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

Optimal. Leaf size=140 \[ \frac {\sqrt {2} \sqrt {c-d} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} f}+\frac {2 \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} f} \]

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

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Rubi [A]  time = 0.46, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3981, 3983, 203, 3980, 206} \[ \frac {\sqrt {2} \sqrt {c-d} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} f}+\frac {2 \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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*Se
c[e + f*x]])])/(Sqrt[a]*f) + (2*Sqrt[d]*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]*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 3981

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

Rubi steps

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

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Mathematica [A]  time = 17.02, size = 187, normalized size = 1.34 \[ \frac {\sqrt {c} \sin (e+f x) \sqrt {c+d \sec (e+f x)} \left (2 \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {c \cos (e+f x)+d}}{\sqrt {d} \sqrt {c-c \cos (e+f x)}}\right )-\sqrt {2} \sqrt {c-d} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c \cos (e+f x)+d}}{\sqrt {c-d} \sqrt {c-c \cos (e+f x)}}\right )\right )}{f \sqrt {a (\sec (e+f x)+1)} \sqrt {c-c \cos (e+f x)} \sqrt {c \cos (e+f x)+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 1048, normalized size = 7.49 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 1.99, size = 503, normalized size = 3.59 \[ \frac {\sqrt {\frac {d +c \cos \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right ) \left (d \ln \left (-\frac {2 \left (\sqrt {2}\, \sqrt {-d}\, \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )-c \sin \left (f x +e \right )-d \sin \left (f x +e \right )-c \cos \left (f x +e \right )+d \cos \left (f x +e \right )+c -d \right )}{-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}\right ) \sqrt {c -d}-d \ln \left (\frac {2 \sqrt {2}\, \sqrt {-d}\, \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )-2 c \sin \left (f x +e \right )-2 d \sin \left (f x +e \right )+2 c \cos \left (f x +e \right )-2 d \cos \left (f x +e \right )-2 c +2 d}{1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}\right ) \sqrt {c -d}+\ln \left (\frac {\sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{1+\cos \left (f x +e \right )}}\, \sqrt {c -d}\, \sin \left (f x +e \right )-c \cos \left (f x +e \right )+d \cos \left (f x +e \right )+c -d}{\sin \left (f x +e \right ) \sqrt {c -d}}\right ) \sqrt {2}\, \sqrt {-d}\, c -\ln \left (\frac {\sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{1+\cos \left (f x +e \right )}}\, \sqrt {c -d}\, \sin \left (f x +e \right )-c \cos \left (f x +e \right )+d \cos \left (f x +e \right )+c -d}{\sin \left (f x +e \right ) \sqrt {c -d}}\right ) \sqrt {2}\, \sqrt {-d}\, d \right ) \sqrt {2}}{f \sin \left (f x +e \right )^{2} \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{1+\cos \left (f x +e \right )}}\, a \sqrt {c -d}\, \sqrt {-d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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