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

Optimal. Leaf size=67 \[ -\frac {A+i B}{2 a d (1+i \tan (c+d x))}-\frac {x (-B+i A)}{2 a}+\frac {i B \log (\cos (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3589, 3475, 12, 3526, 8} \[ -\frac {A+i B}{2 a d (1+i \tan (c+d x))}-\frac {x (-B+i A)}{2 a}+\frac {i B \log (\cos (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3526

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

Rule 3589

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

Rubi steps

\begin {align*} \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx &=-\frac {i \int \frac {a (i A-B) \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{a}-\frac {(i B) \int \tan (c+d x) \, dx}{a}\\ &=\frac {i B \log (\cos (c+d x))}{a d}-(-A-i B) \int \frac {\tan (c+d x)}{a+i a \tan (c+d x)} \, dx\\ &=\frac {i B \log (\cos (c+d x))}{a d}-\frac {A+i B}{2 d (a+i a \tan (c+d x))}-\frac {(i A-B) \int 1 \, dx}{2 a}\\ &=-\frac {(i A-B) x}{2 a}+\frac {i B \log (\cos (c+d x))}{a d}-\frac {A+i B}{2 d (a+i a \tan (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 1.08, size = 148, normalized size = 2.21 \[ \frac {\cos (c+d x) (A+B \tan (c+d x)) \left (\tan (c+d x) \left (-2 i A d x+A+2 i B \log \left (\cos ^2(c+d x)\right )-2 B d x+i B\right )-2 A d x+i A+4 B \tan ^{-1}(\tan (d x)) (\tan (c+d x)-i)+2 B \log \left (\cos ^2(c+d x)\right )+2 i B d x-B\right )}{4 a d (\tan (c+d x)-i) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]*(A + B*Tan[c + d*x])*(I*A - B - 2*A*d*x + (2*I)*B*d*x + 2*B*Log[Cos[c + d*x]^2] + (A + I*B - (2*
I)*A*d*x - 2*B*d*x + (2*I)*B*Log[Cos[c + d*x]^2])*Tan[c + d*x] + 4*B*ArcTan[Tan[d*x]]*(-I + Tan[c + d*x])))/(4
*a*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(-I + Tan[c + d*x]))

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fricas [A]  time = 0.78, size = 67, normalized size = 1.00 \[ \frac {{\left ({\left (-2 i \, A + 6 \, B\right )} d x e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i \, B e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - A - i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((-2*I*A + 6*B)*d*x*e^(2*I*d*x + 2*I*c) + 4*I*B*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - A - I*B
)*e^(-2*I*d*x - 2*I*c)/(a*d)

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giac [A]  time = 0.34, size = 82, normalized size = 1.22 \[ -\frac {\frac {{\left (A + 3 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a} - \frac {{\left (A - i \, B\right )} \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a} - \frac {A \tan \left (d x + c\right ) + 3 i \, B \tan \left (d x + c\right ) + i \, A + B}{a {\left (\tan \left (d x + c\right ) - i\right )}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.20, size = 121, normalized size = 1.81 \[ \frac {A \ln \left (\tan \left (d x +c \right )+i\right )}{4 d a}-\frac {i B \ln \left (\tan \left (d x +c \right )+i\right )}{4 d a}-\frac {\ln \left (\tan \left (d x +c \right )-i\right ) A}{4 d a}-\frac {3 i \ln \left (\tan \left (d x +c \right )-i\right ) B}{4 d a}+\frac {i A}{2 d a \left (\tan \left (d x +c \right )-i\right )}-\frac {B}{2 d a \left (\tan \left (d x +c \right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*(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 = 6.24, size = 81, normalized size = 1.21 \[ -\frac {\frac {A}{2\,a}+\frac {B\,1{}\mathrm {i}}{2\,a}}{d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{4\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,3{}\mathrm {i}\right )}{4\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.44, size = 121, normalized size = 1.81 \[ \frac {i B \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a d} + \begin {cases} - \frac {\left (A + i B\right ) e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: 4 a d e^{2 i c} \neq 0 \\x \left (- \frac {- i A + 3 B}{2 a} + \frac {\left (- i A e^{2 i c} + i A + 3 B e^{2 i c} - B\right ) e^{- 2 i c}}{2 a}\right ) & \text {otherwise} \end {cases} - \frac {x \left (i A - 3 B\right )}{2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*B*log(exp(2*I*d*x) + exp(-2*I*c))/(a*d) + Piecewise((-(A + I*B)*exp(-2*I*c)*exp(-2*I*d*x)/(4*a*d), Ne(4*a*d*
exp(2*I*c), 0)), (x*(-(-I*A + 3*B)/(2*a) + (-I*A*exp(2*I*c) + I*A + 3*B*exp(2*I*c) - B)*exp(-2*I*c)/(2*a)), Tr
ue)) - x*(I*A - 3*B)/(2*a)

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