Integrand size = 23, antiderivative size = 87 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\left (a^2 c-b^2 c-2 a b d\right ) x-\frac {\left (2 a b c+a^2 d-b^2 d\right ) \log (\cos (e+f x))}{f}+\frac {b (b c+a d) \tan (e+f x)}{f}+\frac {d (a+b \tan (e+f x))^2}{2 f} \] Output:
(a^2*c-2*a*b*d-b^2*c)*x-(a^2*d+2*a*b*c-b^2*d)*ln(cos(f*x+e))/f+b*(a*d+b*c) *tan(f*x+e)/f+1/2*d*(a+b*tan(f*x+e))^2/f
Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.10 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\frac {(a+i b)^2 (-i c+d) \log (i-\tan (e+f x))+(a-i b)^2 (i c+d) \log (i+\tan (e+f x))+2 b (b c+2 a d) \tan (e+f x)+b^2 d \tan ^2(e+f x)}{2 f} \] Input:
Integrate[(a + b*Tan[e + f*x])^2*(c + d*Tan[e + f*x]),x]
Output:
((a + I*b)^2*((-I)*c + d)*Log[I - Tan[e + f*x]] + (a - I*b)^2*(I*c + d)*Lo g[I + Tan[e + f*x]] + 2*b*(b*c + 2*a*d)*Tan[e + f*x] + b^2*d*Tan[e + f*x]^ 2)/(2*f)
Time = 0.41 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4011, 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.
\(\displaystyle \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x))dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int (a+b \tan (e+f x)) (a c-b d+(b c+a d) \tan (e+f x))dx+\frac {d (a+b \tan (e+f x))^2}{2 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \tan (e+f x)) (a c-b d+(b c+a d) \tan (e+f x))dx+\frac {d (a+b \tan (e+f x))^2}{2 f}\) |
\(\Big \downarrow \) 4008 |
\(\displaystyle \left (a^2 d+2 a b c-b^2 d\right ) \int \tan (e+f x)dx+x \left (a^2 c-2 a b d-b^2 c\right )+\frac {b (a d+b c) \tan (e+f x)}{f}+\frac {d (a+b \tan (e+f x))^2}{2 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \left (a^2 d+2 a b c-b^2 d\right ) \int \tan (e+f x)dx+x \left (a^2 c-2 a b d-b^2 c\right )+\frac {b (a d+b c) \tan (e+f x)}{f}+\frac {d (a+b \tan (e+f x))^2}{2 f}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -\frac {\left (a^2 d+2 a b c-b^2 d\right ) \log (\cos (e+f x))}{f}+x \left (a^2 c-2 a b d-b^2 c\right )+\frac {b (a d+b c) \tan (e+f x)}{f}+\frac {d (a+b \tan (e+f x))^2}{2 f}\) |
Input:
Int[(a + b*Tan[e + f*x])^2*(c + d*Tan[e + f*x]),x]
Output:
(a^2*c - b^2*c - 2*a*b*d)*x - ((2*a*b*c + a^2*d - b^2*d)*Log[Cos[e + f*x]] )/f + (b*(b*c + a*d)*Tan[e + f*x])/f + (d*(a + b*Tan[e + f*x])^2)/(2*f)
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_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03
method | result | size |
norman | \(\left (a^{2} c -2 a b d -b^{2} c \right ) x +\frac {b \left (2 a d +b c \right ) \tan \left (f x +e \right )}{f}+\frac {b^{2} d \tan \left (f x +e \right )^{2}}{2 f}+\frac {\left (a^{2} d +2 a b c -b^{2} d \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f}\) | \(90\) |
derivativedivides | \(\frac {\frac {b^{2} d \tan \left (f x +e \right )^{2}}{2}+2 \tan \left (f x +e \right ) a b d +\tan \left (f x +e \right ) b^{2} c +\frac {\left (a^{2} d +2 a b c -b^{2} d \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}+\left (a^{2} c -2 a b d -b^{2} c \right ) \arctan \left (\tan \left (f x +e \right )\right )}{f}\) | \(97\) |
