\(\int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\) [52]

Optimal result

Integrand size = 34, antiderivative size = 148 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {(i A-7 B) x}{8 a^3}-\frac {i B \log (\cos (c+d x))}{a^3 d}+\frac {(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac {A+7 i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

1/8*(I*A-7*B)*x/a^3-I*B*ln(cos(d*x+c))/a^3/d+1/6*(I*A-B)*tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^3+1/8*(A+3*I*B)*tan
(d*x+c)^2/a/d/(a+I*a*tan(d*x+c))^2+1/8*(A+7*I*B)/d/(a^3+I*a^3*tan(d*x+c))

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3676, 3670, 3556, 12, 3607, 8} \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {A+7 i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {x (-7 B+i A)}{8 a^3}-\frac {i B \log (\cos (c+d x))}{a^3 d}+\frac {(-B+i A) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2} \]

[In]

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

[Out]

((I*A - 7*B)*x)/(8*a^3) - (I*B*Log[Cos[c + d*x]])/(a^3*d) + ((I*A - B)*Tan[c + d*x]^3)/(6*d*(a + I*a*Tan[c + d
*x])^3) + ((A + (3*I)*B)*Tan[c + d*x]^2)/(8*a*d*(a + I*a*Tan[c + d*x])^2) + (A + (7*I)*B)/(8*d*(a^3 + I*a^3*Ta
n[c + d*x]))

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 3556

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

Rule 3607

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 3670

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 3676

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*} \text {integral}& = \frac {(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^2(c+d x) (3 a (i A-B)+6 i a B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\tan (c+d x) \left (-6 a^2 (A+3 i B)-24 a^2 B \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4} \\ & = \frac {(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}-\frac {i \int -\frac {6 a^3 (i A-7 B) \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{24 a^5}+\frac {(i B) \int \tan (c+d x) \, dx}{a^3} \\ & = -\frac {i B \log (\cos (c+d x))}{a^3 d}+\frac {(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}-\frac {(A+7 i B) \int \frac {\tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = -\frac {i B \log (\cos (c+d x))}{a^3 d}+\frac {(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac {A+7 i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(i A-7 B) \int 1 \, dx}{8 a^3} \\ & = \frac {(i A-7 B) x}{8 a^3}-\frac {i B \log (\cos (c+d x))}{a^3 d}+\frac {(i A-B) \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+3 i B) \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac {A+7 i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.13 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.12 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {i \left (3 i ((A+15 i B) \log (i-\tan (c+d x))-(A-i B) \log (i+\tan (c+d x)))+6 (A+7 i B) \tan (c+d x)+24 B \tan ^2(c+d x)-2 (A+7 i B) \tan ^3(c+d x)+\frac {2 \tan ^4(c+d x) \left (-6 (A+i B)+(-3 i A+9 B) \tan (c+d x)+(A+7 i B) \tan ^2(c+d x)\right )}{(-i+\tan (c+d x))^3}\right )}{48 a^3 d} \]

[In]

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

[Out]

((-1/48*I)*((3*I)*((A + (15*I)*B)*Log[I - Tan[c + d*x]] - (A - I*B)*Log[I + Tan[c + d*x]]) + 6*(A + (7*I)*B)*T
an[c + d*x] + 24*B*Tan[c + d*x]^2 - 2*(A + (7*I)*B)*Tan[c + d*x]^3 + (2*Tan[c + d*x]^4*(-6*(A + I*B) + ((-3*I)
*A + 9*B)*Tan[c + d*x] + (A + (7*I)*B)*Tan[c + d*x]^2))/(-I + Tan[c + d*x])^3))/(a^3*d)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.08

method	result	size
risch	\(-\frac {15 x B}{8 a^{3}}+\frac {i x A}{8 a^{3}}+\frac {11 i {\mathrm e}^{-2 i \left (d x +c \right )} B}{16 a^{3} d}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )} A}{16 a^{3} d}-\frac {5 i {\mathrm e}^{-4 i \left (d x +c \right )} B}{32 a^{3} d}-\frac {3 \,{\mathrm e}^{-4 i \left (d x +c \right )} A}{32 a^{3} d}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )} B}{48 a^{3} d}+\frac {{\mathrm e}^{-6 i \left (d x +c \right )} A}{48 a^{3} d}-\frac {2 B c}{a^{3} d}-\frac {i B \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{3} d}\)	\(160\)
derivativedivides	\(\frac {i A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {i B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{3}}-\frac {7 B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {17 B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {7 i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}+\frac {5 A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {7 i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}\)	\(179\)
default	\(\frac {i A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {i B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{3}}-\frac {7 B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {17 B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {7 i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}+\frac {5 A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {7 i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}\)	\(179\)

