3.94 \(\int \frac {\cot (c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=114 \[ \frac {A+i B}{d \sqrt {a+i a \tan (c+d x)}}+\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d} \]

[Out]

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

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Rubi [A]  time = 0.35, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3596, 3600, 3480, 206, 3599, 63, 208} \[ \frac {A+i B}{d \sqrt {a+i a \tan (c+d x)}}+\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d) + ((A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]
/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d) + (A + I*B)/(d*Sqrt[a + I*a*Tan[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 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 3596

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*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*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] && LtQ[m, 0] &&  !GtQ[n,
0]

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

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Mathematica [A]  time = 2.73, size = 208, normalized size = 1.82 \[ \frac {\sqrt {\sec (c+d x)} \left ((A+i B) \sqrt {1+e^{2 i (c+d x)}}+(A-i B) e^{i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )-2 \sqrt {2} A e^{i (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {1+e^{2 i (c+d x)}}}\right )\right )}{2 d \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((A + I*B)*Sqrt[1 + E^((2*I)*(c + d*x))] + (A - I*B)*E^(I*(c + d*x))*ArcSinh[E^(I*(c + d*x))] - 2*Sqrt[2]*A*E
^(I*(c + d*x))*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[1 + E^((2*I)*(c + d*x))]])*Sqrt[Sec[c + d*x]])/(2*d*Sqrt
[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E
^((2*I)*(c + d*x))])

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fricas [B]  time = 0.64, size = 581, normalized size = 5.10 \[ \frac {{\left (a d \sqrt {\frac {2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {{\left ({\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left (2 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - a d \sqrt {\frac {2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {{\left ({\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left (-2 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 2 \, a d \sqrt {\frac {A^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {16 \, {\left (3 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2} + 2 \, \sqrt {2} {\left (a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{2} 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 {A^{2}}{a d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) + 2 \, a d \sqrt {\frac {A^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {16 \, {\left (3 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2} - 2 \, \sqrt {2} {\left (a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{2} 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 {A^{2}}{a d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) + 2 \, \sqrt {2} {\left ({\left (A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*d*sqrt((2*A^2 - 4*I*A*B - 2*B^2)/(a*d^2))*e^(I*d*x + I*c)*log(((4*I*A + 4*B)*a*e^(I*d*x + I*c) + sqrt(2
)*(2*I*a*d*e^(2*I*d*x + 2*I*c) + 2*I*a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((2*A^2 - 4*I*A*B - 2*B^2)/(a*
d^2)))*e^(-I*d*x - I*c)/(I*A + B)) - a*d*sqrt((2*A^2 - 4*I*A*B - 2*B^2)/(a*d^2))*e^(I*d*x + I*c)*log(((4*I*A +
4*B)*a*e^(I*d*x + I*c) + sqrt(2)*(-2*I*a*d*e^(2*I*d*x + 2*I*c) - 2*I*a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*s
qrt((2*A^2 - 4*I*A*B - 2*B^2)/(a*d^2)))*e^(-I*d*x - I*c)/(I*A + B)) - 2*a*d*sqrt(A^2/(a*d^2))*e^(I*d*x + I*c)*
log(16*(3*A*a^2*e^(2*I*d*x + 2*I*c) + A*a^2 + 2*sqrt(2)*(a^2*d*e^(3*I*d*x + 3*I*c) + a^2*d*e^(I*d*x + I*c))*sq
rt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(A^2/(a*d^2)))*e^(-2*I*d*x - 2*I*c)/A) + 2*a*d*sqrt(A^2/(a*d^2))*e^(I*d*x
+ I*c)*log(16*(3*A*a^2*e^(2*I*d*x + 2*I*c) + A*a^2 - 2*sqrt(2)*(a^2*d*e^(3*I*d*x + 3*I*c) + a^2*d*e^(I*d*x + I
*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(A^2/(a*d^2)))*e^(-2*I*d*x - 2*I*c)/A) + 2*sqrt(2)*((A + I*B)*e^(2*
I*d*x + 2*I*c) + A + I*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 3.72, size = 948, normalized size = 8.32 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(I*A*2^(1/2)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-I*B*cos(d
*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d
*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+2*I*A*cos(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))+A*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+2*I*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))-I*B*2^(1/2)*(-
2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)*2^(1/2))+B*2^(1/2)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(
d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+2*A*cos(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*(-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))+A*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-2*A*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*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
*sin(d*x+c)+2*A*(-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))-4*I*B*cos(
d*x+c)-4*A*cos(d*x+c))/(I*sin(d*x+c)+cos(d*x+c))/a

