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

Optimal. Leaf size=88 \[ \frac {i \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 88, normalized size of antiderivative = 1.00, 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 = {3546, 3544, 205} \[ \frac {i \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Tan[c + d*x]]/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

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

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 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 3546

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(2*b*f*m), x] - Dist[(a*c - b*d)/(2*b^2), Int[(a + b*Tan[e
+ f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), 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] && EqQ[m + n, 0] && LeQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {i \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {i \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{2 a}\\ &=\frac {i \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a \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 ) \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 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.34, size = 132, normalized size = 1.50 \[ \frac {i \sqrt {\tan (c+d x)} \left (\sqrt {-1+e^{2 i (c+d x)}}-e^{i (c+d x)} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{\sqrt {2} d \sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Tan[c + d*x]]/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

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

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fricas [B]  time = 0.55, size = 300, normalized size = 3.41 \[ \frac {{\left (a d \sqrt {\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {1}{4} i \, a d \sqrt {\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - a d \sqrt {\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-\frac {1}{4} i \, a d \sqrt {\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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*d^2))*e^(I*d*x + I*c)*log(1/4*I*a*d*sqrt(2*I/(a*d^2))*e^(I*d*x + I*c) + 1/4*sqrt(2)*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^(2*I*d*x + 2*I*c)
+ 1)) - a*d*sqrt(2*I/(a*d^2))*e^(I*d*x + I*c)*log(-1/4*I*a*d*sqrt(2*I/(a*d^2))*e^(I*d*x + I*c) + 1/4*sqrt(2)*
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^(2*I*d*x + 2
*I*c) + 1)) + sqrt(2)*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))*(2*I*e^(2*I*d*x + 2*I*c) + 2*I))*e^(-I*d*x - I*c)/(a*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\tan \left (d x + c\right )}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(tan(d*x + c))/sqrt(I*a*tan(d*x + c) + a), x)

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maple [B]  time = 0.24, size = 350, normalized size = 3.98 \[ \frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (2 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \tan \left (d x +c \right ) a -\sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a -4 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}+\sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +4 \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{4 d a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(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)/a*(2*I*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-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-4*I*(a*tan(d*x+c)*(1+I*
tan(d*x+c)))^(1/2)*(-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))*a+4*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(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)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\tan \left (d x + c\right )}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(tan(d*x + c))/sqrt(I*a*tan(d*x + c) + a), x)

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mupad [B]  time = 6.71, size = 242, normalized size = 2.75 \[ \frac {\ln \left (2+\frac {\sqrt {a}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4-4{}\mathrm {i}\right )}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}}+\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )}{\sqrt {a}\,d}-\frac {2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )\,\left (d\,1{}\mathrm {i}-\frac {a\,d\,\mathrm {tan}\left (c+d\,x\right )}{{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}\right )}-\frac {\sqrt {\frac {1}{8}{}\mathrm {i}}\,\ln \left (-\frac {a\,\mathrm {tan}\left (c+d\,x\right )}{{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}+\frac {2\,{\left (-1\right )}^{3/4}\,\sqrt {2}\,\sqrt {a}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}}+1{}\mathrm {i}\right )}{\sqrt {a}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(1/2)/(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

(log((a*tan(c + d*x)*2i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (a^(1/2)*tan(c + d*x)^(1/2)*(4 + 4i))/(
(a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2)) + 2)*(1/4 + 1i/4))/(a^(1/2)*d) - (2*tan(c + d*x)^(1/2))/(((a + a*tan(
c + d*x)*1i)^(1/2) - a^(1/2))*(d*1i - (a*d*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2)) - ((1i/
8)^(1/2)*log((2*(-1)^(3/4)*2^(1/2)*a^(1/2)*tan(c + d*x)^(1/2))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2)) - (a*
tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + 1i))/(a^(1/2)*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\tan {\left (c + d x \right )}}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(tan(c + d*x))/sqrt(I*a*(tan(c + d*x) - I)), x)

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