3.1333 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx\)

Optimal. Leaf size=251 \[ \frac {a^{5/2} (40 A+38 B+25 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{8 d}-\frac {a^3 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (8 A-2 B-3 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{4 d \sqrt {\sec (c+d x)}}-\frac {a (6 A-C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}}{d} \]

[Out]

-1/3*a*(6*A-C)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)-1/24*a^3*(24*A-54*B-49*C)*sin(d*x+c)/d/(a+
a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)-1/4*a^2*(8*A-2*B-3*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2
)+2*A*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+1/8*a^(5/2)*(40*A+38*B+25*C)*arcsin(sin(d*x+c)*a^(1
/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.95, antiderivative size = 251, normalized size of antiderivative = 1.00, 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, 3043, 2976, 2981, 2774, 216} \[ \frac {a^{5/2} (40 A+38 B+25 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{8 d}-\frac {a^3 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (8 A-2 B-3 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{4 d \sqrt {\sec (c+d x)}}-\frac {a (6 A-C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}}{d} \]

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)*Sec[c + d*x]^(3/2),x]

[Out]

(a^(5/2)*(40*A + 38*B + 25*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[
Sec[c + d*x]])/(8*d) - (a^3*(24*A - 54*B - 49*C)*Sin[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x
]]) - (a^2*(8*A - 2*B - 3*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(4*d*Sqrt[Sec[c + d*x]]) - (a*(6*A - C)*(a
+ a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c +
d*x]]*Sin[c + d*x])/d

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]

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 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 3043

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

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]

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 ) \sec ^{\frac {3}{2}}(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 )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 A (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {\left (2 \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 (5 A+B)-\frac {1}{2} a (6 A-C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{a}\\ &=-\frac {a (6 A-C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {\left (2 \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 (24 A+6 B+C)-\frac {3}{4} a^2 (8 A-2 B-3 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{3 a}\\ &=-\frac {a^2 (8 A-2 B-3 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}-\frac {a (6 A-C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\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 (72 A+30 B+13 C)-\frac {1}{8} a^3 (24 A-54 B-49 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx}{3 a}\\ &=-\frac {a^3 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {a^2 (8 A-2 B-3 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}-\frac {a (6 A-C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{16} \left (a^2 (40 A+38 B+25 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 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {a^2 (8 A-2 B-3 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}-\frac {a (6 A-C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}-\frac {\left (a^2 (40 A+38 B+25 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 )}{8 d}\\ &=\frac {a^{5/2} (40 A+38 B+25 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)}}{8 d}-\frac {a^3 (24 A-54 B-49 C) \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {a^2 (8 A-2 B-3 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}-\frac {a (6 A-C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.00, size = 156, normalized size = 0.62 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \sqrt {a (\cos (c+d x)+1)} \left (3 \sqrt {2} (40 A+38 B+25 C) \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+2 \sin \left (\frac {1}{2} (c+d x)\right ) (3 (8 A+22 B+27 C) \cos (c+d x)+48 A+(6 B+17 C) \cos (2 (c+d x))+6 B+2 C \cos (3 (c+d x))+17 C)\right )}{48 d} \]

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)*Sec[c + d*x]^(3/2),x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(3*Sqrt[2]*(40*A + 38*B + 25*C)*ArcSin[Sqr
t[2]*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + 2*(48*A + 6*B + 17*C + 3*(8*A + 22*B + 27*C)*Cos[c + d*x] + (6*B +
17*C)*Cos[2*(c + d*x)] + 2*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)

________________________________________________________________________________________

fricas [A]  time = 0.77, size = 178, normalized size = 0.71 \[ -\frac {3 \, {\left ({\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (40 \, A + 38 \, B + 25 \, C\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (8 \, C a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (6 \, B + 17 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (8 \, A + 22 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right ) + 48 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

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)^(3/2),x, algorithm="fricas")

[Out]

-1/24*(3*((40*A + 38*B + 25*C)*a^2*cos(d*x + c) + (40*A + 38*B + 25*C)*a^2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c)
+ a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (8*C*a^2*cos(d*x + c)^3 + 2*(6*B + 17*C)*a^2*cos(d*x + c)^2
+ 3*(8*A + 22*B + 25*C)*a^2*cos(d*x + c) + 48*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c))
)/(d*cos(d*x + c) + d)

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)^(3/2),x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [B]  time = 0.54, size = 513, normalized size = 2.04 \[ \frac {\left (8 C \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+120 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) \cos \left (d x +c \right )+12 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+114 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) \cos \left (d x +c \right )+34 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+75 C \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+24 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+120 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+66 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+114 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+75 C \sin \left (d x +c \right ) \cos \left (d x +c \right )+75 C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+48 A \sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{24 d \left (1+\cos \left (d x +c \right )\right )} \]

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)^(3/2),x)

[Out]

1/24/d*(8*C*sin(d*x+c)*cos(d*x+c)^3+120*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)/cos(d*x+c))*cos(d*x+c)+12*B*sin(d*x+c)*cos(d*x+c)^2+114*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*cos(d*x+c)+34*C*sin(d*x+c)*cos(d*x+c)^2+75*C*c
os(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+24
*A*cos(d*x+c)*sin(d*x+c)+120*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)/cos(d*x+c))+66*B*cos(d*x+c)*sin(d*x+c)+114*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(sin(d*x+c)*(cos(d
*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+75*C*sin(d*x+c)*cos(d*x+c)+75*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arct
an(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+48*A*sin(d*x+c))*cos(d*x+c)*(1/cos(d*x+c))^(3/2)*(
a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))*a^2

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)^(3/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

int((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________