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

Optimal. Leaf size=99 \[ \frac{(A+i B) \sqrt{\tan (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

[Out]

((1/2 - I/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) +
((A + I*B)*Sqrt[Tan[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.193384, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3596, 12, 3544, 205} \[ \frac{(A+i B) \sqrt{\tan (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/2 - I/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) +
((A + I*B)*Sqrt[Tan[c + d*x]])/(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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3544

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 3.06364, size = 123, normalized size = 1.24 \[ \frac{\sqrt{\tan (c+d x)} \left ((A+i B) \sqrt{-1+e^{2 i (c+d x)}}+(A-i B) e^{i (c+d x)} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{d \sqrt{-1+e^{2 i (c+d x)}} \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

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Maple [B]  time = 0.094, size = 639, normalized size = 6.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*(I*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+
I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a-2*I*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(
1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)*a+B*2^(1/2)*ln(-(-2*
2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a-
I*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)
+I))*a+4*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)+2*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^
(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)*a+4*I*B*(a*tan(d*x+
c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)-B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))
^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a-4*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)
+4*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/a/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+
I)^2/(-I*a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.04757, size = 1183, normalized size = 11.95 \begin{align*} \frac{{\left (a d \sqrt{\frac{-2 i \, A^{2} - 4 \, A B + 2 i \, B^{2}}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac{{\left (i \, a d \sqrt{\frac{-2 i \, A^{2} - 4 \, A B + 2 i \, 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}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{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 )}}{4 i \, A + 4 \, B}\right ) - a d \sqrt{\frac{-2 i \, A^{2} - 4 \, A B + 2 i \, B^{2}}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac{{\left (-i \, a d \sqrt{\frac{-2 i \, A^{2} - 4 \, A B + 2 i \, 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}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{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 )}}{4 i \, A + 4 \, B}\right ) + 2 \, \sqrt{2}{\left ({\left (A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{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))/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/4*(a*d*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)/(a*d^2))*e^(2*I*d*x + 2*I*c)*log((I*a*d*sqrt((-2*I*A^2 - 4*A*B + 2*
I*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))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(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*I*A^2 - 4*A*B + 2*I*B^2)/(a*d^2))*e^(2*I*d*x + 2*I*c)*log((-I*a*d*sqrt((-2*I*A^2 -
4*A*B + 2*I*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))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(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*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))*
sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(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 )} \sqrt{\tan{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [B]  time = 1.35289, size = 207, normalized size = 2.09 \begin{align*} \frac{-\left (i + 1\right ) \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} +{\left (\left (2 i - 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a - \left (2 i - 2\right ) \, a^{2}\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} \sqrt{i \, a \tan \left (d x + c\right ) + a} B}{2 \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a - 3 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} + 2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-(I + 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)*a^2 + ((2*I - 2)*(I*a*tan(d*x +
c) + a)*a - (2*I - 2)*a^2)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*tan(d*x + c) + a)*B)/(((I*a*tan
(d*x + c) + a)^3*a - 3*(I*a*tan(d*x + c) + a)^2*a^2 + 2*(I*a*tan(d*x + c) + a)*a^3)*d)