3.217 \(\int \frac{\tan ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=85 \[ \frac{a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{b^{5/2} f (a-b)}-\frac{(a+b) \tan (e+f x)}{b^2 f}-\frac{x}{a-b}+\frac{\tan ^3(e+f x)}{3 b f} \]

[Out]

-(x/(a - b)) + (a^(5/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/((a - b)*b^(5/2)*f) - ((a + b)*Tan[e + f*x])/(
b^2*f) + Tan[e + f*x]^3/(3*b*f)

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Rubi [A]  time = 0.187996, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3670, 479, 582, 522, 203, 205} \[ \frac{a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{b^{5/2} f (a-b)}-\frac{(a+b) \tan (e+f x)}{b^2 f}-\frac{x}{a-b}+\frac{\tan ^3(e+f x)}{3 b f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^6/(a + b*Tan[e + f*x]^2),x]

[Out]

-(x/(a - b)) + (a^(5/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/((a - b)*b^(5/2)*f) - ((a + b)*Tan[e + f*x])/(
b^2*f) + Tan[e + f*x]^3/(3*b*f)

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 479

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 582

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

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

Mathematica [A]  time = 0.781735, size = 92, normalized size = 1.08 \[ \frac{\sqrt{b} \left ((a-b) \tan (e+f x) \left (3 a-b \sec ^2(e+f x)+4 b\right )+3 b^2 (e+f x)\right )-3 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{3 b^{5/2} f (b-a)} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^6/(a + b*Tan[e + f*x]^2),x]

[Out]

(-3*a^(5/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] + Sqrt[b]*(3*b^2*(e + f*x) + (a - b)*(3*a + 4*b - b*Sec[e +
f*x]^2)*Tan[e + f*x]))/(3*b^(5/2)*(-a + b)*f)

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Maple [A]  time = 0.017, size = 102, normalized size = 1.2 \begin{align*}{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{3\,fb}}-{\frac{a\tan \left ( fx+e \right ) }{f{b}^{2}}}-{\frac{\tan \left ( fx+e \right ) }{fb}}+{\frac{{a}^{3}}{f{b}^{2} \left ( a-b \right ) }\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 ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^6/(a+b*tan(f*x+e)^2),x)

[Out]

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

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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)^6/(a+b*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.21668, size = 618, normalized size = 7.27 \begin{align*} \left [-\frac{12 \, b^{2} f x - 4 \,{\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, a^{2} \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 ) + 12 \,{\left (a^{2} - b^{2}\right )} \tan \left (f x + e\right )}{12 \,{\left (a b^{2} - b^{3}\right )} f}, -\frac{6 \, b^{2} f x - 2 \,{\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{3} - 3 \, a^{2} \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 ) + 6 \,{\left (a^{2} - b^{2}\right )} \tan \left (f x + e\right )}{6 \,{\left (a b^{2} - b^{3}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^6/(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

[-1/12*(12*b^2*f*x - 4*(a*b - b^2)*tan(f*x + e)^3 + 3*a^2*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)) + 12*(a^2 - b^2)*tan(f*x + e))/((a*b^2 - b^3)*f), -1/6*(6*b^2*f*x - 2*(a*b - b^2)*tan(f*x + e)^3 - 3*
a^2*sqrt(a/b)*arctan(1/2*(b*tan(f*x + e)^2 - a)*sqrt(a/b)/(a*tan(f*x + e))) + 6*(a^2 - b^2)*tan(f*x + e))/((a*
b^2 - b^3)*f)]

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Sympy [A]  time = 20.8982, size = 685, normalized size = 8.06 \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)**6/(a+b*tan(f*x+e)**2),x)

[Out]

