3.520 \(\int \frac{(A+B \cos (c+d x)) \sec ^{\frac{11}{2}}(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx\)

Optimal. Leaf size=295 \[ -\frac{2 (A-9 B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{63 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (19 A-3 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}-\frac{2 (29 A-93 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (257 A-129 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{315 d \sqrt{a \cos (c+d x)+a}}-\frac{\sqrt{2} (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 A \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x)}{9 d \sqrt{a \cos (c+d x)+a}} \]

[Out]

-((Sqrt[2]*(A - B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[C
os[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d)) + (2*(257*A - 129*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqr
t[a + a*Cos[c + d*x]]) - (2*(29*A - 93*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) +
(2*(19*A - 3*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) - (2*(A - 9*B)*Sec[c + d*x]^
(7/2)*Sin[c + d*x])/(63*d*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(9*d*Sqrt[a + a*Co
s[c + d*x]])

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Rubi [A]  time = 1.05655, antiderivative size = 295, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2961, 2984, 12, 2782, 205} \[ -\frac{2 (A-9 B) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{63 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (19 A-3 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}-\frac{2 (29 A-93 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{315 d \sqrt{a \cos (c+d x)+a}}+\frac{2 (257 A-129 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{315 d \sqrt{a \cos (c+d x)+a}}-\frac{\sqrt{2} (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 A \sin (c+d x) \sec ^{\frac{9}{2}}(c+d x)}{9 d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^(11/2))/Sqrt[a + a*Cos[c + d*x]],x]

[Out]

-((Sqrt[2]*(A - B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[C
os[c + d*x]]*Sqrt[Sec[c + d*x]])/(Sqrt[a]*d)) + (2*(257*A - 129*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(315*d*Sqr
t[a + a*Cos[c + d*x]]) - (2*(29*A - 93*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(315*d*Sqrt[a + a*Cos[c + d*x]]) +
(2*(19*A - 3*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) - (2*(A - 9*B)*Sec[c + d*x]^
(7/2)*Sin[c + d*x])/(63*d*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(9*d*Sqrt[a + a*Co
s[c + d*x]])

Rule 2961

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Sin[e + f*x])^m*(
c + d*Sin[e + f*x])^n)/(g*Sin[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*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]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{(A+B \cos (c+d x)) \sec ^{\frac{11}{2}}(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \cos (c+d x)}{\cos ^{\frac{11}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{2 A \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{2} a (A-9 B)+4 a A \cos (c+d x)}{\cos ^{\frac{9}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{9 a}\\ &=-\frac{2 (A-9 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{3}{4} a^2 (19 A-3 B)-\frac{3}{2} a^2 (A-9 B) \cos (c+d x)}{\cos ^{\frac{7}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{63 a^2}\\ &=\frac{2 (19 A-3 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (A-9 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (8 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{3}{8} a^3 (29 A-93 B)+\frac{3}{2} a^3 (19 A-3 B) \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{315 a^3}\\ &=-\frac{2 (29 A-93 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (19 A-3 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (A-9 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (16 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{3}{16} a^4 (257 A-129 B)-\frac{3}{8} a^4 (29 A-93 B) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{945 a^4}\\ &=\frac{2 (257 A-129 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (29 A-93 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (19 A-3 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (A-9 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (32 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int -\frac{945 a^5 (A-B)}{32 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{945 a^5}\\ &=\frac{2 (257 A-129 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (29 A-93 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (19 A-3 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (A-9 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}-\left ((A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{2 (257 A-129 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (29 A-93 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (19 A-3 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (A-9 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}+\frac{\left (2 a (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{d}\\ &=-\frac{\sqrt{2} (A-B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{\sqrt{a} d}+\frac{2 (257 A-129 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (29 A-93 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{315 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (19 A-3 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}-\frac{2 (A-9 B) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{63 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sec ^{\frac{9}{2}}(c+d x) \sin (c+d x)}{9 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 9.50608, size = 272, normalized size = 0.92 \[ \frac{2 e^{-\frac{1}{2} i (c+d x)} \cos \left (\frac{1}{2} (c+d x)\right ) \left (-315 i (A-B) \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )-\frac{1}{4} \sin \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{9}{2}}(c+d x) \left (\cos \left (\frac{1}{2} (c+d x)\right )+i \sin \left (\frac{1}{2} (c+d x)\right )\right ) ((214 A-918 B) \cos (c+d x)-8 (157 A-69 B) \cos (2 (c+d x))+58 A \cos (3 (c+d x))-257 A \cos (4 (c+d x))-1279 A-186 B \cos (3 (c+d x))+129 B \cos (4 (c+d x))+423 B)\right )}{315 d \sqrt{a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^(11/2))/Sqrt[a + a*Cos[c + d*x]],x]