[In]

int(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-15/8*x/a^3*B+1/8*I*x/a^3*A+11/16*I/a^3/d*exp(-2*I*(d*x+c))*B+3/16/a^3/d*exp(-2*I*(d*x+c))*A-5/32*I/a^3/d*exp(
-4*I*(d*x+c))*B-3/32/a^3/d*exp(-4*I*(d*x+c))*A+1/48*I/a^3/d*exp(-6*I*(d*x+c))*B+1/48/a^3/d*exp(-6*I*(d*x+c))*A
-2*B/a^3/d*c-I*B/a^3/d*ln(exp(2*I*(d*x+c))+1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.70 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\left (12 \, {\left (-i \, A + 15 \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} + 96 i \, B e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 6 \, {\left (3 \, A + 11 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (3 \, A + 5 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, A - 2 i \, B\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

[In]

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/96*(12*(-I*A + 15*B)*d*x*e^(6*I*d*x + 6*I*c) + 96*I*B*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 6*
(3*A + 11*I*B)*e^(4*I*d*x + 4*I*c) + 3*(3*A + 5*I*B)*e^(2*I*d*x + 2*I*c) - 2*A - 2*I*B)*e^(-6*I*d*x - 6*I*c)/(
a^3*d)

Sympy [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.00 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=- \frac {i B \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{3} d} + \begin {cases} \frac {\left (\left (512 A a^{6} d^{2} e^{6 i c} + 512 i B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (- 2304 A a^{6} d^{2} e^{8 i c} - 3840 i B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (4608 A a^{6} d^{2} e^{10 i c} + 16896 i B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {i A - 15 B}{8 a^{3}} + \frac {\left (i A e^{6 i c} - 3 i A e^{4 i c} + 3 i A e^{2 i c} - i A - 15 B e^{6 i c} + 11 B e^{4 i c} - 5 B e^{2 i c} + B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (i A - 15 B\right )}{8 a^{3}} \]

[In]

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

[Out]

-I*B*log(exp(2*I*d*x) + exp(-2*I*c))/(a**3*d) + Piecewise((((512*A*a**6*d**2*exp(6*I*c) + 512*I*B*a**6*d**2*ex
p(6*I*c))*exp(-6*I*d*x) + (-2304*A*a**6*d**2*exp(8*I*c) - 3840*I*B*a**6*d**2*exp(8*I*c))*exp(-4*I*d*x) + (4608
*A*a**6*d**2*exp(10*I*c) + 16896*I*B*a**6*d**2*exp(10*I*c))*exp(-2*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(
a**9*d**3*exp(12*I*c), 0)), (x*(-(I*A - 15*B)/(8*a**3) + (I*A*exp(6*I*c) - 3*I*A*exp(4*I*c) + 3*I*A*exp(2*I*c)
- I*A - 15*B*exp(6*I*c) + 11*B*exp(4*I*c) - 5*B*exp(2*I*c) + B)*exp(-6*I*c)/(8*a**3)), True)) + x*(I*A - 15*B
)/(8*a**3)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.82 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.86 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\frac {6 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} - \frac {6 \, {\left (A + 15 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {11 \, A \tan \left (d x + c\right )^{3} + 165 i \, B \tan \left (d x + c\right )^{3} + 51 i \, A \tan \left (d x + c\right )^{2} + 291 \, B \tan \left (d x + c\right )^{2} + 75 \, A \tan \left (d x + c\right ) - 171 i \, B \tan \left (d x + c\right ) - 29 i \, A - 29 \, B}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]

[In]

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/96*(6*(A - I*B)*log(tan(d*x + c) + I)/a^3 - 6*(A + 15*I*B)*log(tan(d*x + c) - I)/a^3 + (11*A*tan(d*x + c)^3
+ 165*I*B*tan(d*x + c)^3 + 51*I*A*tan(d*x + c)^2 + 291*B*tan(d*x + c)^2 + 75*A*tan(d*x + c) - 171*I*B*tan(d*x
+ c) - 29*I*A - 29*B)/(a^3*(tan(d*x + c) - I)^3))/d

Mupad [B] (verification not implemented)

Time = 7.69 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.99 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {5\,A}{12\,a^3}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {7\,A}{8\,a^3}+\frac {B\,17{}\mathrm {i}}{8\,a^3}\right )+\frac {B\,17{}\mathrm {i}}{12\,a^3}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {27\,B}{8\,a^3}+\frac {A\,9{}\mathrm {i}}{8\,a^3}\right )}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,15{}\mathrm {i}\right )}{16\,a^3\,d} \]

[In]

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

[Out]

((5*A)/(12*a^3) - tan(c + d*x)^2*((7*A)/(8*a^3) + (B*17i)/(8*a^3)) + (B*17i)/(12*a^3) + tan(c + d*x)*((A*9i)/(
8*a^3) - (27*B)/(8*a^3)))/(d*(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1)) - (log(tan(c + d*x)
+ 1i)*(A - B*1i))/(16*a^3*d) + (log(tan(c + d*x) - 1i)*(A + B*15i))/(16*a^3*d)