3.62 \(\int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=143 \[ \frac {A-i B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {x (B+i A)}{16 a^4}+\frac {A-i B}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {A+3 i B}{12 a d (a+i a \tan (c+d x))^3}-\frac {A+i B}{8 d (a+i a \tan (c+d x))^4} \]

[Out]

-1/16*(I*A+B)*x/a^4+1/8*(-A-I*B)/d/(a+I*a*tan(d*x+c))^4+1/12*(A+3*I*B)/a/d/(a+I*a*tan(d*x+c))^3+1/16*(A-I*B)/d
/(a^2+I*a^2*tan(d*x+c))^2+1/16*(A-I*B)/d/(a^4+I*a^4*tan(d*x+c))

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Rubi [A]  time = 0.19, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3590, 3526, 3479, 8} \[ \frac {A-i B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {A-i B}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}-\frac {x (B+i A)}{16 a^4}+\frac {A+3 i B}{12 a d (a+i a \tan (c+d x))^3}-\frac {A+i B}{8 d (a+i a \tan (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^4,x]

[Out]

-((I*A + B)*x)/(16*a^4) - (A + I*B)/(8*d*(a + I*a*Tan[c + d*x])^4) + (A + (3*I)*B)/(12*a*d*(a + I*a*Tan[c + d*
x])^3) + (A - I*B)/(16*d*(a^2 + I*a^2*Tan[c + d*x])^2) + (A - I*B)/(16*d*(a^4 + I*a^4*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))^4} \, dx &=-\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}-\frac {i \int \frac {a (A+i B)+2 a B \tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx}{2 a^2}\\ &=-\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {A+3 i B}{12 a d (a+i a \tan (c+d x))^3}-\frac {(i A+B) \int \frac {1}{(a+i a \tan (c+d x))^2} \, dx}{4 a^2}\\ &=-\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {A+3 i B}{12 a d (a+i a \tan (c+d x))^3}+\frac {A-i B}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}-\frac {(i A+B) \int \frac {1}{a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=-\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {A+3 i B}{12 a d (a+i a \tan (c+d x))^3}+\frac {A-i B}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {A-i B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {(i A+B) \int 1 \, dx}{16 a^4}\\ &=-\frac {(i A+B) x}{16 a^4}-\frac {A+i B}{8 d (a+i a \tan (c+d x))^4}+\frac {A+3 i B}{12 a d (a+i a \tan (c+d x))^3}+\frac {A-i B}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {A-i B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.51, size = 141, normalized size = 0.99 \[ \frac {\sec ^4(c+d x) (-3 (8 i A d x+A+B (8 d x+i)) \cos (4 (c+d x))+32 i A \sin (2 (c+d x))+3 i A \sin (4 (c+d x))+24 A d x \sin (4 (c+d x))+16 A \cos (2 (c+d x))-24 i B d x \sin (4 (c+d x))-3 B \sin (4 (c+d x))+12 i B)}{384 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(Sec[c + d*x]^4*((12*I)*B + 16*A*Cos[2*(c + d*x)] - 3*(A + (8*I)*A*d*x + B*(I + 8*d*x))*Cos[4*(c + d*x)] + (32
*I)*A*Sin[2*(c + d*x)] + (3*I)*A*Sin[4*(c + d*x)] - 3*B*Sin[4*(c + d*x)] + 24*A*d*x*Sin[4*(c + d*x)] - (24*I)*
B*d*x*Sin[4*(c + d*x)]))/(384*a^4*d*(-I + Tan[c + d*x])^4)

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fricas [A]  time = 0.76, size = 79, normalized size = 0.55 \[ \frac {{\left ({\left (-24 i \, A - 24 \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} + 24 \, A e^{\left (6 i \, d x + 6 i \, c\right )} + 12 i \, B e^{\left (4 i \, d x + 4 i \, c\right )} - 8 \, A e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, A - 3 i \, B\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \]

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))^4,x, algorithm="fricas")

[Out]

1/384*((-24*I*A - 24*B)*d*x*e^(8*I*d*x + 8*I*c) + 24*A*e^(6*I*d*x + 6*I*c) + 12*I*B*e^(4*I*d*x + 4*I*c) - 8*A*
e^(2*I*d*x + 2*I*c) - 3*A - 3*I*B)*e^(-8*I*d*x - 8*I*c)/(a^4*d)

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giac [A]  time = 0.99, size = 154, normalized size = 1.08 \[ -\frac {\frac {12 \, {\left (A - i \, B\right )} \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{4}} - \frac {12 \, {\left (A - i \, B\right )} \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{4}} - \frac {25 \, A \tan \left (d x + c\right )^{4} - 25 i \, B \tan \left (d x + c\right )^{4} - 124 i \, A \tan \left (d x + c\right )^{3} - 124 \, B \tan \left (d x + c\right )^{3} - 246 \, A \tan \left (d x + c\right )^{2} + 246 i \, B \tan \left (d x + c\right )^{2} + 252 i \, A \tan \left (d x + c\right ) + 124 \, B \tan \left (d x + c\right ) + 57 \, A - 25 i \, B}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]

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))^4,x, algorithm="giac")

[Out]

