\(\int \tan ^4(e+f x) (a+b \tan ^2(e+f x)) \, dx\) [192]

Optimal result

Integrand size = 21, antiderivative size = 60 \[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=(a-b) x-\frac {(a-b) \tan (e+f x)}{f}+\frac {(a-b) \tan ^3(e+f x)}{3 f}+\frac {b \tan ^5(e+f x)}{5 f} \]

[Out]

(a-b)*x-(a-b)*tan(f*x+e)/f+1/3*(a-b)*tan(f*x+e)^3/f+1/5*b*tan(f*x+e)^5/f

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 60, 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 = {3712, 3554, 8} \[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {(a-b) \tan ^3(e+f x)}{3 f}-\frac {(a-b) \tan (e+f x)}{f}+x (a-b)+\frac {b \tan ^5(e+f x)}{5 f} \]

[In]

Int[Tan[e + f*x]^4*(a + b*Tan[e + f*x]^2),x]

[Out]

(a - b)*x - ((a - b)*Tan[e + f*x])/f + ((a - b)*Tan[e + f*x]^3)/(3*f) + (b*Tan[e + f*x]^5)/(5*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3712

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp
[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[A - C, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[
{a, b, e, f, A, C, m}, x] && NeQ[A*b^2 + a^2*C, 0] &&  !LeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {b \tan ^5(e+f x)}{5 f}+(a-b) \int \tan ^4(e+f x) \, dx \\ & = \frac {(a-b) \tan ^3(e+f x)}{3 f}+\frac {b \tan ^5(e+f x)}{5 f}+(-a+b) \int \tan ^2(e+f x) \, dx \\ & = -\frac {(a-b) \tan (e+f x)}{f}+\frac {(a-b) \tan ^3(e+f x)}{3 f}+\frac {b \tan ^5(e+f x)}{5 f}+(a-b) \int 1 \, dx \\ & = (a-b) x-\frac {(a-b) \tan (e+f x)}{f}+\frac {(a-b) \tan ^3(e+f x)}{3 f}+\frac {b \tan ^5(e+f x)}{5 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.62 \[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {a \arctan (\tan (e+f x))}{f}-\frac {b \arctan (\tan (e+f x))}{f}-\frac {a \tan (e+f x)}{f}+\frac {b \tan (e+f x)}{f}+\frac {a \tan ^3(e+f x)}{3 f}-\frac {b \tan ^3(e+f x)}{3 f}+\frac {b \tan ^5(e+f x)}{5 f} \]

[In]

Integrate[Tan[e + f*x]^4*(a + b*Tan[e + f*x]^2),x]

[Out]

(a*ArcTan[Tan[e + f*x]])/f - (b*ArcTan[Tan[e + f*x]])/f - (a*Tan[e + f*x])/f + (b*Tan[e + f*x])/f + (a*Tan[e +
f*x]^3)/(3*f) - (b*Tan[e + f*x]^3)/(3*f) + (b*Tan[e + f*x]^5)/(5*f)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95

method	result	size
norman	\(\left (a -b \right ) x -\frac {\left (a -b \right ) \tan \left (f x +e \right )}{f}+\frac {\left (a -b \right ) \tan \left (f x +e \right )^{3}}{3 f}+\frac {b \tan \left (f x +e \right )^{5}}{5 f}\)	\(57\)
parallelrisch	\(\frac {3 \tan \left (f x +e \right )^{5} b +5 \tan \left (f x +e \right )^{3} a -5 b \tan \left (f x +e \right )^{3}+15 a f x -15 b f x -15 \tan \left (f x +e \right ) a +15 b \tan \left (f x +e \right )}{15 f}\)	\(68\)
derivativedivides	\(\frac {\frac {\tan \left (f x +e \right )^{5} b}{5}+\frac {\tan \left (f x +e \right )^{3} a}{3}-\frac {b \tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right ) a +b \tan \left (f x +e \right )+\left (a -b \right ) \arctan \left (\tan \left (f x +e \right )\right )}{f}\)	\(69\)
default	\(\frac {\frac {\tan \left (f x +e \right )^{5} b}{5}+\frac {\tan \left (f x +e \right )^{3} a}{3}-\frac {b \tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right ) a +b \tan \left (f x +e \right )+\left (a -b \right ) \arctan \left (\tan \left (f x +e \right )\right )}{f}\)	\(69\)
parts	\(\frac {a \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {b \left (\frac {\tan \left (f x +e \right )^{5}}{5}-\frac {\tan \left (f x +e \right )^{3}}{3}+\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}\)	\(74\)
risch	\(a x -b x -\frac {2 i \left (30 a \,{\mathrm e}^{8 i \left (f x +e \right )}-45 b \,{\mathrm e}^{8 i \left (f x +e \right )}+90 a \,{\mathrm e}^{6 i \left (f x +e \right )}-90 b \,{\mathrm e}^{6 i \left (f x +e \right )}+110 a \,{\mathrm e}^{4 i \left (f x +e \right )}-140 b \,{\mathrm e}^{4 i \left (f x +e \right )}+70 a \,{\mathrm e}^{2 i \left (f x +e \right )}-70 b \,{\mathrm e}^{2 i \left (f x +e \right )}+20 a -23 b \right )}{15 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}\)	\(131\)

