3.118 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=266 \[ -\frac{(47 A+24 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{8 a^{3/2} d}+\frac{(17 A+9 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{3 (7 A+4 C) \tan (c+d x)}{8 a d \sqrt{a \cos (c+d x)+a}}+\frac{(5 A+3 C) \tan (c+d x) \sec ^2(c+d x)}{6 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(13 A+6 C) \tan (c+d x) \sec (c+d x)}{12 a d \sqrt{a \cos (c+d x)+a}} \]

[Out]

-((47*A + 24*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*a^(3/2)*d) + ((17*A + 9*C)*ArcTan
h[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + (3*(7*A + 4*C)*Tan[c + d
*x])/(8*a*d*Sqrt[a + a*Cos[c + d*x]]) - ((13*A + 6*C)*Sec[c + d*x]*Tan[c + d*x])/(12*a*d*Sqrt[a + a*Cos[c + d*
x]]) - ((A + C)*Sec[c + d*x]^2*Tan[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((5*A + 3*C)*Sec[c + d*x]^2*Ta
n[c + d*x])/(6*a*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.88116, antiderivative size = 266, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3042, 2984, 2985, 2649, 206, 2773} \[ -\frac{(47 A+24 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{8 a^{3/2} d}+\frac{(17 A+9 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{3 (7 A+4 C) \tan (c+d x)}{8 a d \sqrt{a \cos (c+d x)+a}}+\frac{(5 A+3 C) \tan (c+d x) \sec ^2(c+d x)}{6 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(13 A+6 C) \tan (c+d x) \sec (c+d x)}{12 a d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

-((47*A + 24*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*a^(3/2)*d) + ((17*A + 9*C)*ArcTan
h[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + (3*(7*A + 4*C)*Tan[c + d
*x])/(8*a*d*Sqrt[a + a*Cos[c + d*x]]) - ((13*A + 6*C)*Sec[c + d*x]*Tan[c + d*x])/(12*a*d*Sqrt[a + a*Cos[c + d*
x]]) - ((A + C)*Sec[c + d*x]^2*Tan[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((5*A + 3*C)*Sec[c + d*x]^2*Ta
n[c + d*x])/(6*a*d*Sqrt[a + a*Cos[c + d*x]])

Rule 3042

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x
])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 2984

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

Rule 2985

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx &=-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\left (a (5 A+3 C)-\frac{1}{2} a (7 A+3 C) \cos (c+d x)\right ) \sec ^4(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A+3 C) \sec ^2(c+d x) \tan (c+d x)}{6 a d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\left (-a^2 (13 A+6 C)+\frac{5}{2} a^2 (5 A+3 C) \cos (c+d x)\right ) \sec ^3(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{6 a^3}\\ &=-\frac{(13 A+6 C) \sec (c+d x) \tan (c+d x)}{12 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A+3 C) \sec ^2(c+d x) \tan (c+d x)}{6 a d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\left (\frac{9}{2} a^3 (7 A+4 C)-\frac{3}{2} a^3 (13 A+6 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{12 a^4}\\ &=\frac{3 (7 A+4 C) \tan (c+d x)}{8 a d \sqrt{a+a \cos (c+d x)}}-\frac{(13 A+6 C) \sec (c+d x) \tan (c+d x)}{12 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A+3 C) \sec ^2(c+d x) \tan (c+d x)}{6 a d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\left (-\frac{3}{4} a^4 (47 A+24 C)+\frac{9}{4} a^4 (7 A+4 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{12 a^5}\\ &=\frac{3 (7 A+4 C) \tan (c+d x)}{8 a d \sqrt{a+a \cos (c+d x)}}-\frac{(13 A+6 C) \sec (c+d x) \tan (c+d x)}{12 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A+3 C) \sec ^2(c+d x) \tan (c+d x)}{6 a d \sqrt{a+a \cos (c+d x)}}+\frac{(17 A+9 C) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{4 a}-\frac{(47 A+24 C) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx}{16 a^2}\\ &=\frac{3 (7 A+4 C) \tan (c+d x)}{8 a d \sqrt{a+a \cos (c+d x)}}-\frac{(13 A+6 C) \sec (c+d x) \tan (c+d x)}{12 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A+3 C) \sec ^2(c+d x) \tan (c+d x)}{6 a d \sqrt{a+a \cos (c+d x)}}-\frac{(17 A+9 C) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{2 a d}+\frac{(47 A+24 C) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{8 a d}\\ &=-\frac{(47 A+24 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{8 a^{3/2} d}+\frac{(17 A+9 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{3 (7 A+4 C) \tan (c+d x)}{8 a d \sqrt{a+a \cos (c+d x)}}-\frac{(13 A+6 C) \sec (c+d x) \tan (c+d x)}{12 a d \sqrt{a+a \cos (c+d x)}}-\frac{(A+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A+3 C) \sec ^2(c+d x) \tan (c+d x)}{6 a d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.86616, size = 205, normalized size = 0.77 \[ \frac{-48 (17 A+9 C) \cos ^5\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 \sqrt{2} (47 A+24 C) \cos ^5\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )-\sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3\left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) (3 (55 A+36 C) \cos (c+d x)+(74 A+48 C) \cos (2 (c+d x))+63 A \cos (3 (c+d x))+106 A+36 C \cos (3 (c+d x))+48 C)}{48 d \left (\sin ^2\left (\frac{1}{2} (c+d x)\right )-1\right ) (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

