3.6.46 \(\int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx\) [546]

Optimal. Leaf size=244 \[ -\frac {(-1)^{3/4} a^{3/2} (12 A-11 i B) \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{4 d}-\frac {(2+2 i) a^{3/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a (4 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}} \]

[Out]

-1/4*(-1)^(3/4)*a^(3/2)*(12*A-11*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-(2+2*I)*a^(3/2)*(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+1/2*I*a*B*(a+I*a*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(3/2)+1/4*a*(4
*I*A+5*B)*(a+I*a*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.56, antiderivative size = 244, 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, 3675, 3678, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} -\frac {(-1)^{3/4} a^{3/2} (12 A-11 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 )}{4 d}-\frac {(2+2 i) a^{3/2} (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 (5 B+4 i A) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((-1)^(3/4)*a^(3/2)*(12*A - (11*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 - ((2 + 2*I)*a^(3/2)*(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 + ((I/2)*a*B*Sqrt[a + I*a*
Tan[c + d*x]])/(d*Cot[c + d*x]^(3/2)) + (a*((4*I)*A + 5*B)*Sqrt[a + I*a*Tan[c + d*x]])/(4*d*Sqrt[Cot[c + d*x]]
)

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 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 3678

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 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 \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\\ &=\frac {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {1}{2} \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (\frac {1}{2} a (4 A-3 i B)+\frac {1}{2} a (4 i A+5 B) \tan (c+d x)\right ) \, dx\\ &=\frac {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a (4 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{4} a^2 (4 i A+5 B)+\frac {1}{4} a^2 (12 A-11 i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{2 a}\\ &=\frac {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a (4 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}-\left (2 a (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}{8} \left ((12 i A+11 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 {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a (4 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {\left (4 i a^3 (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^2 (12 i A+11 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 )}{8 d}\\ &=-\frac {(2-2 i) a^{3/2} (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 {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a (4 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {\left (a^2 (12 i A+11 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 )}{4 d}\\ &=-\frac {(2-2 i) a^{3/2} (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 {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a (4 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {\left (a^2 (12 i A+11 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 )}{4 d}\\ &=-\frac {\sqrt [4]{-1} a^{3/2} (12 i A+11 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)}}{4 d}-\frac {(2-2 i) a^{3/2} (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 {i a B \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {a (4 i A+5 B) \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 4.02, size = 441, normalized size = 1.81 \begin {gather*} \frac {\cos ^2(c+d x) \sqrt {\cot (c+d x)} (\cos (d x)-i \sin (d x)) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \left (-\sqrt {2} \left (\sqrt {2} (12 A-11 i B) \log \left (\frac {2 e^{\frac {5 i c}{2}} \left (\sqrt {2}-i \sqrt {2} e^{i (c+d x)}+2 i \sqrt {-1+e^{2 i (c+d x)}}\right )}{(12 A-11 i B) \left (-i+e^{i (c+d x)}\right )}\right )+\sqrt {2} (-12 A+11 i B) \log \left (\frac {2 e^{\frac {5 i c}{2}} \left (-i \sqrt {2}+\sqrt {2} e^{i (c+d x)}+2 \sqrt {-1+e^{2 i (c+d x)}}\right )}{(12 i A+11 B) \left (i+e^{i (c+d x)}\right )}\right )+32 (A-i B) \log \left ((\cos (c)-i \sin (c)) \left (\cos (c+d x)+i \sin (c+d x)+\sqrt {-1+\cos (2 (c+d x))+i \sin (2 (c+d x))}\right )\right )\right ) \sqrt {i (i+\cot (c+d x)) \sin ^2(c+d x)} (\cos (2 c+d x)-i \sin (2 c+d x))+4 (i \cos (c)+\sin (c)) \tan (c+d x) (4 A-5 i B+2 B \tan (c+d x))\right )}{16 d (A \cos (c+d x)+B \sin (c+d x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[c + d*x]^2*Sqrt[Cot[c + d*x]]*(Cos[d*x] - I*Sin[d*x])*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x])*(
-(Sqrt[2]*(Sqrt[2]*(12*A - (11*I)*B)*Log[(2*E^(((5*I)/2)*c)*(Sqrt[2] - I*Sqrt[2]*E^(I*(c + d*x)) + (2*I)*Sqrt[
-1 + E^((2*I)*(c + d*x))]))/((12*A - (11*I)*B)*(-I + E^(I*(c + d*x))))] + Sqrt[2]*(-12*A + (11*I)*B)*Log[(2*E^
(((5*I)/2)*c)*((-I)*Sqrt[2] + Sqrt[2]*E^(I*(c + d*x)) + 2*Sqrt[-1 + E^((2*I)*(c + d*x))]))/(((12*I)*A + 11*B)*
(I + E^(I*(c + d*x))))] + 32*(A - I*B)*Log[(Cos[c] - I*Sin[c])*(Cos[c + d*x] + I*Sin[c + d*x] + Sqrt[-1 + Cos[
2*(c + d*x)] + I*Sin[2*(c + d*x)]])])*Sqrt[I*(I + Cot[c + d*x])*Sin[c + d*x]^2]*(Cos[2*c + d*x] - I*Sin[2*c +
d*x])) + 4*(I*Cos[c] + Sin[c])*Tan[c + d*x]*(4*A - (5*I)*B + 2*B*Tan[c + d*x])))/(16*d*(A*Cos[c + d*x] + B*Sin
[c + d*x]))

