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

Optimal. Leaf size=112 \[ -\frac{(1+i) \sqrt{a} (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 (-1)^{3/4} \sqrt{a} B \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.323059, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.184, Rules used = {3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{(1+i) \sqrt{a} (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 (-1)^{3/4} \sqrt{a} B \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3601

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

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]

Rule 3599

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*B)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt{\tan (c+d x)}} \, dx &=-\left ((-A+i B) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\right )+\frac{(i B) \int \frac{(a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{a}\\ &=\frac{(i a B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{d}-\frac{\left (2 a^2 (i A+B)\right ) \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{(1+i) \sqrt{a} (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 )}{d}+\frac{(2 i a B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{(1+i) \sqrt{a} (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 )}{d}+\frac{(2 i a B) \operatorname{Subst}\left (\int \frac{1}{1-i a x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac{2 (-1)^{3/4} \sqrt{a} B \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{(1+i) \sqrt{a} (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 )}{d}\\ \end{align*}

Mathematica [B]  time = 3.9432, size = 238, normalized size = 2.12 \[ \frac{\sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos (c+d x) \sqrt{a+i a \tan (c+d x)} \left (4 (A-i B) \log \left (\sqrt{-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )+i \sqrt{2} B \left (\log \left (-2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )-\log \left (2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )\right )\right )}{2 d \sqrt{-1+e^{2 i (c+d x)}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.061, size = 498, normalized size = 4.5 \begin{align*}{\frac{a}{2\,d \left ( -\tan \left ( dx+c \right ) +i \right ) }\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{\tan \left ( dx+c \right ) } \left ( iA\sqrt{ia}\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) \tan \left ( dx+c \right ) -2\,iB\ln \left ({\frac{1}{2} \left ( 2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a \right ){\frac{1}{\sqrt{ia}}}} \right ) \sqrt{-ia}\tan \left ( dx+c \right ) -iB\sqrt{ia}\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) +B\sqrt{ia}\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) \tan \left ( dx+c \right ) +A\sqrt{ia}\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) -2\,B\ln \left ( 1/2\,{\frac{2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a}{\sqrt{ia}}} \right ) \sqrt{-ia} \right ){\frac{1}{\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{-ia}}}{\frac{1}{\sqrt{ia}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [B]  time = 1.84494, size = 1516, normalized size = 13.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt((-2*I*A^2 - 4*A*B + 2*I*B^2)*a/d^2)*log((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) + I*d*sqr
t((-2*I*A^2 - 4*A*B + 2*I*B^2)*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - 1/2*sqrt((-2*I*A^
2 - 4*A*B + 2*I*B^2)*a/d^2)*log((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) - I*d*sqrt((-2*I*A^2 - 4*A
*B + 2*I*B^2)*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - 1/2*sqrt(4*I*B^2*a/d^2)*log((sqrt(
2)*(B*e^(2*I*d*x + 2*I*c) + 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) + I*sqrt(4*I*B^2*a/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/B) + 1/2*s
qrt(4*I*B^2*a/d^2)*log((sqrt(2)*(B*e^(2*I*d*x + 2*I*c) + 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) - I*sqrt(4*I*B^2*a/d^2)*d*e^(2*I*d*x + 2*I*c))*e^
(-2*I*d*x - 2*I*c)/B)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )} \left (A + B \tan{\left (c + d x \right )}\right )}{\sqrt{\tan{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.48671, size = 189, normalized size = 1.69 \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 )}^{2} a - 3 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} + 2 \, a^{3}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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