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

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

[Out]

A*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 = 0.01, antiderivative size = 59, normalized size of antiderivative = 1.00, 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} \[ \frac {A x \sqrt {\cos (c+d x)}}{\sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.05, size = 42, normalized size = 0.71 \[ \frac {\sqrt {\cos (c+d x)} (A (c+d x)+B \sin (c+d x))}{d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.93, size = 187, normalized size = 3.17 \[ \left [-\frac {A \sqrt {-b} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) - 2 \, \sqrt {b \cos \left (d x + c\right )} B \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b d \cos \left (d x + c\right )}, \frac {A \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + \sqrt {b \cos \left (d x + c\right )} B \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((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 = 0.18, size = 39, normalized size = 0.66 \[ \frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (A \left (d x +c \right )+B \sin \left (d x +c \right )\right )}{d \sqrt {b \cos \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 40, normalized size = 0.68 \[ \frac {\frac {2 \, A \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{\sqrt {b}} + \frac {B \sin \left (d x + c\right )}{\sqrt {b}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.54, size = 61, normalized size = 1.03 \[ \frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (B\,\sin \left (2\,c+2\,d\,x\right )+2\,A\,d\,x\,\cos \left (c+d\,x\right )\right )}{b\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 12.68, size = 46, normalized size = 0.78 \[ \begin {cases} \frac {A x}{\sqrt {b}} + \frac {B \sin {\left (c + d x \right )}}{\sqrt {b} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\relax (c )}\right ) \sqrt {\cos {\relax (c )}}}{\sqrt {b \cos {\relax (c )}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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