3.4.30 \(\int \cot (c+d x) (a+i a \tan (c+d x))^m \, dx\) [330]

Optimal. Leaf size=89 \[ \frac {\, _2F_1\left (1,m;1+m;\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^m}{2 d m}-\frac {\, _2F_1(1,m;1+m;1+i \tan (c+d x)) (a+i a \tan (c+d x))^m}{d m} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {3643, 3562, 70, 3680, 67} \begin {gather*} \frac {(a+i a \tan (c+d x))^m \, _2F_1\left (1,m;m+1;\frac {1}{2} (i \tan (c+d x)+1)\right )}{2 d m}-\frac {(a+i a \tan (c+d x))^m \, _2F_1(1,m;m+1;i \tan (c+d x)+1)}{d m} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]*(a + I*a*Tan[c + d*x])^m,x]

[Out]

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

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 3562

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

Rule 3643

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

Rule 3680

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

Rubi steps

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

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Mathematica [F]
time = 10.67, size = 0, normalized size = 0.00 \begin {gather*} \int \cot (c+d x) (a+i a \tan (c+d x))^m \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[Cot[c + d*x]*(a + I*a*Tan[c + d*x])^m,x]

[Out]

Integrate[Cot[c + d*x]*(a + I*a*Tan[c + d*x])^m, x]

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Maple [F]
time = 1.58, size = 0, normalized size = 0.00 \[\int \cot \left (d x +c \right ) \left (a +i a \tan \left (d x +c \right )\right )^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)*(a+I*a*tan(d*x+c))^m,x)

[Out]

int(cot(d*x+c)*(a+I*a*tan(d*x+c))^m,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+I*a*tan(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^m*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+I*a*tan(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^m*(I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c
) - 1), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{m} \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+I*a*tan(d*x+c))**m,x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**m*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+I*a*tan(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^m*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+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)*(a + a*tan(c + d*x)*1i)^m,x)

[Out]

int(cot(c + d*x)*(a + a*tan(c + d*x)*1i)^m, x)

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