3.244 \(\int \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\)

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

[Out]

2/3*A*b^2*sin(d*x+c)/d/(b*cos(d*x+c))^(3/2)+2*b*B*sin(d*x+c)/d/(b*cos(d*x+c))^(1/2)+2/3*b*(A+3*C)*(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))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1
/2)-2*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))*(b*cos(d*x+c))^(
1/2)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {16, 3021, 2748, 2636, 2640, 2639, 2642, 2641} \[ \frac {2 A b^2 \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {2 b (A+3 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b B \sin (c+d x)}{d \sqrt {b \cos (c+d x)}}-\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

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

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, 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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, 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 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.40, size = 90, normalized size = 0.64 \[ \frac {2 b \left (\tan (c+d x) (A+3 B \cos (c+d x))+(A+3 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-3 B \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{3 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

(2*b*(-3*B*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + (A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2,
2] + (A + 3*B*Cos[c + d*x])*Tan[c + d*x]))/(3*d*Sqrt[b*Cos[c + d*x]])

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3*(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sec(d*x + c)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3*(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sec(d*x + c)^3, x)

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maple [B]  time = 3.29, size = 505, normalized size = 3.61 \[ \frac {2 \sqrt {b \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 (2 A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 B \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 C \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-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 )+2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\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 )+6 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 C \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 )\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}}{3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)^3/(4*sin(1/2*d*x+1/2*c)^4-4*s
in(1/2*d*x+1/2*c)^2+1)*(2*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+6*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1
/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-12*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*C*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c
)^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2*A*
cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^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)*Elli
pticE(cos(1/2*d*x+1/2*c),2^(1/2))+6*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*C*(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)))*(-2*sin(1/2*d*x+1/2*c)^4*b+sin(1/2*d*
x+1/2*c)^2*b)^(1/2)/(b*(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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3*(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sec(d*x + c)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)+C*cos(d*x+c)**2)*sec(d*x+c)**3*(b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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