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

Optimal. Leaf size=217 \[ \frac{(35 A-11 B+3 C) \tan (c+d x)}{16 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(115 A-43 B+3 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(5 A-2 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(15 A-7 B-C) \tan (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[Out]

-(((5*A - 2*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(5/2)*d)) + ((115*A - 43*B + 3*C)*
ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Tan[
c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - ((15*A - 7*B - C)*Tan[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2
)) + ((35*A - 11*B + 3*C)*Tan[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.80223, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {3041, 2978, 2984, 2985, 2649, 206, 2773} \[ \frac{(35 A-11 B+3 C) \tan (c+d x)}{16 a^2 d \sqrt{a \cos (c+d x)+a}}+\frac{(115 A-43 B+3 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(5 A-2 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(15 A-7 B-C) \tan (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((5*A - 2*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(5/2)*d)) + ((115*A - 43*B + 3*C)*
ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Tan[
c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - ((15*A - 7*B - C)*Tan[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2
)) + ((35*A - 11*B + 3*C)*Tan[c + d*x])/(16*a^2*d*Sqrt[a + a*Cos[c + d*x]])

Rule 3041

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((a*A - b*B + 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)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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 2978

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

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+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx &=-\frac{(A-B+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\left (a (5 A-B+C)-\frac{1}{2} a (5 A-5 B-3 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-7 B-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\left (\frac{1}{2} a^2 (35 A-11 B+3 C)-\frac{3}{4} a^2 (15 A-7 B-C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-7 B-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(35 A-11 B+3 C) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\left (-4 a^3 (5 A-2 B)+\frac{1}{4} a^3 (35 A-11 B+3 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^5}\\ &=-\frac{(A-B+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-7 B-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(35 A-11 B+3 C) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(5 A-2 B) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx}{2 a^3}+\frac{(115 A-43 B+3 C) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A-B+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-7 B-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(35 A-11 B+3 C) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{(5 A-2 B) \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 )}{a^2 d}-\frac{(115 A-43 B+3 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 )}{16 a^2 d}\\ &=-\frac{(5 A-2 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac{(115 A-43 B+3 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(A-B+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-7 B-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(35 A-11 B+3 C) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.77043, size = 189, normalized size = 0.87 \[ \frac{\sec \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (\sin \left (\frac{1}{2} (c+d x)\right ) (2 (55 A-15 B+7 C) \cos (c+d x)+(35 A-11 B+3 C) \cos (2 (c+d x))+67 A-11 B+3 C)+4 (115 A-43 B+3 C) \cos (c+d x) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-64 \sqrt{2} (5 A-2 B) \cos (c+d x) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{32 a d (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[(c + d*x)/2]*Sec[c + d*x]*(4*(115*A - 43*B + 3*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^4*Cos[c + d*
x] - 64*Sqrt[2]*(5*A - 2*B)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^4*Cos[c + d*x] + (67*A - 11*B +
3*C + 2*(55*A - 15*B + 7*C)*Cos[c + d*x] + (35*A - 11*B + 3*C)*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(32*a*d*(
a*(1 + Cos[c + d*x]))^(3/2))

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Maple [B]  time = 0.317, size = 1327, normalized size = 6.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^(5/2),x)

[Out]

1/16*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(230*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-86*B*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)^6*a+6*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-160*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-160*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)^6*a+64
*B*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+64*B*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-115*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+43*B*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-3*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+70*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+80*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+80*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-22*B*
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)^4-32*B*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-32*B
*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+6*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)^4-15*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+7*B*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+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)^2-2*A*2^(1/2
)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*B*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-2*C*2^(1/2)*(a*sin
(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(7/2)/cos(1/2*d*x+1/2*c)^3/(2*cos(1/2*d*x+1/2*c)-2^(1/2))/(2*cos(1/2*d*x+1
/2*c)+2^(1/2))/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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.83463, size = 1112, normalized size = 5.12 \begin{align*} \frac{\sqrt{2}{\left ({\left (115 \, A - 43 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (115 \, A - 43 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (115 \, A - 43 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (115 \, A - 43 \, B + 3 \, C\right )} \cos \left (d x + c\right )\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 ) - 16 \,{\left ({\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )\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 ({\left (35 \, A - 11 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (55 \, A - 15 \, B + 7 \, C\right )} \cos \left (d x + c\right ) + 16 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/64*(sqrt(2)*((115*A - 43*B + 3*C)*cos(d*x + c)^4 + 3*(115*A - 43*B + 3*C)*cos(d*x + c)^3 + 3*(115*A - 43*B +
3*C)*cos(d*x + c)^2 + (115*A - 43*B + 3*C)*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*co
s(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)) - 16*((5
*A - 2*B)*cos(d*x + c)^4 + 3*(5*A - 2*B)*cos(d*x + c)^3 + 3*(5*A - 2*B)*cos(d*x + c)^2 + (5*A - 2*B)*cos(d*x +
c))*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*((35*A - 11*B + 3*C)*cos(d*x + c)^2 + (55*A - 15*
B + 7*C)*cos(d*x + c) + 16*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x +
c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

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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+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**2/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 5.59757, size = 575, normalized size = 2.65 \begin{align*} \frac{2 \, \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}{\left (\frac{2 \, \sqrt{2}{\left (A a^{5} - B a^{5} + C a^{5}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{8}} + \frac{\sqrt{2}{\left (21 \, A a^{5} - 13 \, B a^{5} + 5 \, C a^{5}\right )}}{a^{8}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{\sqrt{2}{\left (115 \, A \sqrt{a} - 43 \, B \sqrt{a} + 3 \, C \sqrt{a}\right )} \log \left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{a^{3}} - \frac{32 \,{\left (5 \, A \sqrt{a} - 2 \, B \sqrt{a}\right )} \log \left ({\left |{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - a{\left (2 \, \sqrt{2} + 3\right )} \right |}\right )}{a^{3}} + \frac{32 \,{\left (5 \, A \sqrt{a} - 2 \, B \sqrt{a}\right )} \log \left ({\left |{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} + a{\left (2 \, \sqrt{2} - 3\right )} \right |}\right )}{a^{3}} + \frac{128 \, \sqrt{2}{\left (3 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} A \sqrt{a} - A a^{\frac{3}{2}}\right )}}{{\left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )} a^{2}}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/64*(2*sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(A*a^5 - B*a^5 + C*a^5)*tan(1/2*d*x + 1/2*c)^2/a^8 + sqr
t(2)*(21*A*a^5 - 13*B*a^5 + 5*C*a^5)/a^8)*tan(1/2*d*x + 1/2*c) - sqrt(2)*(115*A*sqrt(a) - 43*B*sqrt(a) + 3*C*s
qrt(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^3 - 32*(5*A*sqrt(a) - 2*B
*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^3 + 32*(5*A*sqrt(a) - 2*B*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^3 + 128*sqrt(2)*(3*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2
+ a))^2*A*sqrt(a) - A*a^(3/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sq
rt(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)*a^2))/d