3.374 \(\int \frac{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx\)
Optimal. Leaf size=45 \[ \frac{2 i \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d \sqrt{a+i b}} \]
[Out]
((2*I)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d)
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Rubi [A] time = 0.0520216, antiderivative size = 45, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used =
{3537, 63, 208} \[ \frac{2 i \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d \sqrt{a+i b}} \]
Antiderivative was successfully verified.
[In]
Int[(1 - I*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]],x]
[Out]
((2*I)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d)
Rule 3537
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 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{1-i \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx &=-\frac{i \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{d}\\ &=-\frac{2 \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i a}{b}-\frac{i x^2}{b}} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{b d}\\ &=\frac{2 i \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{\sqrt{a+i b} d}\\ \end{align*}
Mathematica [A] time = 0.0438025, size = 45, normalized size = 1. \[ \frac{2 i \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{d \sqrt{a+i b}} \]
Antiderivative was successfully verified.
[In]
Integrate[(1 - I*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]],x]
[Out]
((2*I)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d)
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Maple [B] time = 0.102, size = 1624, normalized size = 36.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((1-I*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x)
[Out]
I/d/(a^2+b^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/
2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a-1/2*I/d/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)/((a^2+b^2)^(1/2)*a+a^2+b^2)*ln((a+b
*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*b^2+1/2/d/(2*(a^2+b^2)^(1/2)+
2*a)^(1/2)/(a^2+b^2)^(1/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1
/2))*b-I/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a
^2+b^2)^(1/2)-2*a)^(1/2))+1/2*I/d/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a
^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))+1/d/(a^2+b^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan
(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*b+1/2*I/d/(2*(a^2+b^2)^(1/2)+2*a)
^(1/2)/(a^2+b^2)^(1/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))
*a+I/d/((a^2+b^2)^(1/2)*a+a^2+b^2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*
tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*b^2-I/d/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)/(a^2+b^2)^(1/2)/((a^2+
b^2)^(1/2)*a+a^2+b^2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*
a^3-I/d/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)/(a^2+b^2)^(1/2)/((a^2+b^2)^(1/2)*a+a^2+b^2)*ln((a+b*tan(d*x+c))^(1/2)*(2
*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*a*b^2-1/2/d/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)/((a^2+b^
2)^(1/2)*a+a^2+b^2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*a*
b-1/2/d/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)/(a^2+b^2)^(1/2)/((a^2+b^2)^(1/2)*a+a^2+b^2)*ln((a+b*tan(d*x+c))^(1/2)*(2
*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*b*a^2-1/2/d/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)/(a^2+b^2
)^(1/2)/((a^2+b^2)^(1/2)*a+a^2+b^2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^
2+b^2)^(1/2))*b^3-I/d/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)/((a^2+b^2)^(1/2)*a+a^2+b^2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(
a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*a^2-1/d/((a^2+b^2)^(1/2)*a+a^2+b^2)/(2*(a^2+b^2)^(1/
2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a
*b-1/d/(a^2+b^2)^(1/2)/((a^2+b^2)^(1/2)*a+a^2+b^2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*
a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^2*b-1/d/(a^2+b^2)^(1/2)/((a^2+b^2)^(1/2)*a
+a^2+b^2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^
2+b^2)^(1/2)-2*a)^(1/2))*b^3
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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((1-I*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 1.11101, size = 718, normalized size = 15.96 \begin{align*} -\frac{1}{4} \, \sqrt{\frac{4 i}{{\left (-i \, a + b\right )} d^{2}}} \log \left (\frac{{\left ({\left ({\left (i \, a - b\right )} d e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (i \, a - b\right )} d\right )} \sqrt{\frac{{\left (a - i \, b\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + a + i \, b}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{4 i}{{\left (-i \, a + b\right )} d^{2}}} + 2 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, a + 2 i \, b\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, a + b\right )} d}\right ) + \frac{1}{4} \, \sqrt{\frac{4 i}{{\left (-i \, a + b\right )} d^{2}}} \log \left (\frac{{\left ({\left ({\left (-i \, a + b\right )} d e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-i \, a + b\right )} d\right )} \sqrt{\frac{{\left (a - i \, b\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + a + i \, b}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{4 i}{{\left (-i \, a + b\right )} d^{2}}} + 2 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, a + 2 i \, b\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, a + b\right )} d}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1-I*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
-1/4*sqrt(4*I/((-I*a + b)*d^2))*log((((I*a - b)*d*e^(2*I*d*x + 2*I*c) + (I*a - b)*d)*sqrt(((a - I*b)*e^(2*I*d*
x + 2*I*c) + a + I*b)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(4*I/((-I*a + b)*d^2)) + 2*a*e^(2*I*d*x + 2*I*c) + 2*a +
2*I*b)*e^(-2*I*d*x - 2*I*c)/((-I*a + b)*d)) + 1/4*sqrt(4*I/((-I*a + b)*d^2))*log((((-I*a + b)*d*e^(2*I*d*x + 2
*I*c) + (-I*a + b)*d)*sqrt(((a - I*b)*e^(2*I*d*x + 2*I*c) + a + I*b)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(4*I/((-I*
a + b)*d^2)) + 2*a*e^(2*I*d*x + 2*I*c) + 2*a + 2*I*b)*e^(-2*I*d*x - 2*I*c)/((-I*a + b)*d))
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{i \tan{\left (c + d x \right )}}{\sqrt{a + b \tan{\left (c + d x \right )}}}\, dx - \int - \frac{1}{\sqrt{a + b \tan{\left (c + d x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1-I*tan(d*x+c))/(a+b*tan(d*x+c))**(1/2),x)
[Out]
-Integral(I*tan(c + d*x)/sqrt(a + b*tan(c + d*x)), x) - Integral(-1/sqrt(a + b*tan(c + d*x)), x)
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Giac [B] time = 1.38913, size = 203, normalized size = 4.51 \begin{align*} \frac{2 \, \sqrt{2} \arctan \left (\frac{16 i \, \sqrt{b \tan \left (d x + c\right ) + a} a + 16 i \, \sqrt{a^{2} + b^{2}} \sqrt{b \tan \left (d x + c\right ) + a}}{8 \, \sqrt{2} \sqrt{a + \sqrt{a^{2} + b^{2}}} a + 8 i \, \sqrt{2} \sqrt{a + \sqrt{a^{2} + b^{2}}} b + 8 \, \sqrt{2} \sqrt{a^{2} + b^{2}} \sqrt{a + \sqrt{a^{2} + b^{2}}}}\right )}{\sqrt{a + \sqrt{a^{2} + b^{2}}} d{\left (\frac{i \, b}{a + \sqrt{a^{2} + b^{2}}} + 1\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1-I*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")
[Out]
2*sqrt(2)*arctan((16*I*sqrt(b*tan(d*x + c) + a)*a + 16*I*sqrt(a^2 + b^2)*sqrt(b*tan(d*x + c) + a))/(8*sqrt(2)*
sqrt(a + sqrt(a^2 + b^2))*a + 8*I*sqrt(2)*sqrt(a + sqrt(a^2 + b^2))*b + 8*sqrt(2)*sqrt(a^2 + b^2)*sqrt(a + sqr
t(a^2 + b^2))))/(sqrt(a + sqrt(a^2 + b^2))*d*(I*b/(a + sqrt(a^2 + b^2)) + 1))