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

Optimal. Leaf size=221 \[ \frac{(19 A-12 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac{(13 A-9 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(7 A-6 B) \sin (c+d x)}{4 a d \sqrt{a \sec (c+d x)+a}}+\frac{(2 A-B) \sin (c+d x) \cos (c+d x)}{2 a d \sqrt{a \sec (c+d x)+a}}-\frac{(A-B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

((19*A - 12*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*a^(3/2)*d) - ((13*A - 9*B)*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)*Cos[c + d*x]*Sin[c
+ d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((7*A - 6*B)*Sin[c + d*x])/(4*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((2*
A - B)*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 = 0.583225, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4020, 4022, 3920, 3774, 203, 3795} \[ \frac{(19 A-12 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac{(13 A-9 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(7 A-6 B) \sin (c+d x)}{4 a d \sqrt{a \sec (c+d x)+a}}+\frac{(2 A-B) \sin (c+d x) \cos (c+d x)}{2 a d \sqrt{a \sec (c+d x)+a}}-\frac{(A-B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4020

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

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

Mathematica [C]  time = 2.28628, size = 395, normalized size = 1.79 \[ \frac{\sec (c+d x) \left (-40 A \sqrt{1-\sec (c+d x)} (\sin (c+d x)+\tan (c+d x)) \text{Hypergeometric2F1}\left (\frac{1}{2},3,\frac{3}{2},1-\sec (c+d x)\right )+(91 A-48 B) (\sin (c+d x)+\tan (c+d x)) \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )-13 A \sin (c+d x) \sqrt{1-\sec (c+d x)}+\frac{13}{2} A \sin (2 (c+d x)) \sqrt{1-\sec (c+d x)}+18 A \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec (c+d x)}-52 \sqrt{2} A \sin (c+d x) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )-52 \sqrt{2} A \tan (c+d x) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )+24 B \sin (c+d x) \sqrt{1-\sec (c+d x)}+8 B \sin (2 (c+d x)) \sqrt{1-\sec (c+d x)}+36 \sqrt{2} B \sin (c+d x) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )+36 \sqrt{2} B \tan (c+d x) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{16 d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]*(-52*Sqrt[2]*A*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Sin[c + d*x] + 36*Sqrt[2]*B*ArcTanh[Sqrt[
1 - Sec[c + d*x]]/Sqrt[2]]*Sin[c + d*x] - 13*A*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + 24*B*Sqrt[1 - Sec[c + d*x
]]*Sin[c + d*x] + 18*A*Cos[c + d*x]^2*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + (13*A*Sqrt[1 - Sec[c + d*x]]*Sin[2
*(c + d*x)])/2 + 8*B*Sqrt[1 - Sec[c + d*x]]*Sin[2*(c + d*x)] - 52*Sqrt[2]*A*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqr
t[2]]*Tan[c + d*x] + 36*Sqrt[2]*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Tan[c + d*x] + (91*A - 48*B)*ArcTanh
[Sqrt[1 - Sec[c + d*x]]]*(Sin[c + d*x] + Tan[c + d*x]) - 40*A*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]]
*Sqrt[1 - Sec[c + d*x]]*(Sin[c + d*x] + Tan[c + d*x])))/(16*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3
/2))

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Maple [B]  time = 0.333, size = 1075, normalized size = 4.9 \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))/(a+a*sec(d*x+c))^(3/2),x)

[Out]

-1/16/d/a^2*(-1+cos(d*x+c))*(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))-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))+26*A*sin(d*x+c)*cos(d*x+c)^2*(-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))+38*A*sin(d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arc
tanh(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)-18*B*sin(d*x+c)*cos(d*x+c
)^2*(-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)/s
in(d*x+c))-24*B*sin(d*x+c)*cos(d*x+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)+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))+19*A*(-2*cos(d*x+c)/(co
s(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)-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))-12*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arctanh(1/2*2^(1/2)*(-2*co
s(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)+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)-8*A*cos(d*x+c)
^5-18*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)+20*A*cos(d*x+c)^4-16*B*cos(d*x+c)^4+16*A*cos(d*x+c)^3-8*B*cos(d*x+c)^3-28*A*cos(d*x+
c)^2+24*B*cos(d*x+c)^2)*(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 (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))/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((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 = 14.6731, size = 1673, normalized size = 7.57 \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))/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/8*(sqrt(2)*((13*A - 9*B)*cos(d*x + c)^2 + 2*(13*A - 9*B)*cos(d*x + c) + 13*A - 9*B)*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)*cos(d*x + c)^2 + 2*(19*A - 12*B)*cos(d*x +
c) + 19*A - 12*B)*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*cos(d*x + c)^3 - (3*A - 4*B)*cos(d*x +
c)^2 - (7*A - 6*B)*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)*cos(d*x + c)^2 + 2*(13*A - 9*B)*cos(d*x + c) + 13*A
- 9*B)*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)*cos(d*x + c)^2 + 2*(19*A - 12*B)*cos(d*x + c) + 19*A - 12*B)*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)*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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )}\right ) \cos ^{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(cos(d*x+c)**2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Integral((A + B*sec(c + d*x))*cos(c + d*x)**2/(a*(sec(c + d*x) + 1))**(3/2), x)

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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))/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError