3.8.12 \(\int \cot (c+d x) (a+b \tan (c+d x))^n \, dx\) [712]

Optimal. Leaf size=175 \[ \frac {\, _2F_1\left (1,1+n;2+n;\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a-i b) d (1+n)}+\frac {\, _2F_1\left (1,1+n;2+n;\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a+i b) d (1+n)}-\frac {\, _2F_1\left (1,1+n;2+n;1+\frac {b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^{1+n}}{a d (1+n)} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3655, 3620, 3618, 70, 3715, 67} \begin {gather*} \frac {(a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (a-i b)}+\frac {(a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}-\frac {(a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b \tan (c+d x)}{a}+1\right )}{a d (n+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Rule 3620

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3655

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/
(c^2 + d^2), Int[(a + b*Tan[e + f*x])^m*(c - d*Tan[e + f*x]), x], x] + Dist[d^2/(c^2 + d^2), Int[(a + b*Tan[e
+ f*x])^m*((1 + Tan[e + f*x]^2)/(c + d*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rubi steps

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

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Mathematica [A]
time = 0.27, size = 154, normalized size = 0.88 \begin {gather*} \frac {\left (a (a+i b) \, _2F_1\left (1,1+n;2+n;\frac {a+b \tan (c+d x)}{a-i b}\right )+(a-i b) \left (a \, _2F_1\left (1,1+n;2+n;\frac {a+b \tan (c+d x)}{a+i b}\right )-2 (a+i b) \, _2F_1\left (1,1+n;2+n;1+\frac {b \tan (c+d x)}{a}\right )\right )\right ) (a+b \tan (c+d x))^{1+n}}{2 a (a-i b) (a+i b) d (1+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(a + I*b)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - I*b)] + (a - I*b)*(a*Hypergeometric
2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + I*b)] - 2*(a + I*b)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (b*T
an[c + d*x])/a]))*(a + b*Tan[c + d*x])^(1 + n))/(2*a*(a - I*b)*(a + I*b)*d*(1 + n))

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Maple [F]
time = 0.29, size = 0, normalized size = 0.00 \[\int \cot \left (d x +c \right ) \left (a +b \tan \left (d x +c \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (c + d x \right )}\right )^{n} \cot {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(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 \begin {gather*} \int \mathrm {cot}\left (c+d\,x\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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