3.237 \(\int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}} \, dx\)

Optimal. Leaf size=141 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} \sqrt {d} f}-\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} f \sqrt {c-d}} \]

[Out]

-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)+2*arctanh(a^(1/2)*d^(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
)/d^(1/2)

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Rubi [A]  time = 0.53, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {3985, 3983, 203, 3980, 206} \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} \sqrt {d} f}-\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} f \sqrt {c-d}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^2/(Sqrt[a + a*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]]),x]

[Out]

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

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 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 3980

Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)])/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) +
(c_)], x_Symbol] :> Dist[(-2*b)/f, Subst[Int[1/(1 - b*d*x^2), x], x, Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sq
rt[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 3983

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]

Rule 3985

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

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 171, normalized size = 1.21 \[ \frac {2 \cos \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {c \cos (e+f x)+d} \left (\sqrt {2} \sqrt {c-d} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c \cos (e+f x)+d}}\right )-\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {c-d} \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c \cos (e+f x)+d}}\right )\right )}{\sqrt {d} f \sqrt {c-d} \sqrt {a (\sec (e+f x)+1)} \sqrt {c+d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^2/(Sqrt[a + a*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]]),x]

[Out]

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

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fricas [A]  time = 0.81, size = 1100, normalized size = 7.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2/(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*d*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)) + sqrt(a*d)*log(-(8*a*c*d*cos(f*x + e)
+ (a*c^2 - 6*a*c*d + a*d^2)*cos(f*x + e)^3 + 4*((c - d)*cos(f*x + e)^2 + 2*d*cos(f*x + e))*sqrt(a*d)*sqrt((a*c
os(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e))*sin(f*x + e) + 8*a*d^2 + (a*c^2 + 2*a*c
*d - 7*a*d^2)*cos(f*x + e)^2)/(cos(f*x + e)^3 + cos(f*x + e)^2)))/(a*d*f), 1/2*(2*sqrt(2)*a*d*arctan(sqrt(2)*s
qrt((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)*s
in(f*x + e)))/sqrt(a*c - a*d) + sqrt(a*d)*log(-(8*a*c*d*cos(f*x + e) + (a*c^2 - 6*a*c*d + a*d^2)*cos(f*x + e)^
3 + 4*((c - d)*cos(f*x + e)^2 + 2*d*cos(f*x + e))*sqrt(a*d)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*co
s(f*x + e) + d)/cos(f*x + e))*sin(f*x + e) + 8*a*d^2 + (a*c^2 + 2*a*c*d - 7*a*d^2)*cos(f*x + e)^2)/(cos(f*x +
e)^3 + cos(f*x + e)^2)))/(a*d*f), 1/2*(sqrt(2)*a*d*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*sq
rt(-a*d)*arctan(-2*sqrt(-a*d)*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)*cos(f*x + e)^2 + 2*a*d + (a*c + a*d)*cos(f*x + e))))/(a*d*f), (sqrt(2)*
a*d*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) + sqrt(-a*d)*arctan(-2*sqrt(-a*d)*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)*cos(f*x + e)^2 +
2*a*d + (a*c + a*d)*cos(f*x + e))))/(a*d*f)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )^{2}}{\sqrt {a \sec \left (f x + e\right ) + a} \sqrt {d \sec \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 2.03, size = 403, normalized size = 2.86 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {d +c \cos \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right ) \left (-\ln \left (-\frac {\sqrt {c -d}\, \cos \left (f x +e \right )-\sin \left (f x +e \right ) \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{1+\cos \left (f x +e \right )}}-\sqrt {c -d}}{\sin \left (f x +e \right )}\right ) \sqrt {2}\, \sqrt {-d}+\ln \left (-\frac {2 \left (\sqrt {2}\, \sqrt {-d}\, \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )-c \sin \left (f x +e \right )-d \sin \left (f x +e \right )-c \cos \left (f x +e \right )+d \cos \left (f x +e \right )+c -d \right )}{-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}\right ) \sqrt {c -d}-\ln \left (\frac {2 \sqrt {2}\, \sqrt {-d}\, \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )-2 c \sin \left (f x +e \right )-2 d \sin \left (f x +e \right )+2 c \cos \left (f x +e \right )-2 d \cos \left (f x +e \right )-2 c +2 d}{1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}\right ) \sqrt {c -d}\right ) \sqrt {2}}{f \sin \left (f x +e \right )^{2} \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{1+\cos \left (f x +e \right )}}\, a \sqrt {c -d}\, \sqrt {-d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )^{2}}{\sqrt {a \sec \left (f x + e\right ) + a} \sqrt {d \sec \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(f*x + e)^2/(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 \[ \int \frac {1}{{\cos \left (e+f\,x\right )}^2\,\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,\sqrt {c+\frac {d}{\cos \left (e+f\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}{\left (e + f x \right )}}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \sqrt {c + d \sec {\left (e + f x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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