3.49 \(\int (a \sec ^2(x))^{3/2} \, dx\)

Optimal. Leaf size=46 \[ \frac {1}{2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a \sec ^2(x)}}\right )+\frac {1}{2} a \tan (x) \sqrt {a \sec ^2(x)} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4122, 195, 217, 206} \[ \frac {1}{2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a \sec ^2(x)}}\right )+\frac {1}{2} a \tan (x) \sqrt {a \sec ^2(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 4122

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(b*ff)
/f, Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 55, normalized size = 1.20 \[ \frac {1}{2} a \cos (x) \sqrt {a \sec ^2(x)} \left (\tan (x) \sec (x)-\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Cos[x]*Sqrt[a*Sec[x]^2]*(-Log[Cos[x/2] - Sin[x/2]] + Log[Cos[x/2] + Sin[x/2]] + Sec[x]*Tan[x]))/2

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fricas [A]  time = 0.67, size = 39, normalized size = 0.85 \[ -\frac {{\left (a \cos \relax (x)^{2} \log \left (-\frac {\sin \relax (x) - 1}{\sin \relax (x) + 1}\right ) - 2 \, a \sin \relax (x)\right )} \sqrt {\frac {a}{\cos \relax (x)^{2}}}}{4 \, \cos \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a*cos(x)^2*log(-(sin(x) - 1)/(sin(x) + 1)) - 2*a*sin(x))*sqrt(a/cos(x)^2)/cos(x)

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giac [A]  time = 0.60, size = 42, normalized size = 0.91 \[ \frac {1}{4} \, {\left (\log \left (\sin \relax (x) + 1\right ) \mathrm {sgn}\left (\cos \relax (x)\right ) - \log \left (-\sin \relax (x) + 1\right ) \mathrm {sgn}\left (\cos \relax (x)\right ) - \frac {2 \, \mathrm {sgn}\left (\cos \relax (x)\right ) \sin \relax (x)}{\sin \relax (x)^{2} - 1}\right )} a^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(log(sin(x) + 1)*sgn(cos(x)) - log(-sin(x) + 1)*sgn(cos(x)) - 2*sgn(cos(x))*sin(x)/(sin(x)^2 - 1))*a^(3/2)

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maple [A]  time = 0.30, size = 57, normalized size = 1.24 \[ -\frac {\left (\left (\cos ^{2}\relax (x )\right ) \ln \left (-\frac {-1+\cos \relax (x )+\sin \relax (x )}{\sin \relax (x )}\right )-\left (\cos ^{2}\relax (x )\right ) \ln \left (-\frac {-\sin \relax (x )-1+\cos \relax (x )}{\sin \relax (x )}\right )-\sin \relax (x )\right ) \cos \relax (x ) \left (\frac {a}{\cos \relax (x )^{2}}\right )^{\frac {3}{2}}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sec(x)^2)^(3/2),x)

[Out]

-1/2*(cos(x)^2*ln(-(-1+cos(x)+sin(x))/sin(x))-cos(x)^2*ln(-(-sin(x)-1+cos(x))/sin(x))-sin(x))*cos(x)*(a/cos(x)
^2)^(3/2)

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maxima [B]  time = 1.03, size = 324, normalized size = 7.04 \[ -\frac {{\left (8 \, a \cos \left (3 \, x\right ) \sin \left (2 \, x\right ) - 8 \, a \cos \relax (x) \sin \left (2 \, x\right ) + 8 \, a \cos \left (2 \, x\right ) \sin \relax (x) - 4 \, {\left (a \sin \left (3 \, x\right ) - a \sin \relax (x)\right )} \cos \left (4 \, x\right ) - {\left (a \cos \left (4 \, x\right )^{2} + 4 \, a \cos \left (2 \, x\right )^{2} + a \sin \left (4 \, x\right )^{2} + 4 \, a \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 4 \, a \sin \left (2 \, x\right )^{2} + 2 \, {\left (2 \, a \cos \left (2 \, x\right ) + a\right )} \cos \left (4 \, x\right ) + 4 \, a \cos \left (2 \, x\right ) + a\right )} \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \sin \relax (x) + 1\right ) + {\left (a \cos \left (4 \, x\right )^{2} + 4 \, a \cos \left (2 \, x\right )^{2} + a \sin \left (4 \, x\right )^{2} + 4 \, a \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 4 \, a \sin \left (2 \, x\right )^{2} + 2 \, {\left (2 \, a \cos \left (2 \, x\right ) + a\right )} \cos \left (4 \, x\right ) + 4 \, a \cos \left (2 \, x\right ) + a\right )} \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \sin \relax (x) + 1\right ) + 4 \, {\left (a \cos \left (3 \, x\right ) - a \cos \relax (x)\right )} \sin \left (4 \, x\right ) - 4 \, {\left (2 \, a \cos \left (2 \, x\right ) + a\right )} \sin \left (3 \, x\right ) + 4 \, a \sin \relax (x)\right )} \sqrt {a}}{4 \, {\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(8*a*cos(3*x)*sin(2*x) - 8*a*cos(x)*sin(2*x) + 8*a*cos(2*x)*sin(x) - 4*(a*sin(3*x) - a*sin(x))*cos(4*x) -
(a*cos(4*x)^2 + 4*a*cos(2*x)^2 + a*sin(4*x)^2 + 4*a*sin(4*x)*sin(2*x) + 4*a*sin(2*x)^2 + 2*(2*a*cos(2*x) + a)
*cos(4*x) + 4*a*cos(2*x) + a)*log(cos(x)^2 + sin(x)^2 + 2*sin(x) + 1) + (a*cos(4*x)^2 + 4*a*cos(2*x)^2 + a*sin
(4*x)^2 + 4*a*sin(4*x)*sin(2*x) + 4*a*sin(2*x)^2 + 2*(2*a*cos(2*x) + a)*cos(4*x) + 4*a*cos(2*x) + a)*log(cos(x
)^2 + sin(x)^2 - 2*sin(x) + 1) + 4*(a*cos(3*x) - a*cos(x))*sin(4*x) - 4*(2*a*cos(2*x) + a)*sin(3*x) + 4*a*sin(
x))*sqrt(a)/(2*(2*cos(2*x) + 1)*cos(4*x) + cos(4*x)^2 + 4*cos(2*x)^2 + sin(4*x)^2 + 4*sin(4*x)*sin(2*x) + 4*si
n(2*x)^2 + 4*cos(2*x) + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (\frac {a}{{\cos \relax (x)}^2}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a/cos(x)^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \sec ^{2}{\relax (x )}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*sec(x)**2)**(3/2), x)

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