3.61 \(\int \frac{\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=145 \[ -\frac{B+i A}{16 a^4 d (1+i \tan (c+d x))}+\frac{5 B+i A}{16 a^4 d (1+i \tan (c+d x))^2}-\frac{x (A-i B)}{16 a^4}+\frac{(-B+i A) \tan ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}-\frac{B}{6 a d (a+i a \tan (c+d x))^3} \]

[Out]

-((A - I*B)*x)/(16*a^4) + (I*A + 5*B)/(16*a^4*d*(1 + I*Tan[c + d*x])^2) - (I*A + B)/(16*a^4*d*(1 + I*Tan[c + d
*x])) + ((I*A - B)*Tan[c + d*x]^2)/(8*d*(a + I*a*Tan[c + d*x])^4) - B/(6*a*d*(a + I*a*Tan[c + d*x])^3)

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Rubi [A]  time = 0.290671, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {3595, 3590, 3526, 3479, 8} \[ -\frac{B+i A}{16 a^4 d (1+i \tan (c+d x))}+\frac{5 B+i A}{16 a^4 d (1+i \tan (c+d x))^2}-\frac{x (A-i B)}{16 a^4}+\frac{(-B+i A) \tan ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}-\frac{B}{6 a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^2*(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 + 5*B)/(16*a^4*d*(1 + I*Tan[c + d*x])^2) - (I*A + B)/(16*a^4*d*(1 + I*Tan[c + d
*x])) + ((I*A - B)*Tan[c + d*x]^2)/(8*d*(a + I*a*Tan[c + d*x])^4) - B/(6*a*d*(a + I*a*Tan[c + d*x])^3)

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]

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]

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{\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx &=\frac{(i A-B) \tan ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}-\frac{\int \frac{\tan (c+d x) (2 a (i A-B)-2 a (A-3 i B) \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx}{8 a^2}\\ &=\frac{(i A-B) \tan ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}-\frac{B}{6 a d (a+i a \tan (c+d x))^3}+\frac{i \int \frac{-8 a^2 B-4 a^2 (A-3 i B) \tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{16 a^4}\\ &=\frac{i A+5 B}{16 a^4 d (1+i \tan (c+d x))^2}+\frac{(i A-B) \tan ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}-\frac{B}{6 a d (a+i a \tan (c+d x))^3}-\frac{(A-i B) \int \frac{1}{a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=\frac{i A+5 B}{16 a^4 d (1+i \tan (c+d x))^2}+\frac{(i A-B) \tan ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}-\frac{B}{6 a d (a+i a \tan (c+d x))^3}-\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+5 B}{16 a^4 d (1+i \tan (c+d x))^2}+\frac{(i A-B) \tan ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}-\frac{B}{6 a d (a+i a \tan (c+d x))^3}-\frac{i A+B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.49652, size = 144, normalized size = 0.99 \[ -\frac{(\cos (4 (c+d x))-i \sin (4 (c+d x))) (3 (8 A d x+i A-8 i B d x-B) \cos (4 (c+d x))+24 i A d x \sin (4 (c+d x))+3 A \sin (4 (c+d x))-12 i A-32 i B \sin (2 (c+d x))+3 i B \sin (4 (c+d x))+24 B d x \sin (4 (c+d x))-16 B \cos (2 (c+d x)))}{384 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^2*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^4,x]

[Out]

-((Cos[4*(c + d*x)] - I*Sin[4*(c + d*x)])*((-12*I)*A - 16*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*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)

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Maple [A]  time = 0.035, 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{A}{4\,{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{{\frac{5\,i}{12}}B}{{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}+{\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}{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{{\frac{i}{16}}A}{{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{7\,B}{16\,{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\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(tan(d*x+c)^2*(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-1/4/d/a^4/(tan(d*x+c)-I)^3*A-5/12*I/d/a^4/(tan(d*
x+c)-I)^3*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-1/16/d/a^4/(tan(d*x+c)-I)*A+1/16*I/d
/a^4/(tan(d*x+c)-I)*B+1/16*I/d/a^4/(tan(d*x+c)-I)^2*A-7/16/d/a^4/(tan(d*x+c)-I)^2*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(tan(d*x+c)^2*(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 = 1.38764, size = 232, normalized size = 1.6 \begin{align*} -\frac{{\left (24 \,{\left (A - i \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} - 24 \, B e^{\left (6 i \, d x + 6 i \, c\right )} - 12 i \, A e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, B 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(tan(d*x+c)^2*(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) - 24*B*e^(6*I*d*x + 6*I*c) - 12*I*A*e^(4*I*d*x + 4*I*c) + 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 = 6.47366, size = 243, normalized size = 1.68 \begin{align*} \begin{cases} \frac{\left (98304 i A a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + 196608 B a^{12} d^{3} e^{18 i c} e^{- 2 i d x} - 65536 B a^{12} d^{3} e^{14 i c} e^{- 6 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}\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} - 2 A e^{4 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(tan(d*x+c)**2*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**4,x)

[Out]

Piecewise(((98304*I*A*a**12*d**3*exp(16*I*c)*exp(-4*I*d*x) + 196608*B*a**12*d**3*exp(18*I*c)*exp(-2*I*d*x) - 6
5536*B*a**12*d**3*exp(14*I*c)*exp(-6*I*d*x) + (-24576*I*A*a**12*d**3*exp(12*I*c) + 24576*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*((A - I*B)/(
16*a**4) - (A*exp(8*I*c) - 2*A*exp(4*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)*ex
p(-8*I*c)/(16*a**4)), True)) + x*(-A + I*B)/(16*a**4)

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Giac [A]  time = 1.57564, size = 204, normalized size = 1.41 \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} - 54 \, B \tan \left (d x + c\right )^{2} - 124 \, A \tan \left (d x + c\right ) - 4 i \, B \tan \left (d x + c\right ) + 25 i \, A - 7 \, 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(tan(d*x+c)^2*(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 - 54*B*tan(
d*x + c)^2 - 124*A*tan(d*x + c) - 4*I*B*tan(d*x + c) + 25*I*A - 7*B)/(a^4*(tan(d*x + c) - I)^4))/d