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

Optimal. Leaf size=268 \[ -\frac{(47 A-38 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 a^{3/2} d}+\frac{(17 A-13 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{7 (3 A-2 B) \sin (c+d x)}{8 a d \sqrt{a \sec (c+d x)+a}}+\frac{(5 A-3 B) \sin (c+d x) \cos ^2(c+d x)}{6 a d \sqrt{a \sec (c+d x)+a}}-\frac{(A-B) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{(13 A-12 B) \sin (c+d x) \cos (c+d x)}{12 a d \sqrt{a \sec (c+d x)+a}} \]

[Out]

-((47*A - 38*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*a^(3/2)*d) + ((17*A - 13*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]^2*S
in[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + (7*(3*A - 2*B)*Sin[c + d*x])/(8*a*d*Sqrt[a + a*Sec[c + d*x]])
- ((13*A - 12*B)*Cos[c + d*x]*Sin[c + d*x])/(12*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((5*A - 3*B)*Cos[c + d*x]^2*Si
n[c + d*x])/(6*a*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.779661, antiderivative size = 268, 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} \[ -\frac{(47 A-38 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{8 a^{3/2} d}+\frac{(17 A-13 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{7 (3 A-2 B) \sin (c+d x)}{8 a d \sqrt{a \sec (c+d x)+a}}+\frac{(5 A-3 B) \sin (c+d x) \cos ^2(c+d x)}{6 a d \sqrt{a \sec (c+d x)+a}}-\frac{(A-B) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac{(13 A-12 B) \sin (c+d x) \cos (c+d x)}{12 a d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((47*A - 38*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*a^(3/2)*d) + ((17*A - 13*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]^2*S
in[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + (7*(3*A - 2*B)*Sin[c + d*x])/(8*a*d*Sqrt[a + a*Sec[c + d*x]])
- ((13*A - 12*B)*Cos[c + d*x]*Sin[c + d*x])/(12*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((5*A - 3*B)*Cos[c + d*x]^2*Si
n[c + d*x])/(6*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 ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\cos ^3(c+d x) \left (a (5 A-3 B)-\frac{7}{2} a (A-B) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(5 A-3 B) \cos ^2(c+d x) \sin (c+d x)}{6 a d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{\cos ^2(c+d x) \left (-a^2 (13 A-12 B)+\frac{5}{2} a^2 (5 A-3 B) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{6 a^3}\\ &=-\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac{(13 A-12 B) \cos (c+d x) \sin (c+d x)}{12 a d \sqrt{a+a \sec (c+d x)}}+\frac{(5 A-3 B) \cos ^2(c+d x) \sin (c+d x)}{6 a d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{\cos (c+d x) \left (\frac{21}{2} a^3 (3 A-2 B)-\frac{3}{2} a^3 (13 A-12 B) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{12 a^4}\\ &=-\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{7 (3 A-2 B) \sin (c+d x)}{8 a d \sqrt{a+a \sec (c+d x)}}-\frac{(13 A-12 B) \cos (c+d x) \sin (c+d x)}{12 a d \sqrt{a+a \sec (c+d x)}}+\frac{(5 A-3 B) \cos ^2(c+d x) \sin (c+d x)}{6 a d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{-\frac{3}{4} a^4 (47 A-38 B)+\frac{21}{4} a^4 (3 A-2 B) \sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{12 a^5}\\ &=-\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{7 (3 A-2 B) \sin (c+d x)}{8 a d \sqrt{a+a \sec (c+d x)}}-\frac{(13 A-12 B) \cos (c+d x) \sin (c+d x)}{12 a d \sqrt{a+a \sec (c+d x)}}+\frac{(5 A-3 B) \cos ^2(c+d x) \sin (c+d x)}{6 a d \sqrt{a+a \sec (c+d x)}}-\frac{(47 A-38 B) \int \sqrt{a+a \sec (c+d x)} \, dx}{16 a^2}+\frac{(17 A-13 B) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{7 (3 A-2 B) \sin (c+d x)}{8 a d \sqrt{a+a \sec (c+d x)}}-\frac{(13 A-12 B) \cos (c+d x) \sin (c+d x)}{12 a d \sqrt{a+a \sec (c+d x)}}+\frac{(5 A-3 B) \cos ^2(c+d x) \sin (c+d x)}{6 a d \sqrt{a+a \sec (c+d x)}}+\frac{(47 A-38 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 )}{8 a d}-\frac{(17 A-13 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{(47 A-38 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{8 a^{3/2} d}+\frac{(17 A-13 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 ^2(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{7 (3 A-2 B) \sin (c+d x)}{8 a d \sqrt{a+a \sec (c+d x)}}-\frac{(13 A-12 B) \cos (c+d x) \sin (c+d x)}{12 a d \sqrt{a+a \sec (c+d x)}}+\frac{(5 A-3 B) \cos ^2(c+d x) \sin (c+d x)}{6 a d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.14309, size = 502, normalized size = 1.87 \[ -\frac{A (\sec (c+d x)+1)^{3/2} \left (\frac{336 \tan (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},4,\frac{3}{2},1-\sec (c+d x)\right )}{d \sqrt{\sec (c+d x)+1}}+\frac{17 \tan (c+d x) \left (-8 \cos ^3(c+d x) \sqrt{1-\sec (c+d x)}+2 \cos ^2(c+d x) \sqrt{1-\sec (c+d x)}+3 \left (-7 \cos (c+d x) \sqrt{1-\sec (c+d x)}+9 \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )-8 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )\right )}{d \sqrt{1-\sec (c+d x)} \sqrt{\sec (c+d x)+1}}\right )}{96 (a (\sec (c+d x)+1))^{3/2}}-\frac{B (\sec (c+d x)+1)^{3/2} \left (\frac{40 \tan (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},3,\frac{3}{2},1-\sec (c+d x)\right )}{d \sqrt{\sec (c+d x)+1}}-\frac{13 \tan (c+d x) \left (2 \cos ^2(c+d x) \sqrt{1-\sec (c+d x)}-\cos (c+d x) \sqrt{1-\sec (c+d x)}+7 \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )-4 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{d \sqrt{1-\sec (c+d x)} \sqrt{\sec (c+d x)+1}}\right )}{16 (a (\sec (c+d x)+1))^{3/2}}-\frac{A \sin (c+d x) \cos ^2(c+d x)}{2 d (a (\sec (c+d x)+1))^{3/2}}-\frac{B \sin (c+d x) \cos (c+d x)}{2 d (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(B*Cos[c + d*x]*Sin[c + d*x])/(2*d*(a*(1 + Sec[c + d*x]))^(3/2)) - (A*Cos[c + d*x]^2*Sin[c + d*x])/(2*d*(a*(1
+ Sec[c + d*x]))^(3/2)) - (B*(1 + Sec[c + d*x])^(3/2)*((40*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]]*T
an[c + d*x])/(d*Sqrt[1 + Sec[c + d*x]]) - (13*(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]])))/(16*(a*(1 + Sec[c + d*x]))^(3/2)) - (A*(1 + Sec[c
+ d*x])^(3/2)*((336*Hypergeometric2F1[1/2, 4, 3/2, 1 - Sec[c + d*x]]*Tan[c + d*x])/(d*Sqrt[1 + Sec[c + d*x]])
+ (17*(3*(9*ArcTanh[Sqrt[1 - Sec[c + d*x]]] - 8*Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] - 7*Cos[c + d*
x]*Sqrt[1 - Sec[c + d*x]]) + 2*Cos[c + d*x]^2*Sqrt[1 - Sec[c + d*x]] - 8*Cos[c + d*x]^3*Sqrt[1 - Sec[c + d*x]]
)*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]])))/(96*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [B]  time = 0.292, size = 1425, normalized size = 5.3 \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)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x)

