3.2.26 \(\int \frac {(a+a \cos (c+d x)) (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [126]

Optimal. Leaf size=70 \[ \frac {2 a (A+B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a (3 A+B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \]

[Out]

2*a*(A+B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*a*(3*A+B
)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*a*B*sin(d*x+c)*c
os(d*x+c)^(1/2)/d

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3047, 3102, 2827, 2720, 2719} \begin {gather*} \frac {2 a (3 A+B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a (A+B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a B \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + a*Cos[c + d*x])*(A + B*Cos[c + d*x]))/Sqrt[Cos[c + d*x]],x]

[Out]

(2*a*(A + B)*EllipticE[(c + d*x)/2, 2])/d + (2*a*(3*A + B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*B*Sqrt[Cos[
c + d*x]]*Sin[c + d*x])/(3*d)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2827

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 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((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 + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {(a+a \cos (c+d x)) (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx &=\int \frac {a A+(a A+a B) \cos (c+d x)+a B \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2}{3} \int \frac {\frac {1}{2} a (3 A+B)+\frac {3}{2} a (A+B) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+(a (A+B)) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} (a (3 A+B)) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a (A+B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a (3 A+B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 a B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.19, size = 309, normalized size = 4.41 \begin {gather*} \frac {a (1+\cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-6 (A+B) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\text {ArcTan}(\tan (c)))\right ) \sec (c) \sin (d x+\text {ArcTan}(\tan (c)))+\left (9 (A+B) \cos (c-d x-\text {ArcTan}(\tan (c))) \csc (c) \sec (c)+3 A \cos (c+d x+\text {ArcTan}(\tan (c))) \csc (c) \sec (c)+3 B \cos (c+d x+\text {ArcTan}(\tan (c))) \csc (c) \sec (c)-12 A \cos (c+d x) \cot (c) \sqrt {\sec ^2(c)}-12 B \cos (c+d x) \cot (c) \sqrt {\sec ^2(c)}-4 (3 A+B) \cos (c+d x) \sqrt {\cos ^2(d x-\text {ArcTan}(\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sqrt {\sec ^2(c)} \sec (d x-\text {ArcTan}(\cot (c))) \sin (c)+4 B \cos (c+d x) \sqrt {\sec ^2(c)} \sin (c+d x)\right ) \sqrt {\sin ^2(d x+\text {ArcTan}(\tan (c)))}\right )}{12 d \sqrt {\cos (c+d x)} \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\text {ArcTan}(\tan (c)))}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + a*Cos[c + d*x])*(A + B*Cos[c + d*x]))/Sqrt[Cos[c + d*x]],x]

[Out]

(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*(-6*(A + B)*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[T
an[c]]]^2]*Sec[c]*Sin[d*x + ArcTan[Tan[c]]] + (9*(A + B)*Cos[c - d*x - ArcTan[Tan[c]]]*Csc[c]*Sec[c] + 3*A*Cos
[c + d*x + ArcTan[Tan[c]]]*Csc[c]*Sec[c] + 3*B*Cos[c + d*x + ArcTan[Tan[c]]]*Csc[c]*Sec[c] - 12*A*Cos[c + d*x]
*Cot[c]*Sqrt[Sec[c]^2] - 12*B*Cos[c + d*x]*Cot[c]*Sqrt[Sec[c]^2] - 4*(3*A + B)*Cos[c + d*x]*Sqrt[Cos[d*x - Arc
Tan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Sec[c]^2
]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] + 4*B*Cos[c + d*x]*Sqrt[Sec[c]^2]*Sin[c + d*x])*Sqrt[Sin[d*x + ArcTan[Tan[c
]]]^2]))/(12*d*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(320\) vs. \(2(116)=232\).
time = 0.27, size = 321, normalized size = 4.59

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a \left (4 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(321\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))*(A+B*cos(d*x+c))/cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(4*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+3*
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))-3*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))-2*B*cos(1/2*d*
x+1/2*c)*sin(1/2*d*x+1/2*c)^2+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))-3*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)))/(-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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)/sqrt(cos(d*x + c)), x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 142, normalized size = 2.03 \begin {gather*} \frac {2 \, B a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - i \, \sqrt {2} {\left (3 \, A + B\right )} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} {\left (3 \, A + B\right )} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (A + B\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (A + B\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

1/3*(2*B*a*sqrt(cos(d*x + c))*sin(d*x + c) - I*sqrt(2)*(3*A + B)*a*weierstrassPInverse(-4, 0, cos(d*x + c) + I
*sin(d*x + c)) + I*sqrt(2)*(3*A + B)*a*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt(2)
*(A + B)*a*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*(A
+ B)*a*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a \left (\int \frac {A}{\sqrt {\cos {\left (c + d x \right )}}}\, dx + \int A \sqrt {\cos {\left (c + d x \right )}}\, dx + \int B \sqrt {\cos {\left (c + d x \right )}}\, dx + \int B \cos ^{\frac {3}{2}}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))/cos(d*x+c)**(1/2),x)

[Out]

a*(Integral(A/sqrt(cos(c + d*x)), x) + Integral(A*sqrt(cos(c + d*x)), x) + Integral(B*sqrt(cos(c + d*x)), x) +
Integral(B*cos(c + d*x)**(3/2), x))

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)/sqrt(cos(d*x + c)), x)

________________________________________________________________________________________

Mupad [B]
time = 0.53, size = 79, normalized size = 1.13 \begin {gather*} \frac {2\,B\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,B\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,a\,\left (\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x)))/cos(c + d*x)^(1/2),x)

[Out]

(2*B*a*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*B*a*ellipticE(c/2 + (d*x)/2
, 2))/d + (2*A*a*(ellipticE(c/2 + (d*x)/2, 2) + ellipticF(c/2 + (d*x)/2, 2)))/d

________________________________________________________________________________________