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

Optimal. Leaf size=112 \[ \frac {B+i A}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {x (A-i B)}{8 a^3}+\frac {-B+i A}{6 d (a+i a \tan (c+d x))^3}+\frac {B+i A}{8 a d (a+i a \tan (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3526, 3479, 8} \[ \frac {B+i A}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {x (A-i B)}{8 a^3}+\frac {-B+i A}{6 d (a+i a \tan (c+d x))^3}+\frac {B+i A}{8 a d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.94, size = 150, normalized size = 1.34 \[ \frac {\sec ^3(c+d x) ((-27 A+3 i B) \cos (c+d x)+2 (6 i A d x-A+6 B d x-i B) \cos (3 (c+d x))-9 i A \sin (c+d x)+2 i A \sin (3 (c+d x))-12 A d x \sin (3 (c+d x))-9 B \sin (c+d x)-2 B \sin (3 (c+d x))+12 i B d x \sin (3 (c+d x)))}{96 a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*((-27*A + (3*I)*B)*Cos[c + d*x] + 2*(-A - I*B + (6*I)*A*d*x + 6*B*d*x)*Cos[3*(c + d*x)] - (9*I
)*A*Sin[c + d*x] - 9*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*(-I + Tan[c + d*x])^3)

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fricas [A]  time = 0.50, size = 76, normalized size = 0.68 \[ \frac {{\left (12 \, {\left (A - i \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (18 i \, A + 6 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (9 i \, A - 3 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, A - 2 \, B\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/96*(12*(A - I*B)*d*x*e^(6*I*d*x + 6*I*c) + (18*I*A + 6*B)*e^(4*I*d*x + 4*I*c) + (9*I*A - 3*B)*e^(2*I*d*x + 2
*I*c) + 2*I*A - 2*B)*e^(-6*I*d*x - 6*I*c)/(a^3*d)

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giac [A]  time = 0.69, size = 131, normalized size = 1.17 \[ -\frac {\frac {6 \, {\left (i \, A + B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {6 \, {\left (-i \, A - B\right )} \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} + \frac {-11 i \, A \tan \left (d x + c\right )^{3} - 11 \, B \tan \left (d x + c\right )^{3} - 45 \, A \tan \left (d x + c\right )^{2} + 45 i \, B \tan \left (d x + c\right )^{2} + 69 i \, A \tan \left (d x + c\right ) + 69 \, B \tan \left (d x + c\right ) + 51 \, A - 19 i \, B}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/96*(6*(I*A + B)*log(tan(d*x + c) - I)/a^3 + 6*(-I*A - B)*log(I*tan(d*x + c) - 1)/a^3 + (-11*I*A*tan(d*x + c
)^3 - 11*B*tan(d*x + c)^3 - 45*A*tan(d*x + c)^2 + 45*I*B*tan(d*x + c)^2 + 69*I*A*tan(d*x + c) + 69*B*tan(d*x +
c) + 51*A - 19*I*B)/(a^3*(tan(d*x + c) - I)^3))/d

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maple [B]  time = 0.20, size = 203, normalized size = 1.81 \[ \frac {B \ln \left (\tan \left (d x +c \right )+i\right )}{16 d \,a^{3}}+\frac {i A \ln \left (\tan \left (d x +c \right )+i\right )}{16 d \,a^{3}}-\frac {i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i \ln \left (\tan \left (d x +c \right )-i\right ) A}{16 d \,a^{3}}-\frac {\ln \left (\tan \left (d x +c \right )-i\right ) B}{16 d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x)

[Out]

1/16/d/a^3*B*ln(tan(d*x+c)+I)+1/16*I/d/a^3*A*ln(tan(d*x+c)+I)-1/8*I/d/a^3/(tan(d*x+c)-I)^2*A-1/8/d/a^3/(tan(d*
x+c)-I)^2*B+1/8/d/a^3/(tan(d*x+c)-I)*A-1/8*I/d/a^3/(tan(d*x+c)-I)*B-1/6/d/a^3/(tan(d*x+c)-I)^3*A-1/6*I/d/a^3/(
tan(d*x+c)-I)^3*B-1/16*I/d/a^3*ln(tan(d*x+c)-I)*A-1/16/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((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 = 6.46, size = 111, normalized size = 0.99 \[ -\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {B}{8\,a^3}+\frac {A\,1{}\mathrm {i}}{8\,a^3}\right )-\frac {A\,5{}\mathrm {i}}{12\,a^3}-\frac {B}{12\,a^3}+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {3\,A}{8\,a^3}-\frac {B\,3{}\mathrm {i}}{8\,a^3}\right )}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}-\frac {x\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^3} \]

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)^3,x)

[Out]

- (tan(c + d*x)^2*((A*1i)/(8*a^3) + B/(8*a^3)) - (A*5i)/(12*a^3) - B/(12*a^3) + tan(c + d*x)*((3*A)/(8*a^3) -
(B*3i)/(8*a^3)))/(d*(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1)) - (x*(A*1i + B)*1i)/(8*a^3)

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sympy [A]  time = 0.53, size = 264, normalized size = 2.36 \[ \begin {cases} - \frac {\left (\left (- 512 i A a^{6} d^{2} e^{6 i c} + 512 B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (- 2304 i A a^{6} d^{2} e^{8 i c} + 768 B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (- 4608 i A a^{6} d^{2} e^{10 i c} - 1536 B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: 24576 a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {A - i B}{8 a^{3}} + \frac {\left (A e^{6 i c} + 3 A e^{4 i c} + 3 A e^{2 i c} + A - i B e^{6 i c} - i B e^{4 i c} + i B e^{2 i c} + i B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- A + i B\right )}{8 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**3,x)

[Out]

Piecewise((-((-512*I*A*a**6*d**2*exp(6*I*c) + 512*B*a**6*d**2*exp(6*I*c))*exp(-6*I*d*x) + (-2304*I*A*a**6*d**2
*exp(8*I*c) + 768*B*a**6*d**2*exp(8*I*c))*exp(-4*I*d*x) + (-4608*I*A*a**6*d**2*exp(10*I*c) - 1536*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*(-(A - I*B
)/(8*a**3) + (A*exp(6*I*c) + 3*A*exp(4*I*c) + 3*A*exp(2*I*c) + A - I*B*exp(6*I*c) - I*B*exp(4*I*c) + I*B*exp(2
*I*c) + I*B)*exp(-6*I*c)/(8*a**3)), True)) - x*(-A + I*B)/(8*a**3)

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