[In]

int(tan(f*x+e)^4*(a+b*tan(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

(a-b)*x-(a-b)*tan(f*x+e)/f+1/3*(a-b)*tan(f*x+e)^3/f+1/5*b*tan(f*x+e)^5/f

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90 \[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {3 \, b \tan \left (f x + e\right )^{5} + 5 \, {\left (a - b\right )} \tan \left (f x + e\right )^{3} + 15 \, {\left (a - b\right )} f x - 15 \, {\left (a - b\right )} \tan \left (f x + e\right )}{15 \, f} \]

[In]

integrate(tan(f*x+e)^4*(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

1/15*(3*b*tan(f*x + e)^5 + 5*(a - b)*tan(f*x + e)^3 + 15*(a - b)*f*x - 15*(a - b)*tan(f*x + e))/f

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.37 \[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\begin {cases} a x + \frac {a \tan ^{3}{\left (e + f x \right )}}{3 f} - \frac {a \tan {\left (e + f x \right )}}{f} - b x + \frac {b \tan ^{5}{\left (e + f x \right )}}{5 f} - \frac {b \tan ^{3}{\left (e + f x \right )}}{3 f} + \frac {b \tan {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (a + b \tan ^{2}{\left (e \right )}\right ) \tan ^{4}{\left (e \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(tan(f*x+e)**4*(a+b*tan(f*x+e)**2),x)

[Out]

Piecewise((a*x + a*tan(e + f*x)**3/(3*f) - a*tan(e + f*x)/f - b*x + b*tan(e + f*x)**5/(5*f) - b*tan(e + f*x)**
3/(3*f) + b*tan(e + f*x)/f, Ne(f, 0)), (x*(a + b*tan(e)**2)*tan(e)**4, True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95 \[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {3 \, b \tan \left (f x + e\right )^{5} + 5 \, {\left (a - b\right )} \tan \left (f x + e\right )^{3} + 15 \, {\left (f x + e\right )} {\left (a - b\right )} - 15 \, {\left (a - b\right )} \tan \left (f x + e\right )}{15 \, f} \]

[In]

integrate(tan(f*x+e)^4*(a+b*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

1/15*(3*b*tan(f*x + e)^5 + 5*(a - b)*tan(f*x + e)^3 + 15*(f*x + e)*(a - b) - 15*(a - b)*tan(f*x + e))/f

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 589 vs. \(2 (56) = 112\).

Time = 1.11 (sec) , antiderivative size = 589, normalized size of antiderivative = 9.82 \[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {15 \, a f x \tan \left (f x\right )^{5} \tan \left (e\right )^{5} - 15 \, b f x \tan \left (f x\right )^{5} \tan \left (e\right )^{5} - 75 \, a f x \tan \left (f x\right )^{4} \tan \left (e\right )^{4} + 75 \, b f x \tan \left (f x\right )^{4} \tan \left (e\right )^{4} + 15 \, a \tan \left (f x\right )^{5} \tan \left (e\right )^{4} - 15 \, b \tan \left (f x\right )^{5} \tan \left (e\right )^{4} + 15 \, a \tan \left (f x\right )^{4} \tan \left (e\right )^{5} - 15 \, b \tan \left (f x\right )^{4} \tan \left (e\right )^{5} + 150 \, a f x \tan \left (f x\right )^{3} \tan \left (e\right )^{3} - 150 \, b f x \tan \left (f x\right )^{3} \tan \left (e\right )^{3} - 5 \, a \tan \left (f x\right )^{5} \tan \left (e\right )^{2} + 5 \, b \tan \left (f x\right )^{5} \tan \left (e\right )^{2} - 75 \, a \tan \left (f x\right )^{4} \tan \left (e\right )^{3} + 75 \, b \tan \left (f x\right )^{4} \tan \left (e\right )^{3} - 75 \, a \tan \left (f x\right )^{3} \tan \left (e\right )^{4} + 75 \, b \tan \left (f x\right )^{3} \tan \left (e\right )^{4} - 5 \, a \tan \left (f x\right )^{2} \tan \left (e\right )^{5} + 5 \, b \tan \left (f x\right )^{2} \tan \left (e\right )^{5} - 150 \, a f x \tan \left (f x\right )^{2} \tan \left (e\right )^{2} + 150 \, b f x \tan \left (f x\right )^{2} \tan \left (e\right )^{2} - 3 \, b \tan \left (f x\right )^{5} + 10 \, a \tan \left (f x\right )^{4} \tan \left (e\right ) - 25 \, b \tan \left (f x\right )^{4} \tan \left (e\right ) + 120 \, a \tan \left (f x\right )^{3} \tan \left (e\right )^{2} - 150 \, b \tan \left (f x\right )^{3} \tan \left (e\right )^{2} + 120 \, a \tan \left (f x\right )^{2} \tan \left (e\right )^{3} - 150 \, b \tan \left (f x\right )^{2} \tan \left (e\right )^{3} + 10 \, a \tan \left (f x\right ) \tan \left (e\right )^{4} - 25 \, b \tan \left (f x\right ) \tan \left (e\right )^{4} - 3 \, b \tan \left (e\right )^{5} + 75 \, a f x \tan \left (f x\right ) \tan \left (e\right ) - 75 \, b f x \tan \left (f x\right ) \tan \left (e\right ) - 5 \, a \tan \left (f x\right )^{3} + 5 \, b \tan \left (f x\right )^{3} - 75 \, a \tan \left (f x\right )^{2} \tan \left (e\right ) + 75 \, b \tan \left (f x\right )^{2} \tan \left (e\right ) - 75 \, a \tan \left (f x\right ) \tan \left (e\right )^{2} + 75 \, b \tan \left (f x\right ) \tan \left (e\right )^{2} - 5 \, a \tan \left (e\right )^{3} + 5 \, b \tan \left (e\right )^{3} - 15 \, a f x + 15 \, b f x + 15 \, a \tan \left (f x\right ) - 15 \, b \tan \left (f x\right ) + 15 \, a \tan \left (e\right ) - 15 \, b \tan \left (e\right )}{15 \, {\left (f \tan \left (f x\right )^{5} \tan \left (e\right )^{5} - 5 \, f \tan \left (f x\right )^{4} \tan \left (e\right )^{4} + 10 \, f \tan \left (f x\right )^{3} \tan \left (e\right )^{3} - 10 \, f \tan \left (f x\right )^{2} \tan \left (e\right )^{2} + 5 \, f \tan \left (f x\right ) \tan \left (e\right ) - f\right )}} \]

