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

Optimal. Leaf size=123 \[ \frac {A x \sqrt {\cos (c+d x)}}{\sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {b \cos (c+d x)}}+\frac {C x \sqrt {\cos (c+d x)}}{2 \sqrt {b \cos (c+d x)}}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {17, 2637, 2635, 8} \[ \frac {A x \sqrt {\cos (c+d x)}}{\sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {b \cos (c+d x)}}+\frac {C x \sqrt {\cos (c+d x)}}{2 \sqrt {b \cos (c+d x)}}+\frac {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*x*Sqrt[Cos[c + d*x]])/Sqrt[b*Cos[c + d*x]] + (C*x*Sqrt[Cos[c + d*x]])/(2*Sqrt[b*Cos[c + d*x]]) + (B*Sqrt[Co
s[c + d*x]]*Sin[c + d*x])/(d*Sqrt[b*Cos[c + d*x]]) + (C*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[b*Cos[c + d
*x]])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 17

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m + 1/2)*b^(n - 1/2)*Sqrt[b*v])/Sqrt[a*v]
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]

Rule 2635

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 61, normalized size = 0.50 \[ \frac {\sqrt {\cos (c+d x)} (2 (2 A+C) (c+d x)+4 B \sin (c+d x)+C \sin (2 (c+d x)))}{4 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[c + d*x]]*(2*(2*A + C)*(c + d*x) + 4*B*Sin[c + d*x] + C*Sin[2*(c + d*x)]))/(4*d*Sqrt[b*Cos[c + d*x]]
)

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fricas [A]  time = 0.66, size = 218, normalized size = 1.77 \[ \left [-\frac {{\left (2 \, A + C\right )} \sqrt {-b} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) - 2 \, {\left (C \cos \left (d x + c\right ) + 2 \, B\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, b d \cos \left (d x + c\right )}, \frac {{\left (2 \, A + C\right )} \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (C \cos \left (d x + c\right ) + 2 \, B\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((2*A + C)*sqrt(-b)*cos(d*x + c)*log(2*b*cos(d*x + c)^2 + 2*sqrt(b*cos(d*x + c))*sqrt(-b)*sqrt(cos(d*x +
c))*sin(d*x + c) - b) - 2*(C*cos(d*x + c) + 2*B)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b*d*c
os(d*x + c)), 1/2*((2*A + C)*sqrt(b)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*co
s(d*x + c) + (C*cos(d*x + c) + 2*B)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b*d*cos(d*x + c))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.37, size = 63, normalized size = 0.51 \[ \frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (C \sin \left (d x +c \right ) \cos \left (d x +c \right )+2 A \left (d x +c \right )+2 B \sin \left (d x +c \right )+C \left (d x +c \right )\right )}{2 d \sqrt {b \cos \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.64, size = 64, normalized size = 0.52 \[ \frac {\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C}{\sqrt {b}} + \frac {8 \, A \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{\sqrt {b}} + \frac {4 \, B \sin \left (d x + c\right )}{\sqrt {b}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*C/sqrt(b) + 8*A*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/sqrt(b) + 4*B*si
n(d*x + c)/sqrt(b))/d

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mupad [B]  time = 1.01, size = 93, normalized size = 0.76 \[ \frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (C\,\sin \left (c+d\,x\right )+4\,B\,\sin \left (2\,c+2\,d\,x\right )+C\,\sin \left (3\,c+3\,d\,x\right )+8\,A\,d\,x\,\cos \left (c+d\,x\right )+4\,C\,d\,x\,\cos \left (c+d\,x\right )\right )}{4\,b\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(C*sin(c + d*x) + 4*B*sin(2*c + 2*d*x) + C*sin(3*c + 3*d*x) + 8*A*d
*x*cos(c + d*x) + 4*C*d*x*cos(c + d*x)))/(4*b*d*(cos(2*c + 2*d*x) + 1))

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

[Out]

Timed out

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