3.1.63 \(\int (a \sec ^4(x))^{3/2} \, dx\) [63]

Optimal. Leaf size=61 \[ a \cos (x) \sqrt {a \sec ^4(x)} \sin (x)+\frac {2}{3} a \sqrt {a \sec ^4(x)} \sin ^2(x) \tan (x)+\frac {1}{5} a \sqrt {a \sec ^4(x)} \sin ^2(x) \tan ^3(x) \]

[Out]

a*cos(x)*sin(x)*(a*sec(x)^4)^(1/2)+2/3*a*sin(x)^2*(a*sec(x)^4)^(1/2)*tan(x)+1/5*a*sin(x)^2*(a*sec(x)^4)^(1/2)*
tan(x)^3

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Rubi [A]
time = 0.02, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4208, 3852} \begin {gather*} a \sin (x) \cos (x) \sqrt {a \sec ^4(x)}+\frac {1}{5} a \sin ^2(x) \tan ^3(x) \sqrt {a \sec ^4(x)}+\frac {2}{3} a \sin ^2(x) \tan (x) \sqrt {a \sec ^4(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*Sec[x]^4)^(3/2),x]

[Out]

a*Cos[x]*Sqrt[a*Sec[x]^4]*Sin[x] + (2*a*Sqrt[a*Sec[x]^4]*Sin[x]^2*Tan[x])/3 + (a*Sqrt[a*Sec[x]^4]*Sin[x]^2*Tan
[x]^3)/5

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 \left (a \sec ^4(x)\right )^{3/2} \, dx &=\left (a \cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \int \sec ^6(x) \, dx\\ &=-\left (\left (a \cos ^2(x) \sqrt {a \sec ^4(x)}\right ) \text {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (x)\right )\right )\\ &=a \cos (x) \sqrt {a \sec ^4(x)} \sin (x)+\frac {2}{3} a \sqrt {a \sec ^4(x)} \sin ^2(x) \tan (x)+\frac {1}{5} a \sqrt {a \sec ^4(x)} \sin ^2(x) \tan ^3(x)\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 30, normalized size = 0.49 \begin {gather*} \frac {1}{15} \cos (x) (8+6 \cos (2 x)+\cos (4 x)) \left (a \sec ^4(x)\right )^{3/2} \sin (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*Sec[x]^4)^(3/2),x]

[Out]

(Cos[x]*(8 + 6*Cos[2*x] + Cos[4*x])*(a*Sec[x]^4)^(3/2)*Sin[x])/15

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Maple [A]
time = 0.21, size = 29, normalized size = 0.48

method	result	size
default	\(\frac {\left (8 \left (\cos ^{4}\left (x \right )\right )+4 \left (\cos ^{2}\left (x \right )\right )+3\right ) \cos \left (x \right ) \sin \left (x \right ) \left (\frac {a}{\cos \left (x \right )^{4}}\right )^{\frac {3}{2}}}{15}\)	\(29\)
risch	\(\frac {16 i a \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{4}}}\, \left (5+11 \cos \left (2 x \right )+9 i \sin \left (2 x \right )\right )}{15 \left ({\mathrm e}^{2 i x}+1\right )^{3}}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sec(x)^4)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/15*(8*cos(x)^4+4*cos(x)^2+3)*cos(x)*sin(x)*(a/cos(x)^4)^(3/2)

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Maxima [A]
time = 0.53, size = 25, normalized size = 0.41 \begin {gather*} \frac {1}{5} \, a^{\frac {3}{2}} \tan \left (x\right )^{5} + \frac {2}{3} \, a^{\frac {3}{2}} \tan \left (x\right )^{3} + a^{\frac {3}{2}} \tan \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^4)^(3/2),x, algorithm="maxima")

[Out]

1/5*a^(3/2)*tan(x)^5 + 2/3*a^(3/2)*tan(x)^3 + a^(3/2)*tan(x)

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Fricas [A]
time = 2.54, size = 34, normalized size = 0.56 \begin {gather*} \frac {{\left (8 \, a \cos \left (x\right )^{4} + 4 \, a \cos \left (x\right )^{2} + 3 \, a\right )} \sqrt {\frac {a}{\cos \left (x\right )^{4}}} \sin \left (x\right )}{15 \, \cos \left (x\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^4)^(3/2),x, algorithm="fricas")

[Out]

1/15*(8*a*cos(x)^4 + 4*a*cos(x)^2 + 3*a)*sqrt(a/cos(x)^4)*sin(x)/cos(x)^3

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \sec ^{4}{\left (x \right )}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)**4)**(3/2),x)

[Out]

Integral((a*sec(x)**4)**(3/2), x)

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Giac [A]
time = 0.45, size = 22, normalized size = 0.36 \begin {gather*} \frac {1}{15} \, {\left (3 \, \tan \left (x\right )^{5} + 10 \, \tan \left (x\right )^{3} + 15 \, \tan \left (x\right )\right )} a^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(x)^4)^(3/2),x, algorithm="giac")

[Out]

1/15*(3*tan(x)^5 + 10*tan(x)^3 + 15*tan(x))*a^(3/2)

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Mupad [B]
time = 0.56, size = 36, normalized size = 0.59 \begin {gather*} \frac {4\,a^{3/2}\,\sin \left (x\right )}{5\,{\cos \left (x\right )}^3}+\frac {a^{3/2}\,\sin \left (x\right )}{5\,{\cos \left (x\right )}^5}-\frac {8\,a^{3/2}\,{\sin \left (x\right )}^3}{15\,{\cos \left (x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a/cos(x)^4)^(3/2),x)

[Out]

(4*a^(3/2)*sin(x))/(5*cos(x)^3) + (a^(3/2)*sin(x))/(5*cos(x)^5) - (8*a^(3/2)*sin(x)^3)/(15*cos(x)^3)

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