3.355 \(\int \frac {(a+b \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx\)

Optimal. Leaf size=140 \[ \frac {2 \left (3 a^2 A+2 a b B+A b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 \left (5 a (a B+2 A b)+3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sqrt {\cos (c+d x)}}{15 d}+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}{5 d} \]

[Out]

2/5*(3*b^2*B+5*a*(2*A*b+B*a))*(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*(3*A*a^2+A*b^2+2*B*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+2/15*b*(5*A*b+7*B*a)*sin(d*x+c)*cos(d*x+c)^(1/2)/d+2/5*b*B*(a+b*cos(d*x+c))*sin(d*x+c)*cos(d*x+c
)^(1/2)/d

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Rubi [A]  time = 0.27, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2990, 3023, 2748, 2641, 2639} \[ \frac {2 \left (3 a^2 A+2 a b B+A b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 \left (5 a (a B+2 A b)+3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b (7 a B+5 A b) \sin (c+d x) \sqrt {\cos (c+d x)}}{15 d}+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*b^2*B + 5*a*(2*A*b + a*B))*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(3*a^2*A + A*b^2 + 2*a*b*B)*EllipticF[(
c + d*x)/2, 2])/(3*d) + (2*b*(5*A*b + 7*a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*b*B*Sqrt[Cos[c + d*x
]]*(a + b*Cos[c + d*x])*Sin[c + d*x])/(5*d)

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]

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 2990

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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x
])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c -
b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m,
1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3023

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+b \cos (c+d x))^2 (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx &=\frac {2 b B \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \sin (c+d x)}{5 d}+\frac {2}{5} \int \frac {\frac {1}{2} a (5 a A+b B)+\frac {1}{2} \left (3 b^2 B+5 a (2 A b+a B)\right ) \cos (c+d x)+\frac {1}{2} b (5 A b+7 a B) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 b (5 A b+7 a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 b B \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \sin (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {5}{4} \left (3 a^2 A+A b^2+2 a b B\right )+\frac {3}{4} \left (3 b^2 B+5 a (2 A b+a B)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 b (5 A b+7 a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 b B \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \sin (c+d x)}{5 d}+\frac {1}{3} \left (3 a^2 A+A b^2+2 a b B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{5} \left (3 b^2 B+5 a (2 A b+a B)\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {2 \left (3 b^2 B+5 a (2 A b+a B)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (3 a^2 A+A b^2+2 a b B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {2 b (5 A b+7 a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 b B \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \sin (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 106, normalized size = 0.76 \[ \frac {2 \left (5 \left (3 a^2 A+2 a b B+A b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+3 \left (5 a^2 B+10 a A b+3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+b \sin (c+d x) \sqrt {\cos (c+d x)} (10 a B+5 A b+3 b B \cos (c+d x))\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*(10*a*A*b + 5*a^2*B + 3*b^2*B)*EllipticE[(c + d*x)/2, 2] + 5*(3*a^2*A + A*b^2 + 2*a*b*B)*EllipticF[(c +
d*x)/2, 2] + b*Sqrt[Cos[c + d*x]]*(5*A*b + 10*a*B + 3*b*B*Cos[c + d*x])*Sin[c + d*x]))/(15*d)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B b^{2} \cos \left (d x + c\right )^{3} + A a^{2} + {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b^2*cos(d*x + c)^3 + A*a^2 + (2*B*a*b + A*b^2)*cos(d*x + c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c))/sq
rt(cos(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.19, size = 487, normalized size = 3.48 \[ -\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 )}\, \left (-24 B \,b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (20 A \,b^{2}+40 B a b +24 b^{2} 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 (-10 A \,b^{2}-20 B a b -6 b^{2} 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 )+15 a^{2} 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 )+5 A \,b^{2} \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 )-30 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 ) a b +10 B a 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 )-15 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 ) a^{2}-9 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 ) b^{2}\right )}{15 \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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*B*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)
^6+(20*A*b^2+40*B*a*b+24*B*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*A*b^2-20*B*a*b-6*B*b^2)*sin(1/2*d
*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+15*a^2*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)*Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2))+5*A*b^2*(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))-30*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))*a*b+10*B*a*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(co
s(1/2*d*x+1/2*c),2^(1/2))-15*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))*a^2-9*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))*b^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 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.34, size = 177, normalized size = 1.26 \[ \frac {A\,b^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,A\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a\,b\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {4\,A\,a\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,B\,b^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*b^2*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (2*A*a^2*ellipticF(c/2
+ (d*x)/2, 2))/d + (2*B*a^2*ellipticE(c/2 + (d*x)/2, 2))/d + (2*B*a*b*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3
+ (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (4*A*a*b*ellipticE(c/2 + (d*x)/2, 2))/d - (2*B*b^2*cos(c + d*x)^(7/2
)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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