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

Optimal. Leaf size=55 \[ \frac {2 i a B \sqrt {\tan (c+d x)}}{d}-\frac {2 \sqrt [4]{-1} a (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {3592, 3533, 205} \[ \frac {2 i a B \sqrt {\tan (c+d x)}}{d}-\frac {2 \sqrt [4]{-1} a (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3533

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

Rule 3592

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

Rubi steps

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

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Mathematica [A]  time = 1.88, size = 92, normalized size = 1.67 \[ \frac {2 a \sqrt {\tan (c+d x)} \left ((A-i B) \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )+i B \sqrt {i \tan (c+d x)}\right )}{d \sqrt {i \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.16, size = 310, normalized size = 5.64 \[ \frac {8 i \, B a \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} d \log \left (\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} d \log \left (\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Warning, integration of ab
s or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep^2-1)]Discon
tinuities at zeroes of t_nostep^2-1 were not checkedEvaluation time: 0.58int()  Error: Bad Argument Value

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maple [B]  time = 0.10, size = 444, normalized size = 8.07 \[ \frac {2 i a B \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}-\frac {i a B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {i a B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {i a B \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}+\frac {a A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {a A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {a A \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}+\frac {i a A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {i a A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {i a A \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}+\frac {a B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {a B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {a B \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.81, size = 151, normalized size = 2.75 \[ -\frac {-8 i \, B a \sqrt {\tan \left (d x + c\right )} + {\left (2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-8*I*B*a*sqrt(tan(d*x + c)) + (2*sqrt(2)*(-(I + 1)*A + (I - 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(t
an(d*x + c)))) + 2*sqrt(2)*(-(I + 1)*A + (I - 1)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + sq
rt(2)*((I - 1)*A + (I + 1)*B)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*((I - 1)*A + (I + 1
)*B)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))*a)/d

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mupad [B]  time = 7.08, size = 68, normalized size = 1.24 \[ -\frac {2\,{\left (-1\right )}^{1/4}\,B\,a\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {B\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{d}+\frac {\sqrt {2}\,A\,a\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (1+1{}\mathrm {i}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*a*tan(c + d*x)^(1/2)*2i)/d + (2^(1/2)*A*a*atan(2^(1/2)*tan(c + d*x)^(1/2)*(1/2 - 1i/2))*(1 + 1i))/d - (2*(-
1)^(1/4)*B*a*atanh((-1)^(1/4)*tan(c + d*x)^(1/2)))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int A \sqrt {\tan {\left (c + d x \right )}}\, dx + \int B \tan ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int \left (- \frac {i A}{\sqrt {\tan {\left (c + d x \right )}}}\right )\, dx + \int \left (- i B \sqrt {\tan {\left (c + d x \right )}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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