3.182 \(\int \frac{A+B \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=143 \[ \frac{A+i B}{d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{(3 A+i B) \sqrt{a+i a \tan (c+d x)}}{a d \sqrt{\tan (c+d x)}}+\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

[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)/(d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - ((3*A + I*B)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*S
qrt[Tan[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.365383, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {3596, 3598, 12, 3544, 205} \[ \frac{A+i B}{d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{(3 A+i B) \sqrt{a+i a \tan (c+d x)}}{a d \sqrt{\tan (c+d x)}}+\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*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)/(d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - ((3*A + I*B)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*S
qrt[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 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 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)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{A+i B}{d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{1}{2} a (3 A+i B)-a (i A-B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{a^2}\\ &=\frac{A+i B}{d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{(3 A+i B) \sqrt{a+i a \tan (c+d x)}}{a d \sqrt{\tan (c+d x)}}+\frac{2 \int \frac{a^2 (i A+B) \sqrt{a+i a \tan (c+d x)}}{4 \sqrt{\tan (c+d x)}} \, dx}{a^3}\\ &=\frac{A+i B}{d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{(3 A+i B) \sqrt{a+i a \tan (c+d x)}}{a d \sqrt{\tan (c+d x)}}+\frac{(i A+B) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{2 a}\\ &=\frac{A+i B}{d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{(3 A+i B) \sqrt{a+i a \tan (c+d x)}}{a d \sqrt{\tan (c+d x)}}+\frac{(a (A-i 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 ) (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}+\frac{A+i B}{d \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{(3 A+i B) \sqrt{a+i a \tan (c+d x)}}{a d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.19658, size = 181, normalized size = 1.27 \[ \frac{(A+B \tan (c+d x)) \left (\frac{(A-i B) \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )}{\sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}+\frac{-4 A \cos (c+d x)+2 (B-3 i A) \sin (c+d x)}{\sqrt{\tan (c+d x)}}\right )}{2 d \sqrt{a+i a \tan (c+d x)} (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((((A - I*B)*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])/Sqrt[((-I
)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))] + (-4*A*Cos[c + d*x] + 2*((-3*I)*A + B)*Sin[c + d*x])
/Sqrt[Tan[c + d*x]])*(A + B*Tan[c + d*x]))/(2*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

________________________________________________________________________________________

Maple [B]  time = 0.1, size = 701, normalized size = 4.9 \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))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(3/2),x)

[Out]

-1/4/d*(a*(1+I*tan(d*x+c)))^(1/2)*(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)^3*a+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-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)^3*a-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+4*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2+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))*tan(d*x+c)^2*a-20*I*A*(a*tan(d*x
+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)+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+12*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)
*(-I*a)^(1/2)*tan(d*x+c)^2+4*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)-8*A*(a*tan(d*x+c)
*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/a/tan(d*x+c)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I
)^2/(-I*a)^(1/2)

________________________________________________________________________________________

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))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [B]  time = 2.06693, size = 1330, normalized size = 9.3 \begin{align*} \frac{\sqrt{2}{\left ({\left (-10 i \, A + 2 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, A 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}} \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 )} +{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{\frac{2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}}{a d^{2}}} \log \left (\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 )} + \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 ) -{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{\frac{2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}}{a d^{2}}} \log \left (-\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 )} - \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 )}{4 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \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)/tan(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

1/4*(sqrt(2)*((-10*I*A + 2*B)*e^(4*I*d*x + 4*I*c) - 8*I*A*e^(2*I*d*x + 2*I*c) + 2*I*A - 2*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) + (a*d*e^(4*I*d*
x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)/(a*d^2))*log((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*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)/(a*d^2
))*log(-(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*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))

________________________________________________________________________________________

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 )} \tan ^{\frac{3}{2}}{\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))/(a+I*a*tan(d*x+c))**(1/2)/tan(d*x+c)**(3/2),x)

[Out]

Integral((A + B*tan(c + d*x))/(sqrt(a*(I*tan(c + d*x) + 1))*tan(c + d*x)**(3/2)), x)

________________________________________________________________________________________

Giac [A]  time = 1.45652, size = 234, normalized size = 1.64 \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^{3} +{\left (\left (2 i - 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - \left (2 i - 2\right ) \, a^{3}\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}{{\left (-2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a + 8 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} - 10 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} + 4 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4}\right )} 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)/tan(d*x+c)^(3/2),x, algorithm="giac")

[Out]

(-(I + 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)*a^3 + ((2*I - 2)*(I*a*tan(d*x + c)
+ a)*a^2 - (2*I - 2)*a^3)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*tan(d*x + c) + a)*B)/((-2*I*(I*a*
tan(d*x + c) + a)^4*a + 8*I*(I*a*tan(d*x + c) + a)^3*a^2 - 10*I*(I*a*tan(d*x + c) + a)^2*a^3 + 4*I*(I*a*tan(d*
x + c) + a)*a^4)*d)