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

Optimal. Leaf size=185 \[ -\frac {(-7 B+i A) \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}-\frac {-15 B+i A}{16 a^4 d (1+i \tan (c+d x))}+\frac {x (A+15 i B)}{16 a^4}-\frac {B \log (\cos (c+d x))}{a^4 d}+\frac {(-B+i A) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3} \]

[Out]

1/16*(A+15*I*B)*x/a^4-B*ln(cos(d*x+c))/a^4/d+1/16*(-I*A+15*B)/a^4/d/(1+I*tan(d*x+c))-1/16*(I*A-7*B)*tan(d*x+c)
^2/a^4/d/(1+I*tan(d*x+c))^2+1/8*(I*A-B)*tan(d*x+c)^4/d/(a+I*a*tan(d*x+c))^4+1/12*(A+3*I*B)*tan(d*x+c)^3/a/d/(a
+I*a*tan(d*x+c))^3

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Rubi [A]  time = 0.51, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3595, 3589, 3475, 12, 3526, 8} \[ -\frac {(-7 B+i A) \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}-\frac {-15 B+i A}{16 a^4 d (1+i \tan (c+d x))}+\frac {x (A+15 i B)}{16 a^4}-\frac {B \log (\cos (c+d x))}{a^4 d}+\frac {(-B+i A) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + (15*I)*B)*x)/(16*a^4) - (B*Log[Cos[c + d*x]])/(a^4*d) - (I*A - 15*B)/(16*a^4*d*(1 + I*Tan[c + d*x])) - (
(I*A - 7*B)*Tan[c + d*x]^2)/(16*a^4*d*(1 + I*Tan[c + d*x])^2) + ((I*A - B)*Tan[c + d*x]^4)/(8*d*(a + I*a*Tan[c
+ d*x])^4) + ((A + (3*I)*B)*Tan[c + d*x]^3)/(12*a*d*(a + I*a*Tan[c + d*x])^3)

Rule 8

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{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 3589

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[(B*d)/b, Int[Tan[e + f*x], x], x] + Dist[1/b, Int[Simp[A*b*c + (A*b*d + B*(
b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 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 ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx &=\frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}-\frac {\int \frac {\tan ^3(c+d x) (4 a (i A-B)+8 i a B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx}{8 a^2}\\ &=\frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\tan ^2(c+d x) \left (-12 a^2 (A+3 i B)-48 a^2 B \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{48 a^4}\\ &=-\frac {(i A-7 B) \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan (c+d x) \left (-24 a^3 (i A-7 B)-192 i a^3 B \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{192 a^6}\\ &=-\frac {(i A-7 B) \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}+\frac {i \int \frac {24 a^4 (A+15 i B) \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{192 a^7}+\frac {B \int \tan (c+d x) \, dx}{a^4}\\ &=-\frac {B \log (\cos (c+d x))}{a^4 d}-\frac {(i A-7 B) \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}+\frac {(i A-15 B) \int \frac {\tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=-\frac {B \log (\cos (c+d x))}{a^4 d}-\frac {(i A-7 B) \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}-\frac {i A-15 B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {(A+15 i B) \int 1 \, dx}{16 a^4}\\ &=\frac {(A+15 i B) x}{16 a^4}-\frac {B \log (\cos (c+d x))}{a^4 d}-\frac {(i A-7 B) \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}+\frac {(i A-B) \tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(A+3 i B) \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}-\frac {i A-15 B}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.44, size = 195, normalized size = 1.05 \[ \frac {\sec ^4(c+d x) (16 (21 B-4 i A) \cos (2 (c+d x))+3 \cos (4 (c+d x)) (8 A d x+i A-128 B \log (\cos (c+d x))+120 i B d x-B)+32 A \sin (2 (c+d x))+24 i A d x \sin (4 (c+d x))+3 A \sin (4 (c+d x))+36 i A+288 i B \sin (2 (c+d x))+3 i B \sin (4 (c+d x))-360 B d x \sin (4 (c+d x))-384 i B \sin (4 (c+d x)) \log (\cos (c+d x))-96 B)}{384 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^4*((36*I)*A - 96*B + 16*((-4*I)*A + 21*B)*Cos[2*(c + d*x)] + 3*Cos[4*(c + d*x)]*(I*A - B + 8*A*d
*x + (120*I)*B*d*x - 128*B*Log[Cos[c + d*x]]) + 32*A*Sin[2*(c + d*x)] + (288*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)] - 360*B*d*x*Sin[4*(c + d*x)] - (384*I)*
B*Log[Cos[c + 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 = 117, normalized size = 0.63 \[ \frac {{\left (24 \, {\left (A + 31 i \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} - 384 \, B e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + {\left (-48 i \, A + 312 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (36 i \, A - 96 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-16 i \, A + 24 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/384*(24*(A + 31*I*B)*d*x*e^(8*I*d*x + 8*I*c) - 384*B*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + (-48
*I*A + 312*B)*e^(6*I*d*x + 6*I*c) + (36*I*A - 96*B)*e^(4*I*d*x + 4*I*c) + (-16*I*A + 24*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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giac [A]  time = 2.18, size = 154, normalized size = 0.83 \[ -\frac {\frac {12 \, {\left (-i \, A - B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} - \frac {12 \, {\left (-i \, A + 31 \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} - \frac {25 i \, A \tan \left (d x + c\right )^{4} - 775 \, B \tan \left (d x + c\right )^{4} - 260 \, A \tan \left (d x + c\right )^{3} + 1924 i \, B \tan \left (d x + c\right )^{3} + 522 i \, A \tan \left (d x + c\right )^{2} + 1866 \, B \tan \left (d x + c\right )^{2} + 388 \, A \tan \left (d x + c\right ) - 772 i \, B \tan \left (d x + c\right ) - 103 i \, A - 103 \, B}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4*(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 + 31*B)*log(tan(d*x + c) - I)/a^4 - (25*I*A*tan(d*x
+ c)^4 - 775*B*tan(d*x + c)^4 - 260*A*tan(d*x + c)^3 + 1924*I*B*tan(d*x + c)^3 + 522*I*A*tan(d*x + c)^2 + 186
6*B*tan(d*x + c)^2 + 388*A*tan(d*x + c) - 772*I*B*tan(d*x + c) - 103*I*A - 103*B)/(a^4*(tan(d*x + c) - I)^4))/
d

