3.239 \(\int \frac{\tan (e+f x)}{(a+b \tan ^2(e+f x))^3} \, dx\)
Optimal. Leaf size=93 \[ \frac{1}{2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac{1}{4 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac{\log \left (a \cos ^2(e+f x)+b \sin ^2(e+f x)\right )}{2 f (a-b)^3} \]
[Out]
-Log[a*Cos[e + f*x]^2 + b*Sin[e + f*x]^2]/(2*(a - b)^3*f) + 1/(4*(a - b)*f*(a + b*Tan[e + f*x]^2)^2) + 1/(2*(a
- b)^2*f*(a + b*Tan[e + f*x]^2))
________________________________________________________________________________________
Rubi [A] time = 0.0881894, antiderivative size = 93, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used =
{3670, 444, 44} \[ \frac{1}{2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac{1}{4 f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac{\log \left (a \cos ^2(e+f x)+b \sin ^2(e+f x)\right )}{2 f (a-b)^3} \]
Antiderivative was successfully verified.
[In]
Int[Tan[e + f*x]/(a + b*Tan[e + f*x]^2)^3,x]
[Out]
-Log[a*Cos[e + f*x]^2 + b*Sin[e + f*x]^2]/(2*(a - b)^3*f) + 1/(4*(a - b)*f*(a + b*Tan[e + f*x]^2)^2) + 1/(2*(a
- b)^2*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 444
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 44
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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])
Rubi steps
\begin{align*} \int \frac{\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x}{\left (1+x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{(1+x) (a+b x)^3} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{(a-b)^3 (1+x)}-\frac{b}{(a-b) (a+b x)^3}-\frac{b}{(a-b)^2 (a+b x)^2}+\frac{b}{(-a+b)^3 (a+b x)}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac{\log \left (a \cos ^2(e+f x)+b \sin ^2(e+f x)\right )}{2 (a-b)^3 f}+\frac{1}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{1}{2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 0.65667, size = 82, normalized size = 0.88 \[ \frac{\frac{(a-b)^2}{\left (a+b \tan ^2(e+f x)\right )^2}+\frac{2 (a-b)}{a+b \tan ^2(e+f x)}-2 \log \left (a+b \tan ^2(e+f x)\right )-4 \log (\cos (e+f x))}{4 f (a-b)^3} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[e + f*x]/(a + b*Tan[e + f*x]^2)^3,x]
[Out]
(-4*Log[Cos[e + f*x]] - 2*Log[a + b*Tan[e + f*x]^2] + (a - b)^2/(a + b*Tan[e + f*x]^2)^2 + (2*(a - b))/(a + b*
Tan[e + f*x]^2))/(4*(a - b)^3*f)
________________________________________________________________________________________
Maple [B] time = 0.026, size = 190, normalized size = 2. \begin{align*} -{\frac{\ln \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f \left ( a-b \right ) ^{3}}}+{\frac{{a}^{2}}{4\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{ab}{2\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{b}^{2}}{4\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{a}{2\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{b}{2\,f \left ( a-b \right ) ^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f \left ( a-b \right ) ^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(f*x+e)/(a+b*tan(f*x+e)^2)^3,x)
[Out]
-1/2/f/(a-b)^3*ln(a+b*tan(f*x+e)^2)+1/4/f/(a-b)^3*a^2/(a+b*tan(f*x+e)^2)^2-1/2/f/(a-b)^3*a*b/(a+b*tan(f*x+e)^2
)^2+1/4/f*b^2/(a-b)^3/(a+b*tan(f*x+e)^2)^2+1/2/f/(a-b)^3*a/(a+b*tan(f*x+e)^2)-1/2/f/(a-b)^3/(a+b*tan(f*x+e)^2)
*b+1/2/f/(a-b)^3*ln(1+tan(f*x+e)^2)
________________________________________________________________________________________
Maxima [B] time = 1.1013, size = 259, normalized size = 2.78 \begin{align*} \frac{\frac{4 \,{\left (a b - b^{2}\right )} \sin \left (f x + e\right )^{2} - 4 \, a b + b^{2}}{a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3} +{\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} \sin \left (f x + e\right )^{4} - 2 \,{\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} \sin \left (f x + e\right )^{2}} - \frac{2 \, \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}}}{4 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)/(a+b*tan(f*x+e)^2)^3,x, algorithm="maxima")
[Out]
1/4*((4*(a*b - b^2)*sin(f*x + e)^2 - 4*a*b + b^2)/(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3 + (a^5 - 5*a^4*b + 10*a
^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*sin(f*x + e)^4 - 2*(a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*sin(f*
x + e)^2) - 2*log(-(a - b)*sin(f*x + e)^2 + a)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3))/f
________________________________________________________________________________________
Fricas [B] time = 1.58334, size = 458, normalized size = 4.92 \begin{align*} -\frac{3 \, b^{2} \tan \left (f x + e\right )^{4} + 2 \,{\left (2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{2} + 4 \, a b - b^{2} + 2 \,{\left (b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}\right )} \log \left (\frac{b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \,{\left ({\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} f\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)/(a+b*tan(f*x+e)^2)^3,x, algorithm="fricas")
[Out]
-1/4*(3*b^2*tan(f*x + e)^4 + 2*(2*a*b + b^2)*tan(f*x + e)^2 + 4*a*b - b^2 + 2*(b^2*tan(f*x + e)^4 + 2*a*b*tan(
f*x + e)^2 + a^2)*log((b*tan(f*x + e)^2 + a)/(tan(f*x + e)^2 + 1)))/((a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f*t
an(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*f*tan(f*x + e)^2 + (a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*
b^3)*f)
