3.205 \(\int \tan ^4(e+f x) (a+b \tan ^2(e+f x))^2 \, dx\)
Optimal. Leaf size=91 \[ \frac {b (2 a-b) \tan ^5(e+f x)}{5 f}+\frac {(a-b)^2 \tan ^3(e+f x)}{3 f}-\frac {(a-b)^2 \tan (e+f x)}{f}+x (a-b)^2+\frac {b^2 \tan ^7(e+f x)}{7 f} \]
[Out]
(a-b)^2*x-(a-b)^2*tan(f*x+e)/f+1/3*(a-b)^2*tan(f*x+e)^3/f+1/5*(2*a-b)*b*tan(f*x+e)^5/f+1/7*b^2*tan(f*x+e)^7/f
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Rubi [A] time = 0.08, antiderivative size = 91, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used =
{3670, 461, 203} \[ \frac {b (2 a-b) \tan ^5(e+f x)}{5 f}+\frac {(a-b)^2 \tan ^3(e+f x)}{3 f}-\frac {(a-b)^2 \tan (e+f x)}{f}+x (a-b)^2+\frac {b^2 \tan ^7(e+f x)}{7 f} \]
Antiderivative was successfully verified.
[In]
Int[Tan[e + f*x]^4*(a + b*Tan[e + f*x]^2)^2,x]
[Out]
(a - b)^2*x - ((a - b)^2*Tan[e + f*x])/f + ((a - b)^2*Tan[e + f*x]^3)/(3*f) + ((2*a - b)*b*Tan[e + f*x]^5)/(5*
f) + (b^2*Tan[e + f*x]^7)/(7*f)
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 461
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
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]))
Rubi steps
\begin {align*} \int \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 (a+b x^2\right )^2}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \left (-(a-b)^2+(a-b)^2 x^2+(2 a-b) b x^4+b^2 x^6+\frac {a^2-2 a b+b^2}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(a-b)^2 \tan (e+f x)}{f}+\frac {(a-b)^2 \tan ^3(e+f x)}{3 f}+\frac {(2 a-b) b \tan ^5(e+f x)}{5 f}+\frac {b^2 \tan ^7(e+f x)}{7 f}+\frac {(a-b)^2 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=(a-b)^2 x-\frac {(a-b)^2 \tan (e+f x)}{f}+\frac {(a-b)^2 \tan ^3(e+f x)}{3 f}+\frac {(2 a-b) b \tan ^5(e+f x)}{5 f}+\frac {b^2 \tan ^7(e+f x)}{7 f}\\ \end {align*}
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Mathematica [B] time = 0.08, size = 190, normalized size = 2.09 \[ \frac {a^2 \tan ^{-1}(\tan (e+f x))}{f}+\frac {a^2 \tan ^3(e+f x)}{3 f}-\frac {a^2 \tan (e+f x)}{f}-\frac {2 a b \tan ^{-1}(\tan (e+f x))}{f}+\frac {2 a b \tan ^5(e+f x)}{5 f}-\frac {2 a b \tan ^3(e+f x)}{3 f}+\frac {2 a b \tan (e+f x)}{f}+\frac {b^2 \tan ^{-1}(\tan (e+f x))}{f}+\frac {b^2 \tan ^7(e+f x)}{7 f}-\frac {b^2 \tan ^5(e+f x)}{5 f}+\frac {b^2 \tan ^3(e+f x)}{3 f}-\frac {b^2 \tan (e+f x)}{f} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[e + f*x]^4*(a + b*Tan[e + f*x]^2)^2,x]
[Out]
(a^2*ArcTan[Tan[e + f*x]])/f - (2*a*b*ArcTan[Tan[e + f*x]])/f + (b^2*ArcTan[Tan[e + f*x]])/f - (a^2*Tan[e + f*
x])/f + (2*a*b*Tan[e + f*x])/f - (b^2*Tan[e + f*x])/f + (a^2*Tan[e + f*x]^3)/(3*f) - (2*a*b*Tan[e + f*x]^3)/(3
*f) + (b^2*Tan[e + f*x]^3)/(3*f) + (2*a*b*Tan[e + f*x]^5)/(5*f) - (b^2*Tan[e + f*x]^5)/(5*f) + (b^2*Tan[e + f*
x]^7)/(7*f)
