Integrand size = 34, antiderivative size = 142 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {3 (i A-3 B) x}{4 a^2}+\frac {(A+2 i B) \log (\cos (c+d x))}{a^2 d}+\frac {3 (i A-3 B) \tan (c+d x)}{4 a^2 d}+\frac {(A+2 i B) \tan ^2(c+d x)}{2 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
-3/4*(I*A-3*B)*x/a^2+(A+2*I*B)*ln(cos(d*x+c))/a^2/d+3/4*(I*A-3*B)*tan(d*x+ c)/a^2/d+1/2*(A+2*I*B)*tan(d*x+c)^2/a^2/d/(1+I*tan(d*x+c))+1/4*(I*A-B)*tan (d*x+c)^3/d/(a+I*a*tan(d*x+c))^2
Time = 1.51 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.39 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {(7 A+17 i B) \log (i-\tan (c+d x))+(A-i B) \log (i+\tan (c+d x))+2 (-3 i A+9 B+(7 i A-17 B) \log (i-\tan (c+d x))+(i A+B) \log (i+\tan (c+d x))) \tan (c+d x)+(8 A+28 i B-(7 A+17 i B) \log (i-\tan (c+d x))-(A-i B) \log (i+\tan (c+d x))) \tan ^2(c+d x)-8 B \tan ^3(c+d x)}{8 a^2 d (-i+\tan (c+d x))^2} \]
((7*A + (17*I)*B)*Log[I - Tan[c + d*x]] + (A - I*B)*Log[I + Tan[c + d*x]] + 2*((-3*I)*A + 9*B + ((7*I)*A - 17*B)*Log[I - Tan[c + d*x]] + (I*A + B)*L og[I + Tan[c + d*x]])*Tan[c + d*x] + (8*A + (28*I)*B - (7*A + (17*I)*B)*Lo g[I - Tan[c + d*x]] - (A - I*B)*Log[I + Tan[c + d*x]])*Tan[c + d*x]^2 - 8* B*Tan[c + d*x]^3)/(8*a^2*d*(-I + Tan[c + d*x])^2)
Time = 0.73 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {3042, 4078, 3042, 4078, 27, 3042, 4008, 3042, 3956}
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.
\(\displaystyle \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^3 (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2}dx\) |
\(\Big \downarrow \) 4078 |
\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan ^2(c+d x) (3 a (i A-B)+a (A+5 i B) \tan (c+d x))}{i \tan (c+d x) a+a}dx}{4 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan (c+d x)^2 (3 a (i A-B)+a (A+5 i B) \tan (c+d x))}{i \tan (c+d x) a+a}dx}{4 a^2}\) |
\(\Big \downarrow \) 4078 |
\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {-\frac {\int -2 \tan (c+d x) \left (4 a^2 (A+2 i B)-3 a^2 (i A-3 B) \tan (c+d x)\right )dx}{2 a^2}-\frac {2 (A+2 i B) \tan ^2(c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\frac {\int \tan (c+d x) \left (4 a^2 (A+2 i B)-3 a^2 (i A-3 B) \tan (c+d x)\right )dx}{a^2}-\frac {2 (A+2 i B) \tan ^2(c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\frac {\int \tan (c+d x) \left (4 a^2 (A+2 i B)-3 a^2 (i A-3 B) \tan (c+d x)\right )dx}{a^2}-\frac {2 (A+2 i B) \tan ^2(c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}\) |
\(\Big \downarrow \) 4008 |
\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\frac {4 a^2 (A+2 i B) \int \tan (c+d x)dx-\frac {3 a^2 (-3 B+i A) \tan (c+d x)}{d}+3 a^2 x (-3 B+i A)}{a^2}-\frac {2 (A+2 i B) \tan ^2(c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\frac {4 a^2 (A+2 i B) \int \tan (c+d x)dx-\frac {3 a^2 (-3 B+i A) \tan (c+d x)}{d}+3 a^2 x (-3 B+i A)}{a^2}-\frac {2 (A+2 i B) \tan ^2(c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\frac {-\frac {3 a^2 (-3 B+i A) \tan (c+d x)}{d}-\frac {4 a^2 (A+2 i B) \log (\cos (c+d x))}{d}+3 a^2 x (-3 B+i A)}{a^2}-\frac {2 (A+2 i B) \tan ^2(c+d x)}{d (1+i \tan (c+d x))}}{4 a^2}\) |
((I*A - B)*Tan[c + d*x]^3)/(4*d*(a + I*a*Tan[c + d*x])^2) - ((-2*(A + (2*I )*B)*Tan[c + d*x]^2)/(d*(1 + I*Tan[c + d*x])) + (3*a^2*(I*A - 3*B)*x - (4* a^2*(A + (2*I)*B)*Log[Cos[c + d*x]])/d - (3*a^2*(I*A - 3*B)*Tan[c + d*x])/ d)/a^2)/(4*a^2)
3.1.44.3.1 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_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(a*c - b*d)*x, x] + (Simp[b*d*(Tan[e + f*x]/f), x] + Simp[(b*c + a*d) Int[Tan[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 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 = 0.10 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.22
method | result | size |
derivativedivides | \(-\frac {B \tan \left (d x +c \right )}{d \,a^{2}}-\frac {A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}-\frac {3 i A \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {i B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}+\frac {9 B \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {5 i A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {7 B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}\) | \(173\) |
