3.459 \(\int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx\)
Optimal. Leaf size=150 \[ \frac {2 A \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {a+b \sec (c+d x)}} \]
[Out]
-2*(A*b-B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1
/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)*sec(d*x+c)^(1/2)/a/d/(a+b*sec(d*x+c))^(1/2)+2*A*(cos(1/2*d*x+1/2*c)^2)^(1/
2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*(a+b*sec(d*x+c))^(1/2)/a/d/((b+a*c
os(d*x+c))/(a+b))^(1/2)/sec(d*x+c)^(1/2)
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Rubi [A] time = 0.31, antiderivative size = 150, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used =
{4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac {2 A \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {a+b \sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sec[c + d*x])/(Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]),x]
[Out]
(-2*(A*b - a*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(
a*d*Sqrt[a + b*Sec[c + d*x]]) + (2*A*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(a*d*Sqrt
[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]])
Rule 2653
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 2655
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] && !GtQ[a + b, 0]
Rule 2661
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 2663
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && !GtQ[a + b, 0]
Rule 3856
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 3858
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 4035
Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
Rubi steps
\begin {align*} \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx &=\frac {A \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{a}-\frac {(A b-a B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{a}\\ &=-\frac {\left ((A b-a B) \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{a \sqrt {a+b \sec (c+d x)}}+\frac {\left (A \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{a \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ &=-\frac {\left ((A b-a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{a \sqrt {a+b \sec (c+d x)}}+\frac {\left (A \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{a \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}\\ &=-\frac {2 (A b-a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{a d \sqrt {a+b \sec (c+d x)}}+\frac {2 A E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{a d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 3.89, size = 103, normalized size = 0.69 \[ \frac {2 \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \left ((a B-A b) F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )+A (a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )\right )}{a d \sqrt {a+b \sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Sec[c + d*x])/(Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]),x]
[Out]
(2*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*(A*(a + b)*EllipticE[(c + d*x)/2, (2*a)/(a + b)] + (-(A*b) + a*B)*Ellipt
icF[(c + d*x)/2, (2*a)/(a + b)])*Sqrt[Sec[c + d*x]])/(a*d*Sqrt[a + b*Sec[c + d*x]])
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fricas [F] time = 1.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )}}{b \sec \left (d x + c\right )^{2} + a \sec \left (d x + c\right )}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
integral((B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a)*sqrt(sec(d*x + c))/(b*sec(d*x + c)^2 + a*sec(d*x + c)),
x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sec \left (d x + c\right ) + A}{\sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)/(sqrt(b*sec(d*x + c) + a)*sqrt(sec(d*x + c))), x)
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maple [B] time = 2.51, size = 940, normalized size = 6.27 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sec(d*x+c))/sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2),x)
[Out]
-2/d*(-A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-
b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*cos(d*x+c)*a+A*cos(d*x+c)*EllipticE((-1+cos(d*x+c)
)*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2
)*(1/(1+cos(d*x+c)))^(1/2)*a-A*cos(d*x+c)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-
b))^(1/2))*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*b+B*cos(d*x+c)*El
lipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*((b+a*cos(d*x+c))/(1+c
os(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*a-A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d
*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*sin(d*x+c)+A*El
lipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*((b+a*cos(d*x+c))/(1+cos(d*x+c)
)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-A*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),
(-(a+b)/(a-b))^(1/2))*b*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+B*El
lipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*((b+a*cos(d*x+c))/(1+cos(d*x+c)
)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a-A*cos(d*x+c)*((a-b)/(a
+b))^(1/2)*a+A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b-A*b*((a-b)/(a+b))^(1/2))*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)/(
1/cos(d*x+c))^(1/2)/sin(d*x+c)/(b+a*cos(d*x+c))/((a-b)/(a+b))^(1/2)/a
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sec \left (d x + c\right ) + A}{\sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate((B*sec(d*x + c) + A)/(sqrt(b*sec(d*x + c) + a)*sqrt(sec(d*x + c))), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + B/cos(c + d*x))/((a + b/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(1/2)),x)
[Out]
int((A + B/cos(c + d*x))/((a + b/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(1/2)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \sec {\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}} \sqrt {\sec {\left (c + d x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))/sec(d*x+c)**(1/2)/(a+b*sec(d*x+c))**(1/2),x)
[Out]
Integral((A + B*sec(c + d*x))/(sqrt(a + b*sec(c + d*x))*sqrt(sec(c + d*x))), x)
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