3.356 \(\int \cot (c+d x) (a+b \sec (c+d x))^n \, dx\)

Optimal. Leaf size=162 \[ -\frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \sec (c+d x)}{a-b}\right )}{2 d (n+1) (a-b)}-\frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \sec (c+d x)}{a+b}\right )}{2 d (n+1) (a+b)}+\frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b \sec (c+d x)}{a}+1\right )}{a d (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3885, 961, 65, 831, 68} \[ -\frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \sec (c+d x)}{a-b}\right )}{2 d (n+1) (a-b)}-\frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \sec (c+d x)}{a+b}\right )}{2 d (n+1) (a+b)}+\frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b \sec (c+d x)}{a}+1\right )}{a d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 831

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m, (f + g*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !Ration
alQ[m]

Rule 961

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rule 3885

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

Rubi steps

\begin {align*} \int \cot (c+d x) (a+b \sec (c+d x))^n \, dx &=-\frac {b^2 \operatorname {Subst}\left (\int \frac {(a+x)^n}{x \left (b^2-x^2\right )} \, dx,x,b \sec (c+d x)\right )}{d}\\ &=-\frac {b^2 \operatorname {Subst}\left (\int \left (\frac {(a+x)^n}{b^2 x}-\frac {x (a+x)^n}{b^2 \left (-b^2+x^2\right )}\right ) \, dx,x,b \sec (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {(a+x)^n}{x} \, dx,x,b \sec (c+d x)\right )}{d}+\frac {\operatorname {Subst}\left (\int \frac {x (a+x)^n}{-b^2+x^2} \, dx,x,b \sec (c+d x)\right )}{d}\\ &=\frac {\, _2F_1\left (1,1+n;2+n;1+\frac {b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a d (1+n)}+\frac {\operatorname {Subst}\left (\int \left (-\frac {(a+x)^n}{2 (b-x)}+\frac {(a+x)^n}{2 (b+x)}\right ) \, dx,x,b \sec (c+d x)\right )}{d}\\ &=\frac {\, _2F_1\left (1,1+n;2+n;1+\frac {b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a d (1+n)}-\frac {\operatorname {Subst}\left (\int \frac {(a+x)^n}{b-x} \, dx,x,b \sec (c+d x)\right )}{2 d}+\frac {\operatorname {Subst}\left (\int \frac {(a+x)^n}{b+x} \, dx,x,b \sec (c+d x)\right )}{2 d}\\ &=-\frac {\, _2F_1\left (1,1+n;2+n;\frac {a+b \sec (c+d x)}{a-b}\right ) (a+b \sec (c+d x))^{1+n}}{2 (a-b) d (1+n)}-\frac {\, _2F_1\left (1,1+n;2+n;\frac {a+b \sec (c+d x)}{a+b}\right ) (a+b \sec (c+d x))^{1+n}}{2 (a+b) d (1+n)}+\frac {\, _2F_1\left (1,1+n;2+n;1+\frac {b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a d (1+n)}\\ \end {align*}

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Mathematica [A]  time = 1.69, size = 163, normalized size = 1.01 \[ \frac {(a+b \sec (c+d x))^n \left (-2 \, _2F_1\left (1,-n;1-n;\frac {a \cos (c+d x)}{b+a \cos (c+d x)}\right )+\, _2F_1\left (1,-n;1-n;\frac {(a+b) \cos (c+d x)}{b+a \cos (c+d x)}\right )+2^n \left (\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{b}\right )^{-n} \, _2F_1\left (-n,-n;1-n;\frac {(b-a) \cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{2 b}\right )\right )}{2 d n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.36, size = 0, normalized size = 0.00 \[ \int \cot \left (d x +c \right ) \left (a +b \sec \left (d x +c \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {cot}\left (c+d\,x\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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