3.365 \(\int \frac{\cos ^{\frac{5}{2}}(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx\)
Optimal. Leaf size=182 \[ \frac{2 \left (3 a^2+b^2\right ) (A b-a B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^4 d}-\frac{2 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d}-\frac{2 a^3 (A b-a B) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^4 d (a+b)}+\frac{2 (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 b^2 d}+\frac{2 B \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 b d} \]
[Out]
(-2*(5*a*A*b - 5*a^2*B - 3*b^2*B)*EllipticE[(c + d*x)/2, 2])/(5*b^3*d) + (2*(3*a^2 + b^2)*(A*b - a*B)*Elliptic
F[(c + d*x)/2, 2])/(3*b^4*d) - (2*a^3*(A*b - a*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b^4*(a + b)*d) +
(2*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b^2*d) + (2*B*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*b*d)
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Rubi [A] time = 0.819539, antiderivative size = 182, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used =
{2990, 3049, 3059, 2639, 3002, 2641, 2805} \[ \frac{2 \left (3 a^2+b^2\right ) (A b-a B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^4 d}-\frac{2 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d}-\frac{2 a^3 (A b-a B) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^4 d (a+b)}+\frac{2 (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 b^2 d}+\frac{2 B \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 b d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x]),x]
[Out]
(-2*(5*a*A*b - 5*a^2*B - 3*b^2*B)*EllipticE[(c + d*x)/2, 2])/(5*b^3*d) + (2*(3*a^2 + b^2)*(A*b - a*B)*Elliptic
F[(c + d*x)/2, 2])/(3*b^4*d) - (2*a^3*(A*b - a*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b^4*(a + b)*d) +
(2*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b^2*d) + (2*B*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*b*d)
Rule 2990
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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x
])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c -
b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m,
1] && !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Rule 3049
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*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] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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{\cos ^{\frac{5}{2}}(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx &=\frac{2 B \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac{2 \int \frac{\sqrt{\cos (c+d x)} \left (\frac{3 a B}{2}+\frac{3}{2} b B \cos (c+d x)+\frac{5}{2} (A b-a B) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{5 b}\\ &=\frac{2 (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac{2 B \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac{4 \int \frac{\frac{5}{4} a (A b-a B)+\frac{1}{4} b (5 A b+4 a B) \cos (c+d x)-\frac{3}{4} \left (5 a A b-5 a^2 B-3 b^2 B\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 b^2}\\ &=\frac{2 (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac{2 B \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 b d}-\frac{4 \int \frac{-\frac{5}{4} a b (A b-a B)-\frac{5}{4} \left (3 a^2+b^2\right ) (A b-a B) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 b^3}-\frac{\left (5 a A b-5 a^2 B-3 b^2 B\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 b^3}\\ &=-\frac{2 \left (5 a A b-5 a^2 B-3 b^2 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac{2 (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac{2 B \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 b d}-\frac{\left (a^3 (A b-a B)\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b^4}+\frac{\left (\left (3 a^2+b^2\right ) (A b-a B)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 b^4}\\ &=-\frac{2 \left (5 a A b-5 a^2 B-3 b^2 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac{2 \left (3 a^2+b^2\right ) (A b-a B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^4 d}-\frac{2 a^3 (A b-a B) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^4 (a+b) d}+\frac{2 (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac{2 B \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 b d}\\ \end{align*}
Mathematica [A] time = 2.37185, size = 264, normalized size = 1.45 \[ \frac{\frac{2 b^2 \left (5 a^2 B-5 a A b+9 b^2 B\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{6 \left (5 a^2 B-5 a A b+3 b^2 B\right ) \sin (c+d x) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a \sqrt{\sin ^2(c+d x)}}+4 b^2 \sin (c+d x) \sqrt{\cos (c+d x)} (-5 a B+5 A b+3 b B \cos (c+d x))+2 b^2 (4 a B+5 A b) \left (2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-\frac{2 a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{30 b^4 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x]),x]
[Out]
((2*b^2*(-5*a*A*b + 5*a^2*B + 9*b^2*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + 2*b^2*(5*A*b + 4*a
*B)*(2*EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)) + 4*b^2*Sqrt[Cos[c
+ d*x]]*(5*A*b - 5*a*B + 3*b*B*Cos[c + d*x])*Sin[c + d*x] + (6*(-5*a*A*b + 5*a^2*B + 3*b^2*B)*(-2*a*b*Ellipti
cE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (2*a^2 - b^2)*Ell
ipticPi[-(b/a), -ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*Sqrt[Sin[c + d*x]^2]))/(30*b^4*d)
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Maple [B] time = 4.116, size = 1074, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c)),x)
[Out]
-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((-24*B*a*b^3+24*B*b^4)*cos(1/2*d*x+1/2*c)*sin(1
/2*d*x+1/2*c)^6+(20*A*a*b^3-20*A*b^4-20*B*a^2*b^2+44*B*a*b^3-24*B*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)
+(-10*A*a*b^3+10*A*b^4+10*B*a^2*b^2-16*B*a*b^3+6*B*b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+15*A*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b-15*A*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2+5*A*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^3-5*A*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^4+15*A*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2-15*A*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^3-15*A*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a
^3*b-15*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*
a^4+15*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a
^3*b-5*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a
^2*b^2+5*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))
*a*b^3-15*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)
)*a^3*b+15*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2
))*a^2*b^2-9*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1
/2))*a*b^3+9*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1
/2))*b^4+15*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)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b
/(a-b),2^(1/2))*a^4)/b^4/(a-b)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/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{{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{b \cos \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c)),x, algorithm="maxima")
[Out]
integrate((B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(b*cos(d*x + c) + a), 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(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(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(cos(d*x+c)**(5/2)*(A+B*cos(d*x+c))/(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{{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{b \cos \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(b*cos(d*x + c) + a), x)