________________________________________________________________________________________
Sympy [A] time = 171.352, size = 2876, normalized size = 30.92 \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)/(a+b*tan(f*x+e)**2)**3,x)
[Out]
Piecewise((zoo*x/tan(e)**5, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), (log(tan(e + f*x)**2 + 1)/(2*a**3*f), Eq(b, 0)),
(-1/(6*b**3*f*tan(e + f*x)**6 + 18*b**3*f*tan(e + f*x)**4 + 18*b**3*f*tan(e + f*x)**2 + 6*b**3*f), Eq(a, b)),
(x*tan(e)/(a + b*tan(e)**2)**3, Eq(f, 0)), (-2*a**2*log(-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(4*a**5*f + 8*a**
4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3
*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e
+ f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) - 2*a**2*log(I*sqrt(a)*sqrt(1/b) + tan(e +
f*x))/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*ta
n(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3
*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) + 2*a**2*log(tan(e +
f*x)**2 + 1)/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b
**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a
**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) + 3*a**2/(4*
a**5*f + 8*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x
)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a
*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) - 4*a*b*log(-I*sqrt(a)*sqrt(1
/b) + tan(e + f*x))*tan(e + f*x)**2/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e
+ f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f
*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e +
f*x)**4) - 4*a*b*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)**2/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)*
*2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b
**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**
4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) + 4*a*b*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**2/(4*a**5*f + 8
*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*
a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*ta
n(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) + 2*a*b*tan(e + f*x)**2/(4*a**5*f + 8*a
**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a*
*3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(
e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) - 4*a*b/(4*a**5*f + 8*a**4*b*f*tan(e + f*
x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**
2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*
b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) - 2*b**2*log(-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f
*x)**4/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*t
an(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**
3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) - 2*b**2*log(I*sqrt
(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)**4/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**3*b*
*2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 + 24*a
**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*b**5
*f*tan(e + f*x)**4) + 2*b**2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**4/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 -
12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*
f*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*
tan(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) - 2*b**2*tan(e + f*x)**2/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 -
12*a**4*b*f + 4*a**3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f
*tan(e + f*x)**4 + 24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*t
an(e + f*x)**2 - 4*b**5*f*tan(e + f*x)**4) + b**2/(4*a**5*f + 8*a**4*b*f*tan(e + f*x)**2 - 12*a**4*b*f + 4*a**
3*b**2*f*tan(e + f*x)**4 - 24*a**3*b**2*f*tan(e + f*x)**2 + 12*a**3*b**2*f - 12*a**2*b**3*f*tan(e + f*x)**4 +
24*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f + 12*a*b**4*f*tan(e + f*x)**4 - 8*a*b**4*f*tan(e + f*x)**2 - 4*
b**5*f*tan(e + f*x)**4), True))
________________________________________________________________________________________
Giac [B] time = 1.68757, size = 876, normalized size = 9.42 \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)/(a+b*tan(f*x+e)^2)^3,x, algorithm="giac")
[Out]
-1/4*(2*log(a + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos
(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 4*log(-(cos(f*x + e) - 1)/(cos(f*x +
e) + 1) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (3*a^4 + 12*a^4*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 8*a^3*b
*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 24*a^2*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 8*a*b^3*(cos(f*x +
e) - 1)/(cos(f*x + e) + 1) + 18*a^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 16*a^3*b*(cos(f*x + e) - 1)^2
/(cos(f*x + e) + 1)^2 - 48*a^2*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 80*a*b^3*(cos(f*x + e) - 1)^2/(
cos(f*x + e) + 1)^2 - 16*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 12*a^4*(cos(f*x + e) - 1)^3/(cos(f*x
+ e) + 1)^3 - 8*a^3*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 24*a^2*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e
) + 1)^3 + 8*a*b^3*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 3*a^4*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4
)/((a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*(a + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b*(cos(f*x + e) -
1)/(cos(f*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)^2))/f