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

Optimal. Leaf size=113 \[ \frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d} \]

[Out]

-arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)*cos(d*x+c)^(1/2)*sec(
d*x+c)^(1/2)/d/a^(1/2)+2*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4264, 3812, 3808, 206} \[ \frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cos[c + d*x]]/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

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

Rule 206

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

Rule 3808

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

Rule 3812

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

Rule 4264

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

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 100, normalized size = 0.88 \[ \frac {\sin (c+d x) \left (2 \sqrt {1-\sec (c+d x)}+\sqrt {2} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )\right )}{d \sqrt {\cos (c+d x)-1} \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cos[c + d*x]]/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

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

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fricas [A]  time = 0.72, size = 281, normalized size = 2.49 \[ \left [\frac {\frac {\sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}} + 4 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac {\sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) + 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(2)*(a*cos(d*x + c) + a)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sq
rt(cos(d*x + c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a) + 4
*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c) + a*d), (sqrt(2)*(
a*cos(d*x + c) + a)*sqrt(-1/a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*sqrt(cos(d*x
+ c))/sin(d*x + c)) + 2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x
+ c) + a*d)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\cos \left (d x + c\right )}}{\sqrt {a \sec \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.97, size = 98, normalized size = 0.87 \[ \frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-2 \cos \left (d x +c \right )+2\right )}{d \sin \left (d x +c \right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*cos(d*x+c)^(1/2)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(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)/sin(d*x+c)/a

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maxima [A]  time = 1.27, size = 104, normalized size = 0.92 \[ -\frac {\sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - \sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 4 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, \sqrt {a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\cos \left (c+d\,x\right )}}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\cos {\left (c + d x \right )}}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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