3.154 \(\int \frac{A+C \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))} \, dx\)
Optimal. Leaf size=83 \[ \frac{(A-C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}+\frac{(A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}-\frac{(A+C) \sin (c+d x) \sqrt{\cos (c+d x)}}{d (a \cos (c+d x)+a)} \]
[Out]
((A + 3*C)*EllipticE[(c + d*x)/2, 2])/(a*d) + ((A - C)*EllipticF[(c + d*x)/2, 2])/(a*d) - ((A + C)*Sqrt[Cos[c
+ d*x]]*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))
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Rubi [A] time = 0.154196, antiderivative size = 83, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used =
{3042, 2748, 2641, 2639} \[ \frac{(A-C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}+\frac{(A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}-\frac{(A+C) \sin (c+d x) \sqrt{\cos (c+d x)}}{d (a \cos (c+d x)+a)} \]
Antiderivative was successfully verified.
[In]
Int[(A + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])),x]
[Out]
((A + 3*C)*EllipticE[(c + d*x)/2, 2])/(a*d) + ((A - C)*EllipticF[(c + d*x)/2, 2])/(a*d) - ((A + C)*Sqrt[Cos[c
+ d*x]]*Sin[c + d*x])/(d*(a + a*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 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 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{A+C \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))} \, dx &=-\frac{(A+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac{\int \frac{\frac{1}{2} a (A-C)+\frac{1}{2} a (A+3 C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{a^2}\\ &=-\frac{(A+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac{(A-C) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 a}+\frac{(A+3 C) \int \sqrt{\cos (c+d x)} \, dx}{2 a}\\ &=\frac{(A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}+\frac{(A-C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}-\frac{(A+C) \sqrt{\cos (c+d x)} \sin (c+d x)}{d (a+a \cos (c+d x))}\\ \end{align*}
Mathematica [C] time = 6.47345, size = 1095, normalized size = 13.19 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])),x]
[Out]
((I/4)*A*Cos[c/2 + (d*x)/2]^2*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)/4)*C*Cos[c/2 + (d*x)/2]^2*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]) + (Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*((-2*(A + C + 2*C*Cos[c])*Csc[c])/d - (2*Sec[
c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/d))/(a + a*Cos[c + d*x]) - (A*Cos[c/2 + (d*x)/2]^2*
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]]]])/(d*(a + a*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) + (C*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Hyperge
ometricPFQ[{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]]]])/(d*(a + a*Cos[c + d*x])*Sqrt[1 + Cot[c]^2])
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Maple [A] time = 0.118, size = 247, normalized size = 3. \begin{align*}{\frac{1}{ad}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( -\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( A{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) -A{\it EllipticE} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) -C{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) -3\,C{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) + \left ( 2\,A+2\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+ \left ( -A-C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))/cos(d*x+c)^(1/2),x)
[Out]
((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-
C*EllipticF(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)))+(2*A+2*C)*sin(1/2*d*x+1/2*c
)^4+(-A-C)*sin(1/2*d*x+1/2*c)^2)/a/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)/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 (a \cos \left (d x + c\right ) + a\right )} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + A)/((a*cos(d*x + c) + a)*sqrt(cos(d*x + c))), 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 )^{2} + A\right )} \sqrt{\cos \left (d x + c\right )}}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
integral((C*cos(d*x + c)^2 + A)*sqrt(cos(d*x + c))/(a*cos(d*x + c)^2 + a*cos(d*x + c)), 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)/(a+a*cos(d*x+c))/cos(d*x+c)**(1/2),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 )} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)/((a*cos(d*x + c) + a)*sqrt(cos(d*x + c))), x)