3.45 \(\int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=103 \[ \frac {-3 B+i A}{4 a^2 d (1+i \tan (c+d x))}-\frac {x (A+3 i B)}{4 a^2}+\frac {B \log (\cos (c+d x))}{a^2 d}+\frac {(-B+i A) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3595, 3589, 3475, 12, 3526, 8} \[ \frac {-3 B+i A}{4 a^2 d (1+i \tan (c+d x))}-\frac {x (A+3 i B)}{4 a^2}+\frac {B \log (\cos (c+d x))}{a^2 d}+\frac {(-B+i A) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^2*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

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]

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

________________________________________________________________________________________

Mathematica [A]  time = 1.07, size = 185, normalized size = 1.80 \[ \frac {\sec ^2(c+d x) \left (\cos (2 (c+d x)) \left (4 A d x+i A-8 B \log \left (\cos ^2(c+d x)\right )-4 i B d x-B\right )+4 i A d x \sin (2 (c+d x))+A \sin (2 (c+d x))-4 i A+i B \sin (2 (c+d x))+4 B d x \sin (2 (c+d x))-8 i B \sin (2 (c+d x)) \log \left (\cos ^2(c+d x)\right )+16 i B \tan ^{-1}(\tan (d x)) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+8 B\right )}{16 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^2*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(Sec[c + d*x]^2*((-4*I)*A + 8*B + Cos[2*(c + d*x)]*(I*A - B + 4*A*d*x - (4*I)*B*d*x - 8*B*Log[Cos[c + d*x]^2])
+ (16*I)*B*ArcTan[Tan[d*x]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + A*Sin[2*(c + d*x)] + I*B*Sin[2*(c + d*x
)] + (4*I)*A*d*x*Sin[2*(c + d*x)] + 4*B*d*x*Sin[2*(c + d*x)] - (8*I)*B*Log[Cos[c + d*x]^2]*Sin[2*(c + d*x)]))/
(16*a^2*d*(-I + Tan[c + d*x])^2)

________________________________________________________________________________________

fricas [A]  time = 0.69, size = 84, normalized size = 0.82 \[ -\frac {{\left (4 \, {\left (A + 7 i \, B\right )} d x e^{\left (4 i \, d x + 4 i \, c\right )} - 16 \, B e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - {\left (4 i \, A - 8 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - B\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4*(A + 7*I*B)*d*x*e^(4*I*d*x + 4*I*c) - 16*B*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - (4*I*A
- 8*B)*e^(2*I*d*x + 2*I*c) + I*A - B)*e^(-4*I*d*x - 4*I*c)/(a^2*d)

________________________________________________________________________________________

giac [A]  time = 0.76, size = 107, normalized size = 1.04 \[ -\frac {\frac {2 \, {\left (i \, A + B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {2 \, {\left (-i \, A + 7 \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac {3 i \, A \tan \left (d x + c\right )^{2} - 21 \, B \tan \left (d x + c\right )^{2} - 6 \, A \tan \left (d x + c\right ) + 22 i \, B \tan \left (d x + c\right ) + 5 i \, A + 5 \, B}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2*(I*A + B)*log(tan(d*x + c) + I)/a^2 + 2*(-I*A + 7*B)*log(tan(d*x + c) - I)/a^2 + (3*I*A*tan(d*x + c)^
2 - 21*B*tan(d*x + c)^2 - 6*A*tan(d*x + c) + 22*I*B*tan(d*x + c) + 5*I*A + 5*B)/(a^2*(tan(d*x + c) - I)^2))/d

________________________________________________________________________________________

maple [A]  time = 0.21, size = 162, normalized size = 1.57 \[ -\frac {B \ln \left (\tan \left (d x +c \right )+i\right )}{8 d \,a^{2}}-\frac {i A \ln \left (\tan \left (d x +c \right )+i\right )}{8 d \,a^{2}}+\frac {i A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {5 i B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}+\frac {3 A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}+\frac {i \ln \left (\tan \left (d x +c \right )-i\right ) A}{8 d \,a^{2}}-\frac {7 \ln \left (\tan \left (d x +c \right )-i\right ) B}{8 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/d/a^2*B*ln(tan(d*x+c)+I)-1/8*I/d/a^2*A*ln(tan(d*x+c)+I)+1/4*I/d/a^2/(tan(d*x+c)-I)^2*A-1/4/d/a^2/(tan(d*x
+c)-I)^2*B+5/4*I/d/a^2/(tan(d*x+c)-I)*B+3/4/d/a^2/(tan(d*x+c)-I)*A+1/8*I/d/a^2*ln(tan(d*x+c)-I)*A-7/8/d/a^2*ln
(tan(d*x+c)-I)*B

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 6.29, size = 114, normalized size = 1.11 \[ \frac {\frac {A}{2\,a^2}+\frac {B\,1{}\mathrm {i}}{a^2}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {5\,B}{4\,a^2}+\frac {A\,3{}\mathrm {i}}{4\,a^2}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{8\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-7\,B+A\,1{}\mathrm {i}\right )}{8\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A/(2*a^2) + (B*1i)/a^2 + tan(c + d*x)*((A*3i)/(4*a^2) - (5*B)/(4*a^2)))/(d*(2*tan(c + d*x) + tan(c + d*x)^2*1
i - 1i)) - (log(tan(c + d*x) + 1i)*(A*1i + B))/(8*a^2*d) + (log(tan(c + d*x) - 1i)*(A*1i - 7*B))/(8*a^2*d)

________________________________________________________________________________________

sympy [A]  time = 0.86, size = 224, normalized size = 2.17 \[ \frac {B \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} + \begin {cases} \frac {\left (\left (- 4 i A a^{2} d e^{2 i c} + 4 B a^{2} d e^{2 i c}\right ) e^{- 4 i d x} + \left (16 i A a^{2} d e^{4 i c} - 32 B a^{2} d e^{4 i c}\right ) e^{- 2 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: 64 a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (- \frac {- A - 7 i B}{4 a^{2}} + \frac {i \left (i A e^{4 i c} - 2 i A e^{2 i c} + i A - 7 B e^{4 i c} + 4 B e^{2 i c} - B\right ) e^{- 4 i c}}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (A + 7 i B\right )}{4 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*log(exp(2*I*d*x) + exp(-2*I*c))/(a**2*d) + Piecewise((((-4*I*A*a**2*d*exp(2*I*c) + 4*B*a**2*d*exp(2*I*c))*ex
p(-4*I*d*x) + (16*I*A*a**2*d*exp(4*I*c) - 32*B*a**2*d*exp(4*I*c))*exp(-2*I*d*x))*exp(-6*I*c)/(64*a**4*d**2), N
e(64*a**4*d**2*exp(6*I*c), 0)), (x*(-(-A - 7*I*B)/(4*a**2) + I*(I*A*exp(4*I*c) - 2*I*A*exp(2*I*c) + I*A - 7*B*
exp(4*I*c) + 4*B*exp(2*I*c) - B)*exp(-4*I*c)/(4*a**2)), True)) - x*(A + 7*I*B)/(4*a**2)

________________________________________________________________________________________