3.5.64 \(\int \frac {(a+a \cos (c+d x)) (A+B \cos (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [464]

Optimal. Leaf size=172 \[ \frac {6 a (A+B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (7 A+5 B) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a (A+B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a (7 A+5 B) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}} \]

[Out]

2/7*a*B*sin(d*x+c)/d/sec(d*x+c)^(5/2)+2/5*a*(A+B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/21*a*(7*A+5*B)*sin(d*x+c)/d/
sec(d*x+c)^(1/2)+6/5*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2/21*a*(7*A+5*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli
pticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

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Rubi [A]
time = 0.15, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {3039, 4081, 3872, 3854, 3856, 2719, 2720} \begin {gather*} \frac {2 a (A+B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a (7 A+5 B) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {2 a (7 A+5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {6 a (A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + a*Cos[c + d*x])*(A + B*Cos[c + d*x]))/Sec[c + d*x]^(3/2),x]

[Out]

(6*a*(A + B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*a*(7*A + 5*B)*Sqrt[Co
s[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a*B*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2
)) + (2*a*(A + B)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*a*(7*A + 5*B)*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d
*x]])

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 3039

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4081

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \frac {(a+a \cos (c+d x)) (A+B \cos (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx &=\int \frac {(a+a \sec (c+d x)) (B+A \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 a B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {2}{7} \int \frac {-\frac {7}{2} a (A+B)-\frac {1}{2} a (7 A+5 B) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+(a (A+B)) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx+\frac {1}{7} (a (7 A+5 B)) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a (A+B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a (7 A+5 B) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {1}{5} (3 a (A+B)) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{21} (a (7 A+5 B)) \int \sqrt {\sec (c+d x)} \, dx\\ &=\frac {2 a B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a (A+B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a (7 A+5 B) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}+\frac {1}{5} \left (3 a (A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} \left (a (7 A+5 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {6 a (A+B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (7 A+5 B) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 a B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a (A+B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a (7 A+5 B) \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 2.36, size = 182, normalized size = 1.06 \begin {gather*} \frac {a e^{-i d x} \sqrt {\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (20 (7 A+5 B) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-84 i (A+B) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+\cos (c+d x) (252 i (A+B)+5 (28 A+23 B) \sin (c+d x)+42 (A+B) \sin (2 (c+d x))+15 B \sin (3 (c+d x)))\right )}{210 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Cos[c + d*x])*(A + B*Cos[c + d*x]))/Sec[c + d*x]^(3/2),x]

[Out]

(a*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*(20*(7*A + 5*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] - (
84*I)*(A + B)*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*
x))] + Cos[c + d*x]*((252*I)*(A + B) + 5*(28*A + 23*B)*Sin[c + d*x] + 42*(A + B)*Sin[2*(c + d*x)] + 15*B*Sin[3
*(c + d*x)])))/(210*d*E^(I*d*x))

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Maple [A]
time = 0.36, size = 383, normalized size = 2.23

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a \left (240 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-168 A -528 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (308 A +448 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-112 A -122 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+35 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+25 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(383\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))*(A+B*cos(d*x+c))/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(240*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
8+(-168*A-528*B)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(308*A+448*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)
+(-112*A-122*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+35*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))-63*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))+25*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))-63*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)*El
lipticE(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

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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))/sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)/sec(d*x + c)^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 187, normalized size = 1.09 \begin {gather*} \frac {-5 i \, \sqrt {2} {\left (7 \, A + 5 \, B\right )} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (7 \, A + 5 \, B\right )} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 63 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 ) - 63 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 ) + \frac {2 \, {\left (15 \, B a \cos \left (d x + c\right )^{3} + 21 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{2} + 5 \, {\left (7 \, A + 5 \, B\right )} a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, 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))/sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

1/105*(-5*I*sqrt(2)*(7*A + 5*B)*a*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*(7*A
+ 5*B)*a*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 63*I*sqrt(2)*(A + B)*a*weierstrassZeta(-
4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 63*I*sqrt(2)*(A + B)*a*weierstrassZeta(-4,
0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(15*B*a*cos(d*x + c)^3 + 21*(A + B)*a*cos(d*
x + c)^2 + 5*(7*A + 5*B)*a*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d

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

[Out]

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

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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))/sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)/sec(d*x + c)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,\left (a+a\,\cos \left (c+d\,x\right )\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \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)))/(1/cos(c + d*x))^(3/2),x)

[Out]

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

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