3.525 \(\int \frac{\cos ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=232 \[ \frac{(19 A-12 B+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac{(13 A-9 B+5 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(7 A-6 B+2 C) \sin (c+d x)}{4 a d \sqrt{a \sec (c+d x)+a}}+\frac{(2 A-B+C) \sin (c+d x) \cos (c+d x)}{2 a d \sqrt{a \sec (c+d x)+a}}-\frac{(A-B+C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

((19*A - 12*B + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*a^(3/2)*d) - ((13*A - 9*B + 5
*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A - B + C)*Co
s[c + d*x]*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((7*A - 6*B + 2*C)*Sin[c + d*x])/(4*a*d*Sqrt[a + a
*Sec[c + d*x]]) + ((2*A - B + C)*Cos[c + d*x]*Sin[c + d*x])/(2*a*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.622673, antiderivative size = 232, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {4084, 4022, 3920, 3774, 203, 3795} \[ \frac{(19 A-12 B+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac{(13 A-9 B+5 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(7 A-6 B+2 C) \sin (c+d x)}{4 a d \sqrt{a \sec (c+d x)+a}}+\frac{(2 A-B+C) \sin (c+d x) \cos (c+d x)}{2 a d \sqrt{a \sec (c+d x)+a}}-\frac{(A-B+C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

((19*A - 12*B + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*a^(3/2)*d) - ((13*A - 9*B + 5
*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A - B + C)*Co
s[c + d*x]*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((7*A - 6*B + 2*C)*Sin[c + d*x])/(4*a*d*Sqrt[a + a
*Sec[c + d*x]]) + ((2*A - B + C)*Cos[c + d*x]*Sin[c + d*x])/(2*a*d*Sqrt[a + a*Sec[c + d*x]])

Rule 4084

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[e + f*x])^m*(d*Csc[e + f*x])^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)
))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 4022

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1
/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - A*b*(m + n + 1)*Csc[e + f*x
], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

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{\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac{(A-B+C) \cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\cos ^2(c+d x) \left (2 a (2 A-B+C)-\frac{1}{2} a (5 A-5 B+C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B+C) \cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(2 A-B+C) \cos (c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{\cos (c+d x) \left (-a^2 (7 A-6 B+2 C)+3 a^2 (2 A-B+C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a^3}\\ &=-\frac{(A-B+C) \cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(7 A-6 B+2 C) \sin (c+d x)}{4 a d \sqrt{a+a \sec (c+d x)}}+\frac{(2 A-B+C) \cos (c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{\frac{1}{2} a^3 (19 A-12 B+8 C)-\frac{1}{2} a^3 (7 A-6 B+2 C) \sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a^4}\\ &=-\frac{(A-B+C) \cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(7 A-6 B+2 C) \sin (c+d x)}{4 a d \sqrt{a+a \sec (c+d x)}}+\frac{(2 A-B+C) \cos (c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}-\frac{(13 A-9 B+5 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}+\frac{(19 A-12 B+8 C) \int \sqrt{a+a \sec (c+d x)} \, dx}{8 a^2}\\ &=-\frac{(A-B+C) \cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(7 A-6 B+2 C) \sin (c+d x)}{4 a d \sqrt{a+a \sec (c+d x)}}+\frac{(2 A-B+C) \cos (c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}+\frac{(13 A-9 B+5 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{(19 A-12 B+8 C) \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 )}{4 a d}\\ &=\frac{(19 A-12 B+8 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 a^{3/2} d}-\frac{(13 A-9 B+5 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) \cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(7 A-6 B+2 C) \sin (c+d x)}{4 a d \sqrt{a+a \sec (c+d x)}}+\frac{(2 A-B+C) \cos (c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 28.1997, size = 17669, normalized size = 76.16 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[c + d*x]^2*(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 = 0.319, size = 1569, normalized size = 6.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x)

[Out]

-1/16/d/a^2*(-1+cos(d*x+c))*(-12*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(co
s(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*2^(1/2)*sin(d*x+c)+8*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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)*sin(d*x+c)+38*A*2^(1/2)*sin(d*
x+c)*cos(d*x+c)*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*cos(d*x+c)
/(cos(d*x+c)+1))^(3/2)+19*A*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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))-24*B*cos(d*x+c)*sin(d*x+c)*2^(1/2)*(-2*cos(
d*x+c)/(cos(d*x+c)+1))^(3/2)*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))-8
*B*cos(d*x+c)^3+19*A*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))*(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)+20*A*cos(d*x+c)^4+8*C*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)*(-2*cos(d
*x+c)/(cos(d*x+c)+1))^(3/2)*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))+8*
C*cos(d*x+c)^3-8*C*cos(d*x+c)^2+10*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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))*cos(d*x+c)^2*sin(d*x+c)-8*A*cos(d*x+c)^5-16*B*cos(d*x+c)^4+16*A*co
s(d*x+c)^3+24*B*cos(d*x+c)^2+16*C*cos(d*x+c)*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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))+26*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/
2)*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))*sin(d*x+c)-18*B*(-2*cos(d*x
+c)/(cos(d*x+c)+1))^(3/2)*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))*sin(
d*x+c)+10*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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))*sin(d*x+c)+26*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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))*cos(d*x+c)^2*sin(d*x+c)-18*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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))*cos(d*x+c)^2*sin(d*x+c)-12*B*sin
(d*x+c)*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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))-28*A*cos(d*x+c)^2+52*A*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1)
)^(3/2)*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))-36*B*cos(d*x+c)*sin(d*
x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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))+20*C*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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)))*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)^3/cos(d*x+c)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\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(cos(d*x+c)^2*(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)*cos(d*x + c)^2/(a*sec(d*x + c) + a)^(3/2), x)

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Fricas [A]  time = 118.304, size = 1789, normalized size = 7.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[-1/8*(sqrt(2)*((13*A - 9*B + 5*C)*cos(d*x + c)^2 + 2*(13*A - 9*B + 5*C)*cos(d*x + c) + 13*A - 9*B + 5*C)*sqrt
(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) - 3*a*cos(d*x
+ c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + ((19*A - 12*B + 8*C)*cos(d*x + c)^2 +
2*(19*A - 12*B + 8*C)*cos(d*x + c) + 19*A - 12*B + 8*C)*sqrt(-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*
cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) - 2*(2*A*c
os(d*x + c)^3 - (3*A - 4*B)*cos(d*x + c)^2 - (7*A - 6*B + 2*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x
+ c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), 1/4*(sqrt(2)*((13*A - 9*B + 5*C)*c
os(d*x + c)^2 + 2*(13*A - 9*B + 5*C)*cos(d*x + c) + 13*A - 9*B + 5*C)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - ((19*A - 12*B + 8*C)*cos(d*x + c)^2 + 2*(19*A -
12*B + 8*C)*cos(d*x + c) + 19*A - 12*B + 8*C)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x +
c)/(sqrt(a)*sin(d*x + c))) + (2*A*cos(d*x + c)^3 - (3*A - 4*B)*cos(d*x + c)^2 - (7*A - 6*B + 2*C)*cos(d*x + c
))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError