Integrand size = 37, antiderivative size = 72 \[ \int (a+b \tan (d+e x)) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right ) \, dx=-a \left (a^2+b^2\right ) x-\frac {b \left (a^2+b^2\right ) \log (\cos (d+e x))}{e}+\frac {2 a b^2 \tan (d+e x)}{e}+\frac {a^2 (a+b \tan (d+e x))^2}{2 b e} \] Output:
-a*(a^2+b^2)*x-b*(a^2+b^2)*ln(cos(e*x+d))/e+2*a*b^2*tan(e*x+d)/e+1/2*a^2*( a+b*tan(e*x+d))^2/b/e
Result contains complex when optimal does not.
Time = 0.24 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.22 \[ \int (a+b \tan (d+e x)) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right ) \, dx=\frac {\left (a^2+b^2\right ) ((i a+b) \log (i-\tan (d+e x))+(-i a+b) \log (i+\tan (d+e x)))+2 a \left (a^2+2 b^2\right ) \tan (d+e x)+a^2 b \tan ^2(d+e x)}{2 e} \] Input:
Integrate[(a + b*Tan[d + e*x])*(b^2 + 2*a*b*Tan[d + e*x] + a^2*Tan[d + e*x ]^2),x]
Output:
((a^2 + b^2)*((I*a + b)*Log[I - Tan[d + e*x]] + ((-I)*a + b)*Log[I + Tan[d + e*x]]) + 2*a*(a^2 + 2*b^2)*Tan[d + e*x] + a^2*b*Tan[d + e*x]^2)/(2*e)
Time = 0.41 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {3042, 4113, 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.
\(\displaystyle \int (a+b \tan (d+e x)) \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \tan (d+e x)) \left (a^2 \tan (d+e x)^2+2 a b \tan (d+e x)+b^2\right )dx\) |
\(\Big \downarrow \) 4113 |
\(\displaystyle \int (a+b \tan (d+e x)) \left (-a^2+2 b \tan (d+e x) a+b^2\right )dx+\frac {a^2 (a+b \tan (d+e x))^2}{2 b e}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \tan (d+e x)) \left (-a^2+2 b \tan (d+e x) a+b^2\right )dx+\frac {a^2 (a+b \tan (d+e x))^2}{2 b e}\) |
\(\Big \downarrow \) 4008 |
\(\displaystyle b \left (a^2+b^2\right ) \int \tan (d+e x)dx-a x \left (a^2+b^2\right )+\frac {a^2 (a+b \tan (d+e x))^2}{2 b e}+\frac {2 a b^2 \tan (d+e x)}{e}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (a^2+b^2\right ) \int \tan (d+e x)dx-a x \left (a^2+b^2\right )+\frac {a^2 (a+b \tan (d+e x))^2}{2 b e}+\frac {2 a b^2 \tan (d+e x)}{e}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -\frac {b \left (a^2+b^2\right ) \log (\cos (d+e x))}{e}-a x \left (a^2+b^2\right )+\frac {a^2 (a+b \tan (d+e x))^2}{2 b e}+\frac {2 a b^2 \tan (d+e x)}{e}\) |
Input:
Int[(a + b*Tan[d + e*x])*(b^2 + 2*a*b*Tan[d + e*x] + a^2*Tan[d + e*x]^2),x ]
Output:
-(a*(a^2 + b^2)*x) - (b*(a^2 + b^2)*Log[Cos[d + e*x]])/e + (2*a*b^2*Tan[d + e*x])/e + (a^2*(a + b*Tan[d + e*x])^2)/(2*b*e)
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_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && !LeQ[m, -1]
Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.07
method | result | size |
norman | \(\left (-a^{3}-a \,b^{2}\right ) x +\frac {a \left (a^{2}+2 b^{2}\right ) \tan \left (e x +d \right )}{e}+\frac {a^{2} b \tan \left (e x +d \right )^{2}}{2 e}+\frac {b \left (a^{2}+b^{2}\right ) \ln \left (1+\tan \left (e x +d \right )^{2}\right )}{2 e}\) | \(77\) |
derivativedivides | \(\frac {\frac {\tan \left (e x +d \right )^{2} a^{2} b}{2}+\tan \left (e x +d \right ) a^{3}+2 \tan \left (e x +d \right ) a \,b^{2}+\frac {\left (a^{2} b +b^{3}\right ) \ln \left (1+\tan \left (e x +d \right )^{2}\right )}{2}+\left (-a^{3}-a \,b^{2}\right ) \arctan \left (\tan \left (e x +d \right )\right )}{e}\) | \(84\) |