[In]

integrate(tan(f*x+e)^4*(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

1/15*(15*a*f*x*tan(f*x)^5*tan(e)^5 - 15*b*f*x*tan(f*x)^5*tan(e)^5 - 75*a*f*x*tan(f*x)^4*tan(e)^4 + 75*b*f*x*ta
n(f*x)^4*tan(e)^4 + 15*a*tan(f*x)^5*tan(e)^4 - 15*b*tan(f*x)^5*tan(e)^4 + 15*a*tan(f*x)^4*tan(e)^5 - 15*b*tan(
f*x)^4*tan(e)^5 + 150*a*f*x*tan(f*x)^3*tan(e)^3 - 150*b*f*x*tan(f*x)^3*tan(e)^3 - 5*a*tan(f*x)^5*tan(e)^2 + 5*
b*tan(f*x)^5*tan(e)^2 - 75*a*tan(f*x)^4*tan(e)^3 + 75*b*tan(f*x)^4*tan(e)^3 - 75*a*tan(f*x)^3*tan(e)^4 + 75*b*
tan(f*x)^3*tan(e)^4 - 5*a*tan(f*x)^2*tan(e)^5 + 5*b*tan(f*x)^2*tan(e)^5 - 150*a*f*x*tan(f*x)^2*tan(e)^2 + 150*
b*f*x*tan(f*x)^2*tan(e)^2 - 3*b*tan(f*x)^5 + 10*a*tan(f*x)^4*tan(e) - 25*b*tan(f*x)^4*tan(e) + 120*a*tan(f*x)^
3*tan(e)^2 - 150*b*tan(f*x)^3*tan(e)^2 + 120*a*tan(f*x)^2*tan(e)^3 - 150*b*tan(f*x)^2*tan(e)^3 + 10*a*tan(f*x)
*tan(e)^4 - 25*b*tan(f*x)*tan(e)^4 - 3*b*tan(e)^5 + 75*a*f*x*tan(f*x)*tan(e) - 75*b*f*x*tan(f*x)*tan(e) - 5*a*
tan(f*x)^3 + 5*b*tan(f*x)^3 - 75*a*tan(f*x)^2*tan(e) + 75*b*tan(f*x)^2*tan(e) - 75*a*tan(f*x)*tan(e)^2 + 75*b*
tan(f*x)*tan(e)^2 - 5*a*tan(e)^3 + 5*b*tan(e)^3 - 15*a*f*x + 15*b*f*x + 15*a*tan(f*x) - 15*b*tan(f*x) + 15*a*t
an(e) - 15*b*tan(e))/(f*tan(f*x)^5*tan(e)^5 - 5*f*tan(f*x)^4*tan(e)^4 + 10*f*tan(f*x)^3*tan(e)^3 - 10*f*tan(f*
x)^2*tan(e)^2 + 5*f*tan(f*x)*tan(e) - f)

Mupad [B] (verification not implemented)

Time = 11.59 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.88 \[ \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^5}{5}+\left (\frac {a}{3}-\frac {b}{3}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (b-a\right )\,\mathrm {tan}\left (e+f\,x\right )+f\,x\,\left (a-b\right )}{f} \]

[In]

int(tan(e + f*x)^4*(a + b*tan(e + f*x)^2),x)

[Out]

(tan(e + f*x)^3*(a/3 - b/3) - tan(e + f*x)*(a - b) + (b*tan(e + f*x)^5)/5 + f*x*(a - b))/f