3.797 \(\int \frac {(a+i a \tan (c+d x))^n}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=81 \[ \frac {2 (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1\left (\frac {5}{2};1-n,1;\frac {7}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)} \]

[Out]

2/5*AppellF1(5/2,1-n,1,7/2,-I*tan(d*x+c),I*tan(d*x+c))*(a+I*a*tan(d*x+c))^n/d/cot(d*x+c)^(5/2)/((1+I*tan(d*x+c
))^n)

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Rubi [A]  time = 0.16, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {4241, 3564, 130, 511, 510} \[ \frac {2 (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1\left (\frac {5}{2};1-n,1;\frac {7}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^n/Cot[c + d*x]^(3/2),x]

[Out]

(2*AppellF1[5/2, 1 - n, 1, 7/2, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n)/(5*d*Cot[c + d*x]
^(5/2)*(1 + I*Tan[c + d*x])^n)

Rule 130

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + (b*x^k)/e)^m*(c + (d*x^k)/e)^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 3564

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

Rule 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^n}{\cot ^{\frac {3}{2}}(c+d x)} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n \, dx\\ &=\frac {\left (i a^2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {i x}{a}\right )^{3/2} (a+x)^{-1+n}}{-a^2+a x} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {\left (2 a^3 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (a+i a x^2\right )^{-1+n}}{-a^2+i a^2 x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {\left (2 a^2 \sqrt {\cot (c+d x)} (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (1+i x^2\right )^{-1+n}}{-a^2+i a^2 x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {2 F_1\left (\frac {5}{2};1-n,1;\frac {7}{2};-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n}{5 d \cot ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [F]  time = 5.09, size = 0, normalized size = 0.00 \[ \int \frac {(a+i a \tan (c+d x))^n}{\cot ^{\frac {3}{2}}(c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + I*a*Tan[c + d*x])^n/Cot[c + d*x]^(3/2),x]

[Out]

Integrate[(a + I*a*Tan[c + d*x])^n/Cot[c + d*x]^(3/2), x]

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fricas [F]  time = 1.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\left (\frac {2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}{e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^n/cot(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

integral(-(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x +
2*I*c) - 1))*(e^(4*I*d*x + 4*I*c) - 2*e^(2*I*d*x + 2*I*c) + 1)/(e^(4*I*d*x + 4*I*c) + 2*e^(2*I*d*x + 2*I*c) +
1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\cot \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^n/cot(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^n/cot(d*x + c)^(3/2), x)

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maple [F]  time = 1.24, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +i a \tan \left (d x +c \right )\right )^{n}}{\cot \left (d x +c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^n/cot(d*x+c)^(3/2),x)

[Out]

int((a+I*a*tan(d*x+c))^n/cot(d*x+c)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\cot \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^n/cot(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^n/cot(d*x + c)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^n/cot(c + d*x)^(3/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)^n/cot(c + d*x)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**n/cot(d*x+c)**(3/2),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**n/cot(c + d*x)**(3/2), x)

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