3.1234 \(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx\)

Optimal. Leaf size=156 \[ \frac {\sqrt {2} (A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \cos (c+d x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.46, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4221, 3044, 2984, 12, 2782, 205} \[ \frac {\sqrt {2} (A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

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

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(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]

Rule 2984

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

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
])

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 \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx\\ &=\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {a A}{2}+\frac {1}{2} a (2 A+3 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{3 a}\\ &=-\frac {2 A \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {3 a^2 (A+C)}{4 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{3 a^2}\\ &=-\frac {2 A \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\left ((A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx\\ &=-\frac {2 A \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}-\frac {\left (2 a (A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{2 a^2+a x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{d}\\ &=\frac {\sqrt {2} (A+C) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}-\frac {2 A \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 6.78, size = 576, normalized size = 3.69 \[ \frac {2 \sqrt {\frac {1}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \sqrt {1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )} \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {(A+C) \csc ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-12 \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \, _3F_2\left (2,2,\frac {7}{2};1,\frac {9}{2};-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right )-12 \left (3 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )-7 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+4\right ) \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right )+7 \sqrt {-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3 \left (8 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )-20 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+15\right ) \left (\left (3-7 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sqrt {-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}-3 \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \tanh ^{-1}\left (\sqrt {-\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right )\right )\right )}{63 \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{7/2}}-\frac {4 C \sin ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{3/2}}\right )}{d \sqrt {a (\cos (c+d x)+1)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Cos[c/2 + (d*x)/2]*Sqrt[(1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1)]*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]*((-4*C*Sin[c/2
+ (d*x)/2]^3)/(3*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) + ((A + C)*Csc[c/2 + (d*x)/2]^5*(-12*Cos[(c + d*x)/2]^4*H
ypergeometricPFQ[{2, 2, 7/2}, {1, 9/2}, -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)/
2]^8 - 12*Hypergeometric2F1[2, 7/2, 9/2, -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]*Sin[c/2 + (d*x)
/2]^8*(4 - 7*Sin[c/2 + (d*x)/2]^2 + 3*Sin[c/2 + (d*x)/2]^4) + 7*Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (
d*x)/2]^2))]*(1 - 2*Sin[c/2 + (d*x)/2]^2)^3*(15 - 20*Sin[c/2 + (d*x)/2]^2 + 8*Sin[c/2 + (d*x)/2]^4)*((3 - 7*Si
n[c/2 + (d*x)/2]^2)*Sqrt[-(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))] - 3*ArcTanh[Sqrt[-(Sin[c/2 + (d
*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]]*(1 - 2*Sin[c/2 + (d*x)/2]^2))))/(63*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2
))))/(d*Sqrt[a*(1 + Cos[c + d*x])])

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fricas [A]  time = 0.48, size = 135, normalized size = 0.87 \[ -\frac {\frac {3 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{2} + {\left (A + C\right )} a \cos \left (d x + c\right )\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}} + \frac {2 \, {\left (A \cos \left (d x + c\right ) - A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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maple [B]  time = 0.58, size = 366, normalized size = 2.35 \[ \frac {\left (3 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+3 C \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+6 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+6 C \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+3 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+3 C \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+A \cos \left (d x +c \right ) \sqrt {2}\, \sin \left (d x +c \right )-A \sqrt {2}\, \sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{3 d \left (-1+\cos \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/d*(3*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+3*C*arcsin((-1+co
s(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+6*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*co
s(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+6*C*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*(cos(d*x+c)/(1+co
s(d*x+c)))^(3/2)+3*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+3*C*arcsin((-1+cos(d
*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+A*cos(d*x+c)*2^(1/2)*sin(d*x+c)-A*2^(1/2)*sin(d*x+c))*cos
(d*x+c)*sin(d*x+c)^2*(1/cos(d*x+c))^(5/2)*(a*(1+cos(d*x+c)))^(1/2)/(-1+cos(d*x+c))/(1+cos(d*x+c))^2*2^(1/2)/a

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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