3.1334 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sqrt{\sec (c+d x)} \, dx\)

Optimal. Leaf size=253 \[ \frac{a^{5/2} (304 A+200 B+163 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{64 d}+\frac{a^3 (432 A+392 B+299 C) \sin (c+d x)}{192 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{a^2 (16 A+24 B+17 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{32 d \sqrt{\sec (c+d x)}}+\frac{a (8 B+5 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{24 d \sqrt{\sec (c+d x)}}+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d \sqrt{\sec (c+d x)}} \]

[Out]

(a^(5/2)*(304*A + 200*B + 163*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sq
rt[Sec[c + d*x]])/(64*d) + (a^3*(432*A + 392*B + 299*C)*Sin[c + d*x])/(192*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec
[c + d*x]]) + (a^2*(16*A + 24*B + 17*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(32*d*Sqrt[Sec[c + d*x]]) + (a*
(8*B + 5*C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(24*d*Sqrt[Sec[c + d*x]]) + (C*(a + a*Cos[c + d*x])^(5/2)
*Sin[c + d*x])/(4*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.940752, antiderivative size = 253, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4221, 3045, 2976, 2981, 2774, 216} \[ \frac{a^{5/2} (304 A+200 B+163 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{64 d}+\frac{a^3 (432 A+392 B+299 C) \sin (c+d x)}{192 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{a^2 (16 A+24 B+17 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{32 d \sqrt{\sec (c+d x)}}+\frac{a (8 B+5 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{24 d \sqrt{\sec (c+d x)}}+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(304*A + 200*B + 163*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sq
rt[Sec[c + d*x]])/(64*d) + (a^3*(432*A + 392*B + 299*C)*Sin[c + d*x])/(192*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec
[c + d*x]]) + (a^2*(16*A + 24*B + 17*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(32*d*Sqrt[Sec[c + d*x]]) + (a*
(8*B + 5*C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(24*d*Sqrt[Sec[c + d*x]]) + (C*(a + a*Cos[c + d*x])^(5/2)
*Sin[c + d*x])/(4*d*Sqrt[Sec[c + d*x]])

Rule 4221

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rule 3045

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

Rule 2976

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

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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[n, -1]

Rule 2774

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

Rule 216

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

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{4 d \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^{5/2} \left (\frac{1}{2} a (8 A+C)+\frac{1}{2} a (8 B+5 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{4 a}\\ &=\frac{a (8 B+5 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{24 d \sqrt{\sec (c+d x)}}+\frac{C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{4 d \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \cos (c+d x))^{3/2} \left (\frac{1}{4} a^2 (48 A+8 B+11 C)+\frac{3}{4} a^2 (16 A+24 B+17 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{12 a}\\ &=\frac{a^2 (16 A+24 B+17 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{32 d \sqrt{\sec (c+d x)}}+\frac{a (8 B+5 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{24 d \sqrt{\sec (c+d x)}}+\frac{C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{4 d \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{1}{8} a^3 (240 A+104 B+95 C)+\frac{1}{8} a^3 (432 A+392 B+299 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{24 a}\\ &=\frac{a^3 (432 A+392 B+299 C) \sin (c+d x)}{192 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\frac{a^2 (16 A+24 B+17 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{32 d \sqrt{\sec (c+d x)}}+\frac{a (8 B+5 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{24 d \sqrt{\sec (c+d x)}}+\frac{C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{4 d \sqrt{\sec (c+d x)}}+\frac{1}{128} \left (a^2 (304 A+200 B+163 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{a^3 (432 A+392 B+299 C) \sin (c+d x)}{192 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\frac{a^2 (16 A+24 B+17 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{32 d \sqrt{\sec (c+d x)}}+\frac{a (8 B+5 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{24 d \sqrt{\sec (c+d x)}}+\frac{C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{4 d \sqrt{\sec (c+d x)}}-\frac{\left (a^2 (304 A+200 B+163 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a}}} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{64 d}\\ &=\frac{a^{5/2} (304 A+200 B+163 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{64 d}+\frac{a^3 (432 A+392 B+299 C) \sin (c+d x)}{192 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\frac{a^2 (16 A+24 B+17 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{32 d \sqrt{\sec (c+d x)}}+\frac{a (8 B+5 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{24 d \sqrt{\sec (c+d x)}}+\frac{C (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{4 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.46623, size = 166, normalized size = 0.66 \[ \frac{a^2 \sqrt{\cos (c+d x)} \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} \sqrt{a (\cos (c+d x)+1)} \left (3 \sqrt{2} (304 A+200 B+163 C) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \sin \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} ((96 A+272 B+362 C) \cos (c+d x)+528 A+4 (8 B+23 C) \cos (2 (c+d x))+632 B+12 C \cos (3 (c+d x))+581 C)\right )}{384 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(3*Sqrt[2]*(304*A + 200
*B + 163*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(528*A + 632*B + 581*C + (96*A + 272*B + 3
62*C)*Cos[c + d*x] + 4*(8*B + 23*C)*Cos[2*(c + d*x)] + 12*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(384*d)

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Maple [B]  time = 0.203, size = 477, normalized size = 1.9 \begin{align*} -{\frac{{a}^{2} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}-1 \right ) }{192\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}} \left ( 48\,C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+64\,B\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+184\,C\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+96\,A\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+272\,B\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+326\,C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+528\,A\sin \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+600\,B\sin \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+489\,C\sin \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+912\,A\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) +600\,B\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) +489\,C\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2),x)

[Out]

-1/192/d*a^2*(48*C*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+64*B*(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*sin(d*x+c)*cos(d*x+c)^2+184*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+96*A*sin(d*x+c)*
cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+272*B*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+326
*C*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+528*A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+
600*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+489*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+912*A*ar
ctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+600*B*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)/cos(d*x+c))+489*C*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c)))*(1/cos(d*x+c))^(1
/2)*(a*(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^2*(cos(d*x+c)^2-1)

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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((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 4.95067, size = 541, normalized size = 2.14 \begin{align*} -\frac{3 \,{\left ({\left (304 \, A + 200 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right ) +{\left (304 \, A + 200 \, B + 163 \, C\right )} a^{2}\right )} \sqrt{a} \arctan \left (\frac{\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 ) - \frac{{\left (48 \, C a^{2} \cos \left (d x + c\right )^{4} + 8 \,{\left (8 \, B + 23 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \,{\left (48 \, A + 136 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (176 \, A + 200 \, B + 163 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{192 \,{\left (d \cos \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/192*(3*((304*A + 200*B + 163*C)*a^2*cos(d*x + c) + (304*A + 200*B + 163*C)*a^2)*sqrt(a)*arctan(sqrt(a*cos(d
*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (48*C*a^2*cos(d*x + c)^4 + 8*(8*B + 23*C)*a^2*cos(d*
x + c)^3 + 2*(48*A + 136*B + 163*C)*a^2*cos(d*x + c)^2 + 3*(176*A + 200*B + 163*C)*a^2*cos(d*x + c))*sqrt(a*co
s(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + 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((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [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+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

Timed out