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

Optimal. Leaf size=82 \[ \frac {-B+i A}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3526, 3480, 206} \[ \frac {-B+i A}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d)) + (I*A - B)/(d*Sqrt[a
+ I*a*Tan[c + d*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 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 3526

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^m)/(2*a*f*m), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 1.33, size = 129, normalized size = 1.57 \[ \frac {i e^{-2 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left ((A+i B) \left (1+e^{2 i (c+d x)}\right )-(A-i B) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{\sqrt {2} a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.54, size = 339, normalized size = 4.13 \[ -\frac {{\left (a d \sqrt {-\frac {2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {{\left ({\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )} + 2 \, \sqrt {2} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - a d \sqrt {-\frac {2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {{\left ({\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )} - 2 \, \sqrt {2} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \sqrt {2} {\left ({\left (2 i \, A - 2 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, A - 2 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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) + 2*sq
rt(2)*(a*d*e^(2*I*d*x + 2*I*c) + 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) - 2*sqrt(2)*(a*d*e^(2*I*d*x + 2*I*c) + 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)) - sqrt(2)*((2*I*A - 2*B)*e^(2*I*d*x + 2*I*c) + 2*I
*A - 2*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 \[ \int \frac {B \tan \left (d x + c\right ) + A}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.22, size = 71, normalized size = 0.87 \[ \frac {2 i \left (-\frac {\left (-\frac {i B}{2}+\frac {A}{2}\right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}-\frac {-\frac {A}{2}-\frac {i B}{2}}{\sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.68, size = 91, normalized size = 1.11 \[ \frac {i \, {\left (\sqrt {2} {\left (A - i \, B\right )} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (A + i \, B\right )} a}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}}{4 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I*(sqrt(2)*(A - I*B)*sqrt(a)*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))) + 4*(A + I*B)*a/sqrt(I*a*tan(d*x + c) + a))/(a*d)

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mupad [B]  time = 0.76, size = 117, normalized size = 1.43 \[ \frac {A\,1{}\mathrm {i}}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}-\frac {B}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {-a}\,d}-\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{2\,\sqrt {a}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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