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fricas [A] time = 0.41, size = 94, normalized size = 1.03 \[ \frac {15 \, b^{2} \tan \left (f x + e\right )^{7} + 21 \, {\left (2 \, a b - b^{2}\right )} \tan \left (f x + e\right )^{5} + 35 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{3} + 105 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} f x - 105 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )}{105 \, f} \]
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]
1/105*(15*b^2*tan(f*x + e)^7 + 21*(2*a*b - b^2)*tan(f*x + e)^5 + 35*(a^2 - 2*a*b + b^2)*tan(f*x + e)^3 + 105*(
a^2 - 2*a*b + b^2)*f*x - 105*(a^2 - 2*a*b + b^2)*tan(f*x + e))/f
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giac [B] time = 35.41, size = 1684, normalized size = 18.51 \[ \text {result too large to display} \]
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]
1/105*(105*a^2*f*x*tan(f*x)^7*tan(e)^7 - 210*a*b*f*x*tan(f*x)^7*tan(e)^7 + 105*b^2*f*x*tan(f*x)^7*tan(e)^7 - 7
35*a^2*f*x*tan(f*x)^6*tan(e)^6 + 1470*a*b*f*x*tan(f*x)^6*tan(e)^6 - 735*b^2*f*x*tan(f*x)^6*tan(e)^6 + 105*a^2*
tan(f*x)^7*tan(e)^6 - 210*a*b*tan(f*x)^7*tan(e)^6 + 105*b^2*tan(f*x)^7*tan(e)^6 + 105*a^2*tan(f*x)^6*tan(e)^7
- 210*a*b*tan(f*x)^6*tan(e)^7 + 105*b^2*tan(f*x)^6*tan(e)^7 + 2205*a^2*f*x*tan(f*x)^5*tan(e)^5 - 4410*a*b*f*x*
tan(f*x)^5*tan(e)^5 + 2205*b^2*f*x*tan(f*x)^5*tan(e)^5 - 35*a^2*tan(f*x)^7*tan(e)^4 + 70*a*b*tan(f*x)^7*tan(e)
^4 - 35*b^2*tan(f*x)^7*tan(e)^4 - 735*a^2*tan(f*x)^6*tan(e)^5 + 1470*a*b*tan(f*x)^6*tan(e)^5 - 735*b^2*tan(f*x
)^6*tan(e)^5 - 735*a^2*tan(f*x)^5*tan(e)^6 + 1470*a*b*tan(f*x)^5*tan(e)^6 - 735*b^2*tan(f*x)^5*tan(e)^6 - 35*a
^2*tan(f*x)^4*tan(e)^7 + 70*a*b*tan(f*x)^4*tan(e)^7 - 35*b^2*tan(f*x)^4*tan(e)^7 - 3675*a^2*f*x*tan(f*x)^4*tan
(e)^4 + 7350*a*b*f*x*tan(f*x)^4*tan(e)^4 - 3675*b^2*f*x*tan(f*x)^4*tan(e)^4 - 42*a*b*tan(f*x)^7*tan(e)^2 + 21*
b^2*tan(f*x)^7*tan(e)^2 + 140*a^2*tan(f*x)^6*tan(e)^3 - 490*a*b*tan(f*x)^6*tan(e)^3 + 245*b^2*tan(f*x)^6*tan(e
)^3 + 1995*a^2*tan(f*x)^5*tan(e)^4 - 4410*a*b*tan(f*x)^5*tan(e)^4 + 2205*b^2*tan(f*x)^5*tan(e)^4 + 1995*a^2*ta
n(f*x)^4*tan(e)^5 - 4410*a*b*tan(f*x)^4*tan(e)^5 + 2205*b^2*tan(f*x)^4*tan(e)^5 + 140*a^2*tan(f*x)^3*tan(e)^6
- 490*a*b*tan(f*x)^3*tan(e)^6 + 245*b^2*tan(f*x)^3*tan(e)^6 - 42*a*b*tan(f*x)^2*tan(e)^7 + 21*b^2*tan(f*x)^2*t
an(e)^7 + 3675*a^2*f*x*tan(f*x)^3*tan(e)^3 - 7350*a*b*f*x*tan(f*x)^3*tan(e)^3 + 3675*b^2*f*x*tan(f*x)^3*tan(e)
^3 - 15*b^2*tan(f*x)^7 + 84*a*b*tan(f*x)^6*tan(e) - 147*b^2*tan(f*x)^6*tan(e) - 210*a^2*tan(f*x)^5*tan(e)^2 +
840*a*b*tan(f*x)^5*tan(e)^2 - 735*b^2*tan(f*x)^5*tan(e)^2 - 2730*a^2*tan(f*x)^4*tan(e)^3 + 6300*a*b*tan(f*x)^4
*tan(e)^3 - 3675*b^2*tan(f*x)^4*tan(e)^3 - 2730*a^2*tan(f*x)^3*tan(e)^4 + 6300*a*b*tan(f*x)^3*tan(e)^4 - 3675*
