\(\int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [845]

Optimal result

Integrand size = 33, antiderivative size = 59 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {A x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {B \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {17, 2717} \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {A x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {B \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}} \]

[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 2717

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

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {\sqrt {b \cos (c+d x)} (A (c+d x)+B \sin (c+d x))}{d \sqrt {\cos (c+d x)}} \]

[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]])

Maple [A] (verified)

Time = 4.97 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.66

method	result	size
default	\(\frac {\sqrt {\cos \left (d x +c \right ) b}\, \left (A \left (d x +c \right )+B \sin \left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right )}}\)	\(39\)
risch	\(\frac {A x \sqrt {\cos \left (d x +c \right ) b}}{\sqrt {\cos \left (d x +c \right )}}+\frac {B \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{d \sqrt {\cos \left (d x +c \right )}}\)	\(52\)
parts	\(\frac {A \sqrt {\cos \left (d x +c \right ) b}\, \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}}+\frac {B \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{d \sqrt {\cos \left (d x +c \right )}}\)	\(59\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 181, normalized size of antiderivative = 3.07 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\left [\frac {A \sqrt {-b} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} B \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )}, \frac {A \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + \sqrt {b \cos \left (d x + c\right )} B \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )}\right ] \]

[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))]

Sympy [A] (verification not implemented)

Time = 1.35 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.36 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\begin {cases} \frac {A x \sqrt {b \cos {\left (c + d x \right )}}}{\sqrt {\cos {\left (c + d x \right )}}} + \frac {B \sqrt {b \cos {\left (c + d x \right )}} \sin {\left (c + d x \right )}}{d \sqrt {\cos {\left (c + d x \right )}}} & \text {for}\: d \neq 0 \\\frac {x \sqrt {b \cos {\left (c \right )}} \left (A + B \cos {\left (c \right )}\right )}{\sqrt {\cos {\left (c \right )}}} & \text {otherwise} \end {cases} \]

[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*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))), Ne(d, 0)), (x*sqrt(b*cos(c))*(A + B*cos(c))/sqrt(cos(c)), True))

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \, A \sqrt {b} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + B \sqrt {b} \sin \left (d x + c\right )}{d} \]

[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

Giac [F]

\[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.59 \[ \int \frac {\sqrt {b \cos (c+d x)} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (B\,\sin \left (c+d\,x\right )+A\,d\,x\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}} \]

[In]

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

[Out]

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