(-48*(17*A + 9*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5 + 12*Sqrt[2]*(47*A + 24*C)*ArcTanh[Sqrt[2]*Sin[
(c + d*x)/2]]*Cos[(c + d*x)/2]^5 - Cos[(c + d*x)/2]^3*(106*A + 48*C + 3*(55*A + 36*C)*Cos[c + d*x] + (74*A + 4
8*C)*Cos[2*(c + d*x)] + 63*A*Cos[3*(c + d*x)] + 36*C*Cos[3*(c + d*x)])*Sec[c + d*x]^3*Sin[(c + d*x)/2])/(48*d*
(a*(1 + Cos[c + d*x]))^(3/2)*(-1 + Sin[(c + d*x)/2]^2))

________________________________________________________________________________________

Maple [B]  time = 0.133, size = 2028, normalized size = 7.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(3/2),x)

[Out]

1/6*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-576*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a
^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^8*a-1128*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2
^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^
8*a-1128*A*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-2*a)/(2*cos(1/2*
d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^8*a-576*C*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*sin(1
/2*d*x+1/2*c)^2)^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^8*a+72*C*ln(4/(2*cos(1/2*d*x+1/
2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1
/2*c)^2*a+72*C*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-2*a)/(2*cos(
1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^2*a-432*C*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*s
in(1/2*d*x+1/2*c)^2)^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^4*a-432*C*ln(4/(2*cos(1/2*d
*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*
d*x+1/2*c)^4*a+864*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(
1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^6*a+864*C*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)
*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^6*a+1692*A*ln(4/(2*cos
(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*co
s(1/2*d*x+1/2*c)^6*a+1692*A*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)
-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^6*a-846*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1
/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a-846*A*ln(-4
*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/
2)))*cos(1/2*d*x+1/2*c)^4*a+141*A*ln(-4*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)
^(1/2)-2*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2)))*cos(1/2*d*x+1/2*c)^2*a+141*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(
a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^2*a-12*C*
2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-12*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+864*C*2^(1/
2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^8*a+1632*A*2^(1/
2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^8*a+504*A*2^(1/2
)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^6-2448*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^
(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^6*a-1296*C*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^
(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^6*a-608*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(
1/2)*cos(1/2*d*x+1/2*c)^4+1224*A*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c
))*cos(1/2*d*x+1/2*c)^4*a+648*C*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c)
)*cos(1/2*d*x+1/2*c)^4*a+288*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^6-336*C*2^(1/
2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4-204*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^
(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^2*a-108*C*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(
1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^2*a+218*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1
/2)*cos(1/2*d*x+1/2*c)^2+120*C*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^2)/a^(5/2)/co
s(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^3/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^3/sin(1/2*d*x+1/2*c)/(a*cos(1
/2*d*x+1/2*c)^2)^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.33969, size = 980, normalized size = 3.68 \begin{align*} \frac{12 \, \sqrt{2}{\left ({\left (17 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (17 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (17 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 3 \,{\left ({\left (47 \, A + 24 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (47 \, A + 24 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (47 \, A + 24 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \,{\left (9 \,{\left (7 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (37 \, A + 24 \, C\right )} \cos \left (d x + c\right )^{2} - 6 \, A \cos \left (d x + c\right ) + 8 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{96 \,{\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/96*(12*sqrt(2)*((17*A + 9*C)*cos(d*x + c)^5 + 2*(17*A + 9*C)*cos(d*x + c)^4 + (17*A + 9*C)*cos(d*x + c)^3)*s
qrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3
*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 3*((47*A + 24*C)*cos(d*x + c)^5 + 2*(47*A + 24*C)*cos(d*x + c)^4
+ (47*A + 24*C)*cos(d*x + c)^3)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*sqrt(a*cos(d*x + c) + a
)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(9*(7*A + 4*C)*cos(d*x
+ c)^3 + (37*A + 24*C)*cos(d*x + c)^2 - 6*A*cos(d*x + c) + 8*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^2*d
*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3)

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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((A+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+a*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 3.65745, size = 1010, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-1/48*(6*sqrt(2)*(17*A*sqrt(a) + 9*C*sqrt(a))*log((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 + 3*(47*A*sqrt(a) + 24*C*sqrt(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)))/a^2 - 3*(47*A*sqrt(a) + 24*C*sqrt(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)))/a^2 - 12*sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)
*(sqrt(2)*A*a + sqrt(2)*C*a)*tan(1/2*d*x + 1/2*c)/a^3 - 4*sqrt(2)*(339*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*
tan(1/2*d*x + 1/2*c)^2 + a))^10*A*sqrt(a) + 72*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 +
a))^10*C*sqrt(a) - 3165*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^8*A*a^(3/2) - 888
*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^8*C*a^(3/2) + 9198*(sqrt(a)*tan(1/2*d*x +
1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*a^(5/2) + 3024*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/
2*d*x + 1/2*c)^2 + a))^6*C*a^(5/2) - 4938*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^
4*A*a^(7/2) - 1776*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*C*a^(7/2) + 975*(sqrt
(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*a^(9/2) + 360*(sqrt(a)*tan(1/2*d*x + 1/2*c)
- sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*a^(9/2) - 73*A*a^(11/2) - 24*C*a^(11/2))/(((sqrt(a)*tan(1/2*d*x + 1
/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^
2 + a))^2*a + a^2)^3*a))/d