default | \(\frac {\frac {b^{2} d \tan \left (f x +e \right )^{2}}{2}+2 \tan \left (f x +e \right ) a b d +\tan \left (f x +e \right ) b^{2} c +\frac {\left (a^{2} d +2 a b c -b^{2} d \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}+\left (a^{2} c -2 a b d -b^{2} c \right ) \arctan \left (\tan \left (f x +e \right )\right )}{f}\) | \(97\) |
parts | \(a^{2} c x +\frac {\left (2 a b d +b^{2} c \right ) \left (\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {\left (a^{2} d +2 a b c \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f}+\frac {b^{2} d \left (\frac {\tan \left (f x +e \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}\right )}{f}\) | \(98\) |
parallelrisch | \(\frac {2 a^{2} c f x -4 a b d f x -2 b^{2} c f x +b^{2} d \tan \left (f x +e \right )^{2}+\ln \left (1+\tan \left (f x +e \right )^{2}\right ) a^{2} d +2 \ln \left (1+\tan \left (f x +e \right )^{2}\right ) a b c -\ln \left (1+\tan \left (f x +e \right )^{2}\right ) b^{2} d +4 \tan \left (f x +e \right ) a b d +2 \tan \left (f x +e \right ) b^{2} c}{2 f}\) | \(115\) |
risch | \(\frac {2 i b \left (2 a d \,{\mathrm e}^{2 i \left (f x +e \right )}+b c \,{\mathrm e}^{2 i \left (f x +e \right )}-i b d \,{\mathrm e}^{2 i \left (f x +e \right )}+2 a d +b c \right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+2 i a b c x +\frac {4 i a b c e}{f}+a^{2} c x -2 a b d x -b^{2} c x +i a^{2} d x +\frac {2 i a^{2} d e}{f}-\frac {2 i b^{2} d e}{f}-i b^{2} d x -\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a^{2} d}{f}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a b c}{f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) b^{2} d}{f}\) | \(204\) |
Input:
int((a+b*tan(f*x+e))^2*(c+d*tan(f*x+e)),x,method=_RETURNVERBOSE)
Output:
(a^2*c-2*a*b*d-b^2*c)*x+b*(2*a*d+b*c)/f*tan(f*x+e)+1/2*b^2*d/f*tan(f*x+e)^ 2+1/2*(a^2*d+2*a*b*c-b^2*d)/f*ln(1+tan(f*x+e)^2)
Time = 0.09 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.06 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\frac {b^{2} d \tan \left (f x + e\right )^{2} - 2 \, {\left (2 \, a b d - {\left (a^{2} - b^{2}\right )} c\right )} f x - {\left (2 \, a b c + {\left (a^{2} - b^{2}\right )} d\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (b^{2} c + 2 \, a b d\right )} \tan \left (f x + e\right )}{2 \, f} \] Input:
integrate((a+b*tan(f*x+e))^2*(c+d*tan(f*x+e)),x, algorithm="fricas")
Output:
1/2*(b^2*d*tan(f*x + e)^2 - 2*(2*a*b*d - (a^2 - b^2)*c)*f*x - (2*a*b*c + ( a^2 - b^2)*d)*log(1/(tan(f*x + e)^2 + 1)) + 2*(b^2*c + 2*a*b*d)*tan(f*x + e))/f
Time = 0.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.64 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\begin {cases} a^{2} c x + \frac {a^{2} d \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {a b c \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{f} - 2 a b d x + \frac {2 a b d \tan {\left (e + f x \right )}}{f} - b^{2} c x + \frac {b^{2} c \tan {\left (e + f x \right )}}{f} - \frac {b^{2} d \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {b^{2} d \tan ^{2}{\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \left (a + b \tan {\left (e \right )}\right )^{2} \left (c + d \tan {\left (e \right )}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((a+b*tan(f*x+e))**2*(c+d*tan(f*x+e)),x)
