3.1311 \(\int \sqrt{a+a \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac{9}{2}}(c+d x) \, dx\)
Optimal. Leaf size=178 \[ \frac{2 a (24 A+28 B+35 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{4 a (24 A+28 B+35 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a (A+7 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{35 d \sqrt{a \cos (c+d x)+a}}+\frac{2 A \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{7 d} \]
[Out]
(4*a*(24*A + 28*B + 35*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(24*A + 28*
B + 35*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(A + 7*B)*Sec[c + d*x]^(5/2
)*Sin[c + d*x])/(35*d*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(7/2)*Sin[c + d*x
])/(7*d)
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Rubi [A] time = 0.592377, antiderivative size = 178, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used =
{4221, 3043, 2980, 2772, 2771} \[ \frac{2 a (24 A+28 B+35 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{4 a (24 A+28 B+35 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{105 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a (A+7 B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{35 d \sqrt{a \cos (c+d x)+a}}+\frac{2 A \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{7 d} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(9/2),x]
[Out]
(4*a*(24*A + 28*B + 35*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(24*A + 28*
B + 35*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(A + 7*B)*Sec[c + d*x]^(5/2
)*Sin[c + d*x])/(35*d*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(7/2)*Sin[c + d*x
])/(7*d)
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 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 2980
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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] && LtQ[n, -1]
Rule 2772
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
Rule 2771
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[(-2*b^2*Cos[e + f*x])/(f*(b*c + a*d)*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]
Rubi steps
\begin{align*} \int \sqrt{a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac{9}{2}}(c+d x) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac{9}{2}}(c+d x)} \, dx\\ &=\frac{2 A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{1}{2} a (A+7 B)+\frac{1}{2} a (4 A+7 C) \cos (c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx}{7 a}\\ &=\frac{2 a (A+7 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{35} \left ((24 A+28 B+35 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a (24 A+28 B+35 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (A+7 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{105} \left (2 (24 A+28 B+35 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{4 a (24 A+28 B+35 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (24 A+28 B+35 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a (A+7 B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sqrt{a+a \cos (c+d x)} \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end{align*}
Mathematica [A] time = 0.710306, size = 121, normalized size = 0.68 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{7}{2}}(c+d x) \sqrt{a (\cos (c+d x)+1)} (3 (36 A+42 B+35 C) \cos (c+d x)+(24 A+28 B+35 C) \cos (2 (c+d x))+24 A \cos (3 (c+d x))+54 A+28 B \cos (3 (c+d x))+28 B+35 C \cos (3 (c+d x))+35 C)}{105 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(9/2),x]
[Out]
(Sqrt[a*(1 + Cos[c + d*x])]*(54*A + 28*B + 35*C + 3*(36*A + 42*B + 35*C)*Cos[c + d*x] + (24*A + 28*B + 35*C)*C
os[2*(c + d*x)] + 24*A*Cos[3*(c + d*x)] + 28*B*Cos[3*(c + d*x)] + 35*C*Cos[3*(c + d*x)])*Sec[c + d*x]^(7/2)*Ta
n[(c + d*x)/2])/(105*d)
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Maple [A] time = 0.193, size = 138, normalized size = 0.8 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 48\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+56\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+70\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+24\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+28\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+35\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+18\,A\cos \left ( dx+c \right ) +21\,B\cos \left ( dx+c \right ) +15\,A \right ) \cos \left ( dx+c \right ) }{105\,d\sin \left ( dx+c \right ) } \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{9}{2}}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) }} \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)^(9/2)*(a+a*cos(d*x+c))^(1/2),x)
[Out]
-2/105/d*(-1+cos(d*x+c))*(48*A*cos(d*x+c)^3+56*B*cos(d*x+c)^3+70*C*cos(d*x+c)^3+24*A*cos(d*x+c)^2+28*B*cos(d*x
+c)^2+35*C*cos(d*x+c)^2+18*A*cos(d*x+c)+21*B*cos(d*x+c)+15*A)*cos(d*x+c)*(1/cos(d*x+c))^(9/2)*(a*(1+cos(d*x+c)
))^(1/2)/sin(d*x+c)
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Maxima [B] time = 1.90256, size = 1145, normalized size = 6.43 \begin{align*} \text{result too large to display} \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)^(9/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
2/105*(3*A*(35*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 70*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 84*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 58*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*
x + c) + 1)^7 + 9*sqrt(2)*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(4*sin(d*x + c)
^2/(cos(d*x + c) + 1)^2 + 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + sin(
d*x + c)^8/(cos(d*x + c) + 1)^8 + 1)) + 7*B*(15*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 40*sqrt(2)*s
qrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 42*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 24*sqrt(
2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 7*sqrt(2)*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*(sin(d
*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c
) + 1) + 1)^(9/2)*(4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*sin(d*x +
c)^6/(cos(d*x + c) + 1)^6 + sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 1)) + 35*C*(3*sqrt(2)*sqrt(a)*sin(d*x + c)/
(cos(d*x + c) + 1) - 10*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 12*sqrt(2)*sqrt(a)*sin(d*x + c)^
5/(cos(d*x + c) + 1)^5 - 6*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + sqrt(2)*sqrt(a)*sin(d*x + c)^
9/(cos(d*x + c) + 1)^9)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/
2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*sin(d*x + c)^4/(cos
(d*x + c) + 1)^4 + 4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 1)))/d
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Fricas [A] time = 1.59926, size = 294, normalized size = 1.65 \begin{align*} \frac{2 \,{\left (2 \,{\left (24 \, A + 28 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (24 \, A + 28 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (6 \, A + 7 \, B\right )} \cos \left (d x + c\right ) + 15 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )} \sqrt{\cos \left (d x + c\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)^(9/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
2/105*(2*(24*A + 28*B + 35*C)*cos(d*x + c)^3 + (24*A + 28*B + 35*C)*cos(d*x + c)^2 + 3*(6*A + 7*B)*cos(d*x + c
) + 15*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^4 + d*cos(d*x + c)^3)*sqrt(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)**(9/2)*(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{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac{9}{2}}\,{d x} \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)^(9/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(a*cos(d*x + c) + a)*sec(d*x + c)^(9/2), x)