Integrand size = 34, antiderivative size = 103 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {(A+3 i B) x}{4 a^2}+\frac {B \log (\cos (c+d x))}{a^2 d}+\frac {i A-3 B}{4 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
-1/4*(A+3*I*B)*x/a^2+B*ln(cos(d*x+c))/a^2/d+1/4*(I*A-3*B)/a^2/d/(1+I*tan(d *x+c))+1/4*(I*A-B)*tan(d*x+c)^2/d/(a+I*a*tan(d*x+c))^2
Time = 0.85 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.53 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {\sec ^2(c+d x) (-2 i A+4 B+\cos (2 (c+d x)) (2 i A-4 B+(-i A+7 B) \log (i-\tan (c+d x))+(i A+B) \log (i+\tan (c+d x)))+(-A-3 i B+(A+7 i B) \log (i-\tan (c+d x))-(A-i B) \log (i+\tan (c+d x))) \sin (2 (c+d x)))}{8 a^2 d (-i+\tan (c+d x))^2} \]
(Sec[c + d*x]^2*((-2*I)*A + 4*B + Cos[2*(c + d*x)]*((2*I)*A - 4*B + ((-I)* A + 7*B)*Log[I - Tan[c + d*x]] + (I*A + B)*Log[I + Tan[c + d*x]]) + (-A - (3*I)*B + (A + (7*I)*B)*Log[I - Tan[c + d*x]] - (A - I*B)*Log[I + Tan[c + d*x]])*Sin[2*(c + d*x)]))/(8*a^2*d*(-I + Tan[c + d*x])^2)
Time = 0.60 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {3042, 4078, 27, 3042, 4072, 25, 27, 3042, 3956, 4009, 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.
\(\displaystyle \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^2 (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2}dx\) |
\(\Big \downarrow \) 4078 |
\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {2 \tan (c+d x) (a (i A-B)+2 i a B \tan (c+d x))}{i \tan (c+d x) a+a}dx}{4 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan (c+d x) (a (i A-B)+2 i a B \tan (c+d x))}{i \tan (c+d x) a+a}dx}{2 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan (c+d x) (a (i A-B)+2 i a B \tan (c+d x))}{i \tan (c+d x) a+a}dx}{2 a^2}\) |
\(\Big \downarrow \) 4072 |
\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {2 B \int \tan (c+d x)dx-\frac {i \int -\frac {a^2 (A+3 i B) \tan (c+d x)}{i \tan (c+d x) a+a}dx}{a}}{2 a^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {2 B \int \tan (c+d x)dx+\frac {i \int \frac {a^2 (A+3 i B) \tan (c+d x)}{i \tan (c+d x) a+a}dx}{a}}{2 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {2 B \int \tan (c+d x)dx+i a (A+3 i B) \int \frac {\tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {2 B \int \tan (c+d x)dx+i a (A+3 i B) \int \frac {\tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {-\frac {2 B \log (\cos (c+d x))}{d}+i a (A+3 i B) \int \frac {\tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}\) |
\(\Big \downarrow \) 4009 |
\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {-\frac {2 B \log (\cos (c+d x))}{d}+i a (A+3 i B) \left (-\frac {i \int 1dx}{2 a}-\frac {1}{2 d (a+i a \tan (c+d x))}\right )}{2 a^2}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {(-B+i A) \tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {-\frac {2 B \log (\cos (c+d x))}{d}+i a (A+3 i B) \left (-\frac {1}{2 d (a+i a \tan (c+d x))}-\frac {i x}{2 a}\right )}{2 a^2}\) |
((I*A - B)*Tan[c + d*x]^2)/(4*d*(a + I*a*Tan[c + d*x])^2) - ((-2*B*Log[Cos [c + d*x]])/d + I*a*(A + (3*I)*B)*(((-1/2*I)*x)/a - 1/(2*d*(a + I*a*Tan[c + d*x]))))/(2*a^2)
3.1.45.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
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_)])*((c_.) + (d_.)*tan[(e_.) + (f_ .)*(x_)]))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*(d/ b) Int[Tan[e + f*x], x], x] + Simp[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]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-(A*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), x] + Simp[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]
Time = 0.08 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.18
method | result | size |
risch | \(-\frac {7 i x B}{4 a^{2}}-\frac {x A}{4 a^{2}}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} B}{2 a^{2} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} A}{4 a^{2} d}+\frac {{\mathrm e}^{-4 i \left (d x +c \right )} B}{16 a^{2} d}-\frac {i {\mathrm e}^{-4 i \left (d x +c \right )} A}{16 a^{2} d}-\frac {2 i B c}{a^{2} d}+\frac {B \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{2} d}\) | \(122\) |
