3.9.75 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx\) [875]

Optimal. Leaf size=65 \[ \frac {A x \sqrt {\cos (c+d x)}}{b \sqrt {b \cos (c+d x)}}+\frac {B \sqrt {\cos (c+d x)} \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 65, 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, 2717} \begin {gather*} \frac {A x \sqrt {\cos (c+d x)}}{b \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x) \sqrt {\cos (c+d x)}}{b 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]))/(b*Cos[c + d*x])^(3/2),x]

[Out]

(A*x*Sqrt[Cos[c + d*x]])/(b*Sqrt[b*Cos[c + d*x]]) + (B*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(b*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 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*} \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx &=\frac {\sqrt {\cos (c+d x)} \int (A+B \cos (c+d x)) \, dx}{b \sqrt {b \cos (c+d x)}}\\ &=\frac {A x \sqrt {\cos (c+d x)}}{b \sqrt {b \cos (c+d x)}}+\frac {\left (B \sqrt {\cos (c+d x)}\right ) \int \cos (c+d x) \, dx}{b \sqrt {b \cos (c+d x)}}\\ &=\frac {A x \sqrt {\cos (c+d x)}}{b \sqrt {b \cos (c+d x)}}+\frac {B \sqrt {\cos (c+d x)} \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 42, normalized size = 0.65 \begin {gather*} \frac {\cos ^{\frac {3}{2}}(c+d x) (A (c+d x)+B \sin (c+d x))}{d (b \cos (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.20, size = 39, normalized size = 0.60

method	result	size
default	\(\frac {\left (\cos ^{\frac {3}{2}}\left (d x +c \right )\right ) \left (A \left (d x +c \right )+B \sin \left (d x +c \right )\right )}{d \left (b \cos \left (d x +c \right )\right )^{\frac {3}{2}}}\)	\(39\)
risch	\(\frac {A x \left (\sqrt {\cos }\left (d x +c \right )\right )}{b \sqrt {b \cos \left (d x +c \right )}}+\frac {B \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{b d \sqrt {b \cos \left (d x +c \right )}}\)	\(58\)

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))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/d*cos(d*x+c)^(3/2)*(A*(d*x+c)+B*sin(d*x+c))/(b*cos(d*x+c))^(3/2)

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Maxima [A]
time = 0.58, size = 40, normalized size = 0.62 \begin {gather*} \frac {\frac {2 \, A \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{b^{\frac {3}{2}}} + \frac {B \sin \left (d x + c\right )}{b^{\frac {3}{2}}}}{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))^(3/2),x, algorithm="maxima")

[Out]

(2*A*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/b^(3/2) + B*sin(d*x + c)/b^(3/2))/d

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Fricas [A]
time = 0.40, size = 187, normalized size = 2.88 \begin {gather*} \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^{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 )}{b^{2} 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))^(3/2),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^2*d*cos(d*x + c)), (A*sqrt(b)*ar
ctan(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*sqr
t(cos(d*x + c))*sin(d*x + c))/(b^2*d*cos(d*x + c))]

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Sympy [A]
time = 50.58, size = 80, normalized size = 1.23 \begin {gather*} \begin {cases} \frac {A x \cos ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}} + \frac {B \sin {\left (c + d x \right )} \cos ^{\frac {3}{2}}{\left (c + d x \right )}}{d \left (b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\left (c \right )}\right ) \cos ^{\frac {3}{2}}{\left (c \right )}}{\left (b \cos {\left (c \right )}\right )^{\frac {3}{2}}} & \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))**(3/2),x)

[Out]

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

[Out]

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

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Mupad [B]
time = 1.02, size = 61, normalized size = 0.94 \begin {gather*} \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^2\,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))^(3/2),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^2*d*(cos(2*c + 2*d*
x) + 1))

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