^3}{d (1-\cos (c+d x))}-\frac {a^3 \log (1-\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3879, 43} \[ -\frac {2 a^3}{d (1-\cos (c+d x))}-\frac {a^3 \log (1-\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3879

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

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 46, normalized size = 1.15 \[ -\frac {a^3 \left (\cot ^2\left (\frac {1}{2} (c+d x)\right )+2 \left (\log \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*(Cot[(c + d*x)/2]^2 + 2*(Log[Cos[(c + d*x)/2]] + Log[Tan[(c + d*x)/2]])))/d)

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fricas [A]  time = 0.73, size = 50, normalized size = 1.25 \[ \frac {2 \, a^{3} - {\left (a^{3} \cos \left (d x + c\right ) - a^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{d \cos \left (d x + c\right ) - d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.37, size = 109, normalized size = 2.72 \[ -\frac {a^{3} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - a^{3} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - \frac {{\left (a^{3} + \frac {a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{\cos \left (d x + c\right ) - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^3*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) +
1)) - (a^3 + a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/(cos(d*x + c) - 1))/d

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maple [A]  time = 0.73, size = 51, normalized size = 1.28 \[ \frac {a^{3} \ln \left (\sec \left (d x +c \right )\right )}{d}-\frac {2 a^{3}}{d \left (-1+\sec \left (d x +c \right )\right )}-\frac {a^{3} \ln \left (-1+\sec \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.61, size = 34, normalized size = 0.85 \[ -\frac {a^{3} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac {2 \, a^{3}}{\cos \left (d x + c\right ) - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.23, size = 48, normalized size = 1.20 \[ -\frac {a^3\,\left ({\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^3*(a + a/cos(c + d*x))^3,x)

[Out]

-(a^3*(2*log(tan(c/2 + (d*x)/2)) - log(tan(c/2 + (d*x)/2)^2 + 1) + cot(c/2 + (d*x)/2)^2))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int 3 \cot ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 \cot ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \cot ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \cot ^{3}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*(Integral(3*cot(c + d*x)**3*sec(c + d*x), x) + Integral(3*cot(c + d*x)**3*sec(c + d*x)**2, x) + Integral(
cot(c + d*x)**3*sec(c + d*x)**3, x) + Integral(cot(c + d*x)**3, x))

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