3.573 \(\int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\cot ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=383 \[ \frac {2 (B+i A) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1\left (\frac {1}{2};1-n,1;\frac {3}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d \sqrt {\cot (c+d x)}}-\frac {2 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};-i \tan (c+d x)\right )}{d (2 n+1) (2 n+3) (2 n+5) \sqrt {\cot (c+d x)}}-\frac {2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) (a+i a \tan (c+d x))^n}{d (2 n+1) (2 n+3) (2 n+5) \sqrt {\cot (c+d x)}}-\frac {2 (-A (2 n+5)+2 i B n) (a+i a \tan (c+d x))^n}{d (2 n+3) (2 n+5) \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 B (a+i a \tan (c+d x))^n}{d (2 n+5) \cot ^{\frac {5}{2}}(c+d x)} \]

[Out]

2*B*(a+I*a*tan(d*x+c))^n/d/(5+2*n)/cot(d*x+c)^(5/2)-2*(2*I*B*n-A*(5+2*n))*(a+I*a*tan(d*x+c))^n/d/(3+2*n)/(5+2*
n)/cot(d*x+c)^(3/2)-2*(2*I*A*n*(5+2*n)+B*(4*n^2+10*n+15))*(a+I*a*tan(d*x+c))^n/d/(5+2*n)/(4*n^2+8*n+3)/cot(d*x
+c)^(1/2)+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*(4*B*n*(2*n^2+8*n+9)+I*A*(8*n^3+32*n^2+36*n+15))*hypergeom([1/2, 1-n],[3/2],-I*tan(d*x
+c))*(a+I*a*tan(d*x+c))^n/d/(5+2*n)/(4*n^2+8*n+3)/cot(d*x+c)^(1/2)/((1+I*tan(d*x+c))^n)

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Rubi [A]  time = 1.26, antiderivative size = 383, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {4241, 3597, 3601, 3564, 130, 430, 429, 3599, 66, 64} \[ \frac {2 (B+i A) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n F_1\left (\frac {1}{2};1-n,1;\frac {3}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d \sqrt {\cot (c+d x)}}-\frac {2 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};-i \tan (c+d x)\right )}{d (2 n+1) (2 n+3) (2 n+5) \sqrt {\cot (c+d x)}}-\frac {2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) (a+i a \tan (c+d x))^n}{d (2 n+1) (2 n+3) (2 n+5) \sqrt {\cot (c+d x)}}-\frac {2 (-A (2 n+5)+2 i B n) (a+i a \tan (c+d x))^n}{d (2 n+3) (2 n+5) \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 B (a+i a \tan (c+d x))^n}{d (2 n+5) \cot ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]))/Cot[c + d*x]^(5/2),x]

[Out]

(2*B*(a + I*a*Tan[c + d*x])^n)/(d*(5 + 2*n)*Cot[c + d*x]^(5/2)) - (2*((2*I)*B*n - A*(5 + 2*n))*(a + I*a*Tan[c
+ d*x])^n)/(d*(3 + 2*n)*(5 + 2*n)*Cot[c + d*x]^(3/2)) - (2*((2*I)*A*n*(5 + 2*n) + B*(15 + 10*n + 4*n^2))*(a +
I*a*Tan[c + d*x])^n)/(d*(1 + 2*n)*(3 + 2*n)*(5 + 2*n)*Sqrt[Cot[c + d*x]]) + (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*(4*B*n*(9 + 8*n + 2*n^2) + I*A*(15 + 36*n + 32*n^2 + 8*n^3))*Hypergeometric2F1[1/2, 1 - n, 3/2, (-I)
*Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n)/(d*(1 + 2*n)*(3 + 2*n)*(5 + 2*n)*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 3597

