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

Optimal. Leaf size=236 \[ -\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {(-B+i A) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d} \]

[Out]

(1/8-1/8*I)*(A+(2-I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a/d*2^(1/2)+(1/8-1/8*I)*(A+(2-I)*B)*arctan(1+2^(1/
2)*tan(d*x+c)^(1/2))/a/d*2^(1/2)+(1/16+1/16*I)*(A-(2+I)*B)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a/d*2^(1/
2)-(1/16+1/16*I)*(A-(2+I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a/d*2^(1/2)+1/2*(I*A-B)*tan(d*x+c)^(1/2
)/d/(a+I*a*tan(d*x+c))

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Rubi [A]  time = 0.29, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3595, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {(-B+i A) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/4 + I/4)*(A + (2 - I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a*d) + ((1/4 - I/4)*(A + (2 - I
)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a*d) + ((1/8 + I/8)*(A - (2 + I)*B)*Log[1 - Sqrt[2]*Sqrt
[Tan[c + d*x]] + Tan[c + d*x]])/(Sqrt[2]*a*d) - ((1/8 + I/8)*(A - (2 + I)*B)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]
] + Tan[c + d*x]])/(Sqrt[2]*a*d) + ((I*A - B)*Sqrt[Tan[c + d*x]])/(2*d*(a + I*a*Tan[c + d*x]))

Rule 204

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx &=\frac {(i A-B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}-\frac {\int \frac {\frac {1}{2} a (i A-B)-\frac {1}{2} a (A-3 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{2 a^2}\\ &=\frac {(i A-B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} a (i A-B)-\frac {1}{2} a (A-3 i B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d}\\ &=\frac {(i A-B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}-\frac {\left (\left (\frac {1}{4}+\frac {i}{4}\right ) (A-(2+i) B)\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a d}--\frac {\left (\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B)\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a d}\\ &=\frac {(i A-B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}--\frac {\left (\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}--\frac {\left (\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}--\frac {\left (\left (\frac {1}{8}-\frac {i}{8}\right ) (A+(2-i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a d}--\frac {\left (\left (\frac {1}{8}-\frac {i}{8}\right ) (A+(2-i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a d}\\ &=\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}--\frac {\left (\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}\\ &=-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.83, size = 198, normalized size = 0.84 \[ \frac {(\cos (d x)+i \sin (d x)) (A+B \tan (c+d x)) \left (4 (A+i B) \sin (c+d x) (\sin (d x)+i \cos (d x))+(1+i) (-\sin (c)+i \cos (c)) \sqrt {\sin (2 (c+d x))} \sec (c+d x) \left ((A+(2-i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+i (A-(2+i) B) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{8 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x)) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[d*x] + I*Sin[d*x])*(4*(A + I*B)*(I*Cos[d*x] + Sin[d*x])*Sin[c + d*x] + (1 + I)*((A + (2 - I)*B)*ArcSin[C
os[c + d*x] - Sin[c + d*x]] + I*(A - (2 + I)*B)*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]])*Sec
[c + d*x]*(I*Cos[c] - Sin[c])*Sqrt[Sin[2*(c + d*x)]])*(A + B*Tan[c + d*x]))/(8*d*(A*Cos[c + d*x] + B*Sin[c + d
*x])*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x]))

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fricas [B]  time = 0.72, size = 570, normalized size = 2.42 \[ -\frac {{\left (a d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \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 {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} + {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - a d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \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 {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} - {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - 2 \, a d \sqrt {\frac {i \, B^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \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 {i \, B^{2}}{a^{2} d^{2}}} + i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 2 \, a d \sqrt {\frac {i \, B^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \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 {i \, B^{2}}{a^{2} d^{2}}} - i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) - 2 \, {\left ({\left (i \, A - B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - B\right )} \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 )}}{8 \, a d} \]

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)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.46, size = 192, normalized size = 0.81 \[ -\frac {i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) B}{d a \left (\sqrt {2}+i \sqrt {2}\right )}+\frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) A}{d a \left (\sqrt {2}+i \sqrt {2}\right )}+\frac {i \left (\sqrt {\tan }\left (d x +c \right )\right ) B}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) A}{2 d a \left (\tan \left (d x +c \right )-i\right )}-\frac {2 i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) B}{d a \left (\sqrt {2}-i \sqrt {2}\right )} \]

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)),x)

[Out]

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

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

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)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 8.51, size = 184, normalized size = 0.78 \[ -\mathrm {atan}\left (\frac {2\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}-\frac {2\,\sqrt {\frac {1}{16}{}\mathrm {i}}\,A\,\mathrm {atanh}\left (4\,\sqrt {\frac {1}{16}{}\mathrm {i}}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{a\,d}+\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}-\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*tan(c + d*x)^(1/2)*1i)/(2*a*d*(tan(c + d*x)*1i + 1)) - atan((4*a*d*tan(c + d*x)^(1/2)*(-(B^2*1i)/(16*a^2*d^
2))^(1/2))/B)*(-(B^2*1i)/(16*a^2*d^2))^(1/2)*2i - (2*(1i/16)^(1/2)*A*atanh(4*(1i/16)^(1/2)*tan(c + d*x)^(1/2))
)/(a*d) - atan((2*a*d*tan(c + d*x)^(1/2)*((B^2*1i)/(4*a^2*d^2))^(1/2))/B)*((B^2*1i)/(4*a^2*d^2))^(1/2)*2i - (B
*tan(c + d*x)^(1/2))/(2*a*d*(tan(c + d*x)*1i + 1))

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

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)),x)

[Out]

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

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