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

Optimal. Leaf size=213 \[ \frac{(8 a A-4 a B-4 A b+7 b B) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 \sqrt{a} d}+\frac{(4 a B+4 A b-b B) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 d \sqrt{a \cos (c+d x)+a}}-\frac{\sqrt{2} (a-b) (A-B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{b B \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{a \cos (c+d x)+a}} \]

[Out]

((8*a*A - 4*A*b - 4*a*B + 7*b*B)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*Sqrt[a]*d) - (Sqr
t[2]*(a - b)*(A - B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(Sq
rt[a]*d) + ((4*A*b + 4*a*B - b*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*d*Sqrt[a + a*Cos[c + d*x]]) + (b*B*Cos[c
+ d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.753215, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 54, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3045, 2983, 2982, 2782, 205, 2774, 216} \[ \frac{(8 a A-4 a B-4 A b+7 b B) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 \sqrt{a} d}+\frac{(4 a B+4 A b-b B) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 d \sqrt{a \cos (c+d x)+a}}-\frac{\sqrt{2} (a-b) (A-B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{b B \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a*A - 4*A*b - 4*a*B + 7*b*B)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*Sqrt[a]*d) - (Sqr
t[2]*(a - b)*(A - B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(Sq
rt[a]*d) + ((4*A*b + 4*a*B - b*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*d*Sqrt[a + a*Cos[c + d*x]]) + (b*B*Cos[c
+ d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]])

Rule 3045

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x
])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*
d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]

Rule 2983

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

Rule 2982

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

Rule 2782

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2774

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{\sqrt{\cos (c+d x)} \left (a A+(A b+a B) \cos (c+d x)+b B \cos ^2(c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx &=\frac{b B \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\sqrt{\cos (c+d x)} \left (\frac{1}{2} a (4 a A+3 b B)+\frac{1}{2} a (4 A b+4 a B-b B) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a}\\ &=\frac{(4 A b+4 a B-b B) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d \sqrt{a+a \cos (c+d x)}}+\frac{b B \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\frac{1}{4} a^2 (4 A b+4 a B-b B)+\frac{1}{4} a^2 (8 a A-4 A b-4 a B+7 b B) \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=\frac{(4 A b+4 a B-b B) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d \sqrt{a+a \cos (c+d x)}}+\frac{b B \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)}}-((a-b) (A-B)) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx+\frac{(8 a A-4 A b-4 a B+7 b B) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx}{8 a}\\ &=\frac{(4 A b+4 a B-b B) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d \sqrt{a+a \cos (c+d x)}}+\frac{b B \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)}}+\frac{(2 a (a-b) (A-B)) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{d}-\frac{(8 a A-4 A b-4 a B+7 b B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a}}} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{4 a d}\\ &=\frac{(8 a A-4 A b-4 a B+7 b B) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{4 \sqrt{a} d}-\frac{\sqrt{2} (a-b) (A-B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{\sqrt{a} d}+\frac{(4 A b+4 a B-b B) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d \sqrt{a+a \cos (c+d x)}}+\frac{b B \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 2.92695, size = 540, normalized size = 2.54 \[ \frac{\cos \left (\frac{1}{2} (c+d x)\right ) \left (\frac{4 \sin \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} (4 a B+4 A b+2 b B \cos (c+d x)-b B)}{d}+\frac{\sqrt{2} e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (8 i \sqrt{2} (a-b) (A-B) \log \left (1+e^{i (c+d x)}\right )-i (8 a A-4 a B-4 A b+7 b B) \sinh ^{-1}\left (e^{i (c+d x)}\right )+8 i a A \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )-8 i \sqrt{2} a A \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )+8 a A d x-4 i a B \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )+8 i \sqrt{2} a B \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-4 a B d x-4 i A b \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )+8 i \sqrt{2} A b \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-4 A b d x+7 i b B \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )-8 i \sqrt{2} b B \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )+7 b B d x\right )}{d \sqrt{1+e^{2 i (c+d x)}}}\right )}{8 \sqrt{a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*((Sqrt[2]*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*(8*a*A*d*x - 4
*A*b*d*x - 4*a*B*d*x + 7*b*B*d*x - I*(8*a*A - 4*A*b - 4*a*B + 7*b*B)*ArcSinh[E^(I*(c + d*x))] + (8*I)*Sqrt[2]*
(a - b)*(A - B)*Log[1 + E^(I*(c + d*x))] + (8*I)*a*A*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]] - (4*I)*A*b*Log[1
+ Sqrt[1 + E^((2*I)*(c + d*x))]] - (4*I)*a*B*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]] + (7*I)*b*B*Log[1 + Sqrt[1
+ E^((2*I)*(c + d*x))]] - (8*I)*Sqrt[2]*a*A*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]]
+ (8*I)*Sqrt[2]*A*b*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]] + (8*I)*Sqrt[2]*a*B*Log[1
- E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]] - (8*I)*Sqrt[2]*b*B*Log[1 - E^(I*(c + d*x)) + Sqrt
[2]*Sqrt[1 + E^((2*I)*(c + d*x))]]))/(d*Sqrt[1 + E^((2*I)*(c + d*x))]) + (4*Sqrt[Cos[c + d*x]]*(4*A*b + 4*a*B
- b*B + 2*b*B*Cos[c + d*x])*Sin[(c + d*x)/2])/d))/(8*Sqrt[a*(1 + Cos[c + d*x])])

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Maple [B]  time = 0.104, size = 571, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d*(-1+cos(d*x+c))^3*(4*A*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)*b+4*B*cos(d*x+c)*(cos(d*
x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)*a+4*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)*b+2*B*cos(d*x+c)^2*(c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*b+4*B*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)*a+4*A*cos(d*x+c)
*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*a-4*A*cos(d*x+c)*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*b-4*B*
cos(d*x+c)*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*a+4*B*cos(d*x+c)*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+
c))*b-B*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*b+8*A*cos(d*x+c)*arctan(sin(d*x+c)*(cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*a-4*A*cos(d*x+c)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*
x+c))*b-4*B*cos(d*x+c)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*a+7*B*cos(d*x+c)*arctan
(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*b)*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^(3/2)/(cos(d*
x+c)/(1+cos(d*x+c)))^(5/2)/sin(d*x+c)^6/a

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B b \cos \left (d x + c\right )^{2} + A a +{\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a \cos \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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