3.2.87 \(\int \frac {\sqrt {c+d \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}} \, dx\) [187]

Optimal. Leaf size=141 \[ \frac {2 \sqrt {c} \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {c} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} f}-\frac {\sqrt {2} \sqrt {c-d} \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {c-d} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} f} \]

[Out]

2*arctan(a^(1/2)*c^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e))^(1/2))*c^(1/2)/f/a^(1/2)-arctan(1/
2*a^(1/2)*(c-d)^(1/2)*tan(f*x+e)*2^(1/2)/(a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e))^(1/2))*2^(1/2)*(c-d)^(1/2)/f/
a^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.23, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {4020, 4019, 209, 4068} \begin {gather*} \frac {2 \sqrt {c} \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {c} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} f}-\frac {\sqrt {2} \sqrt {c-d} \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {c-d} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

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

Rule 4019

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)], x_Symbol] :> Dist[-
2*(a/f), Subst[Int[1/(1 + a*c*x^2), x], x, Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*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 4020

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

Rule 4068

Int[csc[(e_.) + (f_.)*(x_)]/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (
c_)]), x_Symbol] :> Dist[-2*(a/(b*f)), Subst[Int[1/(2 + (a*c - b*d)*x^2), x], x, Cot[e + f*x]/(Sqrt[a + b*Csc[
e + f*x]]*Sqrt[c + d*Csc[e + f*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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 14.46, size = 184, normalized size = 1.30 \begin {gather*} \frac {2 \cos \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {-c+d} \tanh ^{-1}\left (\frac {\sqrt {-c+d} \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {d+c \cos (e+f x)}}\right )+\frac {\sqrt {2} \sqrt {c} \sqrt {c+d} \text {ArcSin}\left (\frac {\sqrt {2} \sqrt {c} \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right ) \sqrt {\frac {d+c \cos (e+f x)}{c+d}}}{\sqrt {d+c \cos (e+f x)}}\right ) \sqrt {c+d \sec (e+f x)}}{f \sqrt {d+c \cos (e+f x)} \sqrt {a (1+\sec (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(481\) vs. \(2(114)=228\).
time = 1.96, size = 482, normalized size = 3.42

method	result	size
default	\(-\frac {2 \sqrt {\frac {d +c \cos \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right ) \left (\sqrt {2}\, \sqrt {-\left (c -d \right )^{4} c}\, \arctan \left (\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (c -d \right )^{2} c \sqrt {2}}{\sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {-\left (c -d \right )^{4} c}}\right ) \sqrt {c -d}+\ln \left (\frac {\sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )-\sqrt {c -d}\, \cos \left (f x +e \right )+\sqrt {c -d}}{\sin \left (f x +e \right )}\right ) c^{3}-3 \ln \left (\frac {\sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )-\sqrt {c -d}\, \cos \left (f x +e \right )+\sqrt {c -d}}{\sin \left (f x +e \right )}\right ) c^{2} d +3 \ln \left (\frac {\sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )-\sqrt {c -d}\, \cos \left (f x +e \right )+\sqrt {c -d}}{\sin \left (f x +e \right )}\right ) c \,d^{2}-\ln \left (\frac {\sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )-\sqrt {c -d}\, \cos \left (f x +e \right )+\sqrt {c -d}}{\sin \left (f x +e \right )}\right ) d^{3}\right )}{f \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )^{2} a \sqrt {c -d}\, \left (c^{2}-2 c d +d^{2}\right )}\)	\(482\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/f*((d+c*cos(f*x+e))/cos(f*x+e))^(1/2)*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*cos(f*x+e)*(-1+cos(f*x+e))*(2^(1/
2)*(-(c-d)^4*c)^(1/2)*arctan((-1+cos(f*x+e))/(-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)/sin(f*x+e)*(c-d)^2*c*2
^(1/2)/(-(c-d)^4*c)^(1/2))*(c-d)^(1/2)+ln(((-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-(c-d)^(1/2)*c
os(f*x+e)+(c-d)^(1/2))/sin(f*x+e))*c^3-3*ln(((-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-(c-d)^(1/2)
*cos(f*x+e)+(c-d)^(1/2))/sin(f*x+e))*c^2*d+3*ln(((-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-(c-d)^(
1/2)*cos(f*x+e)+(c-d)^(1/2))/sin(f*x+e))*c*d^2-ln(((-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-(c-d)
^(1/2)*cos(f*x+e)+(c-d)^(1/2))/sin(f*x+e))*d^3)/(-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)/sin(f*x+e)^2/a/(c-d
)^(1/2)/(c^2-2*c*d+d^2)