[Out]

(2*Cos[(c + d*x)/2]*((-315*I)*(A - B)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d
*x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] - ((-1279*A + 423*B + (214*A - 91
8*B)*Cos[c + d*x] - 8*(157*A - 69*B)*Cos[2*(c + d*x)] + 58*A*Cos[3*(c + d*x)] - 186*B*Cos[3*(c + d*x)] - 257*A
*Cos[4*(c + d*x)] + 129*B*Cos[4*(c + d*x)])*Sec[c + d*x]^(9/2)*(Cos[(c + d*x)/2] + I*Sin[(c + d*x)/2])*Sin[(c
+ d*x)/2])/4))/(315*d*E^((I/2)*(c + d*x))*Sqrt[a*(1 + Cos[c + d*x])])

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Maple [B]  time = 0.641, size = 793, normalized size = 2.7 \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))*sec(d*x+c)^(11/2)/(a+cos(d*x+c)*a)^(1/2),x)

[Out]

1/315/d*2^(1/2)/a*(315*A*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-315
*B*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+1575*A*cos(d*x+c)^4*(cos(
d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-1575*B*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c))
)^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+3150*A*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+co
s(d*x+c))/sin(d*x+c))-3150*B*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))
+3150*A*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-3150*B*cos(d*x+c)^2*
(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+1575*A*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+
c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-1575*B*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+c
os(d*x+c))/sin(d*x+c))+315*A*(cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-315*B*(cos(d
*x+c)/(1+cos(d*x+c)))^(9/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+257*A*cos(d*x+c)^4*2^(1/2)*sin(d*x+c)-129*B*cos
(d*x+c)^4*2^(1/2)*sin(d*x+c)-29*A*cos(d*x+c)^3*2^(1/2)*sin(d*x+c)+93*B*cos(d*x+c)^3*2^(1/2)*sin(d*x+c)+57*A*co
s(d*x+c)^2*2^(1/2)*sin(d*x+c)-9*B*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)-5*A*cos(d*x+c)*2^(1/2)*sin(d*x+c)+45*B*cos(d
*x+c)*2^(1/2)*sin(d*x+c)+35*A*2^(1/2)*sin(d*x+c))*cos(d*x+c)*(1/cos(d*x+c))^(11/2)*(a*(1+cos(d*x+c)))^(1/2)*si
n(d*x+c)^8/(-1+cos(d*x+c))^4/(1+cos(d*x+c))^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(11/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.69998, size = 539, normalized size = 1.83 \begin{align*} \frac{\frac{315 \, \sqrt{2}{\left ({\left (A - B\right )} a \cos \left (d x + c\right )^{5} +{\left (A - B\right )} a \cos \left (d x + c\right )^{4}\right )} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right )}{\sqrt{a}} + \frac{2 \,{\left ({\left (257 \, A - 129 \, B\right )} \cos \left (d x + c\right )^{4} -{\left (29 \, A - 93 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (19 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} - 5 \,{\left (A - 9 \, B\right )} \cos \left (d x + c\right ) + 35 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{315 \,{\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(11/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/315*(315*sqrt(2)*((A - B)*a*cos(d*x + c)^5 + (A - B)*a*cos(d*x + c)^4)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) +
a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)))/sqrt(a) + 2*((257*A - 129*B)*cos(d*x + c)^4 - (29*A - 93*B)*cos(
d*x + c)^3 + 3*(19*A - 3*B)*cos(d*x + c)^2 - 5*(A - 9*B)*cos(d*x + c) + 35*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x
+ c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c)^5 + a*d*cos(d*x + c)^4)

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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))*sec(d*x+c)**(11/2)/(a+a*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{11}{2}}}{\sqrt{a \cos \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(11/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(11/2)/sqrt(a*cos(d*x + c) + a), x)