3.3.70 \(\int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {1+\sec (c+d x)}} \, dx\) [270]

Optimal. Leaf size=98 \[ \frac {\sqrt {2} \sinh ^{-1}\left (\frac {\tan (c+d x)}{1+\sec (c+d x)}\right )}{d}+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {2 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {1+\sec (c+d x)}} \]

[Out]

arcsinh(tan(d*x+c)/(1+sec(d*x+c)))*2^(1/2)/d+2/3*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(1+sec(d*x+c))^(1/2)-2/3*sin(d*
x+c)*sec(d*x+c)^(1/2)/d/(1+sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3908, 4098, 3892, 221} \begin {gather*} -\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {\sec (c+d x)+1}}+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x)+1}}+\frac {\sqrt {2} \sinh ^{-1}\left (\frac {\tan (c+d x)}{\sec (c+d x)+1}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sec[c + d*x]^(3/2)*Sqrt[1 + Sec[c + d*x]]),x]

[Out]

(Sqrt[2]*ArcSinh[Tan[c + d*x]/(1 + Sec[c + d*x])])/d + (2*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c
+ d*x]]) - (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[1 + Sec[c + d*x]])

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 3892

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-Sqrt[2
])*(Sqrt[a]/(b*f)), Subst[Int[1/Sqrt[1 + x^2], x], x, b*(Cot[e + f*x]/(a + b*Csc[e + f*x]))], x] /; FreeQ[{a,
b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && EqQ[d - a/b, 0] && GtQ[a, 0]

Rule 3908

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[Cot[e +
f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Dist[1/(2*b*d*n), Int[(d*Csc[e + f*x])^(n + 1)
*((a + b*(2*n + 1)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b
^2, 0] && LtQ[n, 0] && IntegerQ[2*n]

Rule 4098

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dist[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {1+\sec (c+d x)}} \, dx &=\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {1}{3} \int \frac {1-2 \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}} \, dx\\ &=\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {2 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {1+\sec (c+d x)}}+\int \frac {\sqrt {\sec (c+d x)}}{\sqrt {1+\sec (c+d x)}} \, dx\\ &=\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {2 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {1+\sec (c+d x)}}-\frac {\sqrt {2} \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,-\frac {\tan (c+d x)}{1+\sec (c+d x)}\right )}{d}\\ &=\frac {\sqrt {2} \sinh ^{-1}\left (\frac {\tan (c+d x)}{1+\sec (c+d x)}\right )}{d}+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {2 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {1+\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 118, normalized size = 1.20 \begin {gather*} \frac {\left (2 (-1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}-3 \sqrt {2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sqrt {\sec (c+d x)}\right ) \tan (c+d x)}{3 d \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sqrt {1+\sec (c+d x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sec[c + d*x]^(3/2)*Sqrt[1 + Sec[c + d*x]]),x]

[Out]

((2*(-1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]] - 3*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c
+ d*x]]]*Sqrt[Sec[c + d*x]])*Tan[c + d*x])/(3*d*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])]*Sqrt[1 + Sec[c + d*x
]])

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Maple [A]
time = 0.14, size = 116, normalized size = 1.18

method	result	size
default	\(-\frac {\sqrt {\frac {1+\cos \left (d x +c \right )}{\cos \left (d x +c \right )}}\, \left (3 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+2 \left (\cos ^{2}\left (d x +c \right )\right )-4 \cos \left (d x +c \right )+2\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{2}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/d*((1+cos(d*x+c))/cos(d*x+c))^(1/2)*(3*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*(-2/(1+cos(d*x+c)
))^(1/2)*sin(d*x+c)+2*cos(d*x+c)^2-4*cos(d*x+c)+2)*(1/cos(d*x+c))^(3/2)*cos(d*x+c)^2/sin(d*x+c)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (84) = 168\).
time = 0.55, size = 279, normalized size = 2.85 \begin {gather*} -\frac {3 \, \sqrt {2} \cos \left (\frac {2}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 3 \, \sqrt {2} \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) \sin \left (\frac {2}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) - 3 \, \sqrt {2} \log \left (\cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 1\right ) + 3 \, \sqrt {2} \log \left (\cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 1\right ) - 2 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sqrt {2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*sqrt(2)*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(3/2*d*x + 3/2*c) - 3*sqrt(2)*
cos(3/2*d*x + 3/2*c)*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 3*sqrt(2)*log(cos(1/3*arct
an2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c
)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + 3*sqrt(2)*log(cos(1/3*arctan2(si
n(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2
- 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - 2*sqrt(2)*sin(3/2*d*x + 3/2*c) + 3*sqr
t(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))/d

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Fricas [A]
time = 2.95, size = 163, normalized size = 1.66 \begin {gather*} \frac {3 \, {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt {\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(sqrt(2)*cos(d*x + c) + sqrt(2))*log((2*sqrt(2)*sqrt((cos(d*x + c) + 1)/cos(d*x + c))*sqrt(cos(d*x + c)
)*sin(d*x + c) - cos(d*x + c)^2 + 2*cos(d*x + c) + 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(cos(d*x + c)
^2 - cos(d*x + c))*sqrt((cos(d*x + c) + 1)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {\sec {\left (c + d x \right )} + 1} \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**(3/2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(sec(d*x + c) + 1)*sec(d*x + c)^(3/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}+1}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((1/cos(c + d*x) + 1)^(1/2)*(1/cos(c + d*x))^(3/2)),x)

[Out]

int(1/((1/cos(c + d*x) + 1)^(1/2)*(1/cos(c + d*x))^(3/2)), x)

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