\(\int (1+\sec ^2(x))^{3/2} \, dx\) [255]

Optimal result

Integrand size = 10, antiderivative size = 42 \[ \int \left (1+\sec ^2(x)\right )^{3/2} \, dx=2 \text {arcsinh}\left (\frac {\tan (x)}{\sqrt {2}}\right )+\arctan \left (\frac {\tan (x)}{\sqrt {2+\tan ^2(x)}}\right )+\frac {1}{2} \tan (x) \sqrt {2+\tan ^2(x)} \]

[Out]

2*arcsinh(1/2*2^(1/2)*tan(x))+arctan(tan(x)/(2+tan(x)^2)^(1/2))+1/2*(2+tan(x)^2)^(1/2)*tan(x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4213, 427, 537, 221, 385, 209} \[ \int \left (1+\sec ^2(x)\right )^{3/2} \, dx=2 \text {arcsinh}\left (\frac {\tan (x)}{\sqrt {2}}\right )+\arctan \left (\frac {\tan (x)}{\sqrt {\tan ^2(x)+2}}\right )+\frac {1}{2} \tan (x) \sqrt {\tan ^2(x)+2} \]

[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 209

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

Rule 221

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 385

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 427

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 537

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 4213

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]

Rubi steps \begin{align*} \text {integral}& = \text {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} \text {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 \text {Subst}\left (\int \frac {1}{\sqrt {2+x^2}} \, dx,x,\tan (x)\right )+\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {2+x^2}} \, dx,x,\tan (x)\right ) \\ & = 2 \text {arcsinh}\left (\frac {\tan (x)}{\sqrt {2}}\right )+\frac {1}{2} \tan (x) \sqrt {2+\tan ^2(x)}+\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\tan (x)}{\sqrt {2+\tan ^2(x)}}\right ) \\ & = 2 \text {arcsinh}\left (\frac {\tan (x)}{\sqrt {2}}\right )+\arctan \left (\frac {\tan (x)}{\sqrt {2+\tan ^2(x)}}\right )+\frac {1}{2} \tan (x) \sqrt {2+\tan ^2(x)} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 109, normalized size of antiderivative = 2.60 \[ \int \left (1+\sec ^2(x)\right )^{3/2} \, dx=\frac {\left (1+\cos ^2(x)\right ) \sec (x) \sqrt {1+\sec ^2(x)} \left (4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sin (x)}{\sqrt {3+\cos (2 x)}}\right ) \cos ^2(x)-2 i \sqrt {2} \cos ^2(x) \log \left (\sqrt {3+\cos (2 x)}+i \sqrt {2} \sin (x)\right )+\sqrt {3+\cos (2 x)} \sin (x)\right )}{(3+\cos (2 x))^{3/2}} \]

[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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(188\) vs. \(2(35)=70\).

Time = 8.05 (sec) , antiderivative size = 189, normalized size of antiderivative = 4.50

method	result	size
default	\(\frac {\sqrt {2}\, \cos \left (x \right ) \left (\sin \left (x \right ) \cos \left (x \right ) \sqrt {\frac {\cos \left (x \right )^{2}+1}{\left (\cos \left (x \right )+1\right )^{2}}}+2 \cos \left (x \right )^{2} \arctan \left (\frac {\sin \left (x \right )}{\left (\cos \left (x \right )+1\right ) \sqrt {\frac {\cos \left (x \right )^{2}+1}{\left (\cos \left (x \right )+1\right )^{2}}}}\right )-2 \cos \left (x \right )^{2} \operatorname {arctanh}\left (\frac {\sin \left (x \right )-2}{\left (\cos \left (x \right )+1\right ) \sqrt {\frac {\cos \left (x \right )^{2}+1}{\left (\cos \left (x \right )+1\right )^{2}}}}\right )-2 \cos \left (x \right )^{2} \operatorname {arctanh}\left (\frac {\sin \left (x \right )+2}{\left (\cos \left (x \right )+1\right ) \sqrt {\frac {\cos \left (x \right )^{2}+1}{\left (\cos \left (x \right )+1\right )^{2}}}}\right )+\sqrt {\frac {\cos \left (x \right )^{2}+1}{\left (\cos \left (x \right )+1\right )^{2}}}\, \sin \left (x \right )\right ) \sqrt {2+2 \sec \left (x \right )^{2}}\, \left (1+\sec \left (x \right )^{2}\right )}{4 \sqrt {\frac {\cos \left (x \right )^{2}+1}{\left (\cos \left (x \right )+1\right )^{2}}}\, \left (\cos \left (x \right )+1\right ) \left (\cos \left (x \right )^{2}+1\right )}\)	\(189\)

