\(\int \frac {\sqrt {b \cos (c+d x)} (A+C \cos ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [92]

Optimal result

Integrand size = 35, antiderivative size = 90 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {A x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {C x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {C \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {17, 2715, 8} \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {A x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {C x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{2 d} \]

[In]

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

[Out]

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

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 2715

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] && Integ
erQ[2*n]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.58 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {\sqrt {b \cos (c+d x)} (2 (2 A+C) (c+d x)+C \sin (2 (c+d x)))}{4 d \sqrt {\cos (c+d x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 6.90 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {\sqrt {\cos \left (d x +c \right ) b}\, \left (C \cos \left (d x +c \right ) \sin \left (d x +c \right )+2 A \left (d x +c \right )+C \left (d x +c \right )\right )}{2 d \sqrt {\cos \left (d x +c \right )}}\)	\(54\)
risch	\(\frac {\sqrt {\cos \left (d x +c \right ) b}\, x \left (4 A +2 C \right )}{4 \sqrt {\cos \left (d x +c \right )}}+\frac {\sqrt {\cos \left (d x +c \right ) b}\, C \sin \left (2 d x +2 c \right )}{4 \sqrt {\cos \left (d x +c \right )}\, d}\)	\(63\)
parts	\(\frac {C \sqrt {\cos \left (d x +c \right ) b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \sqrt {\cos \left (d x +c \right )}}+\frac {A \sqrt {\cos \left (d x +c \right ) b}\, \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}}\)	\(72\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.80 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\left [\frac {2 \, \sqrt {b \cos \left (d x + c\right )} C \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (2 \, A + C\right )} \sqrt {-b} \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{4 \, d}, \frac {\sqrt {b \cos \left (d x + c\right )} C \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (2 \, A + C\right )} \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right )}{2 \, d}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 13.72 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.62 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\begin {cases} \frac {A x \sqrt {b \cos {\left (c + d x \right )}}}{\sqrt {\cos {\left (c + d x \right )}}} + \frac {C x \sqrt {b \cos {\left (c + d x \right )}} \sin ^{2}{\left (c + d x \right )}}{2 \sqrt {\cos {\left (c + d x \right )}}} + \frac {C x \sqrt {b \cos {\left (c + d x \right )}} \cos ^{\frac {3}{2}}{\left (c + d x \right )}}{2} + \frac {C \sqrt {b \cos {\left (c + d x \right )}} \sin {\left (c + d x \right )} \sqrt {\cos {\left (c + d x \right )}}}{2 d} & \text {for}\: d \neq 0 \\\frac {x \sqrt {b \cos {\left (c \right )}} \left (A + C \cos ^{2}{\left (c \right )}\right )}{\sqrt {\cos {\left (c \right )}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((A*x*sqrt(b*cos(c + d*x))/sqrt(cos(c + d*x)) + C*x*sqrt(b*cos(c + d*x))*sin(c + d*x)**2/(2*sqrt(cos(
c + d*x))) + C*x*sqrt(b*cos(c + d*x))*cos(c + d*x)**(3/2)/2 + C*sqrt(b*cos(c + d*x))*sin(c + d*x)*sqrt(cos(c +
d*x))/(2*d), Ne(d, 0)), (x*sqrt(b*cos(c))*(A + C*cos(c)**2)/sqrt(cos(c)), True))

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.58 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C \sqrt {b} + 8 \, A \sqrt {b} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{4 \, d} \]

[In]

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

Giac [F]

\[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \cos \left (d x + c\right )}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.50 \[ \int \frac {\sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (C\,\sin \left (2\,c+2\,d\,x\right )+4\,A\,d\,x+2\,C\,d\,x\right )}{4\,d\,\sqrt {\cos \left (c+d\,x\right )}} \]

[In]

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

[Out]

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