________________________________________________________________________________________

Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 4489 vs. \(2 (194 ) = 388\).
time = 63.26, size = 4490, normalized size = 18.40

method	result	size
default	\(\text {Expression too large to display}\)	\(4490\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/16/d*2^(1/2)*a*(-24*I*A*2^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))*cos(d*x+c)^2*sin(d*x+c)-14*B*2^(
1/2)*cos(d*x+c)^2*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+11*B*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(
1/2)+1)*cos(d*x+c)^2*sin(d*x+c)+8*A*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+8*A*2^(
1/2)*cos(d*x+c)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-24*I*A*2^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c
))^(1/2))*cos(d*x+c)^2+32*I*B*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)*cos(d*x+c)^2*sin(d*x+c)+32*
I*B*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)*cos(d*x+c)^2*sin(d*x+c)+16*I*B*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))*cos(d*x+c)^2*sin(d*x+c)+8*I*A*2^(1/2)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))
^(1/2)+4*I*B*2^(1/2)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+24*A*2^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*
x+c))^(1/2))*cos(d*x+c)^2*sin(d*x+c)+12*A*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)^2*sin(d*
x+c)-12*A*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)^2*sin(d*x+c)-11*B*2^(1/2)*ln(((-1+cos(d*
x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)^2*sin(d*x+c)-22*B*2^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))*cos
(d*x+c)^2*sin(d*x+c)+12*I*A*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)^2-12*I*A*2^(1/2)*ln(((
-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)^2+32*I*A*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)*c
os(d*x+c)^2*sin(d*x+c)+32*I*A*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)*cos(d*x+c)^2*sin(d*x+c)+16*
I*A*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))*cos(d*x+c)^2*sin(d*x+c)-22*I*B*2^(1/2)*arctan(((-1+
cos(d*x+c))/sin(d*x+c))^(1/2))*cos(d*x+c)^2-11*I*B*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)
^2+11*I*B*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)^2+16*A*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))*cos(d*x+c)^3-32*B*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)*cos(d*x+c)^3-32*B
*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)*cos(d*x+c)^3-16*B*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))*cos(d*x+c)^3+12*I*A*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)^2*sin(d*x+c)-
12*I*A*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)^2*sin(d*x+c)-22*I*B*2^(1/2)*arctan(((-1+cos
(d*x+c))/sin(d*x+c))^(1/2))*cos(d*x+c)^2*sin(d*x+c)-11*I*B*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*co
s(d*x+c)^2*sin(d*x+c)+11*I*B*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)^2*sin(d*x+c)+8*I*A*2^
(1/2)*cos(d*x+c)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-8*I*A*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*((-1+cos(
d*x+c))/sin(d*x+c))^(1/2)-14*I*B*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-10*I*B*2^(
1/2)*cos(d*x+c)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-11*I*B*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(
1/2)-1)*cos(d*x+c)^3+14*I*B*2^(1/2)*cos(d*x+c)^3*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-4*I*B*2^(1/2)*cos(d*x+c)^2
*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-14*I*B*2^(1/2)*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+24*I*A*2^(1/2
)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))*cos(d*x+c)^3-12*I*A*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)
+1)*cos(d*x+c)^3+12*I*A*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)^3-8*I*A*2^(1/2)*cos(d*x+c)
^3*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+22*I*B*2^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))*cos(d*x+c)^3+1
1*I*B*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)^3-22*B*cos(d*x+c)^2*2^(1/2)*arctan(((-1+cos(
d*x+c))/sin(d*x+c))^(1/2))+4*B*2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-4*B*cos(d*x+c)^2*2^(1/2)*
((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+14*B*cos(d*x+c)*2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+12*A*2^(1/2)*ln(
((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)^3+22*B*2^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))*cos
(d*x+c)^3-11*B*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)^3+11*B*2^(1/2)*ln(((-1+cos(d*x+c))/
sin(d*x+c))^(1/2)-1)*cos(d*x+c)^3-14*B*2^(1/2)*cos(d*x+c)^3*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-32*A*arctan(((-
1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)*cos(d*x+c)^2*sin(d*x+c)-32*A*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1
/2)*2^(1/2)+1)*cos(d*x+c)^2*sin(d*x+c)-32*I*A*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)*cos(d*x+c)^
3-32*I*A*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)*cos(d*x+c)^3-16*I*A*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))*cos(d*x+c)^3-32*I*B*...

