\(\int \frac {\tan ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) [218]

Optimal result

Integrand size = 23, antiderivative size = 63 \[ \int \frac {\tan ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {x}{a-b}-\frac {a^{3/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{(a-b) b^{3/2} f}+\frac {\tan (e+f x)}{b f} \]

[Out]

x/(a-b)-a^(3/2)*arctan(b^(1/2)*tan(f*x+e)/a^(1/2))/(a-b)/b^(3/2)/f+tan(f*x+e)/b/f

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3751, 490, 536, 209, 211} \[ \int \frac {\tan ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {a^{3/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{b^{3/2} f (a-b)}+\frac {x}{a-b}+\frac {\tan (e+f x)}{b f} \]

[In]

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

[Out]

x/(a - b) - (a^(3/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/((a - b)*b^(3/2)*f) + Tan[e + f*x]/(b*f)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 490

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(2*n -
1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*d*(m + n*(p + q) + 1))), x] - Dist[e^(2*n)
/(b*d*(m + n*(p + q) + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m +
n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d
, 0] && IGtQ[n, 0] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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^4}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\tan (e+f x)}{b f}-\frac {\text {Subst}\left (\int \frac {a+(a+b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{b f} \\ & = \frac {\tan (e+f x)}{b f}+\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{(a-b) f}-\frac {a^2 \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{(a-b) b f} \\ & = \frac {x}{a-b}-\frac {a^{3/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{(a-b) b^{3/2} f}+\frac {\tan (e+f x)}{b f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.11 \[ \int \frac {\tan ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {e+f x}{(a-b) f}-\frac {a^{3/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{(a-b) b^{3/2} f}+\frac {\tan (e+f x)}{b f} \]

[In]

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

[Out]

(e + f*x)/((a - b)*f) - (a^(3/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/((a - b)*b^(3/2)*f) + Tan[e + f*x]/(b
*f)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (f x +e \right )}{b}-\frac {a^{2} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{b \left (a -b \right ) \sqrt {a b}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a -b}}{f}\)	\(65\)
default	\(\frac {\frac {\tan \left (f x +e \right )}{b}-\frac {a^{2} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{b \left (a -b \right ) \sqrt {a b}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a -b}}{f}\)	\(65\)
risch	\(\frac {x}{a -b}+\frac {2 i}{f b \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {\sqrt {-a b}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{2 b^{2} \left (a -b \right ) f}-\frac {\sqrt {-a b}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{2 b^{2} \left (a -b \right ) f}\)	\(144\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 220, normalized size of antiderivative = 3.49 \[ \int \frac {\tan ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\left [\frac {4 \, b f x - a \sqrt {-\frac {a}{b}} \log \left (\frac {b^{2} \tan \left (f x + e\right )^{4} - 6 \, a b \tan \left (f x + e\right )^{2} + a^{2} + 4 \, {\left (b^{2} \tan \left (f x + e\right )^{3} - a b \tan \left (f x + e\right )\right )} \sqrt {-\frac {a}{b}}}{b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}}\right ) + 4 \, {\left (a - b\right )} \tan \left (f x + e\right )}{4 \, {\left (a b - b^{2}\right )} f}, \frac {2 \, b f x - a \sqrt {\frac {a}{b}} \arctan \left (\frac {{\left (b \tan \left (f x + e\right )^{2} - a\right )} \sqrt {\frac {a}{b}}}{2 \, a \tan \left (f x + e\right )}\right ) + 2 \, {\left (a - b\right )} \tan \left (f x + e\right )}{2 \, {\left (a b - b^{2}\right )} f}\right ] \]

[In]

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

[Out]

[1/4*(4*b*f*x - a*sqrt(-a/b)*log((b^2*tan(f*x + e)^4 - 6*a*b*tan(f*x + e)^2 + a^2 + 4*(b^2*tan(f*x + e)^3 - a*
b*tan(f*x + e))*sqrt(-a/b))/(b^2*tan(f*x + e)^4 + 2*a*b*tan(f*x + e)^2 + a^2)) + 4*(a - b)*tan(f*x + e))/((a*b
- b^2)*f), 1/2*(2*b*f*x - a*sqrt(a/b)*arctan(1/2*(b*tan(f*x + e)^2 - a)*sqrt(a/b)/(a*tan(f*x + e))) + 2*(a -
b)*tan(f*x + e))/((a*b - b^2)*f)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (48) = 96\).

