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]
-((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
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Rubi [A] time = 0.384709, antiderivative size = 123, normalized size of antiderivative = 1.,
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 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 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 \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*}
Mathematica [B] time = 4.61775, 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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Maple [B] time = 0.553, size = 1179, normalized size = 9.6 \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)^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)^2-2*I*B*(-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))*cos(d*x+c)*sin(d*x+c)-2*I*B*(-2*cos(d*x+c)/(cos(d*x+c)+1
))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)*sin(d*x+c)*2^(1/2)+2*A*(-2*cos(d*
x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)*sin(d*x+c)*2^(1
/2)+2*I*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*
x+c)/cos(d*x+c))*sin(d*x+c)*2^(1/2)+2*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*cos(d*x+c)*sin(d*x+c)*2^(1/2)+2*I*A*(-2*cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*cos(d*x+c)*sin(d*x+c
)*2^(1/2)-2*I*A*cos(d*x+c)+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
))*cos(d*x+c)*sin(d*x+c)+2*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(1/2))*sin(d*x+c)*2^(1/2)-2*I*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(
cos(d*x+c)+1))^(1/2))*sin(d*x+c)*2^(1/2)+2*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(
d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*2^(1/2)+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))*sin(d*x+c)+2*B*(-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))*cos(d*
x+c)*sin(d*x+c)+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))*sin(d*x+
c)-2*I*B*(-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))*sin(d*x+c)+2*B*(-
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)+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))*cos(d*x+c)*sin(d*x+c)+2*A*cos(d*x+c)*sin(d*x+c))/(I*sin(d*x+c)+cos(d*x+c)-1)/(cos(d
*x+c)+1)
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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)^2*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 1.79056, size = 1644, normalized size = 13.37 \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)^2*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")
[Out]
1/2*(sqrt(2)*(-2*I*A*e^(2*I*d*x + 2*I*c) - 2*I*A)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - (d*e^(2*
I*d*x + 2*I*c) - d)*sqrt(-(A^2 - 4*I*A*B - 4*B^2)*a/d^2)*log((sqrt(2)*((I*A + 2*B)*e^(2*I*d*x + 2*I*c) + I*A +
2*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + 2*d*sqrt(-(A^2 - 4*I*A*B - 4*B^2)*a/d^2)*e^(2*I*d*x
+ 2*I*c))*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((sqrt(2)*((I*A + 2*B)*e^(2*I*d*x + 2*I*c) + I*A + 2*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c)
- 2*d*sqrt(-(A^2 - 4*I*A*B - 4*B^2)*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + 2*B)) + (d*e^(2*I*
d*x + 2*I*c) - d)*sqrt(-(2*A^2 - 4*I*A*B - 2*B^2)*a/d^2)*log((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) + d*sqrt(-(2*A^2 - 4*I*A*B - 2*B^2)*a/d^2)*e^(2*I*d*x + 2*
I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - (d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-(2*A^2 - 4*I*A*B - 2*B^2)*a/d^2)*log
((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) - d*sqrt
(-(2*A^2 - 4*I*A*B - 2*B^2)*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)))/(d*e^(2*I*d*x + 2*I*c
) - d)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
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)