3.874 \(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=107 \[ \frac {A \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \sqrt {b \cos (c+d x)}}+\frac {B x \sqrt {\cos (c+d x)}}{2 b \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \sqrt {b \cos (c+d x)}} \]

[Out]

1/2*B*cos(d*x+c)^(3/2)*sin(d*x+c)/b/d/(b*cos(d*x+c))^(1/2)+1/2*B*x*cos(d*x+c)^(1/2)/b/(b*cos(d*x+c))^(1/2)+A*s
in(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 = 0.02, antiderivative size = 107, 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, 2734} \[ \frac {A \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \sqrt {b \cos (c+d x)}}+\frac {B x \sqrt {\cos (c+d x)}}{2 b \sqrt {b \cos (c+d x)}}+\frac {B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x]))/(b*Cos[c + d*x])^(3/2),x]

[Out]

(B*x*Sqrt[Cos[c + d*x]])/(2*b