3.9.53 \(\int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [853]

Optimal. Leaf size=61 \[ \frac {A b x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {b B \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 61, 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 b x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {b B \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.07, size = 42, normalized size = 0.69 \begin {gather*} \frac {(b \cos (c+d x))^{3/2} (A (c+d x)+B \sin (c+d x))}{d \cos ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.19, size = 39, normalized size = 0.64

method	result	size
default	\(\frac {\left (b \cos \left (d x +c \right )\right )^{\frac {3}{2}} \left (A \left (d x +c \right )+B \sin \left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{\frac {3}{2}}}\)	\(39\)
risch	\(\frac {A b x \sqrt {b \cos \left (d x +c \right )}}{\sqrt {\cos \left (d x +c \right )}}+\frac {b B \sin \left (d x +c \right ) \sqrt {b \cos \left (d x +c \right )}}{d \sqrt {\cos \left (d x +c \right )}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.55, size = 40, normalized size = 0.66 \begin {gather*} \frac {2 \, A b^{\frac {3}{2}} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + B b^{\frac {3}{2}} \sin \left (d x + c\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 184, normalized size = 3.02 \begin {gather*} \left [\frac {A \sqrt {-b} 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 b \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )}, \frac {A b^{\frac {3}{2}} \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 b \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(A*sqrt(-b)*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))*si
n(d*x + c) - b) + 2*sqrt(b*cos(d*x + c))*B*b*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), (A*b^(3/2)*arc
tan(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*b*sq
rt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]

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Sympy [A]
time = 43.34, size = 80, normalized size = 1.31 \begin {gather*} \begin {cases} \frac {A x \left (b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}} + \frac {B \left (b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}} \sin {\left (c + d x \right )}}{d \cos ^{\frac {3}{2}}{\left (c + d x \right )}} & \text {for}\: d \neq 0 \\\frac {x \left (b \cos {\left (c \right )}\right )^{\frac {3}{2}} \left (A + B \cos {\left (c \right )}\right )}{\cos ^{\frac {3}{2}}{\left (c \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [B]
time = 0.85, size = 36, normalized size = 0.59 \begin {gather*} \frac {b\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (B\,\sin \left (c+d\,x\right )+A\,d\,x\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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