3.9.67 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{\sqrt {b \cos (c+d x)}} \, dx\) [867]

Optimal. Leaf size=98 \[ \frac {B x \sqrt {\cos (c+d x)}}{2 \sqrt {b \cos (c+d x)}}+\frac {A \sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {b \cos (c+d x)}}+\frac {B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {b \cos (c+d x)}} \]

[Out]

1/2*B*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(b*cos(d*x+c))^(1/2)+1/2*B*x*cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(1/2)+A*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 = 0.02, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {17, 2813} \begin {gather*} \frac {A \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {b \cos (c+d x)}}+\frac {B x \sqrt {\cos (c+d x)}}{2 \sqrt {b \cos (c+d x)}}+\frac {B \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[(Cos[c + d*x]^(3/2)*(A + B*Cos[c + d*x]))/Sqrt[b*Cos[c + d*x]],x]

[Out]

(B*x*Sqrt[Cos[c + d*x]])/(2*Sqrt[b*Cos[c + d*x]]) + (A*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[b*Cos[c + d*x]
]) + (B*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*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 2813

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*a*c +
b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Cos[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{\sqrt {b \cos (c+d x)}} \, dx &=\frac {\sqrt {\cos (c+d x)} \int \cos (c+d x) (A+B \cos (c+d x)) \, dx}{\sqrt {b \cos (c+d x)}}\\ &=\frac {B x \sqrt {\cos (c+d x)}}{2 \sqrt {b \cos (c+d x)}}+\frac {A \sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {b \cos (c+d x)}}+\frac {B \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 = 0.10, size = 57, normalized size = 0.58 \begin {gather*} \frac {\sqrt {\cos (c+d x)} (4 A \sin (c+d x)+B (2 (c+d x)+\sin (2 (c+d x))))}{4 d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^(3/2)*(A + B*Cos[c + d*x]))/Sqrt[b*Cos[c + d*x]],x]

[Out]

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

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Maple [A]
time = 0.26, size = 55, normalized size = 0.56

method	result	size
default	\(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (B \sin \left (d x +c \right ) \cos \left (d x +c \right )+2 A \sin \left (d x +c \right )+B \left (d x +c \right )\right )}{2 d \sqrt {b \cos \left (d x +c \right )}}\)	\(55\)
risch	\(\frac {B x \left (\sqrt {\cos }\left (d x +c \right )\right )}{2 \sqrt {b \cos \left (d x +c \right )}}+\frac {A \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{d \sqrt {b \cos \left (d x +c \right )}}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) B \sin \left (2 d x +2 c \right )}{4 \sqrt {b \cos \left (d x +c \right )}\, d}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.61, size = 40, normalized size = 0.41 \begin {gather*} \frac {\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B}{\sqrt {b}} + \frac {4 \, A \sin \left (d x + c\right )}{\sqrt {b}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*B/sqrt(b) + 4*A*sin(d*x + c)/sqrt(b))/d

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Fricas [A]
time = 0.46, size = 210, normalized size = 2.14 \begin {gather*} \left [-\frac {B \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 \, {\left (B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, b d \cos \left (d x + c\right )}, \frac {B \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 ) + {\left (B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b d \cos \left (d x + c\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 51.80, size = 151, normalized size = 1.54 \begin {gather*} \begin {cases} \frac {A \sin {\left (c + d x \right )} \sqrt {\cos {\left (c + d x \right )}}}{d \sqrt {b \cos {\left (c + d x \right )}}} + \frac {B 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 {B x \cos ^{\frac {5}{2}}{\left (c + d x \right )}}{2 \sqrt {b \cos {\left (c + d x \right )}}} + \frac {B \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 + B \cos {\left (c \right )}\right ) \cos ^{\frac {3}{2}}{\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(cos(d*x+c)**(3/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))**(1/2),x)

[Out]

Piecewise((A*sin(c + d*x)*sqrt(cos(c + d*x))/(d*sqrt(b*cos(c + d*x))) + B*x*sin(c + d*x)**2*sqrt(cos(c + d*x))
/(2*sqrt(b*cos(c + d*x))) + B*x*cos(c + d*x)**(5/2)/(2*sqrt(b*cos(c + d*x))) + B*sin(c + d*x)*cos(c + d*x)**(3
/2)/(2*d*sqrt(b*cos(c + d*x))), Ne(d, 0)), (x*(A + B*cos(c))*cos(c)**(3/2)/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(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^(3/2)/sqrt(b*cos(d*x + c)), x)

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Mupad [B]
time = 1.37, size = 82, normalized size = 0.84 \begin {gather*} \frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (B\,\sin \left (c+d\,x\right )+4\,A\,\sin \left (2\,c+2\,d\,x\right )+B\,\sin \left (3\,c+3\,d\,x\right )+4\,B\,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)^(3/2)*(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(c + d*x) + 4*A*sin(2*c + 2*d*x) + B*sin(3*c + 3*d*x) + 4*B*d
*x*cos(c + d*x)))/(4*b*d*(cos(2*c + 2*d*x) + 1))

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