3.889 \(\int \sec (x) (5-11 \sec ^5(x))^2 \tan (x) \, dx\)

Optimal. Leaf size=19 \[ 11 \sec ^{11}(x)-\frac{55 \sec ^6(x)}{3}+25 \sec (x) \]

[Out]

25*Sec[x] - (55*Sec[x]^6)/3 + 11*Sec[x]^11

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Rubi [A]  time = 0.0396507, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4339, 270} \[ 11 \sec ^{11}(x)-\frac{55 \sec ^6(x)}{3}+25 \sec (x) \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]*(5 - 11*Sec[x]^5)^2*Tan[x],x]

[Out]

25*Sec[x] - (55*Sec[x]^6)/3 + 11*Sec[x]^11

Rule 4339

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dist[(b*
c)^(-1), Subst[Int[SubstFor[1/x, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[
c*(a + b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Tan] || EqQ[F, tan])

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]

Rubi steps

\begin{align*} \int \sec (x) \left (5-11 \sec ^5(x)\right )^2 \tan (x) \, dx &=-\operatorname{Subst}\left (\int \frac{\left (11-5 x^5\right )^2}{x^{12}} \, dx,x,\cos (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (\frac{121}{x^{12}}-\frac{110}{x^7}+\frac{25}{x^2}\right ) \, dx,x,\cos (x)\right )\\ &=25 \sec (x)-\frac{55 \sec ^6(x)}{3}+11 \sec ^{11}(x)\\ \end{align*}

Mathematica [A]  time = 0.013386, size = 19, normalized size = 1. \[ 11 \sec ^{11}(x)-\frac{55 \sec ^6(x)}{3}+25 \sec (x) \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]*(5 - 11*Sec[x]^5)^2*Tan[x],x]

[Out]

25*Sec[x] - (55*Sec[x]^6)/3 + 11*Sec[x]^11

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Maple [A]  time = 0.017, size = 18, normalized size = 1. \begin{align*} 25\,\sec \left ( x \right ) -{\frac{55\, \left ( \sec \left ( x \right ) \right ) ^{6}}{3}}+11\, \left ( \sec \left ( x \right ) \right ) ^{11} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)*(5-11*sec(x)^5)^2*tan(x),x)

[Out]

25*sec(x)-55/3*sec(x)^6+11*sec(x)^11

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Maxima [A]  time = 0.957646, size = 27, normalized size = 1.42 \begin{align*} \frac{75 \, \cos \left (x\right )^{10} - 55 \, \cos \left (x\right )^{5} + 33}{3 \, \cos \left (x\right )^{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*(5-11*sec(x)^5)^2*tan(x),x, algorithm="maxima")

[Out]

1/3*(75*cos(x)^10 - 55*cos(x)^5 + 33)/cos(x)^11

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Fricas [A]  time = 2.52757, size = 66, normalized size = 3.47 \begin{align*} \frac{75 \, \cos \left (x\right )^{10} - 55 \, \cos \left (x\right )^{5} + 33}{3 \, \cos \left (x\right )^{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*(5-11*sec(x)^5)^2*tan(x),x, algorithm="fricas")

[Out]

1/3*(75*cos(x)^10 - 55*cos(x)^5 + 33)/cos(x)^11

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Sympy [A]  time = 33.4383, size = 19, normalized size = 1. \begin{align*} 11 \sec ^{11}{\left (x \right )} - \frac{55 \sec ^{6}{\left (x \right )}}{3} + 25 \sec{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*(5-11*sec(x)**5)**2*tan(x),x)

[Out]

11*sec(x)**11 - 55*sec(x)**6/3 + 25*sec(x)

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Giac [A]  time = 1.07273, size = 27, normalized size = 1.42 \begin{align*} \frac{75 \, \cos \left (x\right )^{10} - 55 \, \cos \left (x\right )^{5} + 33}{3 \, \cos \left (x\right )^{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)*(5-11*sec(x)^5)^2*tan(x),x, algorithm="giac")

[Out]

1/3*(75*cos(x)^10 - 55*cos(x)^5 + 33)/cos(x)^11