3.168 \(\int \frac{c+d \sec (e+f x)}{\sqrt{a+a \sec (e+f x)}} \, dx\)

Optimal. Leaf size=91 \[ \frac{2 c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} f}-\frac{\sqrt{2} (c-d) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} f} \]

[Out]

(2*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(Sqrt[a]*f) - (Sqrt[2]*(c - d)*ArcTan[(Sqrt[a]*T
an[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[a]*f)

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Rubi [A]  time = 0.109533, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3920, 3774, 203, 3795} \[ \frac{2 c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} f}-\frac{\sqrt{2} (c-d) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{\sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Sec[e + f*x])/Sqrt[a + a*Sec[e + f*x]],x]

[Out]

(2*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(Sqrt[a]*f) - (Sqrt[2]*(c - d)*ArcTan[(Sqrt[a]*T
an[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[a]*f)

Rule 3920

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*Csc[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]

Rule 3774

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

Rule 203

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

Rule 3795

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

Rubi steps

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

Mathematica [A]  time = 0.28695, size = 92, normalized size = 1.01 \[ \frac{2 \cos \left (\frac{1}{2} (e+f x)\right ) \left ((d-c) \tan ^{-1}\left (\frac{\sin \left (\frac{1}{2} (e+f x)\right )}{\sqrt{\cos (e+f x)}}\right )+\sqrt{2} c \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (e+f x)\right )\right )\right )}{f \sqrt{\cos (e+f x)} \sqrt{a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Sec[e + f*x])/Sqrt[a + a*Sec[e + f*x]],x]

[Out]

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

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Maple [B]  time = 0.217, size = 194, normalized size = 2.1 \begin{align*} -{\frac{1}{af}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}} \left ( c\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( fx+e \right ) }{2\,\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) +c\ln \left ( -{\frac{1}{\sin \left ( fx+e \right ) } \left ( -\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sin \left ( fx+e \right ) +\cos \left ( fx+e \right ) -1 \right ) } \right ) -d\ln \left ( -{\frac{1}{\sin \left ( fx+e \right ) } \left ( -\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\sin \left ( fx+e \right ) +\cos \left ( fx+e \right ) -1 \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f/a*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(c*2^(1/2)*arctanh(1/2*2^(1/
2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))+c*ln(-(-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*si
n(f*x+e)+cos(f*x+e)-1)/sin(f*x+e))-d*ln(-(-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/sin(f
*x+e)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.14631, size = 824, normalized size = 9.05 \begin{align*} \left [-\frac{\sqrt{2}{\left (a c - a d\right )} \sqrt{-\frac{1}{a}} \log \left (-\frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt{-\frac{1}{a}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, \cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 2 \, \sqrt{-a} c \log \left (\frac{2 \, a \cos \left (f x + e\right )^{2} + 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right )}{2 \, a f}, -\frac{2 \, \sqrt{a} c \arctan \left (\frac{\sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt{a} \sin \left (f x + e\right )}\right ) - \frac{\sqrt{2}{\left (a c - a d\right )} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt{a} \sin \left (f x + e\right )}\right )}{\sqrt{a}}}{a f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out