Time = 3.65 (sec) , antiderivative size = 427, normalized size of antiderivative = 6.78 \[ \int \frac {\tan ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\begin {cases} \tilde {\infty } x \tan ^{2}{\left (e \right )} & \text {for}\: a = 0 \wedge b = 0 \wedge f = 0 \\\frac {x + \frac {\tan ^{3}{\left (e + f x \right )}}{3 f} - \frac {\tan {\left (e + f x \right )}}{f}}{a} & \text {for}\: b = 0 \\\frac {- x + \frac {\tan {\left (e + f x \right )}}{f}}{b} & \text {for}\: a = 0 \\- \frac {3 f x \tan ^{2}{\left (e + f x \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} - \frac {3 f x}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} + \frac {2 \tan ^{3}{\left (e + f x \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} + \frac {3 \tan {\left (e + f x \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} & \text {for}\: a = b \\\frac {x \tan ^{4}{\left (e \right )}}{a + b \tan ^{2}{\left (e \right )}} & \text {for}\: f = 0 \\- \frac {a^{2} \log {\left (- \sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a b^{2} f \sqrt {- \frac {a}{b}} - 2 b^{3} f \sqrt {- \frac {a}{b}}} + \frac {a^{2} \log {\left (\sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a b^{2} f \sqrt {- \frac {a}{b}} - 2 b^{3} f \sqrt {- \frac {a}{b}}} + \frac {2 a b \sqrt {- \frac {a}{b}} \tan {\left (e + f x \right )}}{2 a b^{2} f \sqrt {- \frac {a}{b}} - 2 b^{3} f \sqrt {- \frac {a}{b}}} + \frac {2 b^{2} f x \sqrt {- \frac {a}{b}}}{2 a b^{2} f \sqrt {- \frac {a}{b}} - 2 b^{3} f \sqrt {- \frac {a}{b}}} - \frac {2 b^{2} \sqrt {- \frac {a}{b}} \tan {\left (e + f x \right )}}{2 a b^{2} f \sqrt {- \frac {a}{b}} - 2 b^{3} f \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo*x*tan(e)**2, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), ((x + tan(e + f*x)**3/(3*f) - tan(e + f*x)/f)/a,
Eq(b, 0)), ((-x + tan(e + f*x)/f)/b, Eq(a, 0)), (-3*f*x*tan(e + f*x)**2/(2*b*f*tan(e + f*x)**2 + 2*b*f) - 3*f*
x/(2*b*f*tan(e + f*x)**2 + 2*b*f) + 2*tan(e + f*x)**3/(2*b*f*tan(e + f*x)**2 + 2*b*f) + 3*tan(e + f*x)/(2*b*f*
tan(e + f*x)**2 + 2*b*f), Eq(a, b)), (x*tan(e)**4/(a + b*tan(e)**2), Eq(f, 0)), (-a**2*log(-sqrt(-a/b) + tan(e
+ f*x))/(2*a*b**2*f*sqrt(-a/b) - 2*b**3*f*sqrt(-a/b)) + a**2*log(sqrt(-a/b) + tan(e + f*x))/(2*a*b**2*f*sqrt(
-a/b) - 2*b**3*f*sqrt(-a/b)) + 2*a*b*sqrt(-a/b)*tan(e + f*x)/(2*a*b**2*f*sqrt(-a/b) - 2*b**3*f*sqrt(-a/b)) + 2
*b**2*f*x*sqrt(-a/b)/(2*a*b**2*f*sqrt(-a/b) - 2*b**3*f*sqrt(-a/b)) - 2*b**2*sqrt(-a/b)*tan(e + f*x)/(2*a*b**2*
f*sqrt(-a/b) - 2*b**3*f*sqrt(-a/b)), True))

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.03 \[ \int \frac {\tan ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {a^{2} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a b - b^{2}\right )} \sqrt {a b}} - \frac {f x + e}{a - b} - \frac {\tan \left (f x + e\right )}{b}}{f} \]

[In]

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

[Out]

-(a^2*arctan(b*tan(f*x + e)/sqrt(a*b))/((a*b - b^2)*sqrt(a*b)) - (f*x + e)/(a - b) - tan(f*x + e)/b)/f

Giac [A] (verification not implemented)

none

Time = 0.98 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.30 \[ \int \frac {\tan ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} a^{2}}{{\left (a b - b^{2}\right )} \sqrt {a b}} - \frac {f x + e}{a - b} - \frac {\tan \left (f x + e\right )}{b}}{f} \]

[In]

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

[Out]

