3.2.88 \(\int \frac {1}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}} \, dx\) [188]

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

[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))/f/a^(1/2)/c^(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)/f/a^(1/2)/(c-d
)^(1/2)

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Rubi [A]
time = 0.21, 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 = {4023, 4019, 209, 4068} \begin {gather*} \frac {2 \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} \sqrt {c} f}-\frac {\sqrt {2} \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 \sqrt {c-d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + a*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]]),x]

[Out]

(2*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]*Sqrt[c
]*f) - (Sqrt[2]*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]*Sqrt[c - d]*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 4023

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

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 {1}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}} \, dx &=\frac {\int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx}{a}-\int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}} \, dx\\ &=-\frac {2 \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 \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 \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} \sqrt {c} f}-\frac {\sqrt {2} \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} \sqrt {c-d} f}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 171, normalized size = 1.21 \begin {gather*} \frac {2 \left (\sqrt {2} \sqrt {c-d} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {c} \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {d+c \cos (e+f x)}}\right )-\sqrt {c} \text {ArcTan}\left (\frac {\sqrt {c-d} \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {d+c \cos (e+f x)}}\right )\right ) \cos \left (\frac {1}{2} (e+f x)\right ) \sqrt {d+c \cos (e+f x)} \sec (e+f x)}{\sqrt {c} \sqrt {c-d} f \sqrt {a (1+\sec (e+f x))} \sqrt {c+d \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + a*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]]),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(414\) vs. \(2(114)=228\).
time = 2.02, size = 415, normalized size = 2.94

method	result	size
default	\(-\frac {2 \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {d +c \cos \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right ) \left (\ln \left (\frac {\sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )-\sqrt {c -d}\, \cos \left (f x +e \right )+\sqrt {c -d}}{\sin \left (f x +e \right )}\right ) c^{3}-2 \ln \left (\frac {\sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )-\sqrt {c -d}\, \cos \left (f x +e \right )+\sqrt {c -d}}{\sin \left (f x +e \right )}\right ) c^{2} d +\ln \left (\frac {\sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )-\sqrt {c -d}\, \cos \left (f x +e \right )+\sqrt {c -d}}{\sin \left (f x +e \right )}\right ) c \,d^{2}+\sqrt {2}\, \sqrt {-\left (c -d \right )^{4} c}\, \arctan \left (\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (c -d \right )^{2} c \sqrt {2}}{\sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {-\left (c -d \right )^{4} c}}\right ) \sqrt {c -d}\right )}{f \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )^{2} a \sqrt {c -d}\, \left (c^{2}-2 c d +d^{2}\right ) c}\)	\(415\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*((d+c*cos(f*x+e))/cos(f*x+e))^(1/2)*cos(f*x+e)*(-1+cos(f*x+e))*(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^3-2*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))*c^2*d+l
n(((-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
+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))/(-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)/c

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 4.74, size = 975, normalized size = 6.91 \begin {gather*} \left [\frac {\sqrt {2} a c \sqrt {-\frac {1}{a c - a d}} \log \left (\frac {2 \, \sqrt {2} {\left (c - d\right )} \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 )}} \sqrt {-\frac {1}{a c - a d}} \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 {-a c} \log \left (\frac {2 \, a c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-a c} \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 ) - a c + a d + {\left (a c + a d\right )} \cos \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{2 \, a c f}, \frac {\sqrt {2} a c \sqrt {-\frac {1}{a c - a d}} \log \left (\frac {2 \, \sqrt {2} {\left (c - d\right )} \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 )}} \sqrt {-\frac {1}{a c - a d}} \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 {a c} \arctan \left (\frac {\sqrt {a c} \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 )}{a c \sin \left (f x + e\right )}\right )}{2 \, a c f}, \frac {\frac {\sqrt {2} a c \arctan \left (\frac {\sqrt {2} \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 )}{\sqrt {a c - a d} \sin \left (f x + e\right )}\right )}{\sqrt {a c - a d}} - \sqrt {-a c} \log \left (\frac {2 \, a c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-a c} \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 ) - a c + a d + {\left (a c + a d\right )} \cos \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a c f}, \frac {\frac {\sqrt {2} a c \arctan \left (\frac {\sqrt {2} \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 )}{\sqrt {a c - a d} \sin \left (f x + e\right )}\right )}{\sqrt {a c - a d}} - 2 \, \sqrt {a c} \arctan \left (\frac {\sqrt {a c} \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 )}{a c \sin \left (f x + e\right )}\right )}{a c f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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