3.2.18 \(\int \frac {\sqrt {\cos (c+d x)} (A+C \cos ^2(c+d x))}{\sqrt {b \cos (c+d x)}} \, dx\) [118]

Optimal. Leaf size=90 \[ \frac {A x \sqrt {\cos (c+d x)}}{\sqrt {b \cos (c+d x)}}+\frac {C x \sqrt {\cos (c+d x)}}{2 \sqrt {b \cos (c+d x)}}+\frac {C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {b \cos (c+d x)}} \]

[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)

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Rubi [A]
time = 0.02, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {17, 2715, 8} \begin {gather*} \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 {C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[Cos[c + d*x]]*(A + 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]]) + (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 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*} \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx &=\frac {\sqrt {\cos (c+d x)} \int \left (A+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 (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 {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 {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 = 0.07, size = 52, normalized size = 0.58 \begin {gather*} \frac {\sqrt {\cos (c+d x)} (2 (2 A+C) (c+d x)+C \sin (2 (c+d x)))}{4 d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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Maple [A]
time = 0.32, size = 54, normalized size = 0.60

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.59, size = 52, normalized size = 0.58 \begin {gather*} \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}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+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))/d

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Fricas [A]
time = 0.42, size = 169, normalized size = 1.88 \begin {gather*} \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 \, b 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 \, b d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [A]
time = 23.83, size = 146, normalized size = 1.62 \begin {gather*} \begin {cases} \frac {A x \sqrt {\cos {\left (c + d x \right )}}}{\sqrt {b \cos {\left (c + d x \right )}}} + \frac {C x \sin ^{2}{\left (c + d x \right )} \sqrt {\cos {\left (c + d x \right )}}}{2 \sqrt {b \cos {\left (c + d x \right )}}} + \frac {C x \cos ^{\frac {5}{2}}{\left (c + d x \right )}}{2 \sqrt {b \cos {\left (c + d x \right )}}} + \frac {C \sin {\left (c + d x \right )} \cos ^{\frac {3}{2}}{\left (c + d x \right )}}{2 d \sqrt {b \cos {\left (c + d x \right )}}} & \text {for}\: d \neq 0 \\\frac {x \left (A + C \cos ^{2}{\left (c \right )}\right ) \sqrt {\cos {\left (c \right )}}}{\sqrt {b \cos {\left (c \right )}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((A*x*sqrt(cos(c + d*x))/sqrt(b*cos(c + d*x)) + C*x*sin(c + d*x)**2*sqrt(cos(c + d*x))/(2*sqrt(b*cos(
c + d*x))) + C*x*cos(c + d*x)**(5/2)/(2*sqrt(b*cos(c + d*x))) + C*sin(c + d*x)*cos(c + d*x)**(3/2)/(2*d*sqrt(b
*cos(c + d*x))), Ne(d, 0)), (x*(A + C*cos(c)**2)*sqrt(cos(c))/sqrt(b*cos(c)), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [B]
time = 1.49, size = 81, normalized size = 0.90 \begin {gather*} \frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (C\,\sin \left (c+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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^(1/2)*(A + 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) + 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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