-1/384*(12*(A - I*B)*log(I*tan(d*x + c) + 1)/a^4 - 12*(A - I*B)*log(I*tan(d*x + c) - 1)/a^4 - (25*A*tan(d*x +
c)^4 - 25*I*B*tan(d*x + c)^4 - 124*I*A*tan(d*x + c)^3 - 124*B*tan(d*x + c)^3 - 246*A*tan(d*x + c)^2 + 246*I*B*
tan(d*x + c)^2 + 252*I*A*tan(d*x + c) + 124*B*tan(d*x + c) + 57*A - 25*I*B)/(a^4*(tan(d*x + c) - I)^4))/d

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maple [A]  time = 0.23, size = 244, normalized size = 1.71 \[ \frac {A \ln \left (\tan \left (d x +c \right )+i\right )}{32 d \,a^{4}}-\frac {i B \ln \left (\tan \left (d x +c \right )+i\right )}{32 d \,a^{4}}-\frac {A}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {i B}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {i A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {B}{4 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {i A}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {\ln \left (\tan \left (d x +c \right )-i\right ) A}{32 d \,a^{4}}+\frac {i \ln \left (\tan \left (d x +c \right )-i\right ) B}{32 d \,a^{4}} \]

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))^4,x)

[Out]

1/32/d/a^4*A*ln(tan(d*x+c)+I)-1/32*I/d/a^4*B*ln(tan(d*x+c)+I)-1/8/d/a^4/(tan(d*x+c)-I)^4*A-1/8*I/d/a^4/(tan(d*
x+c)-I)^4*B-1/16*I/d/a^4/(tan(d*x+c)-I)*A-1/16/d/a^4/(tan(d*x+c)-I)*B-1/16/d/a^4/(tan(d*x+c)-I)^2*A+1/16*I/d/a
^4/(tan(d*x+c)-I)^2*B-1/4/d/a^4/(tan(d*x+c)-I)^3*B+1/12*I/d/a^4/(tan(d*x+c)-I)^3*A-1/32/d/a^4*ln(tan(d*x+c)-I)
*A+1/32*I/d/a^4*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))^4,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 = 6.51, size = 172, normalized size = 1.20 \[ -\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {A}{4\,a^4}-\frac {B\,1{}\mathrm {i}}{4\,a^4}\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {B}{16\,a^4}+\frac {A\,1{}\mathrm {i}}{16\,a^4}\right )-\frac {A}{12\,a^4}-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B}{16\,a^4}+\frac {A\,19{}\mathrm {i}}{48\,a^4}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,a^4\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{32\,a^4\,d} \]

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)^4,x)

[Out]

(log(tan(c + d*x) - 1i)*(A*1i + B)*1i)/(32*a^4*d) - (tan(c + d*x)^2*(A/(4*a^4) - (B*1i)/(4*a^4)) + tan(c + d*x
)^3*((A*1i)/(16*a^4) + B/(16*a^4)) - A/(12*a^4) - tan(c + d*x)*((A*19i)/(48*a^4) + B/(16*a^4)))/(d*(tan(c + d*
x)*4i - 6*tan(c + d*x)^2 - tan(c + d*x)^3*4i + tan(c + d*x)^4 + 1)) + (log(tan(c + d*x) + 1i)*(A - B*1i))/(32*
a^4*d)

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sympy [A]  time = 0.68, size = 246, normalized size = 1.72 \[ \begin {cases} \frac {\left (196608 A a^{12} d^{3} e^{18 i c} e^{- 2 i d x} - 65536 A a^{12} d^{3} e^{14 i c} e^{- 6 i d x} + 98304 i B a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + \left (- 24576 A a^{12} d^{3} e^{12 i c} - 24576 i B a^{12} d^{3} e^{12 i c}\right ) e^{- 8 i d x}\right ) e^{- 20 i c}}{3145728 a^{16} d^{4}} & \text {for}\: 3145728 a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (- \frac {- i A - B}{16 a^{4}} + \frac {\left (- i A e^{8 i c} - 2 i A e^{6 i c} + 2 i A e^{2 i c} + i A - B e^{8 i c} + 2 B e^{4 i c} - B\right ) e^{- 8 i c}}{16 a^{4}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (i A + B\right )}{16 a^{4}} \]

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))**4,x)

[Out]

Piecewise(((196608*A*a**12*d**3*exp(18*I*c)*exp(-2*I*d*x) - 65536*A*a**12*d**3*exp(14*I*c)*exp(-6*I*d*x) + 983
04*I*B*a**12*d**3*exp(16*I*c)*exp(-4*I*d*x) + (-24576*A*a**12*d**3*exp(12*I*c) - 24576*I*B*a**12*d**3*exp(12*I
*c))*exp(-8*I*d*x))*exp(-20*I*c)/(3145728*a**16*d**4), Ne(3145728*a**16*d**4*exp(20*I*c), 0)), (x*(-(-I*A - B)
/(16*a**4) + (-I*A*exp(8*I*c) - 2*I*A*exp(6*I*c) + 2*I*A*exp(2*I*c) + I*A - B*exp(8*I*c) + 2*B*exp(4*I*c) - B)
*exp(-8*I*c)/(16*a**4)), True)) - x*(I*A + B)/(16*a**4)

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