3.231 \(\int \frac{\tan ^4(e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\)
Optimal. Leaf size=95 \[ \frac{\sqrt{a} (a-3 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 b^{3/2} f (a-b)^2}-\frac{a \tan (e+f x)}{2 b f (a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac{x}{(a-b)^2} \]
[Out]
x/(a - b)^2 + (Sqrt[a]*(a - 3*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(2*(a - b)^2*b^(3/2)*f) - (a*Tan[e +
f*x])/(2*(a - b)*b*f*(a + b*Tan[e + f*x]^2))
________________________________________________________________________________________
Rubi [A] time = 0.117503, antiderivative size = 95, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used =
{3670, 470, 522, 203, 205} \[ \frac{\sqrt{a} (a-3 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 b^{3/2} f (a-b)^2}-\frac{a \tan (e+f x)}{2 b f (a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac{x}{(a-b)^2} \]
Antiderivative was successfully verified.
[In]
Int[Tan[e + f*x]^4/(a + b*Tan[e + f*x]^2)^2,x]
[Out]
x/(a - b)^2 + (Sqrt[a]*(a - 3*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(2*(a - b)^2*b^(3/2)*f) - (a*Tan[e +
f*x])/(2*(a - b)*b*f*(a + b*Tan[e + f*x]^2))
Rule 3670
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]))
Rule 470
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 522
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 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{\tan ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a \tan (e+f x)}{2 (a-b) b f \left (a+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{a+(a-2 b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 (a-b) b f}\\ &=-\frac{a \tan (e+f x)}{2 (a-b) b f \left (a+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{(a-b)^2 f}+\frac{(a (a-3 b)) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{2 (a-b)^2 b f}\\ &=\frac{x}{(a-b)^2}+\frac{\sqrt{a} (a-3 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 (a-b)^2 b^{3/2} f}-\frac{a \tan (e+f x)}{2 (a-b) b f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 0.751754, size = 94, normalized size = 0.99 \[ \frac{\frac{\sqrt{a} (a-3 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{b^{3/2}}-\frac{a (a-b) \sin (2 (e+f x))}{b ((a-b) \cos (2 (e+f x))+a+b)}+2 (e+f x)}{2 f (a-b)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[e + f*x]^4/(a + b*Tan[e + f*x]^2)^2,x]
[Out]
(2*(e + f*x) + (Sqrt[a]*(a - 3*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/b^(3/2) - (a*(a - b)*Sin[2*(e + f*x)
])/(b*(a + b + (a - b)*Cos[2*(e + f*x)])))/(2*(a - b)^2*f)
________________________________________________________________________________________
Maple [A] time = 0.023, size = 160, normalized size = 1.7 \begin{align*} -{\frac{{a}^{2}\tan \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{2}b \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{a\tan \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{2} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{{a}^{2}}{2\,f \left ( a-b \right ) ^{2}b}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{3\,a}{2\,f \left ( a-b \right ) ^{2}}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{f \left ( a-b \right ) ^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^2,x)
[Out]
