3.169 \(\int \frac{A+C \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx\)
Optimal. Leaf size=219 \[ -\frac{(13 A-C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac{(49 A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(49 A-C) \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{(13 A-C) \sin (c+d x)}{6 d \sqrt{\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}-\frac{2 (4 A-C) \sin (c+d x)}{15 a d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^2}-\frac{(A+C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^3} \]
[Out]
-((49*A - C)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) - ((13*A - C)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) + ((49*A
- C)*Sin[c + d*x])/(10*a^3*d*Sqrt[Cos[c + d*x]]) - ((A + C)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[
c + d*x])^3) - (2*(4*A - C)*Sin[c + d*x])/(15*a*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2) - ((13*A - C)*Sin
[c + d*x])/(6*d*Sqrt[Cos[c + d*x]]*(a^3 + a^3*Cos[c + d*x]))
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Rubi [A] time = 0.523739, antiderivative size = 219, 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, 2978, 2748, 2636, 2639, 2641} \[ -\frac{(13 A-C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac{(49 A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(49 A-C) \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{(13 A-C) \sin (c+d x)}{6 d \sqrt{\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}-\frac{2 (4 A-C) \sin (c+d x)}{15 a d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^2}-\frac{(A+C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^3} \]
Antiderivative was successfully verified.
[In]
Int[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^3),x]
[Out]
-((49*A - C)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) - ((13*A - C)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) + ((49*A
- C)*Sin[c + d*x])/(10*a^3*d*Sqrt[Cos[c + d*x]]) - ((A + C)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[
c + d*x])^3) - (2*(4*A - C)*Sin[c + d*x])/(15*a*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2) - ((13*A - C)*Sin
[c + d*x])/(6*d*Sqrt[Cos[c + d*x]]*(a^3 + a^3*Cos[c + d*x]))
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 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+C \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx &=-\frac{(A+C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac{\int \frac{\frac{1}{2} a (11 A+C)-\frac{5}{2} a (A-C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^3}-\frac{2 (4 A-C) \sin (c+d x)}{15 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac{\int \frac{\frac{1}{2} a^2 (41 A+C)-3 a^2 (4 A-C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{15 a^4}\\ &=-\frac{(A+C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^3}-\frac{2 (4 A-C) \sin (c+d x)}{15 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2}-\frac{(13 A-C) \sin (c+d x)}{6 d \sqrt{\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \frac{\frac{3}{4} a^3 (49 A-C)-\frac{5}{4} a^3 (13 A-C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^6}\\ &=-\frac{(A+C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^3}-\frac{2 (4 A-C) \sin (c+d x)}{15 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2}-\frac{(13 A-C) \sin (c+d x)}{6 d \sqrt{\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )}-\frac{(13 A-C) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{12 a^3}+\frac{(49 A-C) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx}{20 a^3}\\ &=-\frac{(13 A-C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{(49 A-C) \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{(A+C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^3}-\frac{2 (4 A-C) \sin (c+d x)}{15 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2}-\frac{(13 A-C) \sin (c+d x)}{6 d \sqrt{\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )}-\frac{(49 A-C) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=-\frac{(49 A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(13 A-C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{(49 A-C) \sin (c+d x)}{10 a^3 d \sqrt{\cos (c+d x)}}-\frac{(A+C) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^3}-\frac{2 (4 A-C) \sin (c+d x)}{15 a d \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2}-\frac{(13 A-C) \sin (c+d x)}{6 d \sqrt{\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}
Mathematica [C] time = 6.88519, size = 1301, normalized size = 5.94 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^3),x]
[Out]
(((-49*I)/10)*A*Cos[c/2 + (d*x)/2]^6*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])^3 + ((I/10)*C*Cos[c/2 + (d*x)/2]^6*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])^3 + (26*A*Cos[c/2 + (d*x)/2]^6*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])^3
*Sqrt[1 + Cot[c]^2]) - (2*C*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTa
n[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])^3*Sqrt[1 +
Cot[c]^2]) + (Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*((2*(20*A + 29*A*Cos[c] - C*Cos[c])*Csc[c/2]*Sec[c/2]*Se
c[c])/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(29*A*Sin[(d*x)/2] - C*Sin[(d*x)/2]))/(5*d) + (2*Sec[c/2]*Sec[c/2
+ (d*x)/2]^5*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(11*A*Sin[(d*x)/2] +
C*Sin[(d*x)/2]))/(15*d) + (16*A*Sec[c]*Sec[c + d*x]*Sin[d*x])/d + (4*(11*A + C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2]
)/(15*d) + (2*(A + C)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d)))/(a + a*Cos[c + d*x])^3
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Maple [B] time = 0.17, size = 685, normalized size = 3.1 \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)^(3/2)/(a+a*cos(d*x+c))^3,x)
[Out]
-1/60*(-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)*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*C*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(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)^4+4*(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)*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*C*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(cos(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*(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)*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*(65*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*C*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(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)*(49*A-C)*sin(1/2*d*x+1/2*c)^8-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*
x+1/2*c)^2)^(1/2)*(817*A-13*C)*sin(1/2*d*x+1/2*c)^6+12*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(1
24*A-C)*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)*(439*A-C)*sin(1/2*d*x+1/2*c)
^2)/a^3/cos(1/2*d*x+1/2*c)^5/(-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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \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)^(3/2)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
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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 )^{2} + A\right )} \sqrt{\cos \left (d x + c\right )}}{a^{3} \cos \left (d x + c\right )^{5} + 3 \, a^{3} \cos \left (d x + c\right )^{4} + 3 \, a^{3} \cos \left (d x + c\right )^{3} + a^{3} \cos \left (d x + c\right )^{2}}, x\right ) \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)^(3/2)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")
[Out]
integral((C*cos(d*x + c)^2 + A)*sqrt(cos(d*x + c))/(a^3*cos(d*x + c)^5 + 3*a^3*cos(d*x + c)^4 + 3*a^3*cos(d*x
+ c)^3 + a^3*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+C*cos(d*x+c)**2)/cos(d*x+c)**(3/2)/(a+a*cos(d*x+c))**3,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 (a \cos \left (d x + c\right ) + a\right )}^{3} \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+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^3,x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)/((a*cos(d*x + c) + a)^3*cos(d*x + c)^(3/2)), x)