3.166 \(\int \frac{\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=264 \[ -\frac{7 (9 A-5 B) \sin (c+d x)}{16 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(39 A-20 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac{(219 A-115 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(31 A-15 B) \sin (c+d x) \cos (c+d x)}{16 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(19 A-11 B) \sin (c+d x) \cos (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[Out]

((39*A - 20*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*a^(5/2)*d) - ((219*A - 115*B)*ArcTa
n[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B)*Cos[c + d*x]*S
in[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((19*A - 11*B)*Cos[c + d*x]*Sin[c + d*x])/(16*a*d*(a + a*Sec[c
+ d*x])^(3/2)) - (7*(9*A - 5*B)*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Sec[c + d*x]]) + ((31*A - 15*B)*Cos[c + d*
x]*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.790338, antiderivative size = 264, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ -\frac{7 (9 A-5 B) \sin (c+d x)}{16 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(39 A-20 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac{(219 A-115 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(31 A-15 B) \sin (c+d x) \cos (c+d x)}{16 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(19 A-11 B) \sin (c+d x) \cos (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

((39*A - 20*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*a^(5/2)*d) - ((219*A - 115*B)*ArcTa
n[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B)*Cos[c + d*x]*S
in[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((19*A - 11*B)*Cos[c + d*x]*Sin[c + d*x])/(16*a*d*(a + a*Sec[c
+ d*x])^(3/2)) - (7*(9*A - 5*B)*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Sec[c + d*x]]) + ((31*A - 15*B)*Cos[c + d*
x]*Sin[c + d*x])/(16*a^2*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))^{5/2}} \, dx &=-\frac{(A-B) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{\int \frac{\cos ^2(c+d x) \left (2 a (3 A-B)-\frac{7}{2} a (A-B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(19 A-11 B) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\cos ^2(c+d x) \left (a^2 (31 A-15 B)-\frac{5}{4} a^2 (19 A-11 B) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(19 A-11 B) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(31 A-15 B) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{\cos (c+d x) \left (-7 a^3 (9 A-5 B)+\frac{3}{2} a^3 (31 A-15 B) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{16 a^5}\\ &=-\frac{(A-B) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(19 A-11 B) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{7 (9 A-5 B) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(31 A-15 B) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{2 a^4 (39 A-20 B)-\frac{7}{2} a^4 (9 A-5 B) \sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{16 a^6}\\ &=-\frac{(A-B) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(19 A-11 B) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{7 (9 A-5 B) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(31 A-15 B) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(219 A-115 B) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}+\frac{(39 A-20 B) \int \sqrt{a+a \sec (c+d x)} \, dx}{8 a^3}\\ &=-\frac{(A-B) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(19 A-11 B) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{7 (9 A-5 B) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(31 A-15 B) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(219 A-115 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 )}{16 a^2 d}-\frac{(39 A-20 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^2 d}\\ &=\frac{(39 A-20 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 a^{5/2} d}-\frac{(219 A-115 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(A-B) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(19 A-11 B) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{7 (9 A-5 B) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(31 A-15 B) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.16367, size = 512, normalized size = 1.94 \[ -\frac{A (\sec (c+d x)+1)^{5/2} \left (\frac{760 \tan (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},3,\frac{3}{2},1-\sec (c+d x)\right )}{d \sqrt{\sec (c+d x)+1}}+\frac{152 \sin (c+d x) \cos (c+d x)}{d (\sec (c+d x)+1)^{3/2}}-\frac{219 \tan (c+d x) \left (2 \cos ^2(c+d x) \sqrt{1-\sec (c+d x)}-\cos (c+d x) \sqrt{1-\sec (c+d x)}+7 \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )-4 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{d \sqrt{1-\sec (c+d x)} \sqrt{\sec (c+d x)+1}}\right )}{128 (a (\sec (c+d x)+1))^{5/2}}-\frac{A \sin (c+d x) \cos (c+d x)}{4 d (a (\sec (c+d x)+1))^{5/2}}-\frac{B \sin (c+d x)}{4 d (a (\sec (c+d x)+1))^{5/2}}-\frac{5 B (\sec (c+d x)+1)^{5/2} \left (\frac{6 \sin (c+d x)}{d (\sec (c+d x)+1)^{3/2}}+\frac{9 \tan (c+d x) \left (\cos (c+d x)+\frac{\tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )}{\sqrt{1-\sec (c+d x)}}\right )}{d \sqrt{\sec (c+d x)+1}}+\frac{23 \tan (c+d x) \left (-\cos (c+d x) \sqrt{1-\sec (c+d x)}+\tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )-\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{d \sqrt{1-\sec (c+d x)} \sqrt{\sec (c+d x)+1}}\right )}{32 (a (\sec (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

