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

Optimal. Leaf size=145 \[ \frac{B+i A}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{B+i A}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac{x (A-i B)}{16 a^4}+\frac{-B+i A}{8 d (a+i a \tan (c+d x))^4}+\frac{B+i A}{12 a d (a+i a \tan (c+d x))^3} \]

[Out]

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

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Rubi [A]  time = 0.106414, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3526, 3479, 8} \[ \frac{B+i A}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{B+i A}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac{x (A-i B)}{16 a^4}+\frac{-B+i A}{8 d (a+i a \tan (c+d x))^4}+\frac{B+i A}{12 a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Rubi steps

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

Mathematica [A]  time = 0.817563, size = 160, normalized size = 1.1 \[ \frac{\sec ^4(c+d x) (16 (B+4 i A) \cos (2 (c+d x))+3 (8 A d x+i A-8 i B d x-B) \cos (4 (c+d x))-32 A \sin (2 (c+d x))+24 i A d x \sin (4 (c+d x))+3 A \sin (4 (c+d x))+36 i A+32 i B \sin (2 (c+d x))+3 i B \sin (4 (c+d x))+24 B d x \sin (4 (c+d x)))}{384 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^4*((36*I)*A + 16*((4*I)*A + B)*Cos[2*(c + d*x)] + 3*(I*A - B + 8*A*d*x - (8*I)*B*d*x)*Cos[4*(c +
d*x)] - 32*A*Sin[2*(c + d*x)] + (32*I)*B*Sin[2*(c + d*x)] + 3*A*Sin[4*(c + d*x)] + (3*I)*B*Sin[4*(c + d*x)] +
(24*I)*A*d*x*Sin[4*(c + d*x)] + 24*B*d*x*Sin[4*(c + d*x)]))/(384*a^4*d*(-I + Tan[c + d*x])^4)

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Maple [A]  time = 0.031, size = 244, normalized size = 1.7 \begin{align*}{\frac{{\frac{i}{8}}A}{{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{4}}}-{\frac{B}{8\,{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{4}}}-{\frac{{\frac{i}{32}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{{a}^{4}d}}-{\frac{\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{32\,{a}^{4}d}}-{\frac{A}{12\,{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}+{\frac{{\frac{i}{12}}B}{{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{{\frac{i}{16}}A}{{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{B}{16\,{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{A}{16\,{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{{\frac{i}{16}}B}{{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{B\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{32\,{a}^{4}d}}+{\frac{{\frac{i}{32}}A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{{a}^{4}d}} \end{align*}

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))^4,x)

[Out]

1/8*I/d/a^4/(tan(d*x+c)-I)^4*A-1/8/d/a^4/(tan(d*x+c)-I)^4*B-1/32*I/d/a^4*ln(tan(d*x+c)-I)*A-1/32/d/a^4*ln(tan(
d*x+c)-I)*B-1/12/d/a^4/(tan(d*x+c)-I)^3*A+1/12*I/d/a^4/(tan(d*x+c)-I)^3*B-1/16*I/d/a^4/(tan(d*x+c)-I)^2*A-1/16
/d/a^4/(tan(d*x+c)-I)^2*B+1/16/d/a^4/(tan(d*x+c)-I)*A-1/16*I/d/a^4/(tan(d*x+c)-I)*B+1/32/d/a^4*B*ln(tan(d*x+c)
+I)+1/32*I/d/a^4*A*ln(tan(d*x+c)+I)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

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))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.42516, size = 261, normalized size = 1.8 \begin{align*} \frac{{\left (24 \,{\left (A - i \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (48 i \, A + 24 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 36 i \, A e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (16 i \, A - 8 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \end{align*}

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))^4,x, algorithm="fricas")

[Out]

1/384*(24*(A - I*B)*d*x*e^(8*I*d*x + 8*I*c) + (48*I*A + 24*B)*e^(6*I*d*x + 6*I*c) + 36*I*A*e^(4*I*d*x + 4*I*c)
+ (16*I*A - 8*B)*e^(2*I*d*x + 2*I*c) + 3*I*A - 3*B)*e^(-8*I*d*x - 8*I*c)/(a^4*d)

