3.412 \(\int \sqrt {\sec ^4(x) \tan (x)} \, dx\)

Optimal. Leaf size=19 \[ \frac {2}{3} \sin (x) \cos (x) \sqrt {\tan (x) \sec ^4(x)} \]

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Rubi [A]  time = 0.12, antiderivative size = 29, normalized size of antiderivative = 1.53, number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {1999, 1954, 1250, 30} \[ \frac {2 \tan ^2(x) \sec ^2(x)}{3 \sqrt {\tan ^5(x)+2 \tan ^3(x)+\tan (x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sec[x]^4*Tan[x]],x]

[Out]

(2*Sec[x]^2*Tan[x]^2)/(3*Sqrt[Tan[x] + 2*Tan[x]^3 + Tan[x]^5])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 1250

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dis
t[(a + b*x^2 + c*x^4)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(f*x)^m*(d + e*x^2)^q*(b/2
+ c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rule 1954

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(k_.) + (c_.)*(x_)^(n_.))^(p_)*((A_) + (B_.)*(x_)^(q_)), x_Symbo
l] :> Dist[(a*x^j + b*x^k + c*x^n)^p/(x^(j*p)*(a + b*x^(k - j) + c*x^(2*(k - j)))^p), Int[x^(m + j*p)*(A + B*x
^(k - j))*(a + b*x^(k - j) + c*x^(2*(k - j)))^p, x], x] /; FreeQ[{a, b, c, A, B, j, k, m, p}, x] && EqQ[q, k -
j] && EqQ[n, 2*k - j] &&  !IntegerQ[p] && PosQ[k - j]

Rule 1999

Int[(u_)^(p_.)*((f_.)*(x_))^(m_.)*(z_), x_Symbol] :> Int[(f*x)^m*ExpandToSum[z, x]*ExpandToSum[u, x]^p, x] /;
FreeQ[{f, m, p}, x] && BinomialQ[z, x] && GeneralizedTrinomialQ[u, x] && EqQ[BinomialDegree[z, x] - Generalize
dTrinomialDegree[u, x], 0] &&  !(BinomialMatchQ[z, x] && GeneralizedTrinomialMatchQ[u, x])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.00 \[ \frac {2}{3} \sin (x) \cos (x) \sqrt {\tan (x) \sec ^4(x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sec[x]^4*Tan[x]],x]

[Out]

(2*Cos[x]*Sin[x]*Sqrt[Sec[x]^4*Tan[x]])/3

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\sec ^4(x) \tan (x)} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Sqrt[Sec[x]^4*Tan[x]],x]

[Out]

Could not integrate

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fricas [A]  time = 0.80, size = 15, normalized size = 0.79 \[ \frac {2}{3} \, \sqrt {\frac {\sin \relax (x)}{\cos \relax (x)^{5}}} \cos \relax (x) \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)/cos(x)^5)^(1/2),x, algorithm="fricas")

[Out]

2/3*sqrt(sin(x)/cos(x)^5)*cos(x)*sin(x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {\sin \relax (x)}{\cos \relax (x)^{5}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)/cos(x)^5)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(sin(x)/cos(x)^5), x)

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maple [A]  time = 0.36, size = 16, normalized size = 0.84

method	result	size
default	\(\frac {2 \cos \relax (x ) \sin \relax (x ) \sqrt {\frac {\sin \relax (x )}{\cos \relax (x )^{5}}}}{3}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(x)/cos(x)^5)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/3*cos(x)*sin(x)*(sin(x)/cos(x)^5)^(1/2)

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maxima [A]  time = 1.05, size = 6, normalized size = 0.32 \[ \frac {2}{3} \, \tan \relax (x)^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)/cos(x)^5)^(1/2),x, algorithm="maxima")

[Out]

2/3*tan(x)^(3/2)

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mupad [B]  time = 0.51, size = 15, normalized size = 0.79 \[ \frac {\sin \left (2\,x\right )\,\sqrt {\frac {\sin \relax (x)}{{\cos \relax (x)}^5}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(x)/cos(x)^5)^(1/2),x)

[Out]

(sin(2*x)*(sin(x)/cos(x)^5)^(1/2))/3

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sin(x)/cos(x)**5)**(1/2),x)

[Out]

Timed out

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