-1/2/f*a^2/(a-b)^2/b*tan(f*x+e)/(a+b*tan(f*x+e)^2)+1/2/f*a/(a-b)^2*tan(f*x+e)/(a+b*tan(f*x+e)^2)+1/2/f*a^2/(a-
b)^2/b/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))-3/2/f*a/(a-b)^2/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2
))+1/f/(a-b)^2*arctan(tan(f*x+e))
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.25805, size = 865, normalized size = 9.11 \begin{align*} \left [\frac{8 \, b^{2} f x \tan \left (f x + e\right )^{2} + 8 \, a b f x -{\left ({\left (a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} - 3 \, a b\right )} \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^{2} - a b\right )} \tan \left (f x + e\right )}{8 \,{\left ({\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} f\right )}}, \frac{4 \, b^{2} f x \tan \left (f x + e\right )^{2} + 4 \, a b f x +{\left ({\left (a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} - 3 \, a b\right )} \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^{2} - a b\right )} \tan \left (f x + e\right )}{4 \,{\left ({\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} f\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")
[Out]
[1/8*(8*b^2*f*x*tan(f*x + e)^2 + 8*a*b*f*x - ((a*b - 3*b^2)*tan(f*x + e)^2 + a^2 - 3*a*b)*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^2 - a*b)*tan(f*x + e))/((a^2*b^2 - 2*a*b^3 + b^4)*f*tan(f*x +
e)^2 + (a^3*b - 2*a^2*b^2 + a*b^3)*f), 1/4*(4*b^2*f*x*tan(f*x + e)^2 + 4*a*b*f*x + ((a*b - 3*b^2)*tan(f*x + e)
^2 + a^2 - 3*a*b)*sqrt(a/b)*arctan(1/2*(b*tan(f*x + e)^2 - a)*sqrt(a/b)/(a*tan(f*x + e))) - 2*(a^2 - a*b)*tan(
f*x + e))/((a^2*b^2 - 2*a*b^3 + b^4)*f*tan(f*x + e)^2 + (a^3*b - 2*a^2*b^2 + a*b^3)*f)]
________________________________________________________________________________________
Sympy [A] time = 52.0489, size = 2179, normalized size = 22.94 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)**4/(a+b*tan(f*x+e)**2)**2,x)
[Out]
Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), ((x + tan(e + f*x)**3/(3*f) - tan(e + f*x)/f)/a**2, Eq(b, 0
)), (x/b**2, Eq(a, 0)), (x*tan(e)**4/(a + b*tan(e)**2)**2, Eq(f, 0)), (-2*I*a**(5/2)*b*sqrt(1/b)*tan(e + f*x)/
(4*I*a**(7/2)*b**2*f*sqrt(1/b) + 4*I*a**(5/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**3*f*sqrt(1/b)
- 8*I*a**(3/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**4*f*sqrt(1/b) + 4*I*sqrt(a)*b**5*f*sqrt(1/b
)*tan(e + f*x)**2) + 4*I*a**(3/2)*b**2*f*x*sqrt(1/b)/(4*I*a**(7/2)*b**2*f*sqrt(1/b) + 4*I*a**(5/2)*b**3*f*sqrt
(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**3*f*sqrt(1/b) - 8*I*a**(3/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a*
*(3/2)*b**4*f*sqrt(1/b) + 4*I*sqrt(a)*b**5*f*sqrt(1/b)*tan(e + f*x)**2) + 2*I*a**(3/2)*b**2*sqrt(1/b)*tan(e +
f*x)/(4*I*a**(7/2)*b**2*f*sqrt(1/b) + 4*I*a**(5/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**3*f*sqrt
(1/b) - 8*I*a**(3/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**4*f*sqrt(1/b) + 4*I*sqrt(a)*b**5*f*sqr
t(1/b)*tan(e + f*x)**2) + 4*I*sqrt(a)*b**3*f*x*sqrt(1/b)*tan(e + f*x)**2/(4*I*a**(7/2)*b**2*f*sqrt(1/b) + 4*I*
a**(5/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**3*f*sqrt(1/b) - 8*I*a**(3/2)*b**4*f*sqrt(1/b)*tan(
e + f*x)**2 + 4*I*a**(3/2)*b**4*f*sqrt(1/b) + 4*I*sqrt(a)*b**5*f*sqrt(1/b)*tan(e + f*x)**2) + a**3*log(-I*sqrt
