3.365 \(\int \frac{1}{\sqrt{1+\cos (c+d x)} \sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=94 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{\cos (c+d x)+1}}\right )}{d}-\frac{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)+1}\right )}{d} \]

[Out]

-((Sqrt[2]*ArcSin[Sin[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/d) + (2*ArcSin[Sin[c
+ d*x]/Sqrt[1 + Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/d

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Rubi [A]  time = 0.153836, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4222, 2777, 2774, 216, 2781} \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{\cos (c+d x)+1}}\right )}{d}-\frac{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)+1}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[1 + Cos[c + d*x]]*Sqrt[Sec[c + d*x]]),x]

[Out]

-((Sqrt[2]*ArcSin[Sin[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/d) + (2*ArcSin[Sin[c
+ d*x]/Sqrt[1 + Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/d

Rule 4222

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rule 2777

Int[Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
d/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] + Dist[(b*c - a*d)/b, Int[1/(Sqrt[a + b*Sin
[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2774

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 216

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

Rule 2781

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

Rubi steps

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

Mathematica [C]  time = 0.56827, size = 171, normalized size = 1.82 \[ \frac{i \sqrt{2} e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \cos \left (\frac{1}{2} (c+d x)\right ) \left (-\sinh ^{-1}\left (e^{i (c+d x)}\right )+\sqrt{2} \tanh ^{-1}\left (\frac{-1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+\tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )}{d \sqrt{\cos (c+d x)+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[1 + Cos[c + d*x]]*Sqrt[Sec[c + d*x]]),x]

[Out]

(I*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(-ArcSinh[E^(I*(c + d
*x))] + Sqrt[2]*ArcTanh[(-1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] + ArcTanh[Sqrt[1 + E^(
(2*I)*(c + d*x))]])*Cos[(c + d*x)/2])/(d*E^((I/2)*(c + d*x))*Sqrt[1 + Cos[c + d*x]])

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Maple [A]  time = 0.36, size = 134, normalized size = 1.4 \begin{align*}{\frac{\sqrt{2}\cos \left ( dx+c \right ) \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}\sqrt{2+2\,\cos \left ( dx+c \right ) } \left ( \sqrt{2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +2\,\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \right ){\frac{1}{\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}}} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.81508, size = 204, normalized size = 2.17 \begin{align*} \frac{\sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{\cos \left (d x + c\right ) + 1} \sqrt{\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - 2 \, \arctan \left (\frac{\sqrt{\cos \left (d x + c\right ) + 1} \sqrt{\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(2)*arctan(sqrt(2)*sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))/sin(d*x + c)) - 2*arctan(sqrt(cos(d*x + c) +
1)*sqrt(cos(d*x + c))/sin(d*x + c)))/d

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(cos(d*x + c) + 1)*sqrt(sec(d*x + c))), x)