default | \(\frac {\frac {\tan \left (e x +d \right )^{2} a^{2} b}{2}+\tan \left (e x +d \right ) a^{3}+2 \tan \left (e x +d \right ) a \,b^{2}+\frac {\left (a^{2} b +b^{3}\right ) \ln \left (1+\tan \left (e x +d \right )^{2}\right )}{2}+\left (-a^{3}-a \,b^{2}\right ) \arctan \left (\tan \left (e x +d \right )\right )}{e}\) | \(84\) |
parallelrisch | \(\frac {-2 a^{3} e x -2 a \,b^{2} e x +\tan \left (e x +d \right )^{2} a^{2} b +\ln \left (1+\tan \left (e x +d \right )^{2}\right ) a^{2} b +\ln \left (1+\tan \left (e x +d \right )^{2}\right ) b^{3}+2 \tan \left (e x +d \right ) a^{3}+4 \tan \left (e x +d \right ) a \,b^{2}}{2 e}\) | \(89\) |
parts | \(a \,b^{2} x +\frac {\left (2 a^{2} b +b^{3}\right ) \ln \left (1+\tan \left (e x +d \right )^{2}\right )}{2 e}+\frac {\left (a^{3}+2 a \,b^{2}\right ) \left (\tan \left (e x +d \right )-\arctan \left (\tan \left (e x +d \right )\right )\right )}{e}+\frac {a^{2} b \left (\frac {\tan \left (e x +d \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (e x +d \right )^{2}\right )}{2}\right )}{e}\) | \(96\) |
risch | \(i a^{2} b x +i b^{3} x -a^{3} x -a \,b^{2} x +\frac {2 i b \,a^{2} d}{e}+\frac {2 i b^{3} d}{e}+\frac {2 i a \left ({\mathrm e}^{2 i \left (e x +d \right )} a^{2}+2 b^{2} {\mathrm e}^{2 i \left (e x +d \right )}-i a b \,{\mathrm e}^{2 i \left (e x +d \right )}+a^{2}+2 b^{2}\right )}{e \left ({\mathrm e}^{2 i \left (e x +d \right )}+1\right )^{2}}-\frac {b \ln \left ({\mathrm e}^{2 i \left (e x +d \right )}+1\right ) a^{2}}{e}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (e x +d \right )}+1\right )}{e}\) | \(162\) |
Input:
int((a+b*tan(e*x+d))*(b^2+2*a*b*tan(e*x+d)+a^2*tan(e*x+d)^2),x,method=_RET URNVERBOSE)
Output:
(-a^3-a*b^2)*x+a*(a^2+2*b^2)*tan(e*x+d)/e+1/2*a^2*b/e*tan(e*x+d)^2+1/2*b*( a^2+b^2)/e*ln(1+tan(e*x+d)^2)
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.03 \[ \int (a+b \tan (d+e x)) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right ) \, dx=\frac {a^{2} b \tan \left (e x + d\right )^{2} - 2 \, {\left (a^{3} + a b^{2}\right )} e x - {\left (a^{2} b + b^{3}\right )} \log \left (\frac {1}{\tan \left (e x + d\right )^{2} + 1}\right ) + 2 \, {\left (a^{3} + 2 \, a b^{2}\right )} \tan \left (e x + d\right )}{2 \, e} \] Input:
integrate((a+b*tan(e*x+d))*(b^2+2*a*b*tan(e*x+d)+a^2*tan(e*x+d)^2),x, algo rithm="fricas")
Output:
1/2*(a^2*b*tan(e*x + d)^2 - 2*(a^3 + a*b^2)*e*x - (a^2*b + b^3)*log(1/(tan (e*x + d)^2 + 1)) + 2*(a^3 + 2*a*b^2)*tan(e*x + d))/e
Time = 0.10 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.69 \[ \int (a+b \tan (d+e x)) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right ) \, dx=\begin {cases} - a^{3} x + \frac {a^{3} \tan {\left (d + e x \right )}}{e} + \frac {a^{2} b \log {\left (\tan ^{2}{\left (d + e x \right )} + 1 \right )}}{2 e} + \frac {a^{2} b \tan ^{2}{\left (d + e x \right )}}{2 e} - a b^{2} x + \frac {2 a b^{2} \tan {\left (d + e x \right )}}{e} + \frac {b^{3} \log {\left (\tan ^{2}{\left (d + e x \right )} + 1 \right )}}{2 e} & \text {for}\: e \neq 0 \\x \left (a + b \tan {\left (d \right )}\right ) \left (a^{2} \tan ^{2}{\left (d \right )} + 2 a b \tan {\left (d \right )} + b^{2}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((a+b*tan(e*x+d))*(b**2+2*a*b*tan(e*x+d)+a**2*tan(e*x+d)**2),x)