[In]

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

[Out]

1/4*2^(1/2)*cos(x)*(sin(x)*cos(x)*((cos(x)^2+1)/(cos(x)+1)^2)^(1/2)+2*cos(x)^2*arctan(sin(x)/(cos(x)+1)/((cos(
x)^2+1)/(cos(x)+1)^2)^(1/2))-2*cos(x)^2*arctanh((sin(x)-2)/(cos(x)+1)/((cos(x)^2+1)/(cos(x)+1)^2)^(1/2))-2*cos
(x)^2*arctanh((sin(x)+2)/(cos(x)+1)/((cos(x)^2+1)/(cos(x)+1)^2)^(1/2))+((cos(x)^2+1)/(cos(x)+1)^2)^(1/2)*sin(x
))*(2+2*sec(x)^2)^(1/2)*(1+sec(x)^2)/((cos(x)^2+1)/(cos(x)+1)^2)^(1/2)/(cos(x)+1)/(cos(x)^2+1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (35) = 70\).

Time = 0.26 (sec) , antiderivative size = 160, normalized size of antiderivative = 3.81 \[ \int \left (1+\sec ^2(x)\right )^{3/2} \, dx=\frac {\arctan \left (\frac {\sqrt {\frac {\cos \left (x\right )^{2} + 1}{\cos \left (x\right )^{2}}} \cos \left (x\right )^{3} \sin \left (x\right ) + \cos \left (x\right ) \sin \left (x\right )}{\cos \left (x\right )^{4} + \cos \left (x\right )^{2} - 1}\right ) \cos \left (x\right ) - \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right ) \cos \left (x\right ) + 2 \, \cos \left (x\right ) \log \left (\cos \left (x\right )^{2} + \cos \left (x\right ) \sin \left (x\right ) + {\left (\cos \left (x\right )^{2} + \cos \left (x\right ) \sin \left (x\right )\right )} \sqrt {\frac {\cos \left (x\right )^{2} + 1}{\cos \left (x\right )^{2}}} + 1\right ) - 2 \, \cos \left (x\right ) \log \left (\cos \left (x\right )^{2} - \cos \left (x\right ) \sin \left (x\right ) + {\left (\cos \left (x\right )^{2} - \cos \left (x\right ) \sin \left (x\right )\right )} \sqrt {\frac {\cos \left (x\right )^{2} + 1}{\cos \left (x\right )^{2}}} + 1\right ) + \sqrt {\frac {\cos \left (x\right )^{2} + 1}{\cos \left (x\right )^{2}}} \sin \left (x\right )}{2 \, \cos \left (x\right )} \]

[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)

Sympy [F]

\[ \int \left (1+\sec ^2(x)\right )^{3/2} \, dx=\int \left (\sec ^{2}{\left (x \right )} + 1\right )^{\frac {3}{2}}\, dx \]

[In]

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

[Out]

Integral((sec(x)**2 + 1)**(3/2), x)

Maxima [F]

\[ \int \left (1+\sec ^2(x)\right )^{3/2} \, dx=\int { {\left (\sec \left (x\right )^{2} + 1\right )}^{\frac {3}{2}} \,d x } \]

[In]

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

[Out]

integrate((sec(x)^2 + 1)^(3/2), x)

Giac [F]

\[ \int \left (1+\sec ^2(x)\right )^{3/2} \, dx=\int { {\left (\sec \left (x\right )^{2} + 1\right )}^{\frac {3}{2}} \,d x } \]

[In]

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

[Out]

integrate((sec(x)^2 + 1)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \left (1+\sec ^2(x)\right )^{3/2} \, dx=\int {\left (\frac {1}{{\cos \left (x\right )}^2}+1\right )}^{3/2} \,d x \]

[In]

int((1/cos(x)^2 + 1)^(3/2),x)

[Out]

int((1/cos(x)^2 + 1)^(3/2), x)