(a)*sqrt(1/b) + tan(e + f*x))/(4*I*a**(7/2)*b**2*f*sqrt(1/b) + 4*I*a**(5/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 -
8*I*a**(5/2)*b**3*f*sqrt(1/b) - 8*I*a**(3/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**4*f*sqrt(1/b)
+ 4*I*sqrt(a)*b**5*f*sqrt(1/b)*tan(e + f*x)**2) - a**3*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(4*I*a**(7/2)*
b**2*f*sqrt(1/b) + 4*I*a**(5/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**3*f*sqrt(1/b) - 8*I*a**(3/2
)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**4*f*sqrt(1/b) + 4*I*sqrt(a)*b**5*f*sqrt(1/b)*tan(e + f*x)
**2) + a**2*b*log(-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)**2/(4*I*a**(7/2)*b**2*f*sqrt(1/b) + 4*I*a*
*(5/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**3*f*sqrt(1/b) - 8*I*a**(3/2)*b**4*f*sqrt(1/b)*tan(e
+ f*x)**2 + 4*I*a**(3/2)*b**4*f*sqrt(1/b) + 4*I*sqrt(a)*b**5*f*sqrt(1/b)*tan(e + f*x)**2) - 3*a**2*b*log(-I*sq
rt(a)*sqrt(1/b) + tan(e + f*x))/(4*I*a**(7/2)*b**2*f*sqrt(1/b) + 4*I*a**(5/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2
- 8*I*a**(5/2)*b**3*f*sqrt(1/b) - 8*I*a**(3/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**4*f*sqrt(1/
b) + 4*I*sqrt(a)*b**5*f*sqrt(1/b)*tan(e + f*x)**2) - a**2*b*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*
x)**2/(4*I*a**(7/2)*b**2*f*sqrt(1/b) + 4*I*a**(5/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**3*f*sqr
t(1/b) - 8*I*a**(3/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**4*f*sqrt(1/b) + 4*I*sqrt(a)*b**5*f*sq
rt(1/b)*tan(e + f*x)**2) + 3*a**2*b*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(4*I*a**(7/2)*b**2*f*sqrt(1/b) + 4
*I*a**(5/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**3*f*sqrt(1/b) - 8*I*a**(3/2)*b**4*f*sqrt(1/b)*t
an(e + f*x)**2 + 4*I*a**(3/2)*b**4*f*sqrt(1/b) + 4*I*sqrt(a)*b**5*f*sqrt(1/b)*tan(e + f*x)**2) - 3*a*b**2*log(
-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)**2/(4*I*a**(7/2)*b**2*f*sqrt(1/b) + 4*I*a**(5/2)*b**3*f*sqrt
(1/b)*tan(e + f*x)**2 - 8*I*a**(5/2)*b**3*f*sqrt(1/b) - 8*I*a**(3/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a*
*(3/2)*b**4*f*sqrt(1/b) + 4*I*sqrt(a)*b**5*f*sqrt(1/b)*tan(e + f*x)**2) + 3*a*b**2*log(I*sqrt(a)*sqrt(1/b) + t
an(e + f*x))*tan(e + f*x)**2/(4*I*a**(7/2)*b**2*f*sqrt(1/b) + 4*I*a**(5/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2 -
8*I*a**(5/2)*b**3*f*sqrt(1/b) - 8*I*a**(3/2)*b**4*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(3/2)*b**4*f*sqrt(1/b)
+ 4*I*sqrt(a)*b**5*f*sqrt(1/b)*tan(e + f*x)**2), True))
________________________________________________________________________________________
Giac [A] time = 2.20204, size = 171, normalized size = 1.8 \begin{align*} \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 )}{\left (a^{2} - 3 \, a b\right )}}{{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \sqrt{a b}} + \frac{2 \,{\left (f x + e\right )}}{a^{2} - 2 \, a b + b^{2}} - \frac{a \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}{\left (a b - b^{2}\right )}}}{2 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")
[Out]
1/2*((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*(a^2 - 3*a*b)/((a^2*b - 2*a*b^2
+ b^3)*sqrt(a*b)) + 2*(f*x + e)/(a^2 - 2*a*b + b^2) - a*tan(f*x + e)/((b*tan(f*x + e)^2 + a)*(a*b - b^2)))/f