3.6.51 \(\int \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [551]

Optimal. Leaf size=236 \[ \frac {(-1)^{3/4} a^{5/2} (2 A-5 i B) \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {a^2 (2 i A-B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}-\frac {2 a A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}{d} \]

[Out]

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

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Rubi [A]
time = 0.54, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {4326, 3674, 3675, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} \frac {(-1)^{3/4} a^{5/2} (2 A-5 i 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 {(4+4 i) a^{5/2} (A-i B) \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 {a^2 (-B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}-\frac {2 a A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

((-1)^(3/4)*a^(5/2)*(2*A - (5*I)*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)*(A - I*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 + (a^2*((2*I)*A - B)*Sqrt[a + I*a
*Tan[c + d*x]])/(d*Sqrt[Cot[c + d*x]]) - (2*a*A*Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2))/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 3675

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*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*
(m + n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[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 {3}{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 {3}{2}}(c+d x)} \, dx\\ &=-\frac {2 a A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}{d}+\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 {1}{2} a (4 i A+B)+\frac {1}{2} a (2 A+i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {a^2 (2 i A-B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}-\frac {2 a A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}{d}+\left (2 \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 (2 i A+B)-\frac {1}{4} a^2 (2 A-5 i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {a^2 (2 i A-B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}-\frac {2 a A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}{d}+\left (4 a^2 (i A+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-\frac {1}{2} \left (a (2 i A+5 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 {a^2 (2 i A-B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}-\frac {2 a A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}{d}-\frac {\left (8 i a^4 (i A+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 (a^3 (2 i A+5 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 )}{2 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 {a^2 (2 i A-B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}-\frac {2 a A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}{d}-\frac {\left (a^3 (2 i A+5 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 {a^2 (2 i A-B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}-\frac {2 a A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}{d}-\frac {\left (a^3 (2 i A+5 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 {\sqrt [4]{-1} a^{5/2} (2 i A+5 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 {a^2 (2 i A-B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}-\frac {2 a A \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}{d}\\ \end {align*}

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Mathematica [A]
time = 5.72, size = 387, normalized size = 1.64 \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 (32 (A-i B) \log \left (e^{i (c+d x)}+\sqrt {-1+e^{2 i (c+d x)}}\right )-\sqrt {2} (2 A-5 i 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 (B+2 A \cot (c+d x)) \sqrt {\sec (c+d x)} (\cos (2 c)-i \sin (2 c))}{\sqrt {\cot (c+d x)} (\cos (d x)+i \sin (d x))^2}\right ) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{8 d \sec ^{\frac {7}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[c + d*x]^(3/2)*(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)))]*(32*(A - I*B)*Log[E^(I*(c + d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]]
- Sqrt[2]*(2*A - (5*I)*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*(B + 2*A*Cot[c + d*x])*Sqrt[Sec[c + d*x]]*(Cos[2*c] - I*Sin[2*c]))/(Sqrt[Cot[c + d*x]]*(Cos[
d*x] + I*Sin[d*x])^2))*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(8*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. 1895 vs. \(2 (193 ) = 386\).
time = 65.30, size = 1896, normalized size = 8.03

method	result	size
default	\(\text {Expression too large to display}\)	\(1896\)

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

[Out]