-((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*a^2/((a*b - b^2)*sqrt(a*b)) - (f*x
+ e)/(a - b) - tan(f*x + e)/b)/f

Mupad [B] (verification not implemented)

Time = 11.62 (sec) , antiderivative size = 1212, normalized size of antiderivative = 19.24 \[ \int \frac {\tan ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

tan(e + f*x)/(b*f) - (2*atan((((((4*a*b^4 - 8*a^2*b^3 + 4*a^3*b^2)/b - (tan(e + f*x)*(4*a*b^5 - 4*b^6 + 4*a^2*
b^4 - 4*a^3*b^3)*2i)/(b*(2*a - 2*b)))*1i)/(2*a - 2*b) + (2*tan(e + f*x)*(a^4 + b^4))/b)/(2*a - 2*b) - ((((4*a*
b^4 - 8*a^2*b^3 + 4*a^3*b^2)/b + (tan(e + f*x)*(4*a*b^5 - 4*b^6 + 4*a^2*b^4 - 4*a^3*b^3)*2i)/(b*(2*a - 2*b)))*
1i)/(2*a - 2*b) - (2*tan(e + f*x)*(a^4 + b^4))/b)/(2*a - 2*b))/((((((4*a*b^4 - 8*a^2*b^3 + 4*a^3*b^2)/b - (tan
(e + f*x)*(4*a*b^5 - 4*b^6 + 4*a^2*b^4 - 4*a^3*b^3)*2i)/(b*(2*a - 2*b)))*1i)/(2*a - 2*b) + (2*tan(e + f*x)*(a^
4 + b^4))/b)*1i)/(2*a - 2*b) - (2*(a^2*b + a^3))/b + (((((4*a*b^4 - 8*a^2*b^3 + 4*a^3*b^2)/b + (tan(e + f*x)*(
4*a*b^5 - 4*b^6 + 4*a^2*b^4 - 4*a^3*b^3)*2i)/(b*(2*a - 2*b)))*1i)/(2*a - 2*b) - (2*tan(e + f*x)*(a^4 + b^4))/b
)*1i)/(2*a - 2*b))))/(f*(2*a - 2*b)) + (atan((((-a^3*b^3)^(1/2)*((((4*a*b^4 - 8*a^2*b^3 + 4*a^3*b^2)/b + (tan(
e + f*x)*(-a^3*b^3)^(1/2)*(4*a*b^5 - 4*b^6 + 4*a^2*b^4 - 4*a^3*b^3))/(b*(a*b^3 - b^4)))*(-a^3*b^3)^(1/2))/(2*(
a*b^3 - b^4)) - (2*tan(e + f*x)*(a^4 + b^4))/b)*1i)/(2*(a*b^3 - b^4)) - ((-a^3*b^3)^(1/2)*((((4*a*b^4 - 8*a^2*
b^3 + 4*a^3*b^2)/b - (tan(e + f*x)*(-a^3*b^3)^(1/2)*(4*a*b^5 - 4*b^6 + 4*a^2*b^4 - 4*a^3*b^3))/(b*(a*b^3 - b^4
)))*(-a^3*b^3)^(1/2))/(2*(a*b^3 - b^4)) + (2*tan(e + f*x)*(a^4 + b^4))/b)*1i)/(2*(a*b^3 - b^4)))/(((-a^3*b^3)^
(1/2)*((((4*a*b^4 - 8*a^2*b^3 + 4*a^3*b^2)/b + (tan(e + f*x)*(-a^3*b^3)^(1/2)*(4*a*b^5 - 4*b^6 + 4*a^2*b^4 - 4
*a^3*b^3))/(b*(a*b^3 - b^4)))*(-a^3*b^3)^(1/2))/(2*(a*b^3 - b^4)) - (2*tan(e + f*x)*(a^4 + b^4))/b))/(2*(a*b^3
- b^4)) - (2*(a^2*b + a^3))/b + ((-a^3*b^3)^(1/2)*((((4*a*b^4 - 8*a^2*b^3 + 4*a^3*b^2)/b - (tan(e + f*x)*(-a^
3*b^3)^(1/2)*(4*a*b^5 - 4*b^6 + 4*a^2*b^4 - 4*a^3*b^3))/(b*(a*b^3 - b^4)))*(-a^3*b^3)^(1/2))/(2*(a*b^3 - b^4))
+ (2*tan(e + f*x)*(a^4 + b^4))/b))/(2*(a*b^3 - b^4))))*(-a^3*b^3)^(1/2)*1i)/(f*(a*b^3 - b^4))