Output:
Piecewise((a**2*c*x + a**2*d*log(tan(e + f*x)**2 + 1)/(2*f) + a*b*c*log(ta n(e + f*x)**2 + 1)/f - 2*a*b*d*x + 2*a*b*d*tan(e + f*x)/f - b**2*c*x + b** 2*c*tan(e + f*x)/f - b**2*d*log(tan(e + f*x)**2 + 1)/(2*f) + b**2*d*tan(e + f*x)**2/(2*f), Ne(f, 0)), (x*(a + b*tan(e))**2*(c + d*tan(e)), True))
Time = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.06 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\frac {b^{2} d \tan \left (f x + e\right )^{2} - 2 \, {\left (2 \, a b d - {\left (a^{2} - b^{2}\right )} c\right )} {\left (f x + e\right )} + {\left (2 \, a b c + {\left (a^{2} - b^{2}\right )} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right ) + 2 \, {\left (b^{2} c + 2 \, a b d\right )} \tan \left (f x + e\right )}{2 \, f} \] Input:
integrate((a+b*tan(f*x+e))^2*(c+d*tan(f*x+e)),x, algorithm="maxima")
Output:
1/2*(b^2*d*tan(f*x + e)^2 - 2*(2*a*b*d - (a^2 - b^2)*c)*(f*x + e) + (2*a*b *c + (a^2 - b^2)*d)*log(tan(f*x + e)^2 + 1) + 2*(b^2*c + 2*a*b*d)*tan(f*x + e))/f
Time = 0.22 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.21 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\frac {{\left (a^{2} c - b^{2} c - 2 \, a b d\right )} {\left (f x + e\right )}}{f} + \frac {{\left (2 \, a b c + a^{2} d - b^{2} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{2 \, f} + \frac {b^{2} d f \tan \left (f x + e\right )^{2} + 2 \, b^{2} c f \tan \left (f x + e\right ) + 4 \, a b d f \tan \left (f x + e\right )}{2 \, f^{2}} \] Input:
integrate((a+b*tan(f*x+e))^2*(c+d*tan(f*x+e)),x, algorithm="giac")
Output:
(a^2*c - b^2*c - 2*a*b*d)*(f*x + e)/f + 1/2*(2*a*b*c + a^2*d - b^2*d)*log( tan(f*x + e)^2 + 1)/f + 1/2*(b^2*d*f*tan(f*x + e)^2 + 2*b^2*c*f*tan(f*x + e) + 4*a*b*d*f*tan(f*x + e))/f^2
Time = 2.22 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (c\,b^2+2\,a\,d\,b\right )}{f}-x\,\left (-c\,a^2+2\,d\,a\,b+c\,b^2\right )+\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )\,\left (\frac {d\,a^2}{2}+c\,a\,b-\frac {d\,b^2}{2}\right )}{f}+\frac {b^2\,d\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,f} \] Input:
int((a + b*tan(e + f*x))^2*(c + d*tan(e + f*x)),x)
Output:
(tan(e + f*x)*(b^2*c + 2*a*b*d))/f - x*(b^2*c - a^2*c + 2*a*b*d) + (log(ta n(e + f*x)^2 + 1)*((a^2*d)/2 - (b^2*d)/2 + a*b*c))/f + (b^2*d*tan(e + f*x) ^2)/(2*f)
Time = 0.21 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.31 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\frac {\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) a^{2} d +2 \,\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) a b c -\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) b^{2} d +\tan \left (f x +e \right )^{2} b^{2} d +4 \tan \left (f x +e \right ) a b d +2 \tan \left (f x +e \right ) b^{2} c +2 a^{2} c f x -4 a b d f x -2 b^{2} c f x}{2 f} \] Input:
int((a+b*tan(f*x+e))^2*(c+d*tan(f*x+e)),x)
Output:
(log(tan(e + f*x)**2 + 1)*a**2*d + 2*log(tan(e + f*x)**2 + 1)*a*b*c - log( tan(e + f*x)**2 + 1)*b**2*d + tan(e + f*x)**2*b**2*d + 4*tan(e + f*x)*a*b* d + 2*tan(e + f*x)*b**2*c + 2*a**2*c*f*x - 4*a*b*d*f*x - 2*b**2*c*f*x)/(2* f)