3.1061 \(\int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=167 \[ -\frac {2 C \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (3 b B-2 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 b C \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3015, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 C \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (3 b B-2 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 b C \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2753

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rule 3015

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

Rubi steps

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

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Mathematica [A]  time = 0.65, size = 144, normalized size = 0.86 \[ \frac {-2 C \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-2 (a+b) (2 a C-3 b B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+2 b C \sin (c+d x) (a+b \cos (c+d x))}{3 d \sqrt {a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b \cos \left (d x + c\right ) - C a + B b\right )} \sqrt {b \cos \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b*cos(d*x + c) - C*a + B*b)*sqrt(b*cos(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 3.11, size = 598, normalized size = 3.58 \[ -\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 C \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}+3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a b -3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) b^{2}+2 C \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a b -6 C \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}-C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{2}+C \,b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )-2 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{2}+2 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a b -2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a b +2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*C*cos(1/2*d*x+1/2*c)^5*b^2+3*B*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)
)*a*b-3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*
c),(-2*b/(a-b))^(1/2))*b^2+2*C*a*b*cos(1/2*d*x+1/2*c)^3-6*C*cos(1/2*d*x+1/2*c)^3*b^2-C*(sin(1/2*d*x+1/2*c)^2)^
(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2+C*b^2*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(
a-b))^(1/2))-2*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d
*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2+2*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2
)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b-2*C*cos(1/2*d*x+1/2*c)*a*b+2*C*cos(1/2*d*x+1/2*c)*b^2)/
(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b
)^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.58, size = 303, normalized size = 1.81 \[ \frac {2\,C\,b\,\sin \left (c+d\,x\right )\,\sqrt {a+b\,\cos \left (c+d\,x\right )}}{3\,d}+\frac {2\,C\,\sqrt {\frac {a+b\,\cos \left (c+d\,x\right )}{a+b}}\,\left (\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |\frac {2\,b}{a+b}\right )\,\left (2\,a^2+b^2\right )-2\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |\frac {2\,b}{a+b}\right )\,\left (a+b\right )\right )}{3\,d\,\sqrt {a+b\,\cos \left (c+d\,x\right )}}-\frac {2\,C\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |\frac {2\,b}{a+b}\right )\,\sqrt {\frac {a+b\,\cos \left (c+d\,x\right )}{a+b}}}{d\,\sqrt {a+b\,\cos \left (c+d\,x\right )}}+\frac {2\,B\,b\,\left (\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |\frac {2\,b}{a+b}\right )\,\left (a+b\right )-a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |\frac {2\,b}{a+b}\right )\right )\,\sqrt {\frac {a+b\,\cos \left (c+d\,x\right )}{a+b}}}{d\,\sqrt {a+b\,\cos \left (c+d\,x\right )}}+\frac {2\,B\,a\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |\frac {2\,b}{a+b}\right )\,\sqrt {\frac {a+b\,\cos \left (c+d\,x\right )}{a+b}}}{d\,\sqrt {a+b\,\cos \left (c+d\,x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*C*b*sin(c + d*x)*(a + b*cos(c + d*x))^(1/2))/(3*d) + (2*C*((a + b*cos(c + d*x))/(a + b))^(1/2)*(ellipticF(c
/2 + (d*x)/2, (2*b)/(a + b))*(2*a^2 + b^2) - 2*a*ellipticE(c/2 + (d*x)/2, (2*b)/(a + b))*(a + b)))/(3*d*(a + b
*cos(c + d*x))^(1/2)) - (2*C*a^2*ellipticF(c/2 + (d*x)/2, (2*b)/(a + b))*((a + b*cos(c + d*x))/(a + b))^(1/2))
/(d*(a + b*cos(c + d*x))^(1/2)) + (2*B*b*(ellipticE(c/2 + (d*x)/2, (2*b)/(a + b))*(a + b) - a*ellipticF(c/2 +
(d*x)/2, (2*b)/(a + b)))*((a + b*cos(c + d*x))/(a + b))^(1/2))/(d*(a + b*cos(c + d*x))^(1/2)) + (2*B*a*b*ellip
ticF(c/2 + (d*x)/2, (2*b)/(a + b))*((a + b*cos(c + d*x))/(a + b))^(1/2))/(d*(a + b*cos(c + d*x))^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- B b \sqrt {a + b \cos {\left (c + d x \right )}}\right )\, dx - \int C a \sqrt {a + b \cos {\left (c + d x \right )}}\, dx - \int \left (- C b \sqrt {a + b \cos {\left (c + d x \right )}} \cos {\left (c + d x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-B*b*sqrt(a + b*cos(c + d*x)), x) - Integral(C*a*sqrt(a + b*cos(c + d*x)), x) - Integral(-C*b*sqrt(a
+ b*cos(c + d*x))*cos(c + d*x), x)

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