3.6.50 \(\int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [550]

Optimal. Leaf size=230 \[ \frac {2 (-1)^{3/4} a^{5/2} B \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(4+4 i) a^{5/2} (i A+B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 (2 i A+B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[Out]

2*(-1)^(3/4)*a^(5/2)*B*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*t
an(d*x+c)^(1/2)/d+(4+4*I)*a^(5/2)*(I*A+B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot
(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d-2*a^2*(2*I*A+B)*cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/d-2/3*a*A*cot(d*x+c
)^(3/2)*(a+I*a*tan(d*x+c))^(3/2)/d

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Rubi [A]
time = 0.51, antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {4326, 3674, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} \frac {(4+4 i) a^{5/2} (B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {2 (-1)^{3/4} a^{5/2} B \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (B+2 i A) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-1)^(3/4)*a^(5/2)*B*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c
+ d*x]]*Sqrt[Tan[c + d*x]])/d + ((4 + 4*I)*a^(5/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt
[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d - (2*a^2*((2*I)*A + B)*Sqrt[Cot[c + d*x]]*Sqr
t[a + I*a*Tan[c + d*x]])/d - (2*a*A*Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(3/2))/(3*d)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3625

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

Rule 3674

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^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x]
)^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c
+ d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m
- 1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && E
qQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

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]

Rule 3682

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 4326

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 {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (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))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {1}{3} \left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^{3/2} \left (\frac {3}{2} a (2 i A+B)+\frac {3}{2} i a B \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 (2 i A+B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {1}{3} \left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{4} a^2 (4 A-3 i B)-\frac {3}{4} a^2 B \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 (2 i A+B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\left (4 a^2 (A-i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx-\left (i a B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 (2 i A+B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {\left (8 i a^4 (A-i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {\left (i a^3 B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {(4+4 i) a^{5/2} (i A+B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 (2 i A+B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {\left (2 i a^3 B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {(4+4 i) a^{5/2} (i A+B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 (2 i A+B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {\left (2 i a^3 B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac {2 (-1)^{3/4} a^{5/2} B \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(4+4 i) a^{5/2} (i A+B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 (2 i A+B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}\\ \end {align*}

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Mathematica [A]
time = 7.57, size = 396, normalized size = 1.72 \begin {gather*} \frac {\left (\sqrt {2} e^{-3 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {\frac {i \left (1+e^{2 i (c+d x)}\right )}{-1+e^{2 i (c+d x)}}} \left (16 (i A+B) \log \left (e^{i (c+d x)}+\sqrt {-1+e^{2 i (c+d x)}}\right )+\sqrt {2} B \left (-\log \left (1-3 e^{2 i (c+d x)}-2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}\right )+\log \left (1-3 e^{2 i (c+d x)}+2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}\right )\right )\right )-\frac {8 \sqrt {\cot (c+d x)} (A \csc (c+d x)+(7 i A+3 B) \sec (c+d x)) (\cos (2 c)-i \sin (2 c))}{\sqrt {\sec (c+d x)} (3 \cos (2 d x)+3 i \sin (2 d x))}\right ) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{4 d \sec ^{\frac {7}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((Sqrt[2]*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[(I*(1 + E^((2*I
)*(c + d*x))))/(-1 + E^((2*I)*(c + d*x)))]*(16*(I*A + B)*Log[E^(I*(c + d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]]
+ Sqrt[2]*B*(-Log[1 - 3*E^((2*I)*(c + d*x)) - 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]] + Log
[1 - 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]])))/E^((3*I)*(c + d*x))
- (8*Sqrt[Cot[c + d*x]]*(A*Csc[c + d*x] + ((7*I)*A + 3*B)*Sec[c + d*x])*(Cos[2*c] - I*Sin[2*c]))/(Sqrt[Sec[c +
d*x]]*(3*Cos[2*d*x] + (3*I)*Sin[2*d*x])))*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(4*d*Sec[c + d*x
]^(7/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2596 vs. \(2 (186 ) = 372\).
time = 71.14, size = 2597, normalized size = 11.29