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

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 \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\cot ^{\frac {5}{2}}(c+d x)} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\\ &=\frac {2 B (a+i a \tan (c+d x))^n}{d (5+2 n) \cot ^{\frac {5}{2}}(c+d x)}+\frac {\left (2 \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 \left (-\frac {5 a B}{2}-\frac {1}{2} a (2 i B n-A (5+2 n)) \tan (c+d x)\right ) \, dx}{a (5+2 n)}\\ &=\frac {2 B (a+i a \tan (c+d x))^n}{d (5+2 n) \cot ^{\frac {5}{2}}(c+d x)}-\frac {2 (2 i B n-A (5+2 n)) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n) \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n \left (\frac {3}{4} a^2 (2 i B n-A (5+2 n))-\frac {1}{4} a^2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) \tan (c+d x)\right ) \, dx}{a^2 (3+2 n) (5+2 n)}\\ &=\frac {2 B (a+i a \tan (c+d x))^n}{d (5+2 n) \cot ^{\frac {5}{2}}(c+d x)}-\frac {2 (2 i B n-A (5+2 n)) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n) \cot ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n) \sqrt {\cot (c+d x)}}+\frac {\left (8 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^n \left (\frac {1}{8} a^3 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right )+\frac {1}{8} a^3 \left (4 i B n \left (9+8 n+2 n^2\right )-A \left (15+36 n+32 n^2+8 n^3\right )\right ) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{a^3 (1+2 n) (3+2 n) (5+2 n)}\\ &=\frac {2 B (a+i a \tan (c+d x))^n}{d (5+2 n) \cot ^{\frac {5}{2}}(c+d x)}-\frac {2 (2 i B n-A (5+2 n)) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n) \cot ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n) \sqrt {\cot (c+d x)}}+\frac {\left ((i A+B) \left (15+46 n+36 n^2+8 n^3\right ) \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}{(1+2 n) (3+2 n) (5+2 n)}-\frac {\left (\left (4 B n \left (9+8 n+2 n^2\right )+i A \left (15+36 n+32 n^2+8 n^3\right )\right ) \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 (1+2 n) (3+2 n) (5+2 n)}\\ &=\frac {2 B (a+i a \tan (c+d x))^n}{d (5+2 n) \cot ^{\frac {5}{2}}(c+d x)}-\frac {2 (2 i B n-A (5+2 n)) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n) \cot ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n) \sqrt {\cot (c+d x)}}+\frac {\left (i a^2 (i A+B) \left (15+46 n+36 n^2+8 n^3\right ) \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 (1+2 n) (3+2 n) (5+2 n)}-\frac {\left (a \left (4 B n \left (9+8 n+2 n^2\right )+i A \left (15+36 n+32 n^2+8 n^3\right )\right ) \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 (1+2 n) (3+2 n) (5+2 n)}\\ &=\frac {2 B (a+i a \tan (c+d x))^n}{d (5+2 n) \cot ^{\frac {5}{2}}(c+d x)}-\frac {2 (2 i B n-A (5+2 n)) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n) \cot ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n) \sqrt {\cot (c+d x)}}-\frac {\left (2 a^3 (i A+B) \left (15+46 n+36 n^2+8 n^3\right ) \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 (1+2 n) (3+2 n) (5+2 n)}-\frac {\left (\left (4 B n \left (9+8 n+2 n^2\right )+i A \left (15+36 n+32 n^2+8 n^3\right )\right ) \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 (1+2 n) (3+2 n) (5+2 n)}\\ &=\frac {2 B (a+i a \tan (c+d x))^n}{d (5+2 n) \cot ^{\frac {5}{2}}(c+d x)}-\frac {2 (2 i B n-A (5+2 n)) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n) \cot ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n) \sqrt {\cot (c+d x)}}-\frac {2 \left (4 B n \left (9+8 n+2 n^2\right )+i A \left (15+36 n+32 n^2+8 n^3\right )\right ) \, _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 (1+2 n) (3+2 n) (5+2 n) \sqrt {\cot (c+d x)}}-\frac {\left (2 a^2 (i A+B) \left (15+46 n+36 n^2+8 n^3\right ) \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 (1+2 n) (3+2 n) (5+2 n)}\\ &=\frac {2 B (a+i a \tan (c+d x))^n}{d (5+2 n) \cot ^{\frac {5}{2}}(c+d x)}-\frac {2 (2 i B n-A (5+2 n)) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n) \cot ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n) \sqrt {\cot (c+d x)}}+\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 \left (4 B n \left (9+8 n+2 n^2\right )+i A \left (15+36 n+32 n^2+8 n^3\right )\right ) \, _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 (1+2 n) (3+2 n) (5+2 n) \sqrt {\cot (c+d x)}}\\ \end {align*}

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Mathematica [F]  time = 22.95, size = 0, normalized size = 0.00 \[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\cot ^{\frac {5}{2}}(c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[((a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]))/Cot[c + d*x]^(5/2),x]

[Out]

Integrate[((a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]))/Cot[c + d*x]^(5/2), x]

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left ({\left (i \, A + B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (-2 i \, A - 4 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, B e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (2 i \, A - 4 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + B\right )} \left (\frac {2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}}{e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c))/cot(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

integral(((I*A + B)*e^(8*I*d*x + 8*I*c) + (-2*I*A - 4*B)*e^(6*I*d*x + 6*I*c) + 6*B*e^(4*I*d*x + 4*I*c) + (2*I*
A - 4*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^(8*I*d*x + 8*I*c) + 4*e^(6*I*d*x + 6*I*c) + 6*e^(4*I*d*x + 4*I*
c) + 4*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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\cot \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c))/cot(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n/cot(d*x + c)^(5/2), x)

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maple [F]  time = 2.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +i a \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )}{\cot \left (d x +c \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c))/cot(d*x+c)^(5/2),x)

[Out]

int((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c))/cot(d*x+c)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\cot \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c))/cot(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n/cot(d*x + c)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^n)/cot(c + d*x)^(5/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^n)/cot(c + d*x)^(5/2), 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((a+I*a*tan(d*x+c))**n*(A+B*tan(d*x+c))/cot(d*x+c)**(5/2),x)

[Out]

Timed out

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