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maple [A]  time = 0.23, size = 244, normalized size = 1.32 \[ \frac {B \ln \left (\tan \left (d x +c \right )+i\right )}{32 d \,a^{4}}+\frac {i A \ln \left (\tan \left (d x +c \right )+i\right )}{32 d \,a^{4}}+\frac {i A}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {B}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {49 i B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {15 A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}+\frac {31 B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {17 i A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {3 i B}{4 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {7 A}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i \ln \left (\tan \left (d x +c \right )-i\right ) A}{32 d \,a^{4}}+\frac {31 \ln \left (\tan \left (d x +c \right )-i\right ) B}{32 d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)+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-49/16*I/d/a^4/(tan(d*x+c)-I)*B-15/16/d/a^4/(tan(d*x+c)-I)*A+31/16/d/a^4/(tan(d*x+c)-I)^2*B-17/16*I
/d/a^4/(tan(d*x+c)-I)^2*A+3/4*I/d/a^4/(tan(d*x+c)-I)^3*B+7/12/d/a^4/(tan(d*x+c)-I)^3*A-1/32*I/d/a^4*ln(tan(d*x
+c)-I)*A+31/32/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)^4*(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 = 6.77, size = 178, normalized size = 0.96 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {29\,B}{4\,a^4}+\frac {A\,7{}\mathrm {i}}{4\,a^4}\right )-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {15\,A}{16\,a^4}+\frac {B\,49{}\mathrm {i}}{16\,a^4}\right )-\frac {A\,1{}\mathrm {i}}{3\,a^4}+\frac {7\,B}{4\,a^4}+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {61\,A}{48\,a^4}+\frac {B\,97{}\mathrm {i}}{16\,a^4}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{32\,a^4\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,31{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c + d*x)^2*((A*7i)/(4*a^4) - (29*B)/(4*a^4)) - tan(c + d*x)^3*((15*A)/(16*a^4) + (B*49i)/(16*a^4)) - (A*1
i)/(3*a^4) + (7*B)/(4*a^4) + tan(c + d*x)*((61*A)/(48*a^4) + (B*97i)/(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*1i + B))/(32*a^4*d) - (log(ta
n(c + d*x) - 1i)*(A + B*31i)*1i)/(32*a^4*d)

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sympy [A]  time = 7.06, size = 366, normalized size = 1.98 \[ - \frac {B \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{4} d} + \begin {cases} \frac {\left (\left (24576 i A a^{12} d^{3} e^{12 i c} - 24576 B a^{12} d^{3} e^{12 i c}\right ) e^{- 8 i d x} + \left (- 131072 i A a^{12} d^{3} e^{14 i c} + 196608 B a^{12} d^{3} e^{14 i c}\right ) e^{- 6 i d x} + \left (294912 i A a^{12} d^{3} e^{16 i c} - 786432 B a^{12} d^{3} e^{16 i c}\right ) e^{- 4 i d x} + \left (- 393216 i A a^{12} d^{3} e^{18 i c} + 2555904 B a^{12} d^{3} e^{18 i c}\right ) e^{- 2 i d x}\right ) e^{- 20 i c}}{3145728 a^{16} d^{4}} & \text {for}\: 3145728 a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (- \frac {A + 31 i B}{16 a^{4}} - \frac {i \left (i A e^{8 i c} - 4 i A e^{6 i c} + 6 i A e^{4 i c} - 4 i A e^{2 i c} + i A - 31 B e^{8 i c} + 26 B e^{6 i c} - 16 B e^{4 i c} + 6 B e^{2 i c} - B\right ) e^{- 8 i c}}{16 a^{4}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- A - 31 i B\right )}{16 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-B*log(exp(2*I*d*x) + exp(-2*I*c))/(a**4*d) + Piecewise((((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) + (-131072*I*A*a**12*d**3*exp(14*I*c) + 196608*B*a**12*d**3*exp(14*I*c))*exp(-6*
I*d*x) + (294912*I*A*a**12*d**3*exp(16*I*c) - 786432*B*a**12*d**3*exp(16*I*c))*exp(-4*I*d*x) + (-393216*I*A*a*
*12*d**3*exp(18*I*c) + 2555904*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 + 31*I*B)/(16*a**4) - I*(I*A*exp(8*I*c) - 4*I*A*exp(6*I*c) + 6*I*
A*exp(4*I*c) - 4*I*A*exp(2*I*c) + I*A - 31*B*exp(8*I*c) + 26*B*exp(6*I*c) - 16*B*exp(4*I*c) + 6*B*exp(2*I*c) -
B)*exp(-8*I*c)/(16*a**4)), True)) - x*(-A - 31*I*B)/(16*a**4)

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