3.255 \(\int (1+\sec ^2(x))^{3/2} \, dx\)

Optimal. Leaf size=42 \[ \tan ^{-1}\left (\frac{\tan (x)}{\sqrt{\tan ^2(x)+2}}\right )+\frac{1}{2} \tan (x) \sqrt{\tan ^2(x)+2}+2 \sinh ^{-1}\left (\frac{\tan (x)}{\sqrt{2}}\right ) \]

[Out]

2*ArcSinh[Tan[x]/Sqrt[2]] + ArcTan[Tan[x]/Sqrt[2 + Tan[x]^2]] + (Tan[x]*Sqrt[2 + Tan[x]^2])/2

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Rubi [A]  time = 0.0368925, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4128, 416, 523, 215, 377, 203} \[ \tan ^{-1}\left (\frac{\tan (x)}{\sqrt{\tan ^2(x)+2}}\right )+\frac{1}{2} \tan (x) \sqrt{\tan ^2(x)+2}+2 \sinh ^{-1}\left (\frac{\tan (x)}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Sec[x]^2)^(3/2),x]

[Out]

2*ArcSinh[Tan[x]/Sqrt[2]] + ArcTan[Tan[x]/Sqrt[2 + Tan[x]^2]] + (Tan[x]*Sqrt[2 + Tan[x]^2])/2

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 215

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

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 203

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

Rubi steps

\begin{align*} \int \left (1+\sec ^2(x)\right )^{3/2} \, dx &=\operatorname{Subst}\left (\int \frac{\left (2+x^2\right )^{3/2}}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\frac{1}{2} \tan (x) \sqrt{2+\tan ^2(x)}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{6+4 x^2}{\left (1+x^2\right ) \sqrt{2+x^2}} \, dx,x,\tan (x)\right )\\ &=\frac{1}{2} \tan (x) \sqrt{2+\tan ^2(x)}+2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+x^2}} \, dx,x,\tan (x)\right )+\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{2+x^2}} \, dx,x,\tan (x)\right )\\ &=2 \sinh ^{-1}\left (\frac{\tan (x)}{\sqrt{2}}\right )+\frac{1}{2} \tan (x) \sqrt{2+\tan ^2(x)}+\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\tan (x)}{\sqrt{2+\tan ^2(x)}}\right )\\ &=2 \sinh ^{-1}\left (\frac{\tan (x)}{\sqrt{2}}\right )+\tan ^{-1}\left (\frac{\tan (x)}{\sqrt{2+\tan ^2(x)}}\right )+\frac{1}{2} \tan (x) \sqrt{2+\tan ^2(x)}\\ \end{align*}

Mathematica [C]  time = 0.172578, size = 109, normalized size = 2.6 \[ \frac{\left (\cos ^2(x)+1\right ) \sec (x) \sqrt{\sec ^2(x)+1} \left (\sin (x) \sqrt{\cos (2 x)+3}-2 i \sqrt{2} \cos ^2(x) \log \left (\sqrt{\cos (2 x)+3}+i \sqrt{2} \sin (x)\right )+4 \sqrt{2} \cos ^2(x) \tanh ^{-1}\left (\frac{\sqrt{2} \sin (x)}{\sqrt{\cos (2 x)+3}}\right )\right )}{(\cos (2 x)+3)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Sec[x]^2)^(3/2),x]

[Out]

((1 + Cos[x]^2)*Sec[x]*Sqrt[1 + Sec[x]^2]*(4*Sqrt[2]*ArcTanh[(Sqrt[2]*Sin[x])/Sqrt[3 + Cos[2*x]]]*Cos[x]^2 - (
2*I)*Sqrt[2]*Cos[x]^2*Log[Sqrt[3 + Cos[2*x]] + I*Sqrt[2]*Sin[x]] + Sqrt[3 + Cos[2*x]]*Sin[x]))/(3 + Cos[2*x])^
(3/2)

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Maple [C]  time = 0.296, size = 429, normalized size = 10.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+sec(x)^2)^(3/2),x)

[Out]

