3.1101 \(\int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx\)

Optimal. Leaf size=234 \[ -\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{5 a^3 d}-\frac{2 b \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a+b)}+\frac{2 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{5 a^3 d \sqrt{\cos (c+d x)}}-\frac{2 (A b-a B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{2 (A b-a B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)} \]

[Out]

(-2*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*a^3*d) - (2*(A*b - a*B)*EllipticF[(c +
d*x)/2, 2])/(3*a^2*d) - (2*b*(A*b^2 - a*(b*B - a*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a^3*(a + b)*
d) + (2*A*Sin[c + d*x])/(5*a*d*Cos[c + d*x]^(5/2)) - (2*(A*b - a*B)*Sin[c + d*x])/(3*a^2*d*Cos[c + d*x]^(3/2))
+ (2*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*Sin[c + d*x])/(5*a^3*d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 1.23611, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {3055, 3059, 2639, 3002, 2641, 2805} \[ -\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{5 a^3 d}-\frac{2 b \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a+b)}+\frac{2 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{5 a^3 d \sqrt{\cos (c+d x)}}-\frac{2 (A b-a B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{2 (A b-a B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + b*Cos[c + d*x])),x]

[Out]

(-2*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*a^3*d) - (2*(A*b - a*B)*EllipticF[(c +
d*x)/2, 2])/(3*a^2*d) - (2*b*(A*b^2 - a*(b*B - a*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a^3*(a + b)*
d) + (2*A*Sin[c + d*x])/(5*a*d*Cos[c + d*x]^(5/2)) - (2*(A*b - a*B)*Sin[c + d*x])/(3*a^2*d*Cos[c + d*x]^(3/2))
+ (2*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*Sin[c + d*x])/(5*a^3*d*Sqrt[Cos[c + d*x]])

Rule 3055

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[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3059

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3002

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rubi steps

\begin{align*} \int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx &=\frac{2 A \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \int \frac{-\frac{5}{2} (A b-a B)+\frac{1}{2} a (3 A+5 C) \cos (c+d x)+\frac{3}{2} A b \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{5 a}\\ &=\frac{2 A \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)}-\frac{2 (A b-a B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{\frac{3}{4} \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right )+\frac{1}{4} a (4 A b+5 a B) \cos (c+d x)-\frac{5}{4} b (A b-a B) \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{15 a^2}\\ &=\frac{2 A \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)}-\frac{2 (A b-a B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 a^3 d \sqrt{\cos (c+d x)}}+\frac{8 \int \frac{-\frac{5}{8} \left (3 A b^3-a^3 B-3 a b^2 B+a^2 b (A+3 C)\right )-\frac{1}{8} a \left (20 A b^2-20 a b B+3 a^2 (3 A+5 C)\right ) \cos (c+d x)-\frac{3}{8} b \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 a^3}\\ &=\frac{2 A \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)}-\frac{2 (A b-a B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 a^3 d \sqrt{\cos (c+d x)}}-\frac{8 \int \frac{\frac{5}{8} b \left (3 A b^3-a^3 B-3 a b^2 B+a^2 b (A+3 C)\right )+\frac{5}{8} a b^2 (A b-a B) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 a^3 b}-\frac{\left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^3}\\ &=-\frac{2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d}+\frac{2 A \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)}-\frac{2 (A b-a B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 a^3 d \sqrt{\cos (c+d x)}}-\frac{(A b-a B) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 a^2}-\frac{\left (b \left (A b^2-a (b B-a C)\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^3}\\ &=-\frac{2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d}-\frac{2 (A b-a B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{2 b \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 (a+b) d}+\frac{2 A \sin (c+d x)}{5 a d \cos ^{\frac{5}{2}}(c+d x)}-\frac{2 (A b-a B) \sin (c+d x)}{3 a^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 a^3 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 4.92415, size = 336, normalized size = 1.44 \[ -\frac{\frac{2 \left (a^2 b (19 A+45 C)-10 a^3 B-45 a b^2 B+45 A b^3\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{4 a \left (3 a^2 (3 A+5 C)-20 a b B+20 A b^2\right ) \left ((a+b) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{b (a+b)}-\frac{2 \left (3 \sin (2 (c+d x)) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )+6 a^2 A \tan (c+d x)+10 a (a B-A b) \sin (c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)}+\frac{6 \sin (c+d x) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt{\sin ^2(c+d x)}}}{30 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + b*Cos[c + d*x])),x]

[Out]

-((2*(45*A*b^3 - 10*a^3*B - 45*a*b^2*B + a^2*b*(19*A + 45*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a +
b) + (4*a*(20*A*b^2 - 20*a*b*B + 3*a^2*(3*A + 5*C))*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a
+ b), (c + d*x)/2, 2]))/(b*(a + b)) + (6*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*(-2*a*b*EllipticE[ArcSin[Sqrt[
Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (2*a^2 - b^2)*EllipticPi[-(b/a),
-ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b*Sqrt[Sin[c + d*x]^2]) - (2*(10*a*(-(A*b) + a*B)*Sin[c +
d*x] + 3*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*Sin[2*(c + d*x)] + 6*a^2*A*Tan[c + d*x]))/Cos[c + d*x]^(3/2))/(
30*a^3*d)

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Maple [B]  time = 3.932, size = 802, normalized size = 3.4 \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)/cos(d*x+c)^(7/2)/(a+b*cos(d*x+c)),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*(A*b^2-B*a*b+C*a^2)*b^2/a^3/(-2*a*b+2*b^2)*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-2/5*A/a/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*s
in(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*Ellip
ticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2
*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*(-2*sin(1/2*d*x+1/2*c)^4+s
in(1/2*d*x+1/2*c)^2)^(1/2)+2*(A*b^2-B*a*b+C*a^2)/a^3*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)
^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*si
n(1/2*d*x+1/2*c)^2-1)+2*(-A*b+B*a)/a^2*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)
^(1/2)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*si
n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*c
os(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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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)/cos(d*x+c)^(7/2)/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [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)/cos(d*x+c)^(7/2)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \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)/cos(d*x+c)^(7/2)/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

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