3.158 \(\int \frac{\cos ^{\frac{5}{2}}(c+d x) (A+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^2} \, dx\)
Optimal. Leaf size=196 \[ -\frac{5 (A+3 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{4 (5 A+14 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{(A+3 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{a^2 d (\cos (c+d x)+1)}+\frac{4 (5 A+14 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15 a^2 d}-\frac{5 (A+3 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a^2 d}-\frac{(A+C) \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]
[Out]
(4*(5*A + 14*C)*EllipticE[(c + d*x)/2, 2])/(5*a^2*d) - (5*(A + 3*C)*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) - (5*
(A + 3*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a^2*d) + (4*(5*A + 14*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(15*a^
2*d) - ((A + 3*C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(a^2*d*(1 + Cos[c + d*x])) - ((A + C)*Cos[c + d*x]^(7/2)*Si
n[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)
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Rubi [A] time = 0.35062, antiderivative size = 196, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used =
{3042, 2977, 2748, 2635, 2641, 2639} \[ -\frac{5 (A+3 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{4 (5 A+14 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{(A+3 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{a^2 d (\cos (c+d x)+1)}+\frac{4 (5 A+14 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{15 a^2 d}-\frac{5 (A+3 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a^2 d}-\frac{(A+C) \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^(5/2)*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^2,x]
[Out]
(4*(5*A + 14*C)*EllipticE[(c + d*x)/2, 2])/(5*a^2*d) - (5*(A + 3*C)*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) - (5*
(A + 3*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a^2*d) + (4*(5*A + 14*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(15*a^
2*d) - ((A + 3*C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(a^2*d*(1 + Cos[c + d*x])) - ((A + C)*Cos[c + d*x]^(7/2)*Si
n[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)
Rule 3042
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*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x
])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]
Rule 2977
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], 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
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Rule 2748
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Rule 2635
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]
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 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]
Rubi steps
\begin{align*} \int \frac{\cos ^{\frac{5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx &=-\frac{(A+C) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{\cos ^{\frac{5}{2}}(c+d x) \left (-\frac{1}{2} a (A+7 C)+\frac{1}{2} a (5 A+11 C) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(A+3 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac{(A+C) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \cos ^{\frac{3}{2}}(c+d x) \left (-\frac{15}{2} a^2 (A+3 C)+2 a^2 (5 A+14 C) \cos (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{(A+3 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac{(A+C) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(5 (A+3 C)) \int \cos ^{\frac{3}{2}}(c+d x) \, dx}{2 a^2}+\frac{(2 (5 A+14 C)) \int \cos ^{\frac{5}{2}}(c+d x) \, dx}{3 a^2}\\ &=-\frac{5 (A+3 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{4 (5 A+14 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a^2 d}-\frac{(A+3 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac{(A+C) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{(5 (A+3 C)) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}+\frac{(2 (5 A+14 C)) \int \sqrt{\cos (c+d x)} \, dx}{5 a^2}\\ &=\frac{4 (5 A+14 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac{5 (A+3 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{5 (A+3 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac{4 (5 A+14 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a^2 d}-\frac{(A+3 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac{(A+C) \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end{align*}
Mathematica [C] time = 6.76216, size = 1248, normalized size = 6.37 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(Cos[c + d*x]^(5/2)*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^2,x]
[Out]
((2*I)*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*
d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)
]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1 +
E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2
*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] +
I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/(a + a*Cos[c
+ d*x])^2 + (((28*I)/5)*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4,
7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*
Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*C
os[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Si
n[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)
*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c]
)))/(a + a*Cos[c + d*x])^2 + (10*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x
- ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Co
t[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])^2*Sq
rt[1 + Cot[c]^2]) + (10*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[
Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*S
in[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(a + a*Cos[c + d*x])^2*Sqrt[1 + Cot[
c]^2]) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*((-8*(5*A + 10*C + 5*A*Cos[c] + 18*C*Cos[c])*Csc[c])/(5*d) -
(16*C*Cos[d*x]*Sin[c])/(3*d) + (4*C*Cos[2*d*x]*Sin[2*c])/(5*d) + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x
)/2] + C*Sin[(d*x)/2]))/(3*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + 2*C*Sin[(d*x)/2]))/d - (16*C*
Cos[c]*Sin[d*x])/(3*d) + (4*C*Cos[2*c]*Sin[2*d*x])/(5*d) + (2*(A + C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/(
a + a*Cos[c + d*x])^2
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Maple [A] time = 0.133, size = 451, normalized size = 2.3 \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+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x)
[Out]
1/30*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-96*C*cos(1/2*d*x+1/2*c)^10+352*C*cos(1/2*d*x+1/
2*c)^8+120*A*cos(1/2*d*x+1/2*c)^6+50*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3+120*A*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-120*C*cos(1/2*d*x+1/2*c)^6+150*C*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c
)^3+336*C*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/
2*d*x+1/2*c),2^(1/2))-190*A*cos(1/2*d*x+1/2*c)^4-266*C*cos(1/2*d*x+1/2*c)^4+75*A*cos(1/2*d*x+1/2*c)^2+135*C*co
s(1/2*d*x+1/2*c)^2-5*A-5*C)/a^2/cos(1/2*d*x+1/2*c)^3/(-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 (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^2, x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \cos \left (d x + c\right )^{4} + A \cos \left (d x + c\right )^{2}\right )} \sqrt{\cos \left (d x + c\right )}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x, algorithm="fricas")
[Out]
integral((C*cos(d*x + c)^4 + A*cos(d*x + c)^2)*sqrt(cos(d*x + c))/(a^2*cos(d*x + c)^2 + 2*a^2*cos(d*x + c) + a
^2), x)
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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+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**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} + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2,x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^2, x)