3.863 \(\int \sec (x) \sqrt{1+\sec (x)} \tan ^3(x) \, dx\)

Optimal. Leaf size=25 \[ \frac{2}{7} (\sec (x)+1)^{7/2}-\frac{4}{5} (\sec (x)+1)^{5/2} \]

[Out]

(-4*(1 + Sec[x])^(5/2))/5 + (2*(1 + Sec[x])^(7/2))/7

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Rubi [A]  time = 0.0853067, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4373, 1570, 1469, 627, 43} \[ \frac{2}{7} (\sec (x)+1)^{7/2}-\frac{4}{5} (\sec (x)+1)^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]*Sqrt[1 + Sec[x]]*Tan[x]^3,x]

[Out]

(-4*(1 + Sec[x])^(5/2))/5 + (2*(1 + Sec[x])^(7/2))/7

Rule 4373

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_), x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dis
t[(b*c*d^(n - 1))^(-1), Subst[Int[SubstFor[(1 - d^2*x^2)^((n - 1)/2)/x^n, Cos[c*(a + b*x)]/d, u, x], x], x, Co
s[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && IntegerQ[(n - 1)/2]
&& NonsumQ[u] && (EqQ[F, Tan] || EqQ[F, tan])

Rule 1570

Int[(x_)^(m_.)*((a_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Int[x^(m - 2*n
*p)*(d + e*x^n)^q*(c + a*x^(2*n))^p, x] /; FreeQ[{a, c, d, e, m, n, q}, x] && EqQ[mn2, -2*n] && IntegerQ[p]

Rule 1469

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[I
nt[(d + e*x)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[Sim
plify[m - n + 1], 0]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \sec (x) \sqrt{1+\sec (x)} \tan ^3(x) \, dx &=-\operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{1}{x}} \left (1-x^2\right )}{x^4} \, dx,x,\cos (x)\right )\\ &=-\operatorname{Subst}\left (\int \frac{\left (-1+\frac{1}{x^2}\right ) \sqrt{1+\frac{1}{x}}}{x^2} \, dx,x,\cos (x)\right )\\ &=\operatorname{Subst}\left (\int \sqrt{1+x} \left (-1+x^2\right ) \, dx,x,\sec (x)\right )\\ &=\operatorname{Subst}\left (\int (-1+x) (1+x)^{3/2} \, dx,x,\sec (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-2 (1+x)^{3/2}+(1+x)^{5/2}\right ) \, dx,x,\sec (x)\right )\\ &=-\frac{4}{5} (1+\sec (x))^{5/2}+\frac{2}{7} (1+\sec (x))^{7/2}\\ \end{align*}

Mathematica [A]  time = 0.189612, size = 30, normalized size = 1.2 \[ -\frac{8}{35} \cos ^4\left (\frac{x}{2}\right ) (9 \cos (x)-5) \sec ^3(x) \sqrt{\sec (x)+1} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]*Sqrt[1 + Sec[x]]*Tan[x]^3,x]

[Out]

(-8*Cos[x/2]^4*(-5 + 9*Cos[x])*Sec[x]^3*Sqrt[1 + Sec[x]])/35

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Maple [A]  time = 0.067, size = 34, normalized size = 1.4 \begin{align*} -{\frac{ \left ( 18\,\cos \left ( x \right ) -10 \right ) \left ( \sin \left ( x \right ) \right ) ^{4}}{35\, \left ( -1+\cos \left ( x \right ) \right ) ^{2} \left ( \cos \left ( x \right ) \right ) ^{3}}\sqrt{{\frac{1+\cos \left ( x \right ) }{\cos \left ( x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(9*cos(x)-5)*((1+cos(x))/cos(x))^(1/2)*sin(x)^4/(-1+cos(x))^2/cos(x)^3

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Maxima [A]  time = 0.957432, size = 28, normalized size = 1.12 \begin{align*} \frac{2}{7} \,{\left (\frac{1}{\cos \left (x\right )} + 1\right )}^{\frac{7}{2}} - \frac{4}{5} \,{\left (\frac{1}{\cos \left (x\right )} + 1\right )}^{\frac{5}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(1/cos(x) + 1)^(7/2) - 4/5*(1/cos(x) + 1)^(5/2)

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Fricas [B]  time = 2.03862, size = 111, normalized size = 4.44 \begin{align*} -\frac{2 \,{\left (9 \, \cos \left (x\right )^{3} + 13 \, \cos \left (x\right )^{2} - \cos \left (x\right ) - 5\right )} \sqrt{\frac{\cos \left (x\right ) + 1}{\cos \left (x\right )}}}{35 \, \cos \left (x\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(9*cos(x)^3 + 13*cos(x)^2 - cos(x) - 5)*sqrt((cos(x) + 1)/cos(x))/cos(x)^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\sec{\left (x \right )} + 1} \tan ^{3}{\left (x \right )} \sec{\left (x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(sec(x) + 1)*tan(x)**3*sec(x), x)

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Giac [B]  time = 1.092, size = 173, normalized size = 6.92 \begin{align*} -\frac{2 \,{\left (35 \,{\left (\sqrt{\cos \left (x\right )^{2} + \cos \left (x\right )} - \cos \left (x\right )\right )}^{6} - 35 \,{\left (\sqrt{\cos \left (x\right )^{2} + \cos \left (x\right )} - \cos \left (x\right )\right )}^{5} - 35 \,{\left (\sqrt{\cos \left (x\right )^{2} + \cos \left (x\right )} - \cos \left (x\right )\right )}^{4} + 105 \,{\left (\sqrt{\cos \left (x\right )^{2} + \cos \left (x\right )} - \cos \left (x\right )\right )}^{3} - 91 \,{\left (\sqrt{\cos \left (x\right )^{2} + \cos \left (x\right )} - \cos \left (x\right )\right )}^{2} + 35 \, \sqrt{\cos \left (x\right )^{2} + \cos \left (x\right )} - 35 \, \cos \left (x\right ) - 5\right )} \mathrm{sgn}\left (\cos \left (x\right )\right )}{35 \,{\left (\sqrt{\cos \left (x\right )^{2} + \cos \left (x\right )} - \cos \left (x\right )\right )}^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(35*(sqrt(cos(x)^2 + cos(x)) - cos(x))^6 - 35*(sqrt(cos(x)^2 + cos(x)) - cos(x))^5 - 35*(sqrt(cos(x)^2 +
cos(x)) - cos(x))^4 + 105*(sqrt(cos(x)^2 + cos(x)) - cos(x))^3 - 91*(sqrt(cos(x)^2 + cos(x)) - cos(x))^2 + 35
*sqrt(cos(x)^2 + cos(x)) - 35*cos(x) - 5)*sgn(cos(x))/(sqrt(cos(x)^2 + cos(x)) - cos(x))^7