3.54 \(\int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=110 \[ \frac {A-i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {x (B+i A)}{8 a^3}+\frac {A+3 i B}{8 a d (a+i a \tan (c+d x))^2}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3} \]

[Out]

-1/8*(I*A+B)*x/a^3+1/6*(-A-I*B)/d/(a+I*a*tan(d*x+c))^3+1/8*(A+3*I*B)/a/d/(a+I*a*tan(d*x+c))^2+1/8*(A-I*B)/d/(a
^3+I*a^3*tan(d*x+c))

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Rubi [A]  time = 0.16, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3590, 3526, 3479, 8} \[ \frac {A-i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {x (B+i A)}{8 a^3}+\frac {A+3 i B}{8 a d (a+i a \tan (c+d x))^2}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

-((I*A + B)*x)/(8*a^3) - (A + I*B)/(6*d*(a + I*a*Tan[c + d*x])^3) + (A + (3*I)*B)/(8*a*d*(a + I*a*Tan[c + d*x]
)^2) + (A - I*B)/(8*d*(a^3 + I*a^3*Tan[c + d*x]))

Rule 8

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

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

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 3590

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

Rubi steps

\begin {align*} \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx &=-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}-\frac {i \int \frac {a (A+i B)+2 a B \tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{2 a^2}\\ &=-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {A+3 i B}{8 a d (a+i a \tan (c+d x))^2}-\frac {(i A+B) \int \frac {1}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {A+3 i B}{8 a d (a+i a \tan (c+d x))^2}+\frac {A-i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {(i A+B) \int 1 \, dx}{8 a^3}\\ &=-\frac {(i A+B) x}{8 a^3}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {A+3 i B}{8 a d (a+i a \tan (c+d x))^2}+\frac {A-i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.55, size = 148, normalized size = 1.35 \[ \frac {(\cos (3 (c+d x))-i \sin (3 (c+d x))) (3 (A+3 i B) \cos (c+d x)-2 (6 i A d x+A+B (6 d x+i)) \cos (3 (c+d x))+9 i A \sin (c+d x)+2 i A \sin (3 (c+d x))+12 A d x \sin (3 (c+d x))-3 B \sin (c+d x)-2 B \sin (3 (c+d x))-12 i B d x \sin (3 (c+d x)))}{96 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

((Cos[3*(c + d*x)] - I*Sin[3*(c + d*x)])*(3*(A + (3*I)*B)*Cos[c + d*x] - 2*(A + (6*I)*A*d*x + B*(I + 6*d*x))*C
os[3*(c + d*x)] + (9*I)*A*Sin[c + d*x] - 3*B*Sin[c + d*x] + (2*I)*A*Sin[3*(c + d*x)] - 2*B*Sin[3*(c + d*x)] +
12*A*d*x*Sin[3*(c + d*x)] - (12*I)*B*d*x*Sin[3*(c + d*x)]))/(96*a^3*d)

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fricas [A]  time = 0.67, size = 75, normalized size = 0.68 \[ \frac {{\left ({\left (-12 i \, A - 12 \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, {\left (A + i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 3 \, {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, A - 2 i \, B\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

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))^3,x, algorithm="fricas")

[Out]

1/96*((-12*I*A - 12*B)*d*x*e^(6*I*d*x + 6*I*c) + 6*(A + I*B)*e^(4*I*d*x + 4*I*c) - 3*(A - I*B)*e^(2*I*d*x + 2*
I*c) - 2*A - 2*I*B)*e^(-6*I*d*x - 6*I*c)/(a^3*d)

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giac [A]  time = 0.74, size = 130, normalized size = 1.18 \[ -\frac {\frac {6 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac {6 \, {\left (A - i \, B\right )} \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} - \frac {11 \, A \tan \left (d x + c\right )^{3} - 11 i \, B \tan \left (d x + c\right )^{3} - 45 i \, A \tan \left (d x + c\right )^{2} - 45 \, B \tan \left (d x + c\right )^{2} - 69 \, A \tan \left (d x + c\right ) + 21 i \, B \tan \left (d x + c\right ) + 19 i \, A + 3 \, B}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]

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))^3,x, algorithm="giac")

[Out]

