3.198 \(\int \tan ^5(e+f x) (a+b \tan ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=105 \[ \frac {b (2 a-b) \tan ^6(e+f x)}{6 f}+\frac {(a-b)^2 \tan ^4(e+f x)}{4 f}-\frac {(a-b)^2 \tan ^2(e+f x)}{2 f}-\frac {(a-b)^2 \log (\cos (e+f x))}{f}+\frac {b^2 \tan ^8(e+f x)}{8 f} \]

[Out]

-(a-b)^2*ln(cos(f*x+e))/f-1/2*(a-b)^2*tan(f*x+e)^2/f+1/4*(a-b)^2*tan(f*x+e)^4/f+1/6*(2*a-b)*b*tan(f*x+e)^6/f+1
/8*b^2*tan(f*x+e)^8/f

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Rubi [A]  time = 0.11, antiderivative size = 105, 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, 446, 88} \[ \frac {b (2 a-b) \tan ^6(e+f x)}{6 f}+\frac {(a-b)^2 \tan ^4(e+f x)}{4 f}-\frac {(a-b)^2 \tan ^2(e+f x)}{2 f}-\frac {(a-b)^2 \log (\cos (e+f x))}{f}+\frac {b^2 \tan ^8(e+f x)}{8 f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^5*(a + b*Tan[e + f*x]^2)^2,x]

[Out]

-(((a - b)^2*Log[Cos[e + f*x]])/f) - ((a - b)^2*Tan[e + f*x]^2)/(2*f) + ((a - b)^2*Tan[e + f*x]^4)/(4*f) + ((2
*a - b)*b*Tan[e + f*x]^6)/(6*f) + (b^2*Tan[e + f*x]^8)/(8*f)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

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 ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^5 \left (a+b x^2\right )^2}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^2 (a+b x)^2}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {\operatorname {Subst}\left (\int \left (-(a-b)^2+(a-b)^2 x+(2 a-b) b x^2+b^2 x^3+\frac {(a-b)^2}{1+x}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac {(a-b)^2 \log (\cos (e+f x))}{f}-\frac {(a-b)^2 \tan ^2(e+f x)}{2 f}+\frac {(a-b)^2 \tan ^4(e+f x)}{4 f}+\frac {(2 a-b) b \tan ^6(e+f x)}{6 f}+\frac {b^2 \tan ^8(e+f x)}{8 f}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 89, normalized size = 0.85 \[ \frac {4 b (2 a-b) \tan ^6(e+f x)+6 (a-b)^2 \tan ^4(e+f x)-12 (a-b)^2 \tan ^2(e+f x)-24 (a-b)^2 \log (\cos (e+f x))+3 b^2 \tan ^8(e+f x)}{24 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^5*(a + b*Tan[e + f*x]^2)^2,x]

[Out]

(-24*(a - b)^2*Log[Cos[e + f*x]] - 12*(a - b)^2*Tan[e + f*x]^2 + 6*(a - b)^2*Tan[e + f*x]^4 + 4*(2*a - b)*b*Ta
n[e + f*x]^6 + 3*b^2*Tan[e + f*x]^8)/(24*f)

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fricas [A]  time = 0.46, size = 107, normalized size = 1.02 \[ \frac {3 \, b^{2} \tan \left (f x + e\right )^{8} + 4 \, {\left (2 \, a b - b^{2}\right )} \tan \left (f x + e\right )^{6} + 6 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{4} - 12 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{2} - 12 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right )}{24 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/24*(3*b^2*tan(f*x + e)^8 + 4*(2*a*b - b^2)*tan(f*x + e)^6 + 6*(a^2 - 2*a*b + b^2)*tan(f*x + e)^4 - 12*(a^2 -
2*a*b + b^2)*tan(f*x + e)^2 - 12*(a^2 - 2*a*b + b^2)*log(1/(tan(f*x + e)^2 + 1)))/f

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.03, size = 198, normalized size = 1.89 \[ \frac {b^{2} \left (\tan ^{8}\left (f x +e \right )\right )}{8 f}+\frac {a b \left (\tan ^{6}\left (f x +e \right )\right )}{3 f}-\frac {b^{2} \left (\tan ^{6}\left (f x +e \right )\right )}{6 f}+\frac {\left (\tan ^{4}\left (f x +e \right )\right ) a^{2}}{4 f}-\frac {\left (\tan ^{4}\left (f x +e \right )\right ) a b}{2 f}+\frac {b^{2} \left (\tan ^{4}\left (f x +e \right )\right )}{4 f}-\frac {\left (\tan ^{2}\left (f x +e \right )\right ) a^{2}}{2 f}+\frac {\left (\tan ^{2}\left (f x +e \right )\right ) a b}{f}-\frac {b^{2} \left (\tan ^{2}\left (f x +e \right )\right )}{2 f}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a^{2}}{2 f}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a b}{f}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) b^{2}}{2 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x)

