3.1317 \(\int \frac{\sqrt{a+a \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=247 \[ \frac{a (48 A+40 B+35 C) \sin (c+d x)}{96 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{a} (48 A+40 B+35 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{64 d}+\frac{a (48 A+40 B+35 C) \sin (c+d x)}{64 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{a (8 B+C) \sin (c+d x)}{24 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{C \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{4 d \sec ^{\frac{5}{2}}(c+d x)} \]

[Out]

(Sqrt[a]*(48*A + 40*B + 35*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]])/(64*d) + (a*(8*B + C)*Sin[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(5/2)) + (C*Sqr
t[a + a*Cos[c + d*x]]*Sin[c + d*x])/(4*d*Sec[c + d*x]^(5/2)) + (a*(48*A + 40*B + 35*C)*Sin[c + d*x])/(96*d*Sqr
t[a + a*Cos[c + d*x]]*Sec[c + d*x]^(3/2)) + (a*(48*A + 40*B + 35*C)*Sin[c + d*x])/(64*d*Sqrt[a + a*Cos[c + d*x
]]*Sqrt[Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.653442, antiderivative size = 247, 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, 2981, 2770, 2774, 216} \[ \frac{a (48 A+40 B+35 C) \sin (c+d x)}{96 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{\sqrt{a} (48 A+40 B+35 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{64 d}+\frac{a (48 A+40 B+35 C) \sin (c+d x)}{64 d \sqrt{\sec (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{a (8 B+C) \sin (c+d x)}{24 d \sec ^{\frac{5}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{C \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{4 d \sec ^{\frac{5}{2}}(c+d x)} \]

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]^(3/2),x]

[Out]

(Sqrt[a]*(48*A + 40*B + 35*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]])/(64*d) + (a*(8*B + C)*Sin[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(5/2)) + (C*Sqr
t[a + a*Cos[c + d*x]]*Sin[c + d*x])/(4*d*Sec[c + d*x]^(5/2)) + (a*(48*A + 40*B + 35*C)*Sin[c + d*x])/(96*d*Sqr
t[a + a*Cos[c + d*x]]*Sec[c + d*x]^(3/2)) + (a*(48*A + 40*B + 35*C)*Sin[c + d*x])/(64*d*Sqrt[a + a*Cos[c + d*x
]]*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 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 2770

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), 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] && GtQ[n, 0] && IntegerQ[2*n]

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 \frac{\sqrt{a+a \cos (c+d x)} \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 \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac{C \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{4 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \left (\frac{1}{2} a (8 A+5 C)+\frac{1}{2} a (8 B+C) \cos (c+d x)\right ) \, dx}{4 a}\\ &=\frac{a (8 B+C) \sin (c+d x)}{24 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{5}{2}}(c+d x)}+\frac{C \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{4 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{1}{48} \left ((48 A+40 B+35 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{a (8 B+C) \sin (c+d x)}{24 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{5}{2}}(c+d x)}+\frac{C \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{4 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{a (48 A+40 B+35 C) \sin (c+d x)}{96 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{64} \left ((48 A+40 B+35 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{a (8 B+C) \sin (c+d x)}{24 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{5}{2}}(c+d x)}+\frac{C \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{4 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{a (48 A+40 B+35 C) \sin (c+d x)}{96 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{a (48 A+40 B+35 C) \sin (c+d x)}{64 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}+\frac{1}{128} \left ((48 A+40 B+35 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 (8 B+C) \sin (c+d x)}{24 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{5}{2}}(c+d x)}+\frac{C \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{4 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{a (48 A+40 B+35 C) \sin (c+d x)}{96 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{a (48 A+40 B+35 C) \sin (c+d x)}{64 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}-\frac{\left ((48 A+40 B+35 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{\sqrt{a} (48 A+40 B+35 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 (8 B+C) \sin (c+d x)}{24 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{5}{2}}(c+d x)}+\frac{C \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{4 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{a (48 A+40 B+35 C) \sin (c+d x)}{96 d \sqrt{a+a \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x)}+\frac{a (48 A+40 B+35 C) \sin (c+d x)}{64 d \sqrt{a+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.00554, size = 164, normalized size = 0.66 \[ \frac{\sqrt{\cos (c+d x)} \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} \sqrt{a (\cos (c+d x)+1)} \left (3 \sqrt{2} (48 A+40 B+35 C) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \sin \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} (2 (48 A+40 B+53 C) \cos (c+d x)+144 A+4 (8 B+7 C) \cos (2 (c+d x))+152 B+12 C \cos (3 (c+d x))+133 C)\right )}{384 d} \]

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]^(3/2),x]

[Out]

(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]*(48*A + 40*B + 3
5*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(144*A + 152*B + 133*C + 2*(48*A + 40*B + 53*C)*C
os[c + d*x] + 4*(8*B + 7*C)*Cos[2*(c + d*x)] + 12*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(384*d)

________________________________________________________________________________________

Maple [B]  time = 0.196, size = 480, normalized size = 1.9 \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)*(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x)

[Out]

-1/192/d*(-1+cos(d*x+c))^3*(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+56*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)+80*B*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)+70*C*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+144*A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)+120*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+105*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)+144*A*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+120*B*arctan(sin(d*x+c)*(cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)/cos(d*x+c))+105*C*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c)))*(a*(1+
cos(d*x+c)))^(1/2)*cos(d*x+c)/(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)/(1/cos(d*x+c))^(3/2)/sin(d*x+c)^6

________________________________________________________________________________________

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+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 4.93754, size = 490, normalized size = 1.98 \begin{align*} -\frac{3 \,{\left ({\left (48 \, A + 40 \, B + 35 \, C\right )} \cos \left (d x + c\right ) + 48 \, A + 40 \, B + 35 \, C\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 \cos \left (d x + c\right )^{4} + 8 \,{\left (8 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (48 \, A + 40 \, B + 35 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (48 \, A + 40 \, B + 35 \, C\right )} \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+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-1/192*(3*((48*A + 40*B + 35*C)*cos(d*x + c) + 48*A + 40*B + 35*C)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqr
t(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (48*C*cos(d*x + c)^4 + 8*(8*B + 7*C)*cos(d*x + c)^3 + 2*(48*A + 40*B
+ 35*C)*cos(d*x + c)^2 + 3*(48*A + 40*B + 35*C)*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(
d*x + c)))/(d*cos(d*x + c) + d)

________________________________________________________________________________________

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)*(a+a*cos(d*x+c))**(1/2)/sec(d*x+c)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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{3}{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)*(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(3/2),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)^(3/2), x)