3.481 \(\int \frac{\sqrt{a+a \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=130 \[ \frac{2 a (8 A+10 B+15 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{2 a (A+5 B) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 A \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]

[Out]

(2*a*(A + 5*B)*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(8*A + 10*B + 15*C)*Sin
[c + d*x])/(15*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])/(5
*d*Cos[c + d*x]^(5/2))

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Rubi [A]  time = 0.368801, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3043, 2980, 2771} \[ \frac{2 a (8 A+10 B+15 C) \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{2 a (A+5 B) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}+\frac{2 A \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]

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))/Cos[c + d*x]^(7/2),x]

[Out]

(2*a*(A + 5*B)*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(8*A + 10*B + 15*C)*Sin
[c + d*x])/(15*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])/(5
*d*Cos[c + d*x]^(5/2))

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 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 \frac{\sqrt{a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 A \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \int \frac{\sqrt{a+a \cos (c+d x)} \left (\frac{1}{2} a (A+5 B)+\frac{1}{2} a (2 A+5 C) \cos (c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx}{5 a}\\ &=\frac{2 a (A+5 B) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 A \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{1}{15} (8 A+10 B+15 C) \int \frac{\sqrt{a+a \cos (c+d x)}}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a (A+5 B) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}}+\frac{2 a (8 A+10 B+15 C) \sin (c+d x)}{15 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)}{5 d \cos ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.370711, size = 85, normalized size = 0.65 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} ((8 A+10 B+15 C) \cos (2 (c+d x))+2 (4 A+5 B) \cos (c+d x)+14 A+10 B+15 C)}{15 d \cos ^{\frac{5}{2}}(c+d x)} \]

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))/Cos[c + d*x]^(7/2),x]

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*(14*A + 10*B + 15*C + 2*(4*A + 5*B)*Cos[c + d*x] + (8*A + 10*B + 15*C)*Cos[2*(c +
d*x)])*Tan[(c + d*x)/2])/(15*d*Cos[c + d*x]^(5/2))

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Maple [A]  time = 0.109, size = 97, normalized size = 0.8 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 8\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+10\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+15\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+4\,A\cos \left ( dx+c \right ) +5\,B\cos \left ( dx+c \right ) +3\,A \right ) }{15\,d\sin \left ( dx+c \right ) }\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}} \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)/cos(d*x+c)^(7/2),x)

[Out]

-2/15/d*(-1+cos(d*x+c))*(8*A*cos(d*x+c)^2+10*B*cos(d*x+c)^2+15*C*cos(d*x+c)^2+4*A*cos(d*x+c)+5*B*cos(d*x+c)+3*
A)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)/cos(d*x+c)^(5/2)

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Maxima [B]  time = 1.78187, size = 707, normalized size = 5.44 \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)*(a+a*cos(d*x+c))^(1/2)/cos(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

2/15*(15*C*(sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1
)^3)/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(3/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(3/2)) + 5*B*(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)*s
qrt(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 + s
in(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)) + A*(15*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 25*sqrt(2)*
sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 17*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 7*sqrt(
2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/((sin(d*x + c)/(co
s(d*x + c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
+ 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)))/d

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Fricas [A]  time = 1.86929, size = 231, normalized size = 1.78 \begin{align*} \frac{2 \,{\left ({\left (8 \, A + 10 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (4 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 3 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\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)/cos(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

2/15*((8*A + 10*B + 15*C)*cos(d*x + c)^2 + (4*A + 5*B)*cos(d*x + c) + 3*A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d
*x + c))*sin(d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)

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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)*(a+a*cos(d*x+c))**(1/2)/cos(d*x+c)**(7/2),x)

[Out]

Timed out

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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}}{\cos \left (d x + c\right )^{\frac{7}{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)/cos(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(a*cos(d*x + c) + a)/cos(d*x + c)^(7/2), x)