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maxima [A]  time = 0.61, size = 133, normalized size = 1.17 \[ -\frac {\frac {\sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} - \frac {4 \, A \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{\sqrt {a}} - \frac {4 \, {\left (A + i \, B\right )}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(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) - 4*A*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))
/sqrt(a) - 4*(A + I*B)/sqrt(I*a*tan(d*x + c) + a))/d

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mupad [B]  time = 7.14, size = 515, normalized size = 4.52 \[ \frac {A+B\,1{}\mathrm {i}}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}-\frac {2\,A\,\mathrm {atanh}\left (\frac {28\,A^3\,a^{3/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{28\,d\,A^3\,a^2+8{}\mathrm {i}\,d\,A^2\,B\,a^2+4\,d\,A\,B^2\,a^2}+\frac {4\,A\,B^2\,a^{3/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{28\,d\,A^3\,a^2+8{}\mathrm {i}\,d\,A^2\,B\,a^2+4\,d\,A\,B^2\,a^2}+\frac {A^2\,B\,a^{3/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,8{}\mathrm {i}}{28\,d\,A^3\,a^2+8{}\mathrm {i}\,d\,A^2\,B\,a^2+4\,d\,A\,B^2\,a^2}\right )}{\sqrt {a}\,d}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,A^3\,{\left (-a\right )}^{3/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,7{}\mathrm {i}}{2\,\left (7\,d\,A^3\,a^2-5{}\mathrm {i}\,d\,A^2\,B\,a^2+3\,d\,A\,B^2\,a^2-1{}\mathrm {i}\,d\,B^3\,a^2\right )}+\frac {\sqrt {2}\,B^3\,{\left (-a\right )}^{3/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\left (7\,d\,A^3\,a^2-5{}\mathrm {i}\,d\,A^2\,B\,a^2+3\,d\,A\,B^2\,a^2-1{}\mathrm {i}\,d\,B^3\,a^2\right )}+\frac {\sqrt {2}\,A\,B^2\,{\left (-a\right )}^{3/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,3{}\mathrm {i}}{2\,\left (7\,d\,A^3\,a^2-5{}\mathrm {i}\,d\,A^2\,B\,a^2+3\,d\,A\,B^2\,a^2-1{}\mathrm {i}\,d\,B^3\,a^2\right )}+\frac {5\,\sqrt {2}\,A^2\,B\,{\left (-a\right )}^{3/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\left (7\,d\,A^3\,a^2-5{}\mathrm {i}\,d\,A^2\,B\,a^2+3\,d\,A\,B^2\,a^2-1{}\mathrm {i}\,d\,B^3\,a^2\right )}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{2\,\sqrt {-a}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A + B*1i)/(d*(a + a*tan(c + d*x)*1i)^(1/2)) - (2*A*atanh((28*A^3*a^(3/2)*d*(a + a*tan(c + d*x)*1i)^(1/2))/(28
*A^3*a^2*d + 4*A*B^2*a^2*d + A^2*B*a^2*d*8i) + (4*A*B^2*a^(3/2)*d*(a + a*tan(c + d*x)*1i)^(1/2))/(28*A^3*a^2*d
+ 4*A*B^2*a^2*d + A^2*B*a^2*d*8i) + (A^2*B*a^(3/2)*d*(a + a*tan(c + d*x)*1i)^(1/2)*8i)/(28*A^3*a^2*d + 4*A*B^
2*a^2*d + A^2*B*a^2*d*8i)))/(a^(1/2)*d) + (2^(1/2)*atanh((2^(1/2)*A^3*(-a)^(3/2)*d*(a + a*tan(c + d*x)*1i)^(1/
2)*7i)/(2*(7*A^3*a^2*d - B^3*a^2*d*1i + 3*A*B^2*a^2*d - A^2*B*a^2*d*5i)) + (2^(1/2)*B^3*(-a)^(3/2)*d*(a + a*ta
n(c + d*x)*1i)^(1/2))/(2*(7*A^3*a^2*d - B^3*a^2*d*1i + 3*A*B^2*a^2*d - A^2*B*a^2*d*5i)) + (2^(1/2)*A*B^2*(-a)^
(3/2)*d*(a + a*tan(c + d*x)*1i)^(1/2)*3i)/(2*(7*A^3*a^2*d - B^3*a^2*d*1i + 3*A*B^2*a^2*d - A^2*B*a^2*d*5i)) +
(5*2^(1/2)*A^2*B*(-a)^(3/2)*d*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(7*A^3*a^2*d - B^3*a^2*d*1i + 3*A*B^2*a^2*d -
A^2*B*a^2*d*5i)))*(A*1i + B))/(2*(-a)^(1/2)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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