3.60 \(\int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=159 \[ \frac {5 A-29 i B}{48 a^4 d (1+i \tan (c+d x))}-\frac {A-13 i B}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {x (B+i A)}{16 a^4}+\frac {(-B+i A) \tan ^3(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+5 i B) \tan ^2(c+d x)}{24 a d (a+i a \tan (c+d x))^3} \]

[Out]

1/16*(I*A+B)*x/a^4+1/48*(-A+13*I*B)/a^4/d/(1+I*tan(d*x+c))^2+1/48*(5*A-29*I*B)/a^4/d/(1+I*tan(d*x+c))+1/8*(I*A
-B)*tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^4+1/24*(A+5*I*B)*tan(d*x+c)^2/a/d/(a+I*a*tan(d*x+c))^3

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Rubi [A]  time = 0.47, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3595, 3590, 3526, 8} \[ \frac {5 A-29 i B}{48 a^4 d (1+i \tan (c+d x))}-\frac {A-13 i B}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {x (B+i A)}{16 a^4}+\frac {(-B+i A) \tan ^3(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+5 i B) \tan ^2(c+d x)}{24 a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((I*A + B)*x)/(16*a^4) - (A - (13*I)*B)/(48*a^4*d*(1 + I*Tan[c + d*x])^2) + (5*A - (29*I)*B)/(48*a^4*d*(1 + I*
Tan[c + d*x])) + ((I*A - B)*Tan[c + d*x]^3)/(8*d*(a + I*a*Tan[c + d*x])^4) + ((A + (5*I)*B)*Tan[c + d*x]^2)/(2
4*a*d*(a + I*a*Tan[c + d*x])^3)

Rule 8

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

Rubi steps

\begin {align*} \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx &=\frac {(i A-B) \tan ^3(c+d x)}{8 d (a+i a \tan (c+d x))^4}-\frac {\int \frac {\tan ^2(c+d x) (3 a (i A-B)-a (A-7 i B) \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx}{8 a^2}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+5 i B) \tan ^2(c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\tan (c+d x) \left (-4 a^2 (A+5 i B)-8 a^2 (i A+4 B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{48 a^4}\\ &=-\frac {A-13 i B}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(i A-B) \tan ^3(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+5 i B) \tan ^2(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac {i \int \frac {4 a^3 (A-13 i B)-16 a^3 (i A+4 B) \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{96 a^6}\\ &=-\frac {A-13 i B}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(i A-B) \tan ^3(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+5 i B) \tan ^2(c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac {5 A-29 i B}{48 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-13 i B}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(i A-B) \tan ^3(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+5 i B) \tan ^2(c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac {5 A-29 i B}{48 d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.53, size = 158, normalized size = 0.99 \[ \frac {\sec ^4(c+d x) (16 (A-4 i B) \cos (2 (c+d x))+3 (8 i A d x+A+8 B d x+i B) \cos (4 (c+d x))+32 i A \sin (2 (c+d x))-3 i A \sin (4 (c+d x))-24 A d x \sin (4 (c+d x))+32 B \sin (2 (c+d x))+24 i B d x \sin (4 (c+d x))+3 B \sin (4 (c+d x))+36 i B)}{384 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(Sec[c + d*x]^4*((36*I)*B + 16*(A - (4*I)*B)*Cos[2*(c + d*x)] + 3*(A + I*B + (8*I)*A*d*x + 8*B*d*x)*Cos[4*(c +
d*x)] + (32*I)*A*Sin[2*(c + d*x)] + 32*B*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 = 0.60, size = 87, normalized size = 0.55 \[ \frac {{\left ({\left (24 i \, A + 24 \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} + 24 \, {\left (A - 2 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 36 i \, B e^{\left (4 i \, d x + 4 i \, c\right )} - 8 \, {\left (A + 2 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, A + 3 i \, B\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(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 - 2*I*B)*e^(6*I*d*x + 6*I*c) + 36*I*B*e^(4*I*d*x + 4*I*
c) - 8*(A + 2*I*B)*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 = 1.63, size = 153, normalized size = 0.96 \[ -\frac {\frac {12 \, {\left (A - i \, B\right )} \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a^{4}} - \frac {12 \, {\left (A - i \, B\right )} \log \left (-i \, \tan \left (d x + c\right ) - 1\right )}{a^{4}} + \frac {25 \, A \tan \left (d x + c\right )^{4} - 25 i \, B \tan \left (d x + c\right )^{4} - 124 i \, A \tan \left (d x + c\right )^{3} + 260 \, B \tan \left (d x + c\right )^{3} - 54 \, A \tan \left (d x + c\right )^{2} - 522 i \, B \tan \left (d x + c\right )^{2} - 4 i \, A \tan \left (d x + c\right ) - 388 \, B \tan \left (d x + c\right ) - 7 \, A + 103 i \, B}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(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 + 260*B*tan(d*x + c)^3 - 54*A*tan(d*x + c)^2 - 522*I*B
*tan(d*x + c)^2 - 4*I*A*tan(d*x + c) - 388*B*tan(d*x + c) - 7*A + 103*I*B)/(a^4*(tan(d*x + c) - I)^4))/d