Output:
Piecewise((-a**3*x + a**3*tan(d + e*x)/e + a**2*b*log(tan(d + e*x)**2 + 1) /(2*e) + a**2*b*tan(d + e*x)**2/(2*e) - a*b**2*x + 2*a*b**2*tan(d + e*x)/e + b**3*log(tan(d + e*x)**2 + 1)/(2*e), Ne(e, 0)), (x*(a + b*tan(d))*(a**2 *tan(d)**2 + 2*a*b*tan(d) + b**2), True))
Time = 0.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.03 \[ \int (a+b \tan (d+e x)) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right ) \, dx=\frac {a^{2} b \tan \left (e x + d\right )^{2} - 2 \, {\left (a^{3} + a b^{2}\right )} {\left (e x + d\right )} + {\left (a^{2} b + b^{3}\right )} \log \left (\tan \left (e x + d\right )^{2} + 1\right ) + 2 \, {\left (a^{3} + 2 \, a b^{2}\right )} \tan \left (e x + d\right )}{2 \, e} \] Input:
integrate((a+b*tan(e*x+d))*(b^2+2*a*b*tan(e*x+d)+a^2*tan(e*x+d)^2),x, algo rithm="maxima")
Output:
1/2*(a^2*b*tan(e*x + d)^2 - 2*(a^3 + a*b^2)*(e*x + d) + (a^2*b + b^3)*log( tan(e*x + d)^2 + 1) + 2*(a^3 + 2*a*b^2)*tan(e*x + d))/e
Time = 0.38 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.25 \[ \int (a+b \tan (d+e x)) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right ) \, dx=-\frac {{\left (a^{3} + a b^{2}\right )} {\left (e x + d\right )}}{e} + \frac {{\left (a^{2} b + b^{3}\right )} \log \left (\tan \left (e x + d\right )^{2} + 1\right )}{2 \, e} + \frac {a^{2} b e \tan \left (e x + d\right )^{2} + 2 \, a^{3} e \tan \left (e x + d\right ) + 4 \, a b^{2} e \tan \left (e x + d\right )}{2 \, e^{2}} \] Input:
integrate((a+b*tan(e*x+d))*(b^2+2*a*b*tan(e*x+d)+a^2*tan(e*x+d)^2),x, algo rithm="giac")
Output:
-(a^3 + a*b^2)*(e*x + d)/e + 1/2*(a^2*b + b^3)*log(tan(e*x + d)^2 + 1)/e + 1/2*(a^2*b*e*tan(e*x + d)^2 + 2*a^3*e*tan(e*x + d) + 4*a*b^2*e*tan(e*x + d))/e^2
Time = 16.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.46 \[ \int (a+b \tan (d+e x)) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right ) \, dx=\frac {\mathrm {tan}\left (d+e\,x\right )\,\left (a^3+2\,a\,b^2\right )}{e}+\frac {\ln \left ({\mathrm {tan}\left (d+e\,x\right )}^2+1\right )\,\left (\frac {a^2\,b}{2}+\frac {b^3}{2}\right )}{e}+\frac {a^2\,b\,{\mathrm {tan}\left (d+e\,x\right )}^2}{2\,e}-\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (d+e\,x\right )\,\left (a^2+b^2\right )}{a^3+a\,b^2}\right )\,\left (a^2+b^2\right )}{e} \] Input:
int((a + b*tan(d + e*x))*(b^2 + a^2*tan(d + e*x)^2 + 2*a*b*tan(d + e*x)),x )
Output:
(tan(d + e*x)*(2*a*b^2 + a^3))/e + (log(tan(d + e*x)^2 + 1)*((a^2*b)/2 + b ^3/2))/e + (a^2*b*tan(d + e*x)^2)/(2*e) - (a*atan((a*tan(d + e*x)*(a^2 + b ^2))/(a*b^2 + a^3))*(a^2 + b^2))/e
Time = 0.15 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.22 \[ \int (a+b \tan (d+e x)) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right ) \, dx=\frac {\mathrm {log}\left (\tan \left (e x +d \right )^{2}+1\right ) a^{2} b +\mathrm {log}\left (\tan \left (e x +d \right )^{2}+1\right ) b^{3}+\tan \left (e x +d \right )^{2} a^{2} b +2 \tan \left (e x +d \right ) a^{3}+4 \tan \left (e x +d \right ) a \,b^{2}-2 a^{3} e x -2 a \,b^{2} e x}{2 e} \] Input:
int((a+b*tan(e*x+d))*(b^2+2*a*b*tan(e*x+d)+a^2*tan(e*x+d)^2),x)
Output:
(log(tan(d + e*x)**2 + 1)*a**2*b + log(tan(d + e*x)**2 + 1)*b**3 + tan(d + e*x)**2*a**2*b + 2*tan(d + e*x)*a**3 + 4*tan(d + e*x)*a*b**2 - 2*a**3*e*x - 2*a*b**2*e*x)/(2*e)