b^2*tan(f*x)^3*tan(e)^4 - 210*a^2*tan(f*x)^2*tan(e)^5 + 840*a*b*tan(f*x)^2*tan(e)^5 - 735*b^2*tan(f*x)^2*tan(e
)^5 + 84*a*b*tan(f*x)*tan(e)^6 - 147*b^2*tan(f*x)*tan(e)^6 - 15*b^2*tan(e)^7 - 2205*a^2*f*x*tan(f*x)^2*tan(e)^
2 + 4410*a*b*f*x*tan(f*x)^2*tan(e)^2 - 2205*b^2*f*x*tan(f*x)^2*tan(e)^2 - 42*a*b*tan(f*x)^5 + 21*b^2*tan(f*x)^
5 + 140*a^2*tan(f*x)^4*tan(e) - 490*a*b*tan(f*x)^4*tan(e) + 245*b^2*tan(f*x)^4*tan(e) + 1995*a^2*tan(f*x)^3*ta
n(e)^2 - 4410*a*b*tan(f*x)^3*tan(e)^2 + 2205*b^2*tan(f*x)^3*tan(e)^2 + 1995*a^2*tan(f*x)^2*tan(e)^3 - 4410*a*b
*tan(f*x)^2*tan(e)^3 + 2205*b^2*tan(f*x)^2*tan(e)^3 + 140*a^2*tan(f*x)*tan(e)^4 - 490*a*b*tan(f*x)*tan(e)^4 +
245*b^2*tan(f*x)*tan(e)^4 - 42*a*b*tan(e)^5 + 21*b^2*tan(e)^5 + 735*a^2*f*x*tan(f*x)*tan(e) - 1470*a*b*f*x*tan
(f*x)*tan(e) + 735*b^2*f*x*tan(f*x)*tan(e) - 35*a^2*tan(f*x)^3 + 70*a*b*tan(f*x)^3 - 35*b^2*tan(f*x)^3 - 735*a
^2*tan(f*x)^2*tan(e) + 1470*a*b*tan(f*x)^2*tan(e) - 735*b^2*tan(f*x)^2*tan(e) - 735*a^2*tan(f*x)*tan(e)^2 + 14
70*a*b*tan(f*x)*tan(e)^2 - 735*b^2*tan(f*x)*tan(e)^2 - 35*a^2*tan(e)^3 + 70*a*b*tan(e)^3 - 35*b^2*tan(e)^3 - 1
05*a^2*f*x + 210*a*b*f*x - 105*b^2*f*x + 105*a^2*tan(f*x) - 210*a*b*tan(f*x) + 105*b^2*tan(f*x) + 105*a^2*tan(
e) - 210*a*b*tan(e) + 105*b^2*tan(e))/(f*tan(f*x)^7*tan(e)^7 - 7*f*tan(f*x)^6*tan(e)^6 + 21*f*tan(f*x)^5*tan(e
)^5 - 35*f*tan(f*x)^4*tan(e)^4 + 35*f*tan(f*x)^3*tan(e)^3 - 21*f*tan(f*x)^2*tan(e)^2 + 7*f*tan(f*x)*tan(e) - f
)
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maple [B] time = 0.03, size = 179, normalized size = 1.97 \[ \frac {b^{2} \left (\tan ^{7}\left (f x +e \right )\right )}{7 f}+\frac {2 \left (\tan ^{5}\left (f x +e \right )\right ) a b}{5 f}-\frac {b^{2} \left (\tan ^{5}\left (f x +e \right )\right )}{5 f}+\frac {\left (\tan ^{3}\left (f x +e \right )\right ) a^{2}}{3 f}-\frac {2 \left (\tan ^{3}\left (f x +e \right )\right ) a b}{3 f}+\frac {b^{2} \left (\tan ^{3}\left (f x +e \right )\right )}{3 f}-\frac {a^{2} \tan \left (f x +e \right )}{f}+\frac {2 a b \tan \left (f x +e \right )}{f}-\frac {b^{2} \tan \left (f x +e \right )}{f}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) a^{2}}{f}-\frac {2 \arctan \left (\tan \left (f x +e \right )\right ) a b}{f}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) b^{2}}{f} \]
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]
1/7*b^2*tan(f*x+e)^7/f+2/5/f*tan(f*x+e)^5*a*b-1/5*b^2*tan(f*x+e)^5/f+1/3/f*tan(f*x+e)^3*a^2-2/3/f*tan(f*x+e)^3
*a*b+1/3*b^2*tan(f*x+e)^3/f-1/f*a^2*tan(f*x+e)+2*a*b*tan(f*x+e)/f-b^2*tan(f*x+e)/f+1/f*arctan(tan(f*x+e))*a^2-
2/f*arctan(tan(f*x+e))*a*b+1/f*arctan(tan(f*x+e))*b^2
