3.156 \(\int \frac{A+B \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx\)
Optimal. Leaf size=168 \[ -\frac{(5 A-2 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{(4 A-B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{(4 A-B) \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}-\frac{(5 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\cos (c+d x)} (\cos (c+d x)+1)}-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^2} \]
[Out]
-(((4*A - B)*EllipticE[(c + d*x)/2, 2])/(a^2*d)) - ((5*A - 2*B)*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) + ((4*A -
B)*Sin[c + d*x])/(a^2*d*Sqrt[Cos[c + d*x]]) - ((5*A - 2*B)*Sin[c + d*x])/(3*a^2*d*Sqrt[Cos[c + d*x]]*(1 + Cos
[c + d*x])) - ((A - B)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2)
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Rubi [A] time = 0.359309, antiderivative size = 168, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used =
{2978, 2748, 2636, 2639, 2641} \[ -\frac{(5 A-2 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{(4 A-B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{(4 A-B) \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}-\frac{(5 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\cos (c+d x)} (\cos (c+d x)+1)}-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^2),x]
[Out]
-(((4*A - B)*EllipticE[(c + d*x)/2, 2])/(a^2*d)) - ((5*A - 2*B)*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) + ((4*A -
B)*Sin[c + d*x])/(a^2*d*Sqrt[Cos[c + d*x]]) - ((5*A - 2*B)*Sin[c + d*x])/(3*a^2*d*Sqrt[Cos[c + d*x]]*(1 + Cos
[c + d*x])) - ((A - B)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2)
Rule 2978
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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 2636
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]
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 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]
Rubi steps
\begin{align*} \int \frac{A+B \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx &=-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac{\int \frac{\frac{1}{2} a (7 A-B)-\frac{3}{2} a (A-B) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{3 a^2}\\ &=-\frac{(5 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\cos (c+d x)} (1+\cos (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac{\int \frac{\frac{3}{2} a^2 (4 A-B)-\frac{1}{2} a^2 (5 A-2 B) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{3 a^4}\\ &=-\frac{(5 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\cos (c+d x)} (1+\cos (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2}-\frac{(5 A-2 B) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}+\frac{(4 A-B) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{2 a^2}\\ &=-\frac{(5 A-2 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(4 A-B) \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}-\frac{(5 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\cos (c+d x)} (1+\cos (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2}-\frac{(4 A-B) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2}\\ &=-\frac{(4 A-B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(5 A-2 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(4 A-B) \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}-\frac{(5 A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\cos (c+d x)} (1+\cos (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2}\\ \end{align*}
Mathematica [C] time = 6.68985, size = 1217, normalized size = 7.24 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(3/2)*(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 + ((I/2)*B*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 + (10*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - Ar
cTan[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]*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*Sqrt[1
+ Cot[c]^2]) - (4*B*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]*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*Sqrt[1 + Cot[c]^
2]) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*((2*(2*A + 2*A*Cos[c] - B*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/d +
(2*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2]))/(3*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(2
*A*Sin[(d*x)/2] - B*Sin[(d*x)/2]))/d + (8*A*Sec[c]*Sec[c + d*x]*Sin[d*x])/d + (2*(A - B)*Sec[c/2 + (d*x)/2]^2*
Tan[c/2])/(3*d)))/(a + a*Cos[c + d*x])^2
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Maple [B] time = 4.243, size = 494, normalized size = 2.9 \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))/cos(d*x+c)^(3/2)/(a+cos(d*x+c)*a)^2,x)
[Out]
-1/6*(-2*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/
2*c)^2)^(1/2)*(12*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*B*Ellipt
icE(cos(1/2*d*x+1/2*c),2^(1/2))+2*B*EllipticF(cos(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+2*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*
c)^2)^(1/2)*(12*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*B*Elliptic
E(cos(1/2*d*x+1/2*c),2^(1/2))+2*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-12*(-2*sin(1/2*d*x
+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*A-B)*sin(1/2*d*x+1/2*c)^6+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*
c)^2)^(1/2)*(43*A-10*B)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(37*A-7*B)*s
in(1/2*d*x+1/2*c)^2)/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(-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))/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^2,x, algorithm="maxima")
[Out]
Timed out
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt{\cos \left (d x + c\right )}}{a^{2} \cos \left (d x + c\right )^{4} + 2 \, a^{2} \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^2,x, algorithm="fricas")
[Out]
integral((B*cos(d*x + c) + A)*sqrt(cos(d*x + c))/(a^2*cos(d*x + c)^4 + 2*a^2*cos(d*x + c)^3 + a^2*cos(d*x + c)
^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((A+B*cos(d*x+c))/cos(d*x+c)**(3/2)/(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{B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^2,x, algorithm="giac")
[Out]
integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^2*cos(d*x + c)^(3/2)), x)