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

Optimal. Leaf size=54 \[ \frac{2 \sinh ^{-1}\left (\frac{\tan (c+d x)}{\sqrt{\sec (c+d x)+1}}\right )}{d}-\frac{\sqrt{2} \sinh ^{-1}\left (\frac{\tan (c+d x)}{\sec (c+d x)+1}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.10941, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3821, 3801, 215, 3807} \[ \frac{2 \sinh ^{-1}\left (\frac{\tan (c+d x)}{\sqrt{\sec (c+d x)+1}}\right )}{d}-\frac{\sqrt{2} \sinh ^{-1}\left (\frac{\tan (c+d x)}{\sec (c+d x)+1}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[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*ArcSinh[Tan[c + d*x]/Sqrt[1 + Sec[c + d*x]]])/d

Rule 3821

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[d/b, I
nt[Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]], x], x] - Dist[(a*d)/b, Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*C
sc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3801

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

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]

Rubi steps

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

Mathematica [A]  time = 0.0823011, size = 76, normalized size = 1.41 \[ \frac{\sqrt{-\tan ^2(c+d x)} \cot (c+d x) \left (2 \sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )-\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sec (c+d x)}}{\sqrt{1-\sec (c+d x)}}\right )\right )}{d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((2*ArcSin[Sqrt[Sec[c + d*x]]] - Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]])*Cot[c +
d*x]*Sqrt[-Tan[c + d*x]^2])/d

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Maple [B]  time = 0.193, size = 180, normalized size = 3.3 \begin{align*}{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}-1 \right ) }{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}} \left ( \sqrt{2}\arctan \left ({\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) }{4}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) -\sqrt{2}\arctan \left ({\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) }{4}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) -2\,\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) +1}{\cos \left ( dx+c \right ) }}} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{3}{2}}}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.18536, size = 639, normalized size = 11.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(2)*log(cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))
)^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 1) - sqrt(2)*log(cos(1/2*arctan2(sin(d*x + c), cos(d*x
+ c)))^2 + sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 1
) - log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 +
2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))
) + 2) + log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))
^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x
+ c))) + 2) - log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x +
c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos
(d*x + c))) + 2) + log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(
d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c)
, cos(d*x + c))) + 2))/d

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Fricas [B]  time = 2.10775, size = 614, normalized size = 11.37 \begin{align*} \frac{\sqrt{2} \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 ) - \log \left (-\frac{\cos \left (d x + c\right )^{2} + 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}{\cos \left (d x + c\right ) + 1}\right ) + \log \left (-\frac{\cos \left (d x + c\right )^{2} - 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}{\cos \left (d x + c\right ) + 1}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(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)) - log(-(cos(d*x + c)^2 + 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)/(cos(d*x + c) + 1)) + log(-(cos(d*x
+ c)^2 - 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)/(cos(d*x
+ c) + 1)))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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