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maxima [A] time = 0.65, size = 97, normalized size = 1.07 \[ \frac {15 \, b^{2} \tan \left (f x + e\right )^{7} + 21 \, {\left (2 \, a b - b^{2}\right )} \tan \left (f x + e\right )^{5} + 35 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{3} + 105 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} {\left (f x + e\right )} - 105 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )}{105 \, f} \]
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]
1/105*(15*b^2*tan(f*x + e)^7 + 21*(2*a*b - b^2)*tan(f*x + e)^5 + 35*(a^2 - 2*a*b + b^2)*tan(f*x + e)^3 + 105*(
a^2 - 2*a*b + b^2)*(f*x + e) - 105*(a^2 - 2*a*b + b^2)*tan(f*x + e))/f
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mupad [B] time = 12.13, size = 127, normalized size = 1.40 \[ \frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,{\left (a-b\right )}^2}{a^2-2\,a\,b+b^2}\right )\,{\left (a-b\right )}^2}{f}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (\frac {2\,a\,b}{5}-\frac {b^2}{5}\right )}{f}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a^2-2\,a\,b+b^2\right )}{f}+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^7}{7\,f}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {a^2}{3}-\frac {2\,a\,b}{3}+\frac {b^2}{3}\right )}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(e + f*x)^4*(a + b*tan(e + f*x)^2)^2,x)
[Out]
(atan((tan(e + f*x)*(a - b)^2)/(a^2 - 2*a*b + b^2))*(a - b)^2)/f + (tan(e + f*x)^5*((2*a*b)/5 - b^2/5))/f - (t
an(e + f*x)*(a^2 - 2*a*b + b^2))/f + (b^2*tan(e + f*x)^7)/(7*f) + (tan(e + f*x)^3*(a^2/3 - (2*a*b)/3 + b^2/3))
/f
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sympy [A] time = 1.08, size = 165, normalized size = 1.81 \[ \begin {cases} a^{2} x + \frac {a^{2} \tan ^{3}{\left (e + f x \right )}}{3 f} - \frac {a^{2} \tan {\left (e + f x \right )}}{f} - 2 a b x + \frac {2 a b \tan ^{5}{\left (e + f x \right )}}{5 f} - \frac {2 a b \tan ^{3}{\left (e + f x \right )}}{3 f} + \frac {2 a b \tan {\left (e + f x \right )}}{f} + b^{2} x + \frac {b^{2} \tan ^{7}{\left (e + f x \right )}}{7 f} - \frac {b^{2} \tan ^{5}{\left (e + f x \right )}}{5 f} + \frac {b^{2} \tan ^{3}{\left (e + f x \right )}}{3 f} - \frac {b^{2} \tan {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (a + b \tan ^{2}{\relax (e )}\right )^{2} \tan ^{4}{\relax (e )} & \text {otherwise} \end {cases} \]
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((a**2*x + a**2*tan(e + f*x)**3/(3*f) - a**2*tan(e + f*x)/f - 2*a*b*x + 2*a*b*tan(e + f*x)**5/(5*f) -
2*a*b*tan(e + f*x)**3/(3*f) + 2*a*b*tan(e + f*x)/f + b**2*x + b**2*tan(e + f*x)**7/(7*f) - b**2*tan(e + f*x)*
*5/(5*f) + b**2*tan(e + f*x)**3/(3*f) - b**2*tan(e + f*x)/f, Ne(f, 0)), (x*(a + b*tan(e)**2)**2*tan(e)**4, Tru
e))
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