\(\int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx\) [224]

Optimal result

Integrand size = 21, antiderivative size = 127 \[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=-\frac {a (4-n) \operatorname {Hypergeometric2F1}\left (1,-1+n,n,\frac {1}{2} (1+\sec (c+d x))\right ) (a+a \sec (c+d x))^{-1+n}}{4 d (1-n)}+\frac {a \operatorname {Hypergeometric2F1}(1,-1+n,n,1+\sec (c+d x)) (a+a \sec (c+d x))^{-1+n}}{d (1-n)}+\frac {a (a+a \sec (c+d x))^{-1+n}}{2 d (1-\sec (c+d x))} \]

[Out]

-1/4*a*(4-n)*hypergeom([1, -1+n],[n],1/2+1/2*sec(d*x+c))*(a+a*sec(d*x+c))^(-1+n)/d/(1-n)+a*hypergeom([1, -1+n]
,[n],1+sec(d*x+c))*(a+a*sec(d*x+c))^(-1+n)/d/(1-n)+1/2*a*(a+a*sec(d*x+c))^(-1+n)/d/(1-sec(d*x+c))

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3965, 105, 162, 67, 70} \[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=-\frac {a (4-n) (a \sec (c+d x)+a)^{n-1} \operatorname {Hypergeometric2F1}\left (1,n-1,n,\frac {1}{2} (\sec (c+d x)+1)\right )}{4 d (1-n)}+\frac {a (a \sec (c+d x)+a)^{n-1} \operatorname {Hypergeometric2F1}(1,n-1,n,\sec (c+d x)+1)}{d (1-n)}+\frac {a (a \sec (c+d x)+a)^{n-1}}{2 d (1-\sec (c+d x))} \]

[In]

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

[Out]

-1/4*(a*(4 - n)*Hypergeometric2F1[1, -1 + n, n, (1 + Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^(-1 + n))/(d*(1 - n
)) + (a*Hypergeometric2F1[1, -1 + n, n, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^(-1 + n))/(d*(1 - n)) + (a*(a +
a*Sec[c + d*x])^(-1 + n))/(2*d*(1 - Sec[c + d*x]))

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 3965

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

Rubi steps \begin{align*} \text {integral}& = \frac {a^4 \text {Subst}\left (\int \frac {(a+a x)^{-2+n}}{x (-a+a x)^2} \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {a (a+a \sec (c+d x))^{-1+n}}{2 d (1-\sec (c+d x))}-\frac {a \text {Subst}\left (\int \frac {(a+a x)^{-2+n} \left (2 a^2+a^2 (2-n) x\right )}{x (-a+a x)} \, dx,x,\sec (c+d x)\right )}{2 d} \\ & = \frac {a (a+a \sec (c+d x))^{-1+n}}{2 d (1-\sec (c+d x))}+\frac {a^2 \text {Subst}\left (\int \frac {(a+a x)^{-2+n}}{x} \, dx,x,\sec (c+d x)\right )}{d}-\frac {\left (a^3 (4-n)\right ) \text {Subst}\left (\int \frac {(a+a x)^{-2+n}}{-a+a x} \, dx,x,\sec (c+d x)\right )}{2 d} \\ & = -\frac {a (4-n) \operatorname {Hypergeometric2F1}\left (1,-1+n,n,\frac {1}{2} (1+\sec (c+d x))\right ) (a+a \sec (c+d x))^{-1+n}}{4 d (1-n)}+\frac {a \operatorname {Hypergeometric2F1}(1,-1+n,n,1+\sec (c+d x)) (a+a \sec (c+d x))^{-1+n}}{d (1-n)}+\frac {a (a+a \sec (c+d x))^{-1+n}}{2 d (1-\sec (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76 \[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=-\frac {a \left (-2+2 n+(-4+n) \operatorname {Hypergeometric2F1}\left (1,-1+n,n,\frac {1}{2} (1+\sec (c+d x))\right ) (-1+\sec (c+d x))+4 \operatorname {Hypergeometric2F1}(1,-1+n,n,1+\sec (c+d x)) (-1+\sec (c+d x))\right ) (a (1+\sec (c+d x)))^{-1+n}}{4 d (-1+n) (-1+\sec (c+d x))} \]

[In]

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

[Out]

-1/4*(a*(-2 + 2*n + (-4 + n)*Hypergeometric2F1[1, -1 + n, n, (1 + Sec[c + d*x])/2]*(-1 + Sec[c + d*x]) + 4*Hyp
ergeometric2F1[1, -1 + n, n, 1 + Sec[c + d*x]]*(-1 + Sec[c + d*x]))*(a*(1 + Sec[c + d*x]))^(-1 + n))/(d*(-1 +
n)*(-1 + Sec[c + d*x]))

Maple [F]

\[\int \cot \left (d x +c \right )^{3} \left (a +a \sec \left (d x +c \right )\right )^{n}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \cot ^{3}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

Integral((a*(sec(c + d*x) + 1))**n*cot(c + d*x)**3, x)

Maxima [F]

\[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]

[In]

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

[Out]

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