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]

-(((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]])

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Rubi [A]  time = 0.073144, antiderivative size = 82, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [A]  time = 0.973927, 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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Maple [A]  time = 0.056, size = 71, normalized size = 0.9 \begin{align*}{\frac{2\,i}{d} \left ( -{ \left ( -{\frac{A}{2}}-{\frac{i}{2}}B \right ){\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}-{\frac{\sqrt{2}}{2} \left ( -{\frac{i}{2}}B+{\frac{A}{2}} \right ){\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ){\frac{1}{\sqrt{a}}}} \right ) } \end{align*}

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

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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((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 = 1.92976, size = 918, normalized size = 11.2 \begin{align*} -\frac{{\left (a d \sqrt{-\frac{2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac{{\left (a d \sqrt{-\frac{2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\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 (2 i \, d x + 2 i \, c\right )} \log \left (-\frac{{\left (a d \sqrt{-\frac{2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \end{align*}

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^(2*I*d*x + 2*I*c)*log((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)/(I*A + 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(-(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)/(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))*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{A + B \tan{\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((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(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{B \tan \left (d x + c\right ) + A}{\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((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)