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Sympy [A]  time = 11.3457, size = 301, normalized size = 2.08 \begin{align*} \begin{cases} \frac{\left (294912 i A a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + \left (24576 i A a^{12} d^{3} e^{12 i c} - 24576 B a^{12} d^{3} e^{12 i c}\right ) e^{- 8 i d x} + \left (131072 i A a^{12} d^{3} e^{14 i c} - 65536 B a^{12} d^{3} e^{14 i c}\right ) e^{- 6 i d x} + \left (393216 i A a^{12} d^{3} e^{18 i c} + 196608 B a^{12} d^{3} e^{18 i c}\right ) e^{- 2 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{A - i B}{16 a^{4}} + \frac{\left (A e^{8 i c} + 4 A e^{6 i c} + 6 A e^{4 i c} + 4 A e^{2 i c} + A - i B e^{8 i c} - 2 i B e^{6 i c} + 2 i B e^{2 i c} + i B\right ) e^{- 8 i c}}{16 a^{4}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (A - i B\right )}{16 a^{4}} \end{align*}

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))**4,x)

[Out]

Piecewise(((294912*I*A*a**12*d**3*exp(16*I*c)*exp(-4*I*d*x) + (24576*I*A*a**12*d**3*exp(12*I*c) - 24576*B*a**1
2*d**3*exp(12*I*c))*exp(-8*I*d*x) + (131072*I*A*a**12*d**3*exp(14*I*c) - 65536*B*a**12*d**3*exp(14*I*c))*exp(-
6*I*d*x) + (393216*I*A*a**12*d**3*exp(18*I*c) + 196608*B*a**12*d**3*exp(18*I*c))*exp(-2*I*d*x))*exp(-20*I*c)/(
3145728*a**16*d**4), Ne(3145728*a**16*d**4*exp(20*I*c), 0)), (x*(-(A - I*B)/(16*a**4) + (A*exp(8*I*c) + 4*A*ex
p(6*I*c) + 6*A*exp(4*I*c) + 4*A*exp(2*I*c) + A - I*B*exp(8*I*c) - 2*I*B*exp(6*I*c) + 2*I*B*exp(2*I*c) + I*B)*e
xp(-8*I*c)/(16*a**4)), True)) + x*(A - I*B)/(16*a**4)

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Giac [A]  time = 1.33658, size = 208, normalized size = 1.43 \begin{align*} -\frac{\frac{12 \,{\left (-i \, A - B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} - \frac{12 \,{\left (-i \, A - B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} - \frac{25 i \, A \tan \left (d x + c\right )^{4} + 25 \, B \tan \left (d x + c\right )^{4} + 124 \, A \tan \left (d x + c\right )^{3} - 124 i \, B \tan \left (d x + c\right )^{3} - 246 i \, A \tan \left (d x + c\right )^{2} - 246 \, B \tan \left (d x + c\right )^{2} - 252 \, A \tan \left (d x + c\right ) + 252 i \, B \tan \left (d x + c\right ) + 153 i \, A + 57 \, B}{a^{4}{\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \end{align*}

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))^4,x, algorithm="giac")

[Out]

-1/384*(12*(-I*A - B)*log(tan(d*x + c) + I)/a^4 - 12*(-I*A - B)*log(tan(d*x + c) - I)/a^4 - (25*I*A*tan(d*x +
c)^4 + 25*B*tan(d*x + c)^4 + 124*A*tan(d*x + c)^3 - 124*I*B*tan(d*x + c)^3 - 246*I*A*tan(d*x + c)^2 - 246*B*ta
n(d*x + c)^2 - 252*A*tan(d*x + c) + 252*I*B*tan(d*x + c) + 153*I*A + 57*B)/(a^4*(tan(d*x + c) - I)^4))/d