-1/4/d*a^2*2^(1/2)*(-5*I*B*sin(d*x+c)*2^(1/2)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(((-1+cos(d*x+c)
)/sin(d*x+c))^(1/2)+1)+5*I*B*sin(d*x+c)*2^(1/2)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(((-1+cos(d*x+
c))/sin(d*x+c))^(1/2)-1)-10*I*B*sin(d*x+c)*2^(1/2)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(((-1+c
os(d*x+c))/sin(d*x+c))^(1/2))+2*I*A*sin(d*x+c)*2^(1/2)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(((-1+c
os(d*x+c))/sin(d*x+c))^(1/2)+1)-2*I*A*sin(d*x+c)*2^(1/2)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(((-1
+cos(d*x+c))/sin(d*x+c))^(1/2)-1)-4*I*A*sin(d*x+c)*2^(1/2)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arcta
n(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))+2*I*B*2^(1/2)-4*A*cos(d*x+c)*2^(1/2)+4*I*A*cos(d*x+c)*2^(1/2)*sin(d*x+c)
-16*A*sin(d*x+c)*cos(d*x+c)*((-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*sin(d*x+c)*cos(d*x+c)*((-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)-8*A*sin(d*x+c)*cos(d*x+c)*((-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)-sin(d*x+c
)-cos(d*x+c)+1))+16*B*sin(d*x+c)*cos(d*x+c)*((-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*B*sin(d*x+c)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(((-1+cos(d*x+c))/si
n(d*x+c))^(1/2)*2^(1/2)-1)+8*B*sin(d*x+c)*cos(d*x+c)*((-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)+sin(d*x+c)+cos(d*x+c)-1))-2*A*sin(d*x+c)*2^(1/2)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(((-1+
cos(d*x+c))/sin(d*x+c))^(1/2)-1)+4*A*sin(d*x+c)*2^(1/2)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan((
(-1+cos(d*x+c))/sin(d*x+c))^(1/2))+5*B*sin(d*x+c)*2^(1/2)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(((-
1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)-5*B*sin(d*x+c)*2^(1/2)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(((-
1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)-10*B*sin(d*x+c)*2^(1/2)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arcta
n(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))+16*I*A*sin(d*x+c)*cos(d*x+c)*((-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*I*A*sin(d*x+c)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*a
rctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)+8*I*A*sin(d*x+c)*cos(d*x+c)*((-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)+sin(d*x+c)+cos(d*x+c)-1))+16*I*B*sin(d*x+c)*cos(d*x+c)*((-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*I*B*sin(d*x+c)*cos(d*x+c)*((-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)+8*I*B*sin(d*x+c)*cos(d*x+c)*((-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)-sin(d*x+c)-cos(d*x+c)+1))+2*A*sin(d*x+c)*2^(1/2)*cos(d
*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)-2*I*B*2^(1/2)*cos(d*x+c)^2+2
*B*2^(1/2)*cos(d*x+c)*sin(d*x+c)-2*2^(1/2)*B*sin(d*x+c)+4*A*2^(1/2)*cos(d*x+c)^2)*sin(d*x+c)*(cos(d*x+c)/sin(d
*x+c))^(3/2)*((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(-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)^(3/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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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. 849 vs. \(2 (182) = 364\).
time = 1.35, size = 849, normalized size = 3.60 \begin {gather*} \frac {8 \, \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 (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, 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} \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 (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, 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 ) - 4 \, \sqrt {2} {\left ({\left (2 \, A - i \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (2 \, A + i \, 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}} - \sqrt {\frac {{\left (4 i \, A^{2} + 20 \, A B - 25 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 {16 \, {\left (3 \, {\left (2 i \, A + 5 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-2 i \, A - 5 \, B\right )} a^{3} + 2 \, \sqrt {2} \sqrt {\frac {{\left (4 i \, A^{2} + 20 \, A B - 25 i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} - 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 )}}{{\left (2 i \, A + 5 \, B\right )} a}\right ) + \sqrt {\frac {{\left (4 i \, A^{2} + 20 \, A B - 25 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 {16 \, {\left (3 \, {\left (2 i \, A + 5 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-2 i \, A - 5 \, B\right )} a^{3} - 2 \, \sqrt {2} \sqrt {\frac {{\left (4 i \, A^{2} + 20 \, A B - 25 i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} - 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 )}}{{\left (2 i \, A + 5 \, B\right )} a}\right )}{4 \, {\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)^(3/2)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(8*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)*(I*d*e^(2*I*d*x + 2*I*c) - I*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*sq
rt(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)*(-I*d*e^(2*I*d*x + 2*I*c) + I*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)) - 4*sqrt(2)*((
2*A - I*B)*a^2*e^(3*I*d*x + 3*I*c) + (2*A + I*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)) - sqrt((4*I*A^2 + 20*A*B - 25*I*B^2)*a^5/d^2)*(d*e^(2*I*
d*x + 2*I*c) + d)*log(-16*(3*(2*I*A + 5*B)*a^3*e^(2*I*d*x + 2*I*c) + (-2*I*A - 5*B)*a^3 + 2*sqrt(2)*sqrt((4*I*
A^2 + 20*A*B - 25*I*B^2)*a^5/d^2)*(d*e^(3*I*d*x + 3*I*c) - 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)/((2*I*A + 5*B)*a)) + sqrt(
(4*I*A^2 + 20*A*B - 25*I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(-16*(3*(2*I*A + 5*B)*a^3*e^(2*I*d*x + 2
*I*c) + (-2*I*A - 5*B)*a^3 - 2*sqrt(2)*sqrt((4*I*A^2 + 20*A*B - 25*I*B^2)*a^5/d^2)*(d*e^(3*I*d*x + 3*I*c) - 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)/((2*I*A + 5*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)**(3/2)*(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)^(3/2)*(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)^(3/2), 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 )}^{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 )}^{5/2} \,d x \end {gather*}

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)^(5/2),x)

[Out]

int(cot(c + d*x)^(3/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2), x)

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