[Out]

-1/192/d/a^2*(-1+cos(d*x+c))*(204*A*sin(d*x+c)*cos(d*x+c)^3*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))^(5/2)-156*B*sin(d*x+c)*cos(d*x+c)^3*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))^(5/2)+423*A*
cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*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)-342*B*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*arct
anh(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)-336*B*cos(d*x+c)^3+204*A*l
n(-(-(-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))^
(5/2)*sin(d*x+c)-156*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))^(5/2)*sin(d*x+c)+612*A*sin(d*x+c)*cos(d*x+c)^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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)-468*B*sin(d*x+c)*cos(d*x+c)^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))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(
5/2)+612*A*sin(d*x+c)*cos(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))^(5/2)-468*B*sin(d*x+c)*cos(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))^(5/2)-208*A*cos(d*x+c)^4+423*A*cos(d*x+c)
*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*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)-342*B*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*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)+112*A*cos(d*x+c)^6-344*A*cos(d*x+c)^5+24
0*B*cos(d*x+c)^5+192*B*cos(d*x+c)^4+504*A*cos(d*x+c)^3-64*A*cos(d*x+c)^7-96*B*cos(d*x+c)^6+141*A*sin(d*x+c)*co
s(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*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))-114*B*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*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))+141*A*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(5/2)*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)-114*B*2
^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*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))*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)^3/cos(d*x+c)^2

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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 )^{3}}{{\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)^3*(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)^3/(a*sec(d*x + c) + a)^(3/2), x)

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Fricas [A]  time = 14.7118, size = 1789, normalized size = 6.68 \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)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/48*(6*sqrt(2)*((17*A - 13*B)*cos(d*x + c)^2 + 2*(17*A - 13*B)*cos(d*x + c) + 17*A - 13*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)) + 3*((47*A - 38*B)*cos(d*x + c)^2 + 2*(47*A - 38*B)*c
os(d*x + c) + 47*A - 38*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*(8*A*cos(d*x + c)^4 - 6*(A - 2*B)*c
os(d*x + c)^3 + (37*A - 18*B)*cos(d*x + c)^2 + 21*(3*A - 2*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), -1/24*(6*sqrt(2)*((17*A - 13*B)*cos
(d*x + c)^2 + 2*(17*A - 13*B)*cos(d*x + c) + 17*A - 13*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))) - 3*((47*A - 38*B)*cos(d*x + c)^2 + 2*(47*A - 38*B)*cos(d*x +
c) + 47*A - 38*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)))
- (8*A*cos(d*x + c)^4 - 6*(A - 2*B)*cos(d*x + c)^3 + (37*A - 18*B)*cos(d*x + c)^2 + 21*(3*A - 2*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(-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)**3*(A+B*sec(d*x+c))/(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)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError