3.351 \(\int \sec (c+d x) \sqrt {a+b \sec (c+d x)} (A+B \sec (c+d x)) \, dx\)
Optimal. Leaf size=256 \[ -\frac {2 (a-b) \sqrt {a+b} (a B+3 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^2 d}+\frac {2 (a-b) \sqrt {a+b} (3 A-B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b d}+\frac {2 B \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d} \]
[Out]
-2/3*(a-b)*(3*A*b+B*a)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2
)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^2/d+2/3*(a-b)*(3*A-B)*cot(d*x+c)*EllipticF(
(a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(
d*x+c))/(a-b))^(1/2)/b/d+2/3*B*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d
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Rubi [A] time = 0.34, antiderivative size = 256, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used =
{4002, 4005, 3832, 4004} \[ -\frac {2 (a-b) \sqrt {a+b} (a B+3 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^2 d}+\frac {2 (a-b) \sqrt {a+b} (3 A-B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b d}+\frac {2 B \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x]),x]
[Out]
(-2*(a - b)*Sqrt[a + b]*(3*A*b + a*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a
+ b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*b^2*d) + (2*(a
- b)*Sqrt[a + b]*(3*A - B)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b
)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*b*d) + (2*B*Sqrt[a + b*Sec
[c + d*x]]*Tan[c + d*x])/(3*d)
Rule 3832
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 4002
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]
Rule 4004
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Rule 4005
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, Int[(Csc[e + f*x]*(1 +
Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]
Rubi steps
\begin {align*} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} (A+B \sec (c+d x)) \, dx &=\frac {2 B \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d}+\frac {2}{3} \int \frac {\sec (c+d x) \left (\frac {1}{2} (3 a A+b B)+\frac {1}{2} (3 A b+a B) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 B \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d}+\frac {1}{3} ((a-b) (3 A-B)) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{3} (3 A b+a B) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\frac {2 (a-b) \sqrt {a+b} (3 A b+a B) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^2 d}+\frac {2 (a-b) \sqrt {a+b} (3 A-B) \cot (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b d}+\frac {2 B \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A] time = 14.65, size = 408, normalized size = 1.59 \[ \frac {\cos (c+d x) \sqrt {a+b \sec (c+d x)} (A+B \sec (c+d x)) \left (\frac {2 (a B+3 A b) \sin (c+d x)}{3 b}+\frac {2}{3} B \tan (c+d x)\right )}{d (A \cos (c+d x)+B)}+\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \sqrt {a+b \sec (c+d x)} (A+B \sec (c+d x)) \left (-\left ((a B+3 A b) \cos (c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)\right )+2 b (a+b) (3 A+B) \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}} \sqrt {\frac {a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 (a+b) (a B+3 A b) \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}} \sqrt {\frac {a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )\right )}{3 b d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b) (A \cos (c+d x)+B)} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x]),x]
[Out]
(2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x])*(-2*(a + b)*(3*A*b + a*
B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[Arc
Sin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(3*A + B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b
+ a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - (3*A*b
+ a*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b*d*(b + a*Cos[c + d*x])*(B
+ A*Cos[c + d*x])*Sqrt[Sec[(c + d*x)/2]^2]*Sec[c + d*x]^(3/2)) + (Cos[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*(A + B
*Sec[c + d*x])*((2*(3*A*b + a*B)*Sin[c + d*x])/(3*b) + (2*B*Tan[c + d*x])/3))/(d*(B + A*Cos[c + d*x]))
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \sec \left (d x + c\right )^{2} + A \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)*(A+B*sec(d*x+c))*(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
integral((B*sec(d*x + c)^2 + A*sec(d*x + c))*sqrt(b*sec(d*x + c) + a), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)*(A+B*sec(d*x+c))*(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)*sec(d*x + c), x)
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maple [B] time = 2.01, size = 1752, normalized size = 6.84 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)*(A+B*sec(d*x+c))*(a+b*sec(d*x+c))^(1/2),x)
[Out]
-2/3/d*(-1+cos(d*x+c))^2*(-B*cos(d*x+c)^2*a^2+B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+
a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b-B*cos(
d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-
1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b+B*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((
b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b-3*A*
cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipt
icE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b+3*A*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c))
)^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2)
)*a*b-B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2
)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b-3*A*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d
*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))
^(1/2))*a*b+3*A*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b)
)^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b+B*cos(d*x+c)^2*b^2-b^2*B+3*A*cos(d*x+c)^
2*b^2-3*A*cos(d*x+c)*b^2-3*A*cos(d*x+c)^2*a*b+B*cos(d*x+c)^3*a*b+B*cos(d*x+c)^2*a*b-2*B*cos(d*x+c)*a*b+3*A*cos
(d*x+c)^3*a*b+B*cos(d*x+c)^3*a^2-3*A*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c
))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^2+3*A*cos(d*x+c)^2*
sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d
*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^2-B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*c
os(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2+B*cos(d*x
+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1
+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^2-3*A*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(
(b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^2+3*A
*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipti
cF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^2-B*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2
+B*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellip
ticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(1+cos(d*x+c))^2
/(b+a*cos(d*x+c))/cos(d*x+c)/sin(d*x+c)^5/b
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)*(A+B*sec(d*x+c))*(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)*sec(d*x + c), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}}{\cos \left (c+d\,x\right )} \,d x \]
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))/cos(c + d*x),x)
[Out]
int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^(1/2))/cos(c + d*x), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \sec {\left (c + d x \right )}\right ) \sqrt {a + b \sec {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)*(A+B*sec(d*x+c))*(a+b*sec(d*x+c))**(1/2),x)
[Out]
Integral((A + B*sec(c + d*x))*sqrt(a + b*sec(c + d*x))*sec(c + d*x), x)
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