3.1.64 \(\int \sqrt {a \sec ^4(x)} \, dx\) [64]

Optimal. Leaf size=15 \[ \cos (x) \sqrt {a \sec ^4(x)} \sin (x) \]

[Out]

cos(x)*sin(x)*(a*sec(x)^4)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4208, 3852, 8} \begin {gather*} \sin (x) \cos (x) \sqrt {a \sec ^4(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Cos[x]*Sqrt[a*Sec[x]^4]*Sin[x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 4208

Int[((b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[b^IntPart[p]*((b*(c*Sec[e + f*x])^n)^
FracPart[p]/(c*Sec[e + f*x])^(n*FracPart[p])), Int[(c*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{b, c, e, f, n, p},
x] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \sqrt {a \sec ^4(x)} \, dx &=\left (\cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \int \sec ^2(x) \, dx\\ &=-\left (\left (\cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \text {Subst}(\int 1 \, dx,x,-\tan (x))\right )\\ &=\cos (x) \sqrt {a \sec ^4(x)} \sin (x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 15, normalized size = 1.00 \begin {gather*} \cos (x) \sqrt {a \sec ^4(x)} \sin (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Cos[x]*Sqrt[a*Sec[x]^4]*Sin[x]

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Maple [A]
time = 0.20, size = 14, normalized size = 0.93

method	result	size
default	\(\cos \left (x \right ) \sin \left (x \right ) \sqrt {\frac {a}{\cos \left (x \right )^{4}}}\)	\(14\)
risch	\(2 i \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}\, \left (1+{\mathrm e}^{-2 i x}\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sec(x)^4)^(1/2),x,method=_RETURNVERBOSE)

[Out]

cos(x)*sin(x)*(a/cos(x)^4)^(1/2)

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Maxima [A]
time = 0.51, size = 6, normalized size = 0.40 \begin {gather*} \sqrt {a} \tan \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^4)^(1/2),x, algorithm="maxima")

[Out]

sqrt(a)*tan(x)

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Fricas [A]
time = 2.90, size = 13, normalized size = 0.87 \begin {gather*} \sqrt {\frac {a}{\cos \left (x\right )^{4}}} \cos \left (x\right ) \sin \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^4)^(1/2),x, algorithm="fricas")

[Out]

sqrt(a/cos(x)^4)*cos(x)*sin(x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \sec ^{4}{\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)**4)**(1/2),x)

[Out]

Integral(sqrt(a*sec(x)**4), x)

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Giac [A]
time = 0.42, size = 6, normalized size = 0.40 \begin {gather*} \sqrt {a} \tan \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^4)^(1/2),x, algorithm="giac")

[Out]

sqrt(a)*tan(x)

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Mupad [B]
time = 0.11, size = 6, normalized size = 0.40 \begin {gather*} \sqrt {a}\,\mathrm {tan}\left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a/cos(x)^4)^(1/2),x)

[Out]

a^(1/2)*tan(x)

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