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

Optimal. Leaf size=194 \[ \frac {2 (B+i A) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1\left (\frac {1}{2};1-n,1;\frac {3}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d \sqrt {\cot (c+d x)}}-\frac {2 i A (1-2 n) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};-i \tan (c+d x)\right )}{d \sqrt {\cot (c+d x)}}-\frac {2 A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^n}{d} \]

[Out]

-2*A*cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^n/d+2*(I*A+B)*AppellF1(1/2,1-n,1,3/2,-I*tan(d*x+c),I*tan(d*x+c))*(a+I
*a*tan(d*x+c))^n/d/cot(d*x+c)^(1/2)/((1+I*tan(d*x+c))^n)-2*I*A*(1-2*n)*hypergeom([1/2, 1-n],[3/2],-I*tan(d*x+c
))*(a+I*a*tan(d*x+c))^n/d/cot(d*x+c)^(1/2)/((1+I*tan(d*x+c))^n)

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Rubi [A]  time = 0.59, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {4241, 3598, 3601, 3564, 130, 430, 429, 3599, 66, 64} \[ \frac {2 (B+i A) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1\left (\frac {1}{2};1-n,1;\frac {3}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d \sqrt {\cot (c+d x)}}-\frac {2 i A (1-2 n) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};-i \tan (c+d x)\right )}{d \sqrt {\cot (c+d x)}}-\frac {2 A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^n}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^n)/d + (2*(I*A + B)*AppellF1[1/2, 1 - n, 1, 3/2, (-I)*Tan[c +
d*x], I*Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n)/(d*Sqrt[Cot[c + d*x]]*(1 + I*Tan[c + d*x])^n) - ((2*I)*A*(1 -
2*n)*Hypergeometric2F1[1/2, 1 - n, 3/2, (-I)*Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n)/(d*Sqrt[Cot[c + d*x]]*(1
+ I*Tan[c + d*x])^n)

Rule 64

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

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c^IntPart[n]*(c + d*x)^FracPart[n])/(1 + (d
*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-(d/(b*c)), 0] && ((RationalQ[m] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0]))
||  !RationalQ[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 429

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

Rule 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[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 3598

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

Rule 3599

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]

Rule 3601

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), 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] && NeQ[A*b + a*B, 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 \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {2 A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^n}{d}+\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^n \left (\frac {1}{2} a (B+2 i A n)-\frac {1}{2} a A (1-2 n) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{a}\\ &=-\frac {2 A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^n}{d}+\left ((i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx-\frac {\left (i A (1-2 n) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a-i a \tan (c+d x)) (a+i a \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx}{a}\\ &=-\frac {2 A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^n}{d}+\frac {\left (i a^2 (i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {(a+x)^{-1+n}}{\sqrt {-\frac {i x}{a}} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d}-\frac {\left (i a A (1-2 n) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {(a+i a x)^{-1+n}}{\sqrt {x}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {2 A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^n}{d}-\frac {\left (2 a^3 (i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {\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 (i A (1-2 n) \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 {(1+i x)^{-1+n}}{\sqrt {x}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {2 A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^n}{d}-\frac {2 i A (1-2 n) \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d \sqrt {\cot (c+d x)}}-\frac {\left (2 a^2 (i A+B) \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 {\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 A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^n}{d}+\frac {2 (i A+B) F_1\left (\frac {1}{2};1-n,1;\frac {3}{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}{d \sqrt {\cot (c+d x)}}-\frac {2 i A (1-2 n) \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d \sqrt {\cot (c+d x)}}\\ \end {align*}

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Mathematica [F]  time = 21.16, size = 0, normalized size = 0.00 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - B\right )} \left (\frac {2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A - B)*(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^(2*I*d*x + 2*I*c) - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 3.75, size = 0, normalized size = 0.00 \[ \int \left (\cot ^{\frac {3}{2}}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*(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 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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