3.175 \(\int \frac {\sqrt {a+a \cos (c+d x)} (A+C \cos ^2(c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=116 \[ \frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a A \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 \sqrt {a} C \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d} \]

[Out]

2*C*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*a^(1/2)/d+2/3*a*A*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos
(d*x+c))^(1/2)+2/3*A*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(3/2)

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Rubi [A]  time = 0.31, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {3044, 2980, 2774, 216} \[ \frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a A \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 \sqrt {a} C \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a]*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (2*a*A*Sin[c + d*x])/(3*d*Sqrt[Cos[c
+ d*x]]*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))

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 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 3044

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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
])

Rubi steps

\begin {align*} \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx &=\frac {2 A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {\sqrt {a+a \cos (c+d x)} \left (\frac {a A}{2}+\frac {3}{2} a C \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{3 a}\\ &=\frac {2 a A \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+C \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a A \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(2 C) \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 )}{d}\\ &=\frac {2 \sqrt {a} C \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {2 a A \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 90, normalized size = 0.78 \[ \frac {\sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} \left (2 A \sin \left (\frac {3}{2} (c+d x)\right )+3 \sqrt {2} C \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {3}{2}}(c+d x)\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(3*Sqrt[2]*C*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(3/2)
+ 2*A*Sin[(3*(c + d*x))/2]))/(3*d*Cos[c + d*x]^(3/2))

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fricas [A]  time = 0.45, size = 123, normalized size = 1.06 \[ \frac {2 \, {\left ({\left (2 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 3 \, {\left (C \cos \left (d x + c\right )^{3} + C \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )\right )}}{3 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*((2*A*cos(d*x + c) + A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) - 3*(C*cos(d*x + c)^3 + C
*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))))/(d*cos(d*
x + c)^3 + d*cos(d*x + c)^2)

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

[Out]

Timed out

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maple [A]  time = 0.39, size = 127, normalized size = 1.09 \[ -\frac {2 \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (-3 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+2 A \left (\cos ^{2}\left (d x +c \right )\right )-A \cos \left (d x +c \right )-A \right )}{3 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/d*(a*(1+cos(d*x+c)))^(1/2)*(-3*C*cos(d*x+c)*sin(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))+2*A*cos(d*x+c)^2-A*cos(d*x+c)-A)/sin(d*x+c)/cos(d*x+c)^(3/2)

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maxima [B]  time = 2.17, size = 339, normalized size = 2.92 \[ \frac {3 \, C \sqrt {a} \arctan \left ({\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + \sin \left (d x + c\right ), {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + \cos \left (d x + c\right )\right ) + \frac {2 \, A {\left (\frac {3 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {4 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (\frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*C*sqrt(a)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arct
an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(
2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + cos(d*x + c)) + 2*A*(3*sqrt
(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 4*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sqrt(2)*
sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((sin(d*x + c)/(cos(d
*x + c) + 1) + 1)^(5/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \left (A + C \cos ^{2}{\left (c + d x \right )}\right )}{\cos ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)**2)*(a+a*cos(d*x+c))**(1/2)/cos(d*x+c)**(5/2),x)

[Out]

Integral(sqrt(a*(cos(c + d*x) + 1))*(A + C*cos(c + d*x)**2)/cos(c + d*x)**(5/2), x)

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