[Out]

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

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maxima [A]  time = 0.56, size = 162, normalized size = 1.54 \[ -\frac {12 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) - \frac {24 \, {\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \sin \left (f x + e\right )^{6} - 6 \, {\left (11 \, a^{2} - 30 \, a b + 18 \, b^{2}\right )} \sin \left (f x + e\right )^{4} + 4 \, {\left (15 \, a^{2} - 38 \, a b + 22 \, b^{2}\right )} \sin \left (f x + e\right )^{2} - 18 \, a^{2} + 44 \, a b - 25 \, b^{2}}{\sin \left (f x + e\right )^{8} - 4 \, \sin \left (f x + e\right )^{6} + 6 \, \sin \left (f x + e\right )^{4} - 4 \, \sin \left (f x + e\right )^{2} + 1}}{24 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

-1/24*(12*(a^2 - 2*a*b + b^2)*log(sin(f*x + e)^2 - 1) - (24*(a^2 - 3*a*b + 2*b^2)*sin(f*x + e)^6 - 6*(11*a^2 -
30*a*b + 18*b^2)*sin(f*x + e)^4 + 4*(15*a^2 - 38*a*b + 22*b^2)*sin(f*x + e)^2 - 18*a^2 + 44*a*b - 25*b^2)/(si
n(f*x + e)^8 - 4*sin(f*x + e)^6 + 6*sin(f*x + e)^4 - 4*sin(f*x + e)^2 + 1))/f

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mupad [B]  time = 11.59, size = 113, normalized size = 1.08 \[ \frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )\,\left (\frac {a^2}{2}-a\,b+\frac {b^2}{2}\right )+{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (\frac {a\,b}{3}-\frac {b^2}{6}\right )+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^8}{8}-{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {a^2}{2}-a\,b+\frac {b^2}{2}\right )+{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (\frac {a^2}{4}-\frac {a\,b}{2}+\frac {b^2}{4}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^5*(a + b*tan(e + f*x)^2)^2,x)

[Out]

(log(tan(e + f*x)^2 + 1)*(a^2/2 - a*b + b^2/2) + tan(e + f*x)^6*((a*b)/3 - b^2/6) + (b^2*tan(e + f*x)^8)/8 - t
an(e + f*x)^2*(a^2/2 - a*b + b^2/2) + tan(e + f*x)^4*(a^2/4 - (a*b)/2 + b^2/4))/f

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sympy [A]  time = 1.34, size = 206, normalized size = 1.96 \[ \begin {cases} \frac {a^{2} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {a^{2} \tan ^{4}{\left (e + f x \right )}}{4 f} - \frac {a^{2} \tan ^{2}{\left (e + f x \right )}}{2 f} - \frac {a b \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{f} + \frac {a b \tan ^{6}{\left (e + f x \right )}}{3 f} - \frac {a b \tan ^{4}{\left (e + f x \right )}}{2 f} + \frac {a b \tan ^{2}{\left (e + f x \right )}}{f} + \frac {b^{2} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {b^{2} \tan ^{8}{\left (e + f x \right )}}{8 f} - \frac {b^{2} \tan ^{6}{\left (e + f x \right )}}{6 f} + \frac {b^{2} \tan ^{4}{\left (e + f x \right )}}{4 f} - \frac {b^{2} \tan ^{2}{\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \left (a + b \tan ^{2}{\relax (e )}\right )^{2} \tan ^{5}{\relax (e )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**5*(a+b*tan(f*x+e)**2)**2,x)

[Out]

Piecewise((a**2*log(tan(e + f*x)**2 + 1)/(2*f) + a**2*tan(e + f*x)**4/(4*f) - a**2*tan(e + f*x)**2/(2*f) - a*b
*log(tan(e + f*x)**2 + 1)/f + a*b*tan(e + f*x)**6/(3*f) - a*b*tan(e + f*x)**4/(2*f) + a*b*tan(e + f*x)**2/f +
b**2*log(tan(e + f*x)**2 + 1)/(2*f) + b**2*tan(e + f*x)**8/(8*f) - b**2*tan(e + f*x)**6/(6*f) + b**2*tan(e + f
*x)**4/(4*f) - b**2*tan(e + f*x)**2/(2*f), Ne(f, 0)), (x*(a + b*tan(e)**2)**2*tan(e)**5, True))

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