3.270 \(\int \frac{1}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{1+\sec (c+d x)}} \, dx\)

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

[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]])

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Rubi [A]  time = 0.151497, antiderivative size = 98, normalized size of antiderivative = 1., 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 = {3823, 4013, 3807, 215} \[ -\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d \sqrt{\sec (c+d x)+1}}+\frac{2 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} \sqrt{\sec (c+d x)+1}}+\frac{\sqrt{2} \sinh ^{-1}\left (\frac{\tan (c+d x)}{\sec (c+d x)+1}\right )}{d} \]

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 3823

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 4013

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]

Rule 3807

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 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]

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} \operatorname{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*}

Mathematica [A]  time = 0.225538, size = 118, normalized size = 1.2 \[ \frac{\tan (c+d x) \left (2 (\cos (c+d x)-1) \sqrt{1-\sec (c+d x)}-3 \sqrt{2} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sec (c+d x)}}{\sqrt{1-\sec (c+d x)}}\right )\right )}{3 d \sqrt{-(\sec (c+d x)-1) \sec (c+d x)} \sqrt{\sec (c+d x)+1}} \]

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 = 0.194, size = 116, normalized size = 1.2 \begin{align*} -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3\,d\sin \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) +1}{\cos \left ( dx+c \right ) }}} \left ( 3\,\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) +2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-4\,\cos \left ( dx+c \right ) +2 \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{3}{2}}}} \end{align*}

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)

[Out]

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

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Maxima [B]  time = 1.89255, size = 377, normalized size = 3.85 \begin{align*} -\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{align*}

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 = 1.97408, size = 448, normalized size = 4.57 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\sec{\left (c + d x \right )} + 1} \sec ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\sec \left (d x + c\right ) + 1} \sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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)