method	result	size
default	\(\text {Expression too large to display}\)	\(2597\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d*a^2*2^(1/2)*(3*I*B*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+
c))^(1/2)-1)-14*A*2^(1/2)-3*I*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)-1)-6*I*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))+3*I*B*((-1+c
os(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)+3*B*cos(d*x+c)^2*((-1+cos(d*x+c)
)/sin(d*x+c))^(1/2)*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)-6*B*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x
+c))^(1/2)*2^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))-3*B*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1
/2)*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)+24*I*A*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ar
ctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)+12*I*A*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln((
((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1)/(-((-1+cos(d*x+c))/sin(d*x+c))^
(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))+24*I*A*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arct
an(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)-24*I*B*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan
(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)-24*I*B*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan((
(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)-12*I*B*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln((-((-1+
cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*
2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1))+16*I*A*cos(d*x+c)*sin(d*x+c)*2^(1/2)-2*A*cos(d*x+c)*2^(1/2)-12*B*
((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln((((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*
x+c)-1)/(-((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))+24*A*cos(d*x+c)^2*((
-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)+6*I*B*2^(1/2)-24*A*((-1+
cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)-24*A*((-1+cos(d*x+c))/sin(d
*x+c))^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)-12*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln((
-((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((-1+cos(d*x+c))/sin(d*x+c))^
(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1))-24*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(((-1+cos(d*x
+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)-24*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^
(1/2)*2^(1/2)-1)+6*B*2^(1/2)*cos(d*x+c)*sin(d*x+c)-6*2^(1/2)*B*sin(d*x+c)+3*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)-14*I*A*sin(d*x+c)*2^(1/2)-24*I*A*((-1+cos(d*x+c))/sin(d*x+
c))^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)-12*I*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(((
(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1)/(-((-1+cos(d*x+c))/sin(d*x+c))^(
1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))-24*I*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(((-1+cos(d*
x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)+16*A*2^(1/2)*cos(d*x+c)^2+24*A*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1
/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)+12*A*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*
ln((-((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((-1+cos(d*x+c))/sin(d*x+
c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1))+24*B*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ar
ctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)+24*B*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan
(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)+12*B*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln((((-1+c
os(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1)/(-((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*
2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))-6*I*B*cos(d*x+c)^2*2^(1/2)+24*I*B*((-1+cos(d*x+c))/sin(d*x+c))^(1
/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)+24*I*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(((-1
+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)+12*I*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln((-((-1+cos(d*x+c))/sin(
d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+
c)+sin(d*x+c)+cos(d*x+c)-1))-3*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1
/2)-1)+6*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))+6*I*B*cos(d*x
+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))-3*I*B*cos(d*x+c)^2
*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1))*sin(d*x+c)*(cos(d*x+c)/s
in(d*x+c))^(5/2)*((I*sin(d*x+c)+cos(d*x+c))*a/cos(d*x+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)^2

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

[Out]

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 780 vs. \(2 (176) = 352\).
time = 1.47, size = 780, normalized size = 3.39 \begin {gather*} -\frac {24 \, \sqrt {2} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {4 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 24 \, \sqrt {2} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {4 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) + 8 \, \sqrt {2} {\left ({\left (8 i \, A + 3 \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 3 \, {\left (-2 i \, A - B\right )} a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}} - 3 \, \sqrt {\frac {4 i \, B^{2} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (-\frac {16 \, {\left (3 \, B a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{3} + \sqrt {2} \sqrt {\frac {4 i \, B^{2} a^{5}}{d^{2}}} {\left (i \, d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{B a}\right ) + 3 \, \sqrt {\frac {4 i \, B^{2} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (-\frac {16 \, {\left (3 \, B a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{3} + \sqrt {2} \sqrt {\frac {4 i \, B^{2} a^{5}}{d^{2}}} {\left (-i \, d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{B a}\right )}{12 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(24*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(4*((A - I*B)*a^3*e^(I
*d*x + I*c) + sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 24*sqrt
(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(4*((A - I*B)*a^3*e^(I*d*x + I*c) -
sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e
^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) + 8*sqrt(2)*((8*I*A + 3
*B)*a^2*e^(3*I*d*x + 3*I*c) + 3*(-2*I*A - B)*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^
(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) - 3*sqrt(4*I*B^2*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(-1
6*(3*B*a^3*e^(2*I*d*x + 2*I*c) - B*a^3 + sqrt(2)*sqrt(4*I*B^2*a^5/d^2)*(I*d*e^(3*I*d*x + 3*I*c) - I*d*e^(I*d*x
+ I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*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)/(B*a)) + 3*sqrt(4*I*B^2*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(-16*(3*B*a^3*e^(2*I*d*x + 2*I*
c) - B*a^3 + sqrt(2)*sqrt(4*I*B^2*a^5/d^2)*(-I*d*e^(3*I*d*x + 3*I*c) + I*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x
+ 2*I*c) + 1))*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)/(B*a)))/(d*e
^(2*I*d*x + 2*I*c) - d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,\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 )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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