3.845 \(\int \frac{\sqrt{b \cos (c+d x)} (A+B \cos (c+d x))}{\sqrt{\cos (c+d x)}} \, dx\)

Optimal. Leaf size=59 \[ \frac{A x \sqrt{b \cos (c+d x)}}{\sqrt{\cos (c+d x)}}+\frac{B \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.012328, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {17, 2637} \[ \frac{A x \sqrt{b \cos (c+d x)}}{\sqrt{\cos (c+d x)}}+\frac{B \sin (c+d x) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*x*Sqrt[b*Cos[c + d*x]])/Sqrt[Cos[c + d*x]] + (B*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*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 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{b \cos (c+d x)} (A+B \cos (c+d x))}{\sqrt{\cos (c+d x)}} \, dx &=\frac{\sqrt{b \cos (c+d x)} \int (A+B \cos (c+d x)) \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{A x \sqrt{b \cos (c+d x)}}{\sqrt{\cos (c+d x)}}+\frac{\left (B \sqrt{b \cos (c+d x)}\right ) \int \cos (c+d x) \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{A x \sqrt{b \cos (c+d x)}}{\sqrt{\cos (c+d x)}}+\frac{B \sqrt{b \cos (c+d x)} \sin (c+d x)}{d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0469873, size = 42, normalized size = 0.71 \[ \frac{\sqrt{b \cos (c+d x)} (A (c+d x)+B \sin (c+d x))}{d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.455, size = 39, normalized size = 0.7 \begin{align*}{\frac{A \left ( dx+c \right ) +B\sin \left ( dx+c \right ) }{d}\sqrt{b\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.99071, size = 54, normalized size = 0.92 \begin{align*} \frac{2 \, A \sqrt{b} \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + B \sqrt{b} \sin \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*A*sqrt(b)*arctan(sin(d*x + c)/(cos(d*x + c) + 1)) + B*sqrt(b)*sin(d*x + c))/d

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Fricas [A]  time = 1.61787, size = 508, normalized size = 8.61 \begin{align*} \left [\frac{A \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 \, \sqrt{b \cos \left (d x + c\right )} B \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )}, \frac{A \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 ) + \sqrt{b \cos \left (d x + c\right )} B \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(A*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*sqrt(b*cos(d*x + c))*B*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), (A*sqrt(b)*arctan(
sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c) + sqrt(b*cos(d*x + c))*B*sqrt(cos
(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]

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Sympy [A]  time = 18.2536, size = 46, normalized size = 0.78 \begin{align*} \begin{cases} A \sqrt{b} x + \frac{B \sqrt{b} \sin{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\\frac{x \sqrt{b \cos{\left (c \right )}} \left (A + B \cos{\left (c \right )}\right )}{\sqrt{\cos{\left (c \right )}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((A*sqrt(b)*x + B*sqrt(b)*sin(c + d*x)/d, Ne(d, 0)), (x*sqrt(b*cos(c))*(A + B*cos(c))/sqrt(cos(c)), T
rue))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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