3.523 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=131 \[ -\frac{(5 A-B-3 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{2 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{3/2} d}-\frac{(A-B+C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]
[Out]
(2*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(3/2)*d) - ((5*A - B - 3*C)*ArcTan[(Sqrt[a]*T
an[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A - B + C)*Tan[c + d*x])/(2*d*(a +
a*Sec[c + d*x])^(3/2))
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Rubi [A] time = 0.194571, antiderivative size = 131, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used =
{4052, 3920, 3774, 203, 3795} \[ -\frac{(5 A-B-3 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{2 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{3/2} d}-\frac{(A-B+C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^(3/2),x]
[Out]
(2*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(3/2)*d) - ((5*A - B - 3*C)*ArcTan[(Sqrt[a]*T
an[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A - B + C)*Tan[c + d*x])/(2*d*(a +
a*Sec[c + d*x])^(3/2))
Rule 4052
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] +
Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*(2*m + 1) + (b*B*(m + 1) - a*(A*(m + 1) - C*
m))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Rule 3920
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[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] && EqQ[a^2 - b^2, 0]
Rule 3774
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
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 3795
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]
Rubi steps
\begin{align*} \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac{(A-B+C) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{\int \frac{-2 a A+\frac{1}{2} a (A-B-3 C) \sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B+C) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{A \int \sqrt{a+a \sec (c+d x)} \, dx}{a^2}-\frac{(5 A-B-3 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A-B+C) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(2 A) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a d}+\frac{(5 A-B-3 C) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{2 a d}\\ &=\frac{2 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac{(5 A-B-3 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A-B+C) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 28.1837, size = 16094, normalized size = 122.85 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^(3/2),x]
[Out]
Result too large to show
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Maple [B] time = 0.249, size = 732, normalized size = 5.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x)
[Out]
-1/4/d/a^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(4*A*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/
2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*2^(1/2)*cos(d*x+c)+5*A*cos(d*x+c)*sin(d*
x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)+4*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1)
)^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)-B*cos(d*x+c)*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*s
in(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-3*C*cos(d*x+c)*sin(d*x+c)*ln(-(-(-2*c
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+5*A*
ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))
^(1/2)*sin(d*x+c)-B*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*
x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-3*C*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/
sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-2*A*cos(d*x+c)^2+2*B*cos(d*x+c)^2-2*C*cos(d*x+c)^2
+2*A*cos(d*x+c)-2*B*cos(d*x+c)+2*C*cos(d*x+c))/(cos(d*x+c)+1)/sin(d*x+c)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/(a*sec(d*x + c) + a)^(3/2), x)
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Fricas [B] time = 24.7142, size = 1625, normalized size = 12.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
[-1/16*(8*(A - B + C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) - sqrt(2)*((5*A - B -
3*C)*cos(d*x + c)^2 + 2*(5*A - B - 3*C)*cos(d*x + c) + 5*A - B - 3*C)*sqrt(-a)*log((17*a*cos(d*x + c)^3 + 4*sq
rt(2)*(3*cos(d*x + c)^2 - cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) + 3*a*co
s(d*x + c)^2 - 13*a*cos(d*x + c) + a)/(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)) + 8*(A*cos(d*x
+ c)^2 + 2*A*cos(d*x + c) + A)*sqrt(-a)*log((8*a*cos(d*x + c)^3 + 4*(2*cos(d*x + c)^2 - cos(d*x + c))*sqrt(-a
)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c) + a)/(cos(d*x + c) + 1)))/(a^2*d*cos
(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), -1/8*(4*(A - B + C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d
*x + c)*sin(d*x + c) - sqrt(2)*((5*A - B - 3*C)*cos(d*x + c)^2 + 2*(5*A - B - 3*C)*cos(d*x + c) + 5*A - B - 3*
C)*sqrt(a)*arctan(1/4*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(3*cos(d*x + c) - 1)/(sqrt(a)*sin(d*x +
c))) + 8*(A*cos(d*x + c)^2 + 2*A*cos(d*x + c) + A)*sqrt(a)*arctan(1/2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*
(2*cos(d*x + c) - 1)/(sqrt(a)*sin(d*x + c))))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(3/2),x)
[Out]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)/(a*(sec(c + d*x) + 1))**(3/2), x)
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Giac [B] time = 10.9264, size = 448, normalized size = 3.42 \begin{align*} -\frac{\frac{\sqrt{2}{\left (5 \, A - B - 3 \, C\right )} \log \left ({\left (\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{\sqrt{-a} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{8 \, A \log \left ({\left |{\left (\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - a{\left (2 \, \sqrt{2} + 3\right )} \right |}\right )}{\sqrt{-a} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{8 \, A \log \left ({\left |{\left (\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} + a{\left (2 \, \sqrt{2} - 3\right )} \right |}\right )}{\sqrt{-a} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{2 \,{\left (\sqrt{2} A a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) - \sqrt{2} B a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) + \sqrt{2} C a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )\right )} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3}}}{8 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")
[Out]
-1/8*(sqrt(2)*(5*A - B - 3*C)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2)/(sq
rt(-a)*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 8*A*log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x +
1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3)))/(sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - 8*A*log(abs((sqrt(-a)*t
an(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3)))/(sqrt(-a)*a*sgn(tan(1/2*d*x
+ 1/2*c)^2 - 1)) - 2*(sqrt(2)*A*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - sqrt(2)*B*a*sgn(tan(1/2*d*x + 1/2*c)^2 -
1) + sqrt(2)*C*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*tan(1/2*d*x + 1/2*c)/a^3
)/d