default | \(-\frac {B \tan \left (d x +c \right )}{d \,a^{2}}-\frac {A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}-\frac {3 i A \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {i B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}+\frac {9 B \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {5 i A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {7 B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}\) | \(173\) |
risch | \(\frac {17 x B}{4 a^{2}}-\frac {7 i x A}{4 a^{2}}-\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} B}{4 a^{2} d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} A}{2 a^{2} d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )} B}{16 a^{2} d}+\frac {{\mathrm e}^{-4 i \left (d x +c \right )} A}{16 a^{2} d}+\frac {4 B c}{a^{2} d}-\frac {2 i A c}{a^{2} d}-\frac {2 i B}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A}{a^{2} d}\) | \(177\) |
-1/d/a^2*B*tan(d*x+c)-1/4/d/a^2/(tan(d*x+c)-I)^2*A-1/4*I/d/a^2/(tan(d*x+c) -I)^2*B-1/2/d/a^2*A*ln(1+tan(d*x+c)^2)-3/4*I/d/a^2*A*arctan(tan(d*x+c))-I/ d/a^2*B*ln(1+tan(d*x+c)^2)+9/4/d/a^2*B*arctan(tan(d*x+c))+5/4*I/d/a^2/(tan (d*x+c)-I)*A-7/4/d/a^2/(tan(d*x+c)-I)*B
Time = 0.26 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {4 \, {\left (7 i \, A - 17 \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} + 4 \, {\left ({\left (7 i \, A - 17 \, B\right )} d x + 2 \, A + 11 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (7 \, A + 11 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 16 \, {\left ({\left (A + 2 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (A + 2 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - A - i \, B}{16 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \]
-1/16*(4*(7*I*A - 17*B)*d*x*e^(6*I*d*x + 6*I*c) + 4*((7*I*A - 17*B)*d*x + 2*A + 11*I*B)*e^(4*I*d*x + 4*I*c) + (7*A + 11*I*B)*e^(2*I*d*x + 2*I*c) - 1 6*((A + 2*I*B)*e^(6*I*d*x + 6*I*c) + (A + 2*I*B)*e^(4*I*d*x + 4*I*c))*log( e^(2*I*d*x + 2*I*c) + 1) - A - I*B)/(a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^( 4*I*d*x + 4*I*c))
Time = 0.44 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.85 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=- \frac {2 i B}{a^{2} d e^{2 i c} e^{2 i d x} + a^{2} d} + \begin {cases} \frac {\left (\left (4 A a^{2} d e^{2 i c} + 4 i B a^{2} d e^{2 i c}\right ) e^{- 4 i d x} + \left (- 32 A a^{2} d e^{4 i c} - 48 i B a^{2} d e^{4 i c}\right ) e^{- 2 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (- \frac {- 7 i A + 17 B}{4 a^{2}} + \frac {\left (- 7 i A e^{4 i c} + 4 i A e^{2 i c} - i A + 17 B e^{4 i c} - 6 B e^{2 i c} + B\right ) e^{- 4 i c}}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- 7 i A + 17 B\right )}{4 a^{2}} + \frac {\left (A + 2 i B\right ) \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} \]
-2*I*B/(a**2*d*exp(2*I*c)*exp(2*I*d*x) + a**2*d) + Piecewise((((4*A*a**2*d *exp(2*I*c) + 4*I*B*a**2*d*exp(2*I*c))*exp(-4*I*d*x) + (-32*A*a**2*d*exp(4 *I*c) - 48*I*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*(-(-7*I*A + 17*B)/(4*a**2) + (-7*I*A*e xp(4*I*c) + 4*I*A*exp(2*I*c) - I*A + 17*B*exp(4*I*c) - 6*B*exp(2*I*c) + B) *exp(-4*I*c)/(4*a**2)), True)) + x*(-7*I*A + 17*B)/(4*a**2) + (A + 2*I*B)* log(exp(2*I*d*x) + exp(-2*I*c))/(a**2*d)
Exception generated. \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.68 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.85 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {2 \, {\left (7 \, A + 17 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac {16 \, B \tan \left (d x + c\right )}{a^{2}} - \frac {21 \, A \tan \left (d x + c\right )^{2} + 51 i \, B \tan \left (d x + c\right )^{2} - 22 i \, A \tan \left (d x + c\right ) + 74 \, B \tan \left (d x + c\right ) - 5 \, A - 27 i \, B}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]
-1/16*(2*(A - I*B)*log(tan(d*x + c) + I)/a^2 + 2*(7*A + 17*I*B)*log(tan(d* x + c) - I)/a^2 + 16*B*tan(d*x + c)/a^2 - (21*A*tan(d*x + c)^2 + 51*I*B*ta n(d*x + c)^2 - 22*I*A*tan(d*x + c) + 74*B*tan(d*x + c) - 5*A - 27*I*B)/(a^ 2*(tan(d*x + c) - I)^2))/d
Time = 7.78 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {\left (A+B\,2{}\mathrm {i}\right )\,1{}\mathrm {i}}{a^2}+\frac {B}{2\,a^2}-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {5\,\left (A+B\,2{}\mathrm {i}\right )}{4\,a^2}-\frac {B\,3{}\mathrm {i}}{4\,a^2}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^2\,d}-\frac {B\,\mathrm {tan}\left (c+d\,x\right )}{a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (7\,A+B\,17{}\mathrm {i}\right )}{8\,a^2\,d} \]