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

Optimal. Leaf size=156 \[ \frac{(-B+i A) \sqrt{\tan (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d}-\frac{2 \sqrt [4]{-1} B \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

[Out]

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

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Rubi [A]  time = 0.480029, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3595, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac{(-B+i A) \sqrt{\tan (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d}-\frac{2 \sqrt [4]{-1} B \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3595

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*b - a*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(2*a*f*
m), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d*n
) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A,
B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]

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

Mathematica [A]  time = 3.56046, size = 183, normalized size = 1.17 \[ \frac{\sqrt{\tan (c+d x)} \left (i (A+i B) \sqrt{-1+e^{2 i (c+d x)}}-i (A-i B) e^{i (c+d x)} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+2 \sqrt{2} B e^{i (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{d \sqrt{-1+e^{2 i (c+d x)}} \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*(A + I*B)*Sqrt[-1 + E^((2*I)*(c + d*x))] - I*(A - I*B)*E^(I*(c + d*x))*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E
^((2*I)*(c + d*x))]] + 2*Sqrt[2]*B*E^(I*(c + d*x))*ArcTanh[(Sqrt[2]*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 = 0.09, size = 900, normalized size = 5.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(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)/a*(I*B*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*t
an(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*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)*a-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*a-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*t
an(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a-8*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*ta
n(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)*a+4*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d
*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+2*B*2^(1/2)*(I*a)^(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*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)^2*a-4*I*A*(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+A*2^(1/2)*(I*a)^(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*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)
^(1/2)*(I*a)^(1/2)*tan(d*x+c)-4*B*(-I*a)^(1/2)*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))*a+4*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(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)/(-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(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.80482, size = 1989, normalized size = 12.75 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)*(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*I*A^2 + 4*A*B - 2*I*B^2)/(a*d^2))*e^(2*I*d*x + 2*I*c)*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*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)/(a*d^2))*e^(2*I*d*x + 2*I*c)*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*sqrt(-4*I*B^2/(a*d^2))*e^(2*I*d*x + 2*I*c)*log(52/605*(4*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) + (3*a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt(-4*I*B^2/(a*d^2)))/(B*e^(2*I*d*x + 2*I*c) + B)) + a*d*sqrt(-4*I*
B^2/(a*d^2))*e^(2*I*d*x + 2*I*c)*log(52/605*(4*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) - (3*a*d*e^(2*I*d*x + 2*I*
c) - a*d)*sqrt(-4*I*B^2/(a*d^2)))/(B*e^(2*I*d*x + 2*I*c) + 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))*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{\left (A + B \tan{\left (c + d x \right )}\right ) \sqrt{\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(tan(d*x+c)**(1/2)*(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(tan(c + d*x))/sqrt(a*(I*tan(c + d*x) + 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError