\(\int \tan (e+f x) (a+b \tan ^2(e+f x))^2 \, dx\) [200]

Optimal result

Integrand size = 21, antiderivative size = 62 \[ \int \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=-\frac {(a-b)^2 \log (\cos (e+f x))}{f}+\frac {(a-b) b \tan ^2(e+f x)}{2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^2}{4 f} \]

[Out]

-(a-b)^2*ln(cos(f*x+e))/f+1/2*(a-b)*b*tan(f*x+e)^2/f+1/4*(a+b*tan(f*x+e)^2)^2/f

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3751, 455, 45} \[ \int \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=\frac {b (a-b) \tan ^2(e+f x)}{2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^2}{4 f}-\frac {(a-b)^2 \log (\cos (e+f x))}{f} \]

[In]

Int[Tan[e + f*x]*(a + b*Tan[e + f*x]^2)^2,x]

[Out]

-(((a - b)^2*Log[Cos[e + f*x]])/f) + ((a - b)*b*Tan[e + f*x]^2)/(2*f) + (a + b*Tan[e + f*x]^2)^2/(4*f)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 3751

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[c*(ff/f), Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2
+ ff^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x \left (a+b x^2\right )^2}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {(a+b x)^2}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f} \\ & = \frac {\text {Subst}\left (\int \left ((a-b) b+\frac {(a-b)^2}{1+x}+b (a+b x)\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f} \\ & = -\frac {(a-b)^2 \log (\cos (e+f x))}{f}+\frac {(a-b) b \tan ^2(e+f x)}{2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^2}{4 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.87 \[ \int \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=\frac {-4 (a-b)^2 \log (\cos (e+f x))+2 (2 a-b) b \tan ^2(e+f x)+b^2 \tan ^4(e+f x)}{4 f} \]

[In]

Integrate[Tan[e + f*x]*(a + b*Tan[e + f*x]^2)^2,x]

[Out]

(-4*(a - b)^2*Log[Cos[e + f*x]] + 2*(2*a - b)*b*Tan[e + f*x]^2 + b^2*Tan[e + f*x]^4)/(4*f)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.06

method	result	size
norman	\(\frac {b^{2} \tan \left (f x +e \right )^{4}}{4 f}+\frac {b \left (2 a -b \right ) \tan \left (f x +e \right )^{2}}{2 f}+\frac {\left (a^{2}-2 a b +b^{2}\right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f}\)	\(66\)
derivativedivides	\(\frac {\frac {b^{2} \tan \left (f x +e \right )^{4}}{4}+\tan \left (f x +e \right )^{2} a b -\frac {b^{2} \tan \left (f x +e \right )^{2}}{2}+\frac {\left (a^{2}-2 a b +b^{2}\right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}}{f}\)	\(67\)
default	\(\frac {\frac {b^{2} \tan \left (f x +e \right )^{4}}{4}+\tan \left (f x +e \right )^{2} a b -\frac {b^{2} \tan \left (f x +e \right )^{2}}{2}+\frac {\left (a^{2}-2 a b +b^{2}\right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}}{f}\)	\(67\)
parallelrisch	\(\frac {b^{2} \tan \left (f x +e \right )^{4}+4 \tan \left (f x +e \right )^{2} a b -2 b^{2} \tan \left (f x +e \right )^{2}+2 \ln \left (1+\tan \left (f x +e \right )^{2}\right ) a^{2}-4 \ln \left (1+\tan \left (f x +e \right )^{2}\right ) a b +2 \ln \left (1+\tan \left (f x +e \right )^{2}\right ) b^{2}}{4 f}\)	\(91\)
parts	\(\frac {a^{2} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f}+\frac {b^{2} \left (\frac {\tan \left (f x +e \right )^{4}}{4}-\frac {\tan \left (f x +e \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}\right )}{f}+\frac {a b \tan \left (f x +e \right )^{2}}{f}-\frac {a b \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f}\)	\(94\)
risch	\(i a^{2} x -2 i a b x +i b^{2} x +\frac {2 i a^{2} e}{f}-\frac {4 i a b e}{f}+\frac {2 i b^{2} e}{f}+\frac {4 b \left (a \,{\mathrm e}^{6 i \left (f x +e \right )}-b \,{\mathrm e}^{6 i \left (f x +e \right )}+2 a \,{\mathrm e}^{4 i \left (f x +e \right )}-b \,{\mathrm e}^{4 i \left (f x +e \right )}+a \,{\mathrm e}^{2 i \left (f x +e \right )}-b \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a^{2}}{f}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a b}{f}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) b^{2}}{f}\)	\(200\)

[In]

int(tan(f*x+e)*(a+b*tan(f*x+e)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/4*b^2/f*tan(f*x+e)^4+1/2*b*(2*a-b)/f*tan(f*x+e)^2+1/2*(a^2-2*a*b+b^2)/f*ln(1+tan(f*x+e)^2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.03 \[ \int \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (2 \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \, f} \]

[In]

integrate(tan(f*x+e)*(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/4*(b^2*tan(f*x + e)^4 + 2*(2*a*b - b^2)*tan(f*x + e)^2 - 2*(a^2 - 2*a*b + b^2)*log(1/(tan(f*x + e)^2 + 1)))/
f

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (49) = 98\).

Time = 0.13 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.81 \[ \int \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=\begin {cases} \frac {a^{2} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} - \frac {a b \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{f} + \frac {a b \tan ^{2}{\left (e + f x \right )}}{f} + \frac {b^{2} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {b^{2} \tan ^{4}{\left (e + f x \right )}}{4 f} - \frac {b^{2} \tan ^{2}{\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \left (a + b \tan ^{2}{\left (e \right )}\right )^{2} \tan {\left (e \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(tan(f*x+e)*(a+b*tan(f*x+e)**2)**2,x)

[Out]

Piecewise((a**2*log(tan(e + f*x)**2 + 1)/(2*f) - a*b*log(tan(e + f*x)**2 + 1)/f + a*b*tan(e + f*x)**2/f + b**2
*log(tan(e + f*x)**2 + 1)/(2*f) + b**2*tan(e + f*x)**4/(4*f) - b**2*tan(e + f*x)**2/(2*f), Ne(f, 0)), (x*(a +
b*tan(e)**2)**2*tan(e), True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.32 \[ \int \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=-\frac {2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) + \frac {4 \, {\left (a b - b^{2}\right )} \sin \left (f x + e\right )^{2} - 4 \, a b + 3 \, b^{2}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1}}{4 \, f} \]

[In]

integrate(tan(f*x+e)*(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

-1/4*(2*(a^2 - 2*a*b + b^2)*log(sin(f*x + e)^2 - 1) + (4*(a*b - b^2)*sin(f*x + e)^2 - 4*a*b + 3*b^2)/(sin(f*x
+ e)^4 - 2*sin(f*x + e)^2 + 1))/f

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1250 vs. \(2 (58) = 116\).

Time = 1.62 (sec) , antiderivative size = 1250, normalized size of antiderivative = 20.16 \[ \int \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=\text {Too large to display} \]

[In]

integrate(tan(f*x+e)*(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")

[Out]

-1/4*(2*a^2*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 +
1))*tan(f*x)^4*tan(e)^4 - 4*a*b*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + ta
n(f*x)^2 + tan(e)^2 + 1))*tan(f*x)^4*tan(e)^4 + 2*b^2*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan
(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)^4*tan(e)^4 - 4*a*b*tan(f*x)^4*tan(e)^4 + 3*b^2*tan(f*x
)^4*tan(e)^4 - 8*a^2*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + t
an(e)^2 + 1))*tan(f*x)^3*tan(e)^3 + 16*a*b*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan
(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)^3*tan(e)^3 - 8*b^2*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e)
+ 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)^3*tan(e)^3 - 4*a*b*tan(f*x)^4*tan(e)^2 + 2*b
^2*tan(f*x)^4*tan(e)^2 + 8*a*b*tan(f*x)^3*tan(e)^3 - 8*b^2*tan(f*x)^3*tan(e)^3 - 4*a*b*tan(f*x)^2*tan(e)^4 + 2
*b^2*tan(f*x)^2*tan(e)^4 + 12*a^2*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + t
an(f*x)^2 + tan(e)^2 + 1))*tan(f*x)^2*tan(e)^2 - 24*a*b*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(t
an(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)^2*tan(e)^2 + 12*b^2*log(4*(tan(f*x)^2*tan(e)^2 - 2*t
an(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)^2*tan(e)^2 - b^2*tan(f*x)^4 +
8*a*b*tan(f*x)^3*tan(e) - 8*b^2*tan(f*x)^3*tan(e) - 8*a*b*tan(f*x)^2*tan(e)^2 + 4*b^2*tan(f*x)^2*tan(e)^2 + 8*
a*b*tan(f*x)*tan(e)^3 - 8*b^2*tan(f*x)*tan(e)^3 - b^2*tan(e)^4 - 8*a^2*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)
*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)*tan(e) + 16*a*b*log(4*(tan(f*x)^2*tan
(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)*tan(e) - 8*b^2*log(
4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 + tan(e)^2 + 1))*tan(f*x)*ta
n(e) - 4*a*b*tan(f*x)^2 + 2*b^2*tan(f*x)^2 + 8*a*b*tan(f*x)*tan(e) - 8*b^2*tan(f*x)*tan(e) - 4*a*b*tan(e)^2 +
2*b^2*tan(e)^2 + 2*a^2*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2 +
tan(e)^2 + 1)) - 4*a*b*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2
+ tan(e)^2 + 1)) + 2*b^2*log(4*(tan(f*x)^2*tan(e)^2 - 2*tan(f*x)*tan(e) + 1)/(tan(f*x)^2*tan(e)^2 + tan(f*x)^2
+ tan(e)^2 + 1)) - 4*a*b + 3*b^2)/(f*tan(f*x)^4*tan(e)^4 - 4*f*tan(f*x)^3*tan(e)^3 + 6*f*tan(f*x)^2*tan(e)^2
- 4*f*tan(f*x)*tan(e) + f)

Mupad [B] (verification not implemented)

Time = 11.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.10 \[ \int \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx=\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )\,\left (\frac {a^2}{2}-a\,b+\frac {b^2}{2}\right )}{f}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (a\,b-\frac {b^2}{2}\right )}{f}+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4}{4\,f} \]

[In]

int(tan(e + f*x)*(a + b*tan(e + f*x)^2)^2,x)

[Out]

(log(tan(e + f*x)^2 + 1)*(a^2/2 - a*b + b^2/2))/f + (tan(e + f*x)^2*(a*b - b^2/2))/f + (b^2*tan(e + f*x)^4)/(4
*f)