Piecewise((zoo*x*tan(e)**4, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), ((-x + tan(e + f*x)**5/(5*f) - tan(e + f*x)**3/(3
*f) + tan(e + f*x)/f)/a, Eq(b, 0)), ((x + tan(e + f*x)**3/(3*f) - tan(e + f*x)/f)/b, Eq(a, 0)), (15*f*x*tan(e
+ f*x)**2/(6*b*f*tan(e + f*x)**2 + 6*b*f) + 15*f*x/(6*b*f*tan(e + f*x)**2 + 6*b*f) + 2*tan(e + f*x)**5/(6*b*f*
tan(e + f*x)**2 + 6*b*f) - 10*tan(e + f*x)**3/(6*b*f*tan(e + f*x)**2 + 6*b*f) - 15*tan(e + f*x)/(6*b*f*tan(e +
f*x)**2 + 6*b*f), Eq(a, b)), (x*tan(e)**6/(a + b*tan(e)**2), Eq(f, 0)), (-6*I*a**(5/2)*b*sqrt(1/b)*tan(e + f*
x)/(6*I*a**(3/2)*b**3*f*sqrt(1/b) - 6*I*sqrt(a)*b**4*f*sqrt(1/b)) + 2*I*a**(3/2)*b**2*sqrt(1/b)*tan(e + f*x)**
3/(6*I*a**(3/2)*b**3*f*sqrt(1/b) - 6*I*sqrt(a)*b**4*f*sqrt(1/b)) - 6*I*sqrt(a)*b**3*f*x*sqrt(1/b)/(6*I*a**(3/2
)*b**3*f*sqrt(1/b) - 6*I*sqrt(a)*b**4*f*sqrt(1/b)) - 2*I*sqrt(a)*b**3*sqrt(1/b)*tan(e + f*x)**3/(6*I*a**(3/2)*
b**3*f*sqrt(1/b) - 6*I*sqrt(a)*b**4*f*sqrt(1/b)) + 6*I*sqrt(a)*b**3*sqrt(1/b)*tan(e + f*x)/(6*I*a**(3/2)*b**3*
f*sqrt(1/b) - 6*I*sqrt(a)*b**4*f*sqrt(1/b)) + 3*a**3*log(-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(6*I*a**(3/2)*b*
*3*f*sqrt(1/b) - 6*I*sqrt(a)*b**4*f*sqrt(1/b)) - 3*a**3*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(6*I*a**(3/2)*
b**3*f*sqrt(1/b) - 6*I*sqrt(a)*b**4*f*sqrt(1/b)), True))

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Giac [B]  time = 4.15437, size = 521, normalized size = 6.13 \begin{align*} \frac{\frac{3 \,{\left (a^{3} b^{4} + b^{7} + a^{2} b{\left | -a b^{3} + b^{4} \right |} + a b^{2}{\left | -a b^{3} + b^{4} \right |} + b^{3}{\left | -a b^{3} + b^{4} \right |}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor + \arctan \left (\frac{8 \, \sqrt{\frac{1}{2}} \tan \left (f x + e\right )}{\sqrt{\frac{16 \, a b^{3} + 16 \, b^{4} + \sqrt{-1024 \, a b^{7} + 256 \,{\left (a b^{3} + b^{4}\right )}^{2}}}{b^{4}}}}\right )\right )}}{a b^{3}{\left | -a b^{3} + b^{4} \right |} + b^{4}{\left | -a b^{3} + b^{4} \right |} +{\left (a b^{3} - b^{4}\right )}^{2}} - \frac{3 \,{\left ({\left (a^{2} + a b + b^{2}\right )} \sqrt{a b}{\left | -a b^{3} + b^{4} \right |}{\left | b \right |} -{\left (a^{3} b^{3} + b^{6}\right )} \sqrt{a b}{\left | b \right |}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor + \arctan \left (\frac{8 \, \sqrt{\frac{1}{2}} \tan \left (f x + e\right )}{\sqrt{\frac{16 \, a b^{3} + 16 \, b^{4} - \sqrt{-1024 \, a b^{7} + 256 \,{\left (a b^{3} + b^{4}\right )}^{2}}}{b^{4}}}}\right )\right )}}{{\left (a b^{3} - b^{4}\right )}^{2} b -{\left (a b^{4} + b^{5}\right )}{\left | -a b^{3} + b^{4} \right |}} + \frac{b^{2} \tan \left (f x + e\right )^{3} - 3 \, a b \tan \left (f x + e\right ) - 3 \, b^{2} \tan \left (f x + e\right )}{b^{3}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^6/(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

1/3*(3*(a^3*b^4 + b^7 + a^2*b*abs(-a*b^3 + b^4) + a*b^2*abs(-a*b^3 + b^4) + b^3*abs(-a*b^3 + b^4))*(pi*floor((
f*x + e)/pi + 1/2) + arctan(8*sqrt(1/2)*tan(f*x + e)/sqrt((16*a*b^3 + 16*b^4 + sqrt(-1024*a*b^7 + 256*(a*b^3 +
b^4)^2))/b^4)))/(a*b^3*abs(-a*b^3 + b^4) + b^4*abs(-a*b^3 + b^4) + (a*b^3 - b^4)^2) - 3*((a^2 + a*b + b^2)*sq
rt(a*b)*abs(-a*b^3 + b^4)*abs(b) - (a^3*b^3 + b^6)*sqrt(a*b)*abs(b))*(pi*floor((f*x + e)/pi + 1/2) + arctan(8*
sqrt(1/2)*tan(f*x + e)/sqrt((16*a*b^3 + 16*b^4 - sqrt(-1024*a*b^7 + 256*(a*b^3 + b^4)^2))/b^4)))/((a*b^3 - b^4
)^2*b - (a*b^4 + b^5)*abs(-a*b^3 + b^4)) + (b^2*tan(f*x + e)^3 - 3*a*b*tan(f*x + e) - 3*b^2*tan(f*x + e))/b^3)
/f