3.705 \(\int \sec ^2(x) \tan ^6(x) (1+\tan ^2(x))^3 \, dx\)

Optimal. Leaf size=33 \[ \frac {\tan ^{13}(x)}{13}+\frac {3 \tan ^{11}(x)}{11}+\frac {\tan ^9(x)}{3}+\frac {\tan ^7(x)}{7} \]

[Out]

1/7*tan(x)^7+1/3*tan(x)^9+3/11*tan(x)^11+1/13*tan(x)^13

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Rubi [A]  time = 0.09, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3657, 2607, 270} \[ \frac {\tan ^{13}(x)}{13}+\frac {3 \tan ^{11}(x)}{11}+\frac {\tan ^9(x)}{3}+\frac {\tan ^7(x)}{7} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^2*Tan[x]^6*(1 + Tan[x]^2)^3,x]

[Out]

Tan[x]^7/7 + Tan[x]^9/3 + (3*Tan[x]^11)/11 + Tan[x]^13/13

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 3657

Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]

Rubi steps

\begin {align*} \int \sec ^2(x) \tan ^6(x) \left (1+\tan ^2(x)\right )^3 \, dx &=\int \sec ^8(x) \tan ^6(x) \, dx\\ &=\operatorname {Subst}\left (\int x^6 \left (1+x^2\right )^3 \, dx,x,\tan (x)\right )\\ &=\operatorname {Subst}\left (\int \left (x^6+3 x^8+3 x^{10}+x^{12}\right ) \, dx,x,\tan (x)\right )\\ &=\frac {\tan ^7(x)}{7}+\frac {\tan ^9(x)}{3}+\frac {3 \tan ^{11}(x)}{11}+\frac {\tan ^{13}(x)}{13}\\ \end {align*}

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Mathematica [B]  time = 0.02, size = 67, normalized size = 2.03 \[ -\frac {16 \tan (x)}{3003}+\frac {1}{13} \tan (x) \sec ^{12}(x)-\frac {27}{143} \tan (x) \sec ^{10}(x)+\frac {53}{429} \tan (x) \sec ^8(x)-\frac {5 \tan (x) \sec ^6(x)}{3003}-\frac {2 \tan (x) \sec ^4(x)}{1001}-\frac {8 \tan (x) \sec ^2(x)}{3003} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^2*Tan[x]^6*(1 + Tan[x]^2)^3,x]

[Out]

(-16*Tan[x])/3003 - (8*Sec[x]^2*Tan[x])/3003 - (2*Sec[x]^4*Tan[x])/1001 - (5*Sec[x]^6*Tan[x])/3003 + (53*Sec[x
]^8*Tan[x])/429 - (27*Sec[x]^10*Tan[x])/143 + (Sec[x]^12*Tan[x])/13

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fricas [A]  time = 0.86, size = 46, normalized size = 1.39 \[ -\frac {{\left (16 \, \cos \relax (x)^{12} + 8 \, \cos \relax (x)^{10} + 6 \, \cos \relax (x)^{8} + 5 \, \cos \relax (x)^{6} - 371 \, \cos \relax (x)^{4} + 567 \, \cos \relax (x)^{2} - 231\right )} \sin \relax (x)}{3003 \, \cos \relax (x)^{13}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2*tan(x)^6*(1+tan(x)^2)^3,x, algorithm="fricas")

[Out]

-1/3003*(16*cos(x)^12 + 8*cos(x)^10 + 6*cos(x)^8 + 5*cos(x)^6 - 371*cos(x)^4 + 567*cos(x)^2 - 231)*sin(x)/cos(
x)^13

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giac [A]  time = 0.14, size = 25, normalized size = 0.76 \[ \frac {1}{13} \, \tan \relax (x)^{13} + \frac {3}{11} \, \tan \relax (x)^{11} + \frac {1}{3} \, \tan \relax (x)^{9} + \frac {1}{7} \, \tan \relax (x)^{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2*tan(x)^6*(1+tan(x)^2)^3,x, algorithm="giac")

[Out]

1/13*tan(x)^13 + 3/11*tan(x)^11 + 1/3*tan(x)^9 + 1/7*tan(x)^7

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maple [A]  time = 0.06, size = 42, normalized size = 1.27 \[ \frac {\sin ^{7}\relax (x )}{7 \cos \relax (x )^{7}}+\frac {\sin ^{9}\relax (x )}{3 \cos \relax (x )^{9}}+\frac {3 \left (\sin ^{11}\relax (x )\right )}{11 \cos \relax (x )^{11}}+\frac {\sin ^{13}\relax (x )}{13 \cos \relax (x )^{13}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^2*tan(x)^6*(1+tan(x)^2)^3,x)

[Out]

1/7*sin(x)^7/cos(x)^7+1/3*sin(x)^9/cos(x)^9+3/11*sin(x)^11/cos(x)^11+1/13*sin(x)^13/cos(x)^13

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maxima [A]  time = 0.39, size = 25, normalized size = 0.76 \[ \frac {1}{13} \, \tan \relax (x)^{13} + \frac {3}{11} \, \tan \relax (x)^{11} + \frac {1}{3} \, \tan \relax (x)^{9} + \frac {1}{7} \, \tan \relax (x)^{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^2*tan(x)^6*(1+tan(x)^2)^3,x, algorithm="maxima")

[Out]

1/13*tan(x)^13 + 3/11*tan(x)^11 + 1/3*tan(x)^9 + 1/7*tan(x)^7

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mupad [B]  time = 2.92, size = 25, normalized size = 0.76 \[ \frac {{\mathrm {tan}\relax (x)}^{13}}{13}+\frac {3\,{\mathrm {tan}\relax (x)}^{11}}{11}+\frac {{\mathrm {tan}\relax (x)}^9}{3}+\frac {{\mathrm {tan}\relax (x)}^7}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(x)^6*(tan(x)^2 + 1)^3)/cos(x)^2,x)

[Out]

tan(x)^7/7 + tan(x)^9/3 + (3*tan(x)^11)/11 + tan(x)^13/13

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sympy [A]  time = 19.28, size = 27, normalized size = 0.82 \[ \frac {\tan ^{13}{\relax (x )}}{13} + \frac {3 \tan ^{11}{\relax (x )}}{11} + \frac {\tan ^{9}{\relax (x )}}{3} + \frac {\tan ^{7}{\relax (x )}}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**2*tan(x)**6*(1+tan(x)**2)**3,x)

[Out]

tan(x)**13/13 + 3*tan(x)**11/11 + tan(x)**9/3 + tan(x)**7/7

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