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(d-c>0)', see `assume?` for mor
e details)Is

________________________________________________________________________________________

Fricas [A]
time = 2.96, size = 945, normalized size = 6.70 \begin {gather*} \left [\frac {\sqrt {2} \sqrt {-\frac {c - d}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c - d}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (3 \, c - d\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (c + d\right )} \cos \left (f x + e\right ) - c + 3 \, d}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 2 \, \sqrt {-\frac {c}{a}} \log \left (-\frac {2 \, \sqrt {-\frac {c}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, c \cos \left (f x + e\right )^{2} - {\left (c + d\right )} \cos \left (f x + e\right ) + c - d}{\cos \left (f x + e\right ) + 1}\right )}{2 \, f}, \frac {\sqrt {2} \sqrt {-\frac {c - d}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c - d}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (3 \, c - d\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (c + d\right )} \cos \left (f x + e\right ) - c + 3 \, d}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) - 4 \, \sqrt {\frac {c}{a}} \arctan \left (\frac {\sqrt {\frac {c}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{c \sin \left (f x + e\right )}\right )}{2 \, f}, -\frac {\sqrt {2} \sqrt {\frac {c - d}{a}} \arctan \left (-\frac {\sqrt {2} \sqrt {\frac {c - d}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{{\left (c - d\right )} \sin \left (f x + e\right )}\right ) - \sqrt {-\frac {c}{a}} \log \left (-\frac {2 \, \sqrt {-\frac {c}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, c \cos \left (f x + e\right )^{2} - {\left (c + d\right )} \cos \left (f x + e\right ) + c - d}{\cos \left (f x + e\right ) + 1}\right )}{f}, -\frac {\sqrt {2} \sqrt {\frac {c - d}{a}} \arctan \left (-\frac {\sqrt {2} \sqrt {\frac {c - d}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{{\left (c - d\right )} \sin \left (f x + e\right )}\right ) + 2 \, \sqrt {\frac {c}{a}} \arctan \left (\frac {\sqrt {\frac {c}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{c \sin \left (f x + e\right )}\right )}{f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(2)*sqrt(-(c - d)/a)*log((2*sqrt(2)*sqrt(-(c - d)/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c
*cos(f*x + e) + d)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (3*c - d)*cos(f*x + e)^2 + 2*(c + d)*cos(f*x + e)
- c + 3*d)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) + 2*sqrt(-c/a)*log(-(2*sqrt(-c/a)*sqrt((a*cos(f*x + e) + a)
/cos(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - 2*c*cos(f*x + e)^2 - (c + d
)*cos(f*x + e) + c - d)/(cos(f*x + e) + 1)))/f, 1/2*(sqrt(2)*sqrt(-(c - d)/a)*log((2*sqrt(2)*sqrt(-(c - d)/a)*
sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (3
*c - d)*cos(f*x + e)^2 + 2*(c + d)*cos(f*x + e) - c + 3*d)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) - 4*sqrt(c/a
)*arctan(sqrt(c/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e))*cos(f*x + e
)/(c*sin(f*x + e))))/f, -(sqrt(2)*sqrt((c - d)/a)*arctan(-sqrt(2)*sqrt((c - d)/a)*sqrt((a*cos(f*x + e) + a)/co
s(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e))*cos(f*x + e)/((c - d)*sin(f*x + e))) - sqrt(-c/a)*log(-(2*
sqrt(-c/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e))*cos(f*x + e)*sin(f*
x + e) - 2*c*cos(f*x + e)^2 - (c + d)*cos(f*x + e) + c - d)/(cos(f*x + e) + 1)))/f, -(sqrt(2)*sqrt((c - d)/a)*
arctan(-sqrt(2)*sqrt((c - d)/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e)
)*cos(f*x + e)/((c - d)*sin(f*x + e))) + 2*sqrt(c/a)*arctan(sqrt(c/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*
sqrt((c*cos(f*x + e) + d)/cos(f*x + e))*cos(f*x + e)/(c*sin(f*x + e))))/f]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c + d \sec {\left (e + f x \right )}}}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c+\frac {d}{\cos \left (e+f\,x\right )}}}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________