________________________________________________________________________________________

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((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 913 vs. \(2 (182) = 364\).
time = 1.53, size = 913, normalized size = 3.74 \begin {gather*} -\frac {16 \, \sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{3}}{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}\right ) - 16 \, \sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{3}}{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}\right ) - 4 \, \sqrt {2} {\left ({\left (4 \, A - 7 i \, B\right )} a e^{\left (5 i \, d x + 5 i \, c\right )} + 4 i \, B a e^{\left (3 i \, d x + 3 i \, c\right )} - {\left (4 \, A - 3 i \, B\right )} a 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 (144 i \, A^{2} + 264 \, A B - 121 i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {16 \, {\left (3 \, {\left (12 i \, A + 11 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-12 i \, A - 11 \, B\right )} a^{2} + 2 \, \sqrt {2} \sqrt {\frac {{\left (144 i \, A^{2} + 264 \, A B - 121 i \, B^{2}\right )} a^{3}}{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 )}}{12 i \, A + 11 \, B}\right ) + \sqrt {\frac {{\left (144 i \, A^{2} + 264 \, A B - 121 i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {16 \, {\left (3 \, {\left (12 i \, A + 11 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-12 i \, A - 11 \, B\right )} a^{2} - 2 \, \sqrt {2} \sqrt {\frac {{\left (144 i \, A^{2} + 264 \, A B - 121 i \, B^{2}\right )} a^{3}}{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 )}}{12 i \, A + 11 \, B}\right )}{16 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/16*(16*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d
)*log(4*((A - I*B)*a^2*e^(I*d*x + I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^3/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)) - 16*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*
d*x + 2*I*c) + d)*log(4*((A - I*B)*a^2*e^(I*d*x + I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^3/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)) - 4*sqrt(2)*((4*A - 7*I*B)*a*e^(5*I*d*x + 5*I*c) + 4*I*B*a*e^(3*I*d*x +
3*I*c) - (4*A - 3*I*B)*a*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((144*I*A^2 + 264*A*B - 121*I*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*
I*d*x + 2*I*c) + d)*log(-16*(3*(12*I*A + 11*B)*a^2*e^(2*I*d*x + 2*I*c) + (-12*I*A - 11*B)*a^2 + 2*sqrt(2)*sqrt
((144*I*A^2 + 264*A*B - 121*I*B^2)*a^3/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)/(12*I*A + 11*B))
+ sqrt((144*I*A^2 + 264*A*B - 121*I*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(-
16*(3*(12*I*A + 11*B)*a^2*e^(2*I*d*x + 2*I*c) + (-12*I*A - 11*B)*a^2 - 2*sqrt(2)*sqrt((144*I*A^2 + 264*A*B - 1
21*I*B^2)*a^3/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)/(12*I*A + 11*B)))/(d*e^(4*I*d*x + 4*I*c)
+ 2*d*e^(2*I*d*x + 2*I*c) + d)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (c + d x \right )}\right )}{\sqrt {\cot {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c))/cot(d*x+c)**(1/2),x)

[Out]

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

________________________________________________________________________________________

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((a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 )}^{3/2}}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}} \,d x \end {gather*}

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)^(3/2))/cot(c + d*x)^(1/2),x)

[Out]

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

________________________________________________________________________________________