-(B*Sin[c + d*x])/(4*d*(a*(1 + Sec[c + d*x]))^(5/2)) - (A*Cos[c + d*x]*Sin[c + d*x])/(4*d*(a*(1 + Sec[c + d*x]
))^(5/2)) - (5*B*(1 + Sec[c + d*x])^(5/2)*((6*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(3/2)) + (9*(Cos[c + d*x] +
ArcTanh[Sqrt[1 - Sec[c + d*x]]]/Sqrt[1 - Sec[c + d*x]])*Tan[c + d*x])/(d*Sqrt[1 + Sec[c + d*x]]) + (23*(ArcTan
h[Sqrt[1 - Sec[c + d*x]]] - Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] - Cos[c + d*x]*Sqrt[1 - Sec[c + d*
x]])*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]])))/(32*(a*(1 + Sec[c + d*x]))^(5/2)) - (A*
(1 + Sec[c + d*x])^(5/2)*((152*Cos[c + d*x]*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(3/2)) + (760*Hypergeometric2F
1[1/2, 3, 3/2, 1 - Sec[c + d*x]]*Tan[c + d*x])/(d*Sqrt[1 + Sec[c + d*x]]) - (219*(7*ArcTanh[Sqrt[1 - Sec[c + d
*x]]] - 4*Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] - Cos[c + d*x]*Sqrt[1 - Sec[c + d*x]] + 2*Cos[c + d*
x]^2*Sqrt[1 - Sec[c + d*x]])*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]])))/(128*(a*(1 + Se
c[c + d*x]))^(5/2))

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Maple [B]  time = 0.388, size = 1427, normalized size = 5.4 \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))^(5/2),x)

[Out]

1/64/d/a^3*(-1+cos(d*x+c))^2*(468*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)*arcta
nh(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))+468*A*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)+219*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)-115*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)-240*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))-240*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)+80*B*cos(d*x+c)^3+300*A*cos(d*x+c)^4+657*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))-34
5*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)*si
n(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-345*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))-32*A*cos(d*x+c)^6+112*A*cos(d*x+c)^5-64*
B*cos(d*x+c)^5-156*B*cos(d*x+c)^4-128*A*cos(d*x+c)^3+140*B*cos(d*x+c)^2+219*A*sin(d*x+c)*cos(d*x+c)^3*(-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))-1
15*B*sin(d*x+c)*cos(d*x+c)^3*(-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)*s
in(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-252*A*cos(d*x+c)^2+657*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))+156*A*sin(d*x+c)*cos
(d*x+c)^3*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)-80*B*sin(d*x+c)*cos(d*x+c)^3*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)+156*A*(-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)-80*B*(-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))*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)^5/cos(d*x+c)

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Maxima [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))/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 28.2047, size = 2059, normalized size = 7.8 \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))^(5/2),x, algorithm="fricas")

[Out]

[1/64*(sqrt(2)*((219*A - 115*B)*cos(d*x + c)^3 + 3*(219*A - 115*B)*cos(d*x + c)^2 + 3*(219*A - 115*B)*cos(d*x
+ c) + 219*A - 115*B)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*si
n(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)) + 8*((39*A - 20
*B)*cos(d*x + c)^3 + 3*(39*A - 20*B)*cos(d*x + c)^2 + 3*(39*A - 20*B)*cos(d*x + c) + 39*A - 20*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)) + 4*(8*A*cos(d*x + c)^4 - 4*(5*A - 4*B)*cos(d*x + c)^3 - 5*(19*A - 11*B)*cos(d
*x + c)^2 - 7*(9*A - 5*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x +
c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), 1/32*(sqrt(2)*((219*A - 115*B)*cos(d*x + c)^3
+ 3*(219*A - 115*B)*cos(d*x + c)^2 + 3*(219*A - 115*B)*cos(d*x + c) + 219*A - 115*B)*sqrt(a)*arctan(sqrt(2)*sq
rt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 8*((39*A - 20*B)*cos(d*x + c)^3 +
3*(39*A - 20*B)*cos(d*x + c)^2 + 3*(39*A - 20*B)*cos(d*x + c) + 39*A - 20*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*(8*A*cos(d*x + c)^4 - 4*(5*A - 4*B)*cos(d*x +
c)^3 - 5*(19*A - 11*B)*cos(d*x + c)^2 - 7*(9*A - 5*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*si
n(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*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))/(a+a*sec(d*x+c))**(5/2),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))/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError