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[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]])

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Rubi [A]  time = 0.347744, antiderivative size = 114, normalized size of antiderivative = 1., 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 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 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]

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 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 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 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 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]

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*}

Mathematica [A]  time = 2.31708, 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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Maple [B]  time = 0.488, size = 948, normalized size = 8.3 \begin{align*} \text{result too large to display} \end{align*}

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*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(I*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*a
rctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)-I*B*(-2
*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos
(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)+2*I*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*ln(-(-(-2*cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))+2*I*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arct
an(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)+A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*arctan
(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)-I*B*(-2*cos(
d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x
+c)+1))^(1/2))*2^(1/2)+B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-si
n(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)+2*I*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))+2*A*(-2*cos(d*x+c)/(cos(d*x+c)
+1))^(1/2)*cos(d*x+c)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-2*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)+A*(-2*cos(d*x+c)/(c
os(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(
1/2))*2^(1/2)+2*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-4*I*B*co
s(d*x+c)-4*A*cos(d*x+c))/(I*sin(d*x+c)+cos(d*x+c))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 2.7198, size = 1543, normalized size = 13.54 \begin{align*} \text{result too large to display} \end{align*}

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^(2*I*d*x + 2*I*c)*log((I*a*d*sqrt((2*A^2 - 4*I*A*B - 2*B^2)
/(a*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*sqrt(a/(e^(2*I*d*x + 2*I*c)
+ 1))*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - a*d*sqrt((2*A^2 - 4*I*A*B - 2*B^2)/(a*d^2))*e^(2*I*d*
x + 2*I*c)*log((-I*a*d*sqrt((2*A^2 - 4*I*A*B - 2*B^2)/(a*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A + B)*e^(2*I
*d*x + 2*I*c) + I*A + B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) -
2*a*d*sqrt(A^2/(a*d^2))*e^(2*I*d*x + 2*I*c)*log(-88/507*(2*sqrt(2)*(A*e^(2*I*d*x + 2*I*c) + A)*sqrt(a/(e^(2*I*
d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + (3*a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt(A^2/(a*d^2)))/(A*e^(2*I*d*x + 2*I*
c) - A)) + 2*a*d*sqrt(A^2/(a*d^2))*e^(2*I*d*x + 2*I*c)*log(-88/507*(2*sqrt(2)*(A*e^(2*I*d*x + 2*I*c) + A)*sqrt
(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - (3*a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt(A^2/(a*d^2)))/(A*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))*e^(-2*I*d*x - 2*I*c)/(a*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \tan{\left (c + d x \right )}\right ) \cot{\left (c + d x \right )}}{\sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

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(a*(I*tan(c + d*x) + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)