-1/96*(6*(A - I*B)*log(tan(d*x + c) - I)/a^3 - 6*(A - I*B)*log(I*tan(d*x + c) - 1)/a^3 - (11*A*tan(d*x + c)^3
- 11*I*B*tan(d*x + c)^3 - 45*I*A*tan(d*x + c)^2 - 45*B*tan(d*x + c)^2 - 69*A*tan(d*x + c) + 21*I*B*tan(d*x + c
) + 19*I*A + 3*B)/(a^3*(tan(d*x + c) - I)^3))/d

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maple [B]  time = 0.22, size = 203, normalized size = 1.85 \[ \frac {A \ln \left (\tan \left (d x +c \right )+i\right )}{16 d \,a^{3}}-\frac {i B \ln \left (\tan \left (d x +c \right )+i\right )}{16 d \,a^{3}}-\frac {A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {3 i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {\ln \left (\tan \left (d x +c \right )-i\right ) A}{16 d \,a^{3}}+\frac {i \ln \left (\tan \left (d x +c \right )-i\right ) B}{16 d \,a^{3}} \]

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))^3,x)

[Out]

1/16/d/a^3*A*ln(tan(d*x+c)+I)-1/16*I/d/a^3*B*ln(tan(d*x+c)+I)-1/8/d/a^3/(tan(d*x+c)-I)^2*A-3/8*I/d/a^3/(tan(d*
x+c)-I)^2*B-1/6*I/d/a^3/(tan(d*x+c)-I)^3*A+1/6/d/a^3/(tan(d*x+c)-I)^3*B-1/8*I/d/a^3/(tan(d*x+c)-I)*A-1/8/d/a^3
/(tan(d*x+c)-I)*B-1/16/d/a^3*ln(tan(d*x+c)-I)*A+1/16*I/d/a^3*ln(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))^3,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 = 6.50, size = 147, normalized size = 1.34 \[ \frac {\frac {A}{12\,a^3}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {A}{8\,a^3}-\frac {B\,1{}\mathrm {i}}{8\,a^3}\right )+\frac {B\,1{}\mathrm {i}}{12\,a^3}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{8\,a^3}+\frac {A\,3{}\mathrm {i}}{8\,a^3}\right )}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{16\,a^3\,d} \]

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)^3,x)

[Out]

(A/(12*a^3) - tan(c + d*x)^2*(A/(8*a^3) - (B*1i)/(8*a^3)) + (B*1i)/(12*a^3) + tan(c + d*x)*((A*3i)/(8*a^3) - B
/(8*a^3)))/(d*(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1)) + (log(tan(c + d*x) - 1i)*(A*1i +
B)*1i)/(16*a^3*d) + (log(tan(c + d*x) + 1i)*(A - B*1i))/(16*a^3*d)

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sympy [A]  time = 0.68, size = 262, normalized size = 2.38 \[ \begin {cases} - \frac {\left (\left (512 A a^{6} d^{2} e^{6 i c} + 512 i B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (768 A a^{6} d^{2} e^{8 i c} - 768 i B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (- 1536 A a^{6} d^{2} e^{10 i c} - 1536 i B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: 24576 a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {- i A - B}{8 a^{3}} + \frac {\left (- i A e^{6 i c} - i A e^{4 i c} + i A e^{2 i c} + i A - B e^{6 i c} + B e^{4 i c} + B e^{2 i c} - B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (i A + B\right )}{8 a^{3}} \]

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))**3,x)

[Out]

Piecewise((-((512*A*a**6*d**2*exp(6*I*c) + 512*I*B*a**6*d**2*exp(6*I*c))*exp(-6*I*d*x) + (768*A*a**6*d**2*exp(
8*I*c) - 768*I*B*a**6*d**2*exp(8*I*c))*exp(-4*I*d*x) + (-1536*A*a**6*d**2*exp(10*I*c) - 1536*I*B*a**6*d**2*exp
(10*I*c))*exp(-2*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(24576*a**9*d**3*exp(12*I*c), 0)), (x*(-(-I*A - B)/
(8*a**3) + (-I*A*exp(6*I*c) - I*A*exp(4*I*c) + I*A*exp(2*I*c) + I*A - B*exp(6*I*c) + B*exp(4*I*c) + B*exp(2*I*
c) - B)*exp(-6*I*c)/(8*a**3)), True)) - x*(I*A + B)/(8*a**3)

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