Integrand size = 32, antiderivative size = 110 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {(i A+B) x}{8 a^3}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}+\frac {A+3 i B}{8 a d (a+i a \tan (c+d x))^2}+\frac {A-i B}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]
-1/8*(I*A+B)*x/a^3+1/6*(-A-I*B)/d/(a+I*a*tan(d*x+c))^3+1/8*(A+3*I*B)/a/d/( a+I*a*tan(d*x+c))^2+1/8*(A-I*B)/d/(a^3+I*a^3*tan(d*x+c))
Time = 1.53 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.35 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {(\cos (3 (c+d x))-i \sin (3 (c+d x))) (3 (A+3 i B) \cos (c+d x)-2 (A+6 i A d x+B (i+6 d x)) \cos (3 (c+d x))+9 i A \sin (c+d x)-3 B \sin (c+d x)+2 i A \sin (3 (c+d x))-2 B \sin (3 (c+d x))+12 A d x \sin (3 (c+d x))-12 i B d x \sin (3 (c+d x)))}{96 a^3 d} \]
((Cos[3*(c + d*x)] - I*Sin[3*(c + d*x)])*(3*(A + (3*I)*B)*Cos[c + d*x] - 2 *(A + (6*I)*A*d*x + B*(I + 6*d*x))*Cos[3*(c + d*x)] + (9*I)*A*Sin[c + d*x] - 3*B*Sin[c + d*x] + (2*I)*A*Sin[3*(c + d*x)] - 2*B*Sin[3*(c + d*x)] + 12 *A*d*x*Sin[3*(c + d*x)] - (12*I)*B*d*x*Sin[3*(c + d*x)]))/(96*a^3*d)
Time = 0.47 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3042, 4073, 3042, 4009, 3042, 3960, 24}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3}dx\) |
\(\Big \downarrow \) 4073 |
\(\displaystyle -\frac {i \int \frac {a (A+i B)+2 a B \tan (c+d x)}{(i \tan (c+d x) a+a)^2}dx}{2 a^2}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {i \int \frac {a (A+i B)+2 a B \tan (c+d x)}{(i \tan (c+d x) a+a)^2}dx}{2 a^2}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 4009 |
\(\displaystyle -\frac {i \left (\frac {1}{2} (A-i B) \int \frac {1}{i \tan (c+d x) a+a}dx+\frac {a (-3 B+i A)}{4 d (a+i a \tan (c+d x))^2}\right )}{2 a^2}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {i \left (\frac {1}{2} (A-i B) \int \frac {1}{i \tan (c+d x) a+a}dx+\frac {a (-3 B+i A)}{4 d (a+i a \tan (c+d x))^2}\right )}{2 a^2}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle -\frac {i \left (\frac {1}{2} (A-i B) \left (\frac {\int 1dx}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))}\right )+\frac {a (-3 B+i A)}{4 d (a+i a \tan (c+d x))^2}\right )}{2 a^2}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle -\frac {i \left (\frac {a (-3 B+i A)}{4 d (a+i a \tan (c+d x))^2}+\frac {1}{2} (A-i B) \left (\frac {x}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))}\right )\right )}{2 a^2}-\frac {A+i B}{6 d (a+i a \tan (c+d x))^3}\) |
-1/6*(A + I*B)/(d*(a + I*a*Tan[c + d*x])^3) - ((I/2)*((a*(I*A - 3*B))/(4*d *(a + I*a*Tan[c + d*x])^2) + ((A - I*B)*(x/(2*a) + (I/2)/(d*(a + I*a*Tan[c + d*x]))))/2))/a^2
3.1.54.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[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]
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] + Simp[(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]
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] + Simp[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]
Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.16
method | result | size |
risch | \(-\frac {x B}{8 a^{3}}-\frac {i x A}{8 a^{3}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B}{16 a^{3} d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} A}{16 a^{3} d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )} B}{32 a^{3} d}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )} A}{32 a^{3} d}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )} B}{48 a^{3} d}-\frac {{\mathrm e}^{-6 i \left (d x +c \right )} A}{48 a^{3} d}\) | \(128\) |
derivativedivides | \(-\frac {A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {3 i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {i A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}\) | \(158\) |
default | \(-\frac {A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {3 i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {i A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}\) | \(158\) |
-1/8*x/a^3*B-1/8*I*x/a^3*A+1/16*I/a^3/d*exp(-2*I*(d*x+c))*B+1/16/a^3/d*exp (-2*I*(d*x+c))*A+1/32*I/a^3/d*exp(-4*I*(d*x+c))*B-1/32/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
Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.67 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\left (12 \, {\left (i \, A + B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} - 6 \, {\left (A + i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, A + 2 i \, B\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]
-1/96*(12*(I*A + B)*d*x*e^(6*I*d*x + 6*I*c) - 6*(A + I*B)*e^(4*I*d*x + 4*I *c) + 3*(A - 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)
Time = 0.29 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.36 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (\left (- 512 A a^{6} d^{2} e^{6 i c} - 512 i B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (- 768 A a^{6} d^{2} e^{8 i c} + 768 i B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (1536 A a^{6} d^{2} e^{10 i c} + 1536 i B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {- i A - B}{8 a^{3}} + \frac {\left (- i A e^{6 i c} - i A e^{4 i c} + i A e^{2 i c} + i A - B e^{6 i c} + B e^{4 i c} + B e^{2 i c} - B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- i A - B\right )}{8 a^{3}} \]
Piecewise((((-512*A*a**6*d**2*exp(6*I*c) - 512*I*B*a**6*d**2*exp(6*I*c))*e xp(-6*I*d*x) + (-768*A*a**6*d**2*exp(8*I*c) + 768*I*B*a**6*d**2*exp(8*I*c) )*exp(-4*I*d*x) + (1536*A*a**6*d**2*exp(10*I*c) + 1536*I*B*a**6*d**2*exp(1 0*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 - B)/(8*a**3) + (-I*A*exp(6*I*c) - I*A*exp(4*I*c) + I*A*exp(2*I*c) + I*A - B*exp(6*I*c) + B*exp(4*I*c) + B*exp(2*I*c) - B)*ex p(-6*I*c)/(8*a**3)), True)) + x*(-I*A - B)/(8*a**3)
Exception generated. \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.56 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.15 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {6 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} - \frac {6 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {11 \, A \tan \left (d x + c\right )^{3} - 11 i \, B \tan \left (d x + c\right )^{3} - 45 i \, A \tan \left (d x + c\right )^{2} - 45 \, B \tan \left (d x + c\right )^{2} - 69 \, A \tan \left (d x + c\right ) + 21 i \, B \tan \left (d x + c\right ) + 19 i \, A + 3 \, B}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]
1/96*(6*(A - I*B)*log(tan(d*x + c) + I)/a^3 - 6*(A - I*B)*log(tan(d*x + c) - I)/a^3 + (11*A*tan(d*x + c)^3 - 11*I*B*tan(d*x + c)^3 - 45*I*A*tan(d*x + c)^2 - 45*B*tan(d*x + c)^2 - 69*A*tan(d*x + c) + 21*I*B*tan(d*x + c) + 1 9*I*A + 3*B)/(a^3*(tan(d*x + c) - I)^3))/d
Time = 7.12 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.34 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {A}{12\,a^3}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {A}{8\,a^3}-\frac {B\,1{}\mathrm {i}}{8\,a^3}\right )+\frac {B\,1{}\mathrm {i}}{12\,a^3}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{8\,a^3}+\frac {A\,3{}\mathrm {i}}{8\,a^3}\right )}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{16\,a^3\,d} \]