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maple [A]  time = 0.24, size = 244, normalized size = 1.53 \[ -\frac {A \ln \left (\tan \left (d x +c \right )+i\right )}{32 d \,a^{4}}+\frac {i B \ln \left (\tan \left (d x +c \right )+i\right )}{32 d \,a^{4}}-\frac {5 i A}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {7 B}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {i A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {15 B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {7 A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {17 i B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {A}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {i B}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {\ln \left (\tan \left (d x +c \right )-i\right ) A}{32 d \,a^{4}}-\frac {i \ln \left (\tan \left (d x +c \right )-i\right ) B}{32 d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3*(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)-5/12*I/d/a^4/(tan(d*x+c)-I)^3*A+7/12/d/a^4/(tan
(d*x+c)-I)^3*B+1/16*I/d/a^4/(tan(d*x+c)-I)*A-15/16/d/a^4/(tan(d*x+c)-I)*B-7/16/d/a^4/(tan(d*x+c)-I)^2*A-17/16*
I/d/a^4/(tan(d*x+c)-I)^2*B+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+1/32/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)^3*(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 = 6.62, size = 178, normalized size = 1.12 \[ \frac {\frac {A}{12\,a^4}+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (-\frac {15\,B}{16\,a^4}+\frac {A\,1{}\mathrm {i}}{16\,a^4}\right )-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {A}{4\,a^4}-\frac {B\,7{}\mathrm {i}}{4\,a^4}\right )-\frac {B\,1{}\mathrm {i}}{3\,a^4}+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {61\,B}{48\,a^4}+\frac {A\,13{}\mathrm {i}}{48\,a^4}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{32\,a^4\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(c + d*x)^3*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

(tan(c + d*x)^3*((A*1i)/(16*a^4) - (15*B)/(16*a^4)) - tan(c + d*x)^2*(A/(4*a^4) - (B*7i)/(4*a^4)) + A/(12*a^4)
- (B*1i)/(3*a^4) + tan(c + d*x)*((A*13i)/(48*a^4) + (61*B)/(48*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) + (log(tan(c + d*x
) + 1i)*(A*1i + B)*1i)/(32*a^4*d)

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sympy [A]  time = 1.00, size = 304, normalized size = 1.91 \[ \begin {cases} \frac {\left (294912 i B a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + \left (24576 A a^{12} d^{3} e^{12 i c} + 24576 i B a^{12} d^{3} e^{12 i c}\right ) e^{- 8 i d x} + \left (- 65536 A a^{12} d^{3} e^{14 i c} - 131072 i B a^{12} d^{3} e^{14 i c}\right ) e^{- 6 i d x} + \left (196608 A a^{12} d^{3} e^{18 i c} - 393216 i B a^{12} d^{3} e^{18 i c}\right ) e^{- 2 i d x}\right ) e^{- 20 i c}}{3145728 a^{16} d^{4}} & \text {for}\: 3145728 a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (- \frac {i A + B}{16 a^{4}} + \frac {\left (i A e^{8 i c} - 2 i A e^{6 i c} + 2 i A e^{2 i c} - i A + B e^{8 i c} - 4 B e^{6 i c} + 6 B e^{4 i c} - 4 B e^{2 i c} + B\right ) e^{- 8 i c}}{16 a^{4}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- i A - B\right )}{16 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**4,x)

[Out]

Piecewise(((294912*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**1
2*d**3*exp(12*I*c))*exp(-8*I*d*x) + (-65536*A*a**12*d**3*exp(14*I*c) - 131072*I*B*a**12*d**3*exp(14*I*c))*exp(
-6*I*d*x) + (196608*A*a**12*d**3*exp(18*I*c) - 393216*I*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*(-(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) - 4*B*exp(6*I*c) + 6*B*exp(4*I*c) - 4*B*exp(2*I*c) + B)*
exp(-8*I*c)/(16*a**4)), True)) - x*(-I*A - B)/(16*a**4)

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