3.545 \(\int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=196 \[ -\frac {(-1)^{3/4} a^{3/2} (3 B+2 i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(2-2 i) a^{3/2} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {i a B \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.65, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {4241, 3594, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac {(-1)^{3/4} a^{3/2} (3 B+2 i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(2-2 i) a^{3/2} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {i a B \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

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 203

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

Rule 205

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

Rule 217

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 3544

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 3594

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

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Mathematica [A]  time = 7.97, size = 360, normalized size = 1.84 \[ \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \left (\sqrt {2} e^{-2 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {\frac {i \left (1+e^{2 i (c+d x)}\right )}{-1+e^{2 i (c+d x)}}} \left (\sqrt {2} (3 B+2 i A) \left (\log \left (-2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )-\log \left (2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )\right )-16 i (A-i B) \log \left (\sqrt {-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )\right )+\frac {8 B (\tan (c+d x)+i)}{\sqrt {\cot (c+d x)} \sqrt {\sec (c+d x)}}\right )}{8 d \sec ^{\frac {5}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.56, size = 818, normalized size = 4.17 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*sqrt(2)*(B*a*e^(3*I*d*x + 3*I*c) - 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((-4*I*A^2 - 12*A*B + 9*I*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2
*I*c) + d)*log(((-96*I*A - 144*B)*a^2*e^(2*I*d*x + 2*I*c) + (32*I*A + 48*B)*a^2 + sqrt(2)*sqrt((-4*I*A^2 - 12*
A*B + 9*I*B^2)*a^3/d^2)*(32*I*d*e^(3*I*d*x + 3*I*c) - 32*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)/(2*I*A + 3*B)) - sqrt((-4*
I*A^2 - 12*A*B + 9*I*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(((-96*I*A - 144*B)*a^2*e^(2*I*d*x + 2*I*c)
+ (32*I*A + 48*B)*a^2 + sqrt(2)*sqrt((-4*I*A^2 - 12*A*B + 9*I*B^2)*a^3/d^2)*(-32*I*d*e^(3*I*d*x + 3*I*c) + 32*
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)/(2*I*A + 3*B)) + sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*
c) + d)*log(-(8*(A - I*B)*a^2*e^(I*d*x + I*c) + sqrt(2)*sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^3/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)/((2*I*A + 2*B)*a)) - sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c
) + d)*log(-(8*(A - I*B)*a^2*e^(I*d*x + I*c) - sqrt(2)*sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^3/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)/((2*I*A + 2*B)*a)))/(d*e^(2*I*d*x + 2*I*c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 4.14, size = 1306, normalized size = 6.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d*(-1+cos(d*x+c))*(-4*A*cos(d*x+c)*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-s
in(d*x+c)+1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))-4*B*cos(d*x+c)*l
n(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(2^(1/2)*((-1+cos(d*x+c))/s
in(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))-8*A*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)
*cos(d*x+c)-8*A*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)-8*B*arctan(2^(1/2)*((-1+cos(d*
x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)-8*B*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)+2*B*
2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+2*I*A*cos(d*x+c)*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)-8
*I*A*cos(d*x+c)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)-8*I*A*cos(d*x+c)*arctan(2^(1/2)*((-1+cos(
d*x+c))/sin(d*x+c))^(1/2)+1)-4*I*A*cos(d*x+c)*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d
*x+c)+sin(d*x+c)-1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))+8*I*B*cos
(d*x+c)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)+8*I*B*cos(d*x+c)*arctan(2^(1/2)*((-1+cos(d*x+c))/
sin(d*x+c))^(1/2)+1)+4*I*B*cos(d*x+c)*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-si
n(d*x+c)+1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))+2*B*cos(d*x+c)*2^
(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+2*A*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*2^(1/2)*cos(d*x+c)-2*A*l
n(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*2^(1/2)*cos(d*x+c)+4*A*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))*2^(1
/2)*cos(d*x+c)+3*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*2^(1/2)*cos(d*x+c)+6*B*arctan(((-1+cos(d*x+c))/sin
(d*x+c))^(1/2))*2^(1/2)*cos(d*x+c)-3*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*2^(1/2)*cos(d*x+c)-2*I*A*cos(d
*x+c)*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)+3*I*B*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*2^(1/2)*
cos(d*x+c)-6*I*B*cos(d*x+c)*2^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))-3*I*B*cos(d*x+c)*2^(1/2)*ln(((-
1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)+4*I*A*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))*2^(1/2)*cos(d*x+c)+2*I*B*2
^(1/2)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(cos(d*x+
c)/sin(d*x+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)/((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*a

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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