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

Optimal. Leaf size=123 \[ -\frac {\sqrt {a} (2 B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {\sqrt {2} \sqrt {a} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {A \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3598, 3600, 3480, 206, 3599, 63, 208} \[ -\frac {\sqrt {a} (2 B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {\sqrt {2} \sqrt {a} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {A \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[a]*(I*A + 2*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d) + (Sqrt[2]*Sqrt[a]*(I*A + B)*ArcTanh[Sq
rt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (A*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d

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 206

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

Rule 208

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

Rule 3480

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

Rule 3598

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*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*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] && 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 3600

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

Rubi steps

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

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Mathematica [B]  time = 4.57, size = 293, normalized size = 2.38 \[ \frac {\sqrt {a+i a \tan (c+d x)} \left (-8 A \cot (c+d x)+e^{-i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \left (\sqrt {2} (2 B+i A) \left (\log \left (\left (-1+e^{i (c+d x)}\right )^2\right )-\log \left (\left (1+e^{i (c+d x)}\right )^2\right )+\log \left (-2 e^{i (c+d x)} \left (1+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )+3 e^{2 i (c+d x)}+2 \sqrt {2} \sqrt {1+e^{2 i (c+d x)}}+3\right )-\log \left (2 e^{i (c+d x)} \left (1+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )+3 e^{2 i (c+d x)}+2 \sqrt {2} \sqrt {1+e^{2 i (c+d x)}}+3\right )\right )+8 (B+i A) \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*A*Cot[c + d*x] + (Sqrt[1 + E^((2*I)*(c + d*x))]*(8*(I*A + B)*ArcSinh[E^(I*(c + d*x))] + Sqrt[2]*(I*A + 2*
B)*(Log[(-1 + E^(I*(c + d*x)))^2] - Log[(1 + E^(I*(c + d*x)))^2] + Log[3 + 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*S
qrt[1 + E^((2*I)*(c + d*x))] - 2*E^(I*(c + d*x))*(1 + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] - Log[3 + 3*E^((
2*I)*(c + d*x)) + 2*Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))] + 2*E^(I*(c + d*x))*(1 + Sqrt[2]*Sqrt[1 + E^((2*I)*(
c + d*x))])])))/E^(I*(c + d*x)))*Sqrt[a + I*a*Tan[c + d*x]])/(8*d)

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fricas [B]  time = 0.67, size = 649, normalized size = 5.28 \[ \frac {{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left ({\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left ({\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (8 \, A^{2} - 16 i \, A B - 8 \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {{\left (A^{2} - 4 i \, A B - 4 \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left ({\left (48 i \, A + 96 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (16 i \, A + 32 \, B\right )} a^{2} + 32 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {{\left (A^{2} - 4 i \, A B - 4 \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + 2 \, B}\right ) + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {{\left (A^{2} - 4 i \, A B - 4 \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left ({\left (48 i \, A + 96 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (16 i \, A + 32 \, B\right )} a^{2} - 32 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {{\left (A^{2} - 4 i \, A B - 4 \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + 2 \, B}\right ) + \sqrt {2} {\left (-4 i \, A e^{\left (3 i \, d x + 3 i \, c\right )} - 4 i \, A e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a/d^2)*log(((4*I*A + 4*B)*a*e^(I*d*x + I*c)
+ sqrt(2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)
))*e^(-I*d*x - I*c)/(I*A + B)) - (d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a/d^2)*log(((4*I
*A + 4*B)*a*e^(I*d*x + I*c) - sqrt(2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a/d^2)*sqrt
(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(I*A + B)) - (d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-(A^2 - 4*I*A*B
- 4*B^2)*a/d^2)*log(((48*I*A + 96*B)*a^2*e^(2*I*d*x + 2*I*c) + (16*I*A + 32*B)*a^2 + 32*sqrt(2)*(a*d*e^(3*I*d*
x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(-(A^2 - 4*I*A*B - 4*B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-
2*I*d*x - 2*I*c)/(I*A + 2*B)) + (d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-(A^2 - 4*I*A*B - 4*B^2)*a/d^2)*log(((48*I*A
+ 96*B)*a^2*e^(2*I*d*x + 2*I*c) + (16*I*A + 32*B)*a^2 - 32*sqrt(2)*(a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I
*c))*sqrt(-(A^2 - 4*I*A*B - 4*B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/(I*A + 2*B))
+ sqrt(2)*(-4*I*A*e^(3*I*d*x + 3*I*c) - 4*I*A*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 4.12, size = 1181, normalized size = 9.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(-2*I*A*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+2*I*A*cos(d
*x+c)^2+2*I*B*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)
))^(1/2))-2*A*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)*2^(1/2))*2^(1/2)-I*A*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-(-2*cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-I*A*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-2*B*cos(d*x+c)*sin(d*x+c)*(
-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(
1/2))*2^(1/2)+2*I*B*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1
/2))-A*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2
))-2*I*A*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(
d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)-2*A*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)-2*B*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(
-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))+2*I*B*sin(d*x+c)*(-2*cos(d*x+c)/(1+
cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)-2*B*sin(d*x+c)*(-2*cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1
/2)+2*I*B*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*2^(1/2))*2^(1/2)-A*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c
)))^(1/2))-2*B*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(
d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-2*A*cos(d*x+c)*sin(d*x+c)+2*I*A*cos(d*x+c))/(I*sin(d*x+c)+cos(d*x+c)-1)/(1+co
s(d*x+c))

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maxima [A]  time = 0.68, size = 145, normalized size = 1.18 \[ -\frac {i \, {\left (\frac {\sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} - \frac {{\left (A - 2 i \, B\right )} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{\sqrt {a}} - \frac {2 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} A}{a \tan \left (d x + c\right )}\right )} a}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*(sqrt(2)*(A - I*B)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(
d*x + c) + a)))/sqrt(a) - (A - 2*I*B)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) +
sqrt(a)))/sqrt(a) - 2*I*sqrt(I*a*tan(d*x + c) + a)*A/(a*tan(d*x + c)))*a/d

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mupad [B]  time = 7.06, size = 168, normalized size = 1.37 \[ -\frac {\mathrm {cot}\left (c+d\,x\right )\,\left (A\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}+A\,\sqrt {a}\,\mathrm {tan}\left (c+d\,x\right )\,\mathrm {atanh}\left (\frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {a}}\right )\,1{}\mathrm {i}+2\,B\,\sqrt {a}\,\mathrm {tan}\left (c+d\,x\right )\,\mathrm {atanh}\left (\frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {a}}\right )-\sqrt {2}\,A\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}-\sqrt {2}\,B\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )\,\mathrm {tan}\left (c+d\,x\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(cot(c + d*x)*(A*(a + a*tan(c + d*x)*1i)^(1/2) + A*a^(1/2)*tan(c + d*x)*atanh((a + a*tan(c + d*x)*1i)^(1/2)/a
^(1/2))*1i + 2*B*a^(1/2)*tan(c + d*x)*atanh((a + a*tan(c + d*x)*1i)^(1/2)/a^(1/2)) - 2^(1/2)*A*a^(1/2)*atanh((
2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^(1/2)))*tan(c + d*x)*1i - 2^(1/2)*B*a^(1/2)*atanh((2^(1/2)*(a + a*
tan(c + d*x)*1i)^(1/2))/(2*a^(1/2)))*tan(c + d*x)))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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