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

Optimal. Leaf size=140 \[ \frac{2 \left (a^2 C+A b^2\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d (a+b)}+\frac{2 A b E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{2 A b \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}+\frac{2 A F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{2 A \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.729576, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3056, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac{2 \left (a^2 C+A b^2\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d (a+b)}+\frac{2 A b E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{2 A b \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}+\frac{2 A F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{2 A \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3056

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[((A*b^2 + 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] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

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

Mathematica [A]  time = 5.72485, size = 223, normalized size = 1.59 \[ \frac{\frac{2 \left (2 a^2 (A+3 C)+9 A b^2\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{6 A \sin (c+d x) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a \sqrt{\sin ^2(c+d x)}}+\frac{4 A \sin (c+d x) (a-3 b \cos (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)}+8 a A \left (2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-\frac{2 a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{6 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(9*A*b^2 + 2*a^2*(A + 3*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + 8*a*A*(2*EllipticF[(c + d
*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)) + (4*A*(a - 3*b*Cos[c + d*x])*Sin[c + d*x
])/Cos[c + d*x]^(3/2) + (6*A*(-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
*Sqrt[Sin[c + d*x]^2]))/(6*a^2*d)

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Maple [B]  time = 1.135, size = 463, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(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+C*a^2)/a^2/(-2*a*b+2*b^2)*b*(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)*Ellipt
icPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+2*A/a*(-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*sin(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)))-2*A/a^2*b*(-(
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*sin(1/2*d*x+1/2*c)^2-1))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x
+1/2*c)^2-1)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(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+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2)/(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} + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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