derivativedivides | \(\frac {i A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {A \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}-\frac {3 i B \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {5 i B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}+\frac {3 A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}\) | \(137\) |
default | \(\frac {i A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {A \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}-\frac {3 i B \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {5 i B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}+\frac {3 A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}\) | \(137\) |
-7/4*I*x/a^2*B-1/4*x/a^2*A-1/2/a^2/d*exp(-2*I*(d*x+c))*B+1/4*I/a^2/d*exp(- 2*I*(d*x+c))*A+1/16/a^2/d*exp(-4*I*(d*x+c))*B-1/16*I/a^2/d*exp(-4*I*(d*x+c ))*A-2*I*B/a^2/d*c+B/a^2/d*ln(exp(2*I*(d*x+c))+1)
Time = 0.25 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {{\left (4 \, {\left (A + 7 i \, B\right )} d x e^{\left (4 i \, d x + 4 i \, c\right )} - 16 \, B e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 4 \, {\left (-i \, A + 2 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - B\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \]
-1/16*(4*(A + 7*I*B)*d*x*e^(4*I*d*x + 4*I*c) - 16*B*e^(4*I*d*x + 4*I*c)*lo g(e^(2*I*d*x + 2*I*c) + 1) + 4*(-I*A + 2*B)*e^(2*I*d*x + 2*I*c) + I*A - B) *e^(-4*I*d*x - 4*I*c)/(a^2*d)
Time = 0.31 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.17 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {B \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} + \begin {cases} \frac {\left (\left (- 4 i A a^{2} d e^{2 i c} + 4 B a^{2} d e^{2 i c}\right ) e^{- 4 i d x} + \left (16 i A a^{2} d e^{4 i c} - 32 B a^{2} d e^{4 i c}\right ) e^{- 2 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (- \frac {- A - 7 i B}{4 a^{2}} + \frac {\left (- A e^{4 i c} + 2 A e^{2 i c} - A - 7 i B e^{4 i c} + 4 i B e^{2 i c} - i B\right ) e^{- 4 i c}}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- A - 7 i B\right )}{4 a^{2}} \]
B*log(exp(2*I*d*x) + exp(-2*I*c))/(a**2*d) + Piecewise((((-4*I*A*a**2*d*ex p(2*I*c) + 4*B*a**2*d*exp(2*I*c))*exp(-4*I*d*x) + (16*I*A*a**2*d*exp(4*I*c ) - 32*B*a**2*d*exp(4*I*c))*exp(-2*I*d*x))*exp(-6*I*c)/(64*a**4*d**2), Ne( a**4*d**2*exp(6*I*c), 0)), (x*(-(-A - 7*I*B)/(4*a**2) + (-A*exp(4*I*c) + 2 *A*exp(2*I*c) - A - 7*I*B*exp(4*I*c) + 4*I*B*exp(2*I*c) - I*B)*exp(-4*I*c) /(4*a**2)), True)) + x*(-A - 7*I*B)/(4*a**2)
Exception generated. \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.55 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.04 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (i \, A + B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {2 \, {\left (-i \, A + 7 \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac {3 i \, A \tan \left (d x + c\right )^{2} - 21 \, B \tan \left (d x + c\right )^{2} - 6 \, A \tan \left (d x + c\right ) + 22 i \, B \tan \left (d x + c\right ) + 5 i \, A + 5 \, B}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]
-1/16*(2*(I*A + B)*log(tan(d*x + c) + I)/a^2 + 2*(-I*A + 7*B)*log(tan(d*x + c) - I)/a^2 + (3*I*A*tan(d*x + c)^2 - 21*B*tan(d*x + c)^2 - 6*A*tan(d*x + c) + 22*I*B*tan(d*x + c) + 5*I*A + 5*B)/(a^2*(tan(d*x + c) - I)^2))/d
Time = 7.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.11 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {A}{2\,a^2}+\frac {B\,1{}\mathrm {i}}{a^2}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {5\,B}{4\,a^2}+\frac {A\,3{}\mathrm {i}}{4\,a^2}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{8\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-7\,B+A\,1{}\mathrm {i}\right )}{8\,a^2\,d} \]