(-1/4+1/4*I)*(8*(-1)^(3/4)*sin(x)*cos(x)^2*(-(I*cos(x)-1-I-cos(x))/(cos(x)+1))^(1/2)*EllipticPi((-1)^(1/4)*(-1
+cos(x))/sin(x),-I,I)*((I*cos(x)+1-I+cos(x))/(cos(x)+1))^(1/2)+4*(-1)^(3/4)*sin(x)*cos(x)^2*(-(I*cos(x)-1-I-co
s(x))/(cos(x)+1))^(1/2)*EllipticPi((-1)^(1/4)*(-1+cos(x))/sin(x),I,I)*((I*cos(x)+1-I+cos(x))/(cos(x)+1))^(1/2)
-8*(-1)^(1/4)*sin(x)*cos(x)^2*(-(I*cos(x)-1-I-cos(x))/(cos(x)+1))^(1/2)*EllipticPi((-1)^(1/4)*(-1+cos(x))/sin(
x),-I,I)*((I*cos(x)+1-I+cos(x))/(cos(x)+1))^(1/2)-4*(-1)^(1/4)*sin(x)*cos(x)^2*(-(I*cos(x)-1-I-cos(x))/(cos(x)
+1))^(1/2)*EllipticPi((-1)^(1/4)*(-1+cos(x))/sin(x),I,I)*((I*cos(x)+1-I+cos(x))/(cos(x)+1))^(1/2)+6*sin(x)*cos
(x)^2*2^(1/2)*(-(I*cos(x)-1-I-cos(x))/(cos(x)+1))^(1/2)*EllipticF((1/2+1/2*I)*2^(1/2)*(-1+cos(x))/sin(x),I)*((
I*cos(x)+1-I+cos(x))/(cos(x)+1))^(1/2)-I*cos(x)^3+I*cos(x)^2-cos(x)^3-I*cos(x)+cos(x)^2+1+I-cos(x))*((cos(x)^2
+1)/cos(x)^2)^(3/2)*sin(x)*cos(x)/(-1+cos(x))/(cos(x)^2+1)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\sec \left (x\right )^{2} + 1\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+sec(x)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((sec(x)^2 + 1)^(3/2), x)

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Fricas [B]  time = 0.542438, size = 536, normalized size = 12.76 \begin{align*} \frac{\arctan \left (\frac{\sqrt{\frac{\cos \left (x\right )^{2} + 1}{\cos \left (x\right )^{2}}} \cos \left (x\right )^{3} \sin \left (x\right ) + \cos \left (x\right ) \sin \left (x\right )}{\cos \left (x\right )^{4} + \cos \left (x\right )^{2} - 1}\right ) \cos \left (x\right ) - \arctan \left (\frac{\sin \left (x\right )}{\cos \left (x\right )}\right ) \cos \left (x\right ) + 2 \, \cos \left (x\right ) \log \left (\cos \left (x\right )^{2} + \cos \left (x\right ) \sin \left (x\right ) +{\left (\cos \left (x\right )^{2} + \cos \left (x\right ) \sin \left (x\right )\right )} \sqrt{\frac{\cos \left (x\right )^{2} + 1}{\cos \left (x\right )^{2}}} + 1\right ) - 2 \, \cos \left (x\right ) \log \left (\cos \left (x\right )^{2} - \cos \left (x\right ) \sin \left (x\right ) +{\left (\cos \left (x\right )^{2} - \cos \left (x\right ) \sin \left (x\right )\right )} \sqrt{\frac{\cos \left (x\right )^{2} + 1}{\cos \left (x\right )^{2}}} + 1\right ) + \sqrt{\frac{\cos \left (x\right )^{2} + 1}{\cos \left (x\right )^{2}}} \sin \left (x\right )}{2 \, \cos \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+sec(x)^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*(arctan((sqrt((cos(x)^2 + 1)/cos(x)^2)*cos(x)^3*sin(x) + cos(x)*sin(x))/(cos(x)^4 + cos(x)^2 - 1))*cos(x)
- arctan(sin(x)/cos(x))*cos(x) + 2*cos(x)*log(cos(x)^2 + cos(x)*sin(x) + (cos(x)^2 + cos(x)*sin(x))*sqrt((cos(
x)^2 + 1)/cos(x)^2) + 1) - 2*cos(x)*log(cos(x)^2 - cos(x)*sin(x) + (cos(x)^2 - cos(x)*sin(x))*sqrt((cos(x)^2 +
1)/cos(x)^2) + 1) + sqrt((cos(x)^2 + 1)/cos(x)^2)*sin(x))/cos(x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\sec ^{2}{\left (x \right )} + 1\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+sec(x)**2)**(3/2),x)

[Out]

Integral((sec(x)**2 + 1)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\sec \left (x\right )^{2} + 1\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+sec(x)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((sec(x)^2 + 1)^(3/2), x)