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]

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

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Rubi [A]  time = 0.09, antiderivative size = 25, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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])

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*}

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Mathematica [A]  time = 0.19, size = 30, normalized size = 1.20 \[ -\frac {8}{35} \cos ^4\left (\frac {x}{2}\right ) (9 \cos (x)-5) \sec ^3(x) \sqrt {\sec (x)+1} \]

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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fricas [B]  time = 0.87, size = 35, normalized size = 1.40 \[ -\frac {2 \, {\left (9 \, \cos \relax (x)^{3} + 13 \, \cos \relax (x)^{2} - \cos \relax (x) - 5\right )} \sqrt {\frac {\cos \relax (x) + 1}{\cos \relax (x)}}}{35 \, \cos \relax (x)^{3}} \]

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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giac [B]  time = 0.17, size = 128, normalized size = 5.12 \[ -\frac {2 \, {\left (35 \, {\left (\sqrt {\cos \relax (x)^{2} + \cos \relax (x)} - \cos \relax (x)\right )}^{6} - 35 \, {\left (\sqrt {\cos \relax (x)^{2} + \cos \relax (x)} - \cos \relax (x)\right )}^{5} - 35 \, {\left (\sqrt {\cos \relax (x)^{2} + \cos \relax (x)} - \cos \relax (x)\right )}^{4} + 105 \, {\left (\sqrt {\cos \relax (x)^{2} + \cos \relax (x)} - \cos \relax (x)\right )}^{3} - 91 \, {\left (\sqrt {\cos \relax (x)^{2} + \cos \relax (x)} - \cos \relax (x)\right )}^{2} + 35 \, \sqrt {\cos \relax (x)^{2} + \cos \relax (x)} - 35 \, \cos \relax (x) - 5\right )} \mathrm {sgn}\left (\cos \relax (x)\right )}{35 \, {\left (\sqrt {\cos \relax (x)^{2} + \cos \relax (x)} - \cos \relax (x)\right )}^{7}} \]

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

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maple [A]  time = 0.17, size = 34, normalized size = 1.36 \[ -\frac {2 \left (9 \cos \relax (x )-5\right ) \sqrt {\frac {1+\cos \relax (x )}{\cos \relax (x )}}\, \left (\sin ^{4}\relax (x )\right )}{35 \left (-1+\cos \relax (x )\right )^{2} \cos \relax (x )^{3}} \]

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 = 0.33, size = 21, normalized size = 0.84 \[ \frac {2}{7} \, {\left (\frac {1}{\cos \relax (x)} + 1\right )}^{\frac {7}{2}} - \frac {4}{5} \, {\left (\frac {1}{\cos \relax (x)} + 1\right )}^{\frac {5}{2}} \]

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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mupad [B]  time = 3.33, size = 24, normalized size = 0.96 \[ -\frac {2\,{\left (\cos \relax (x)+1\right )}^{5/2}\,\sqrt {\frac {1}{\cos \relax (x)}}\,\left (9\,\cos \relax (x)-5\right )}{35\,{\cos \relax (x)}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\sec {\relax (x )} + 1} \tan ^{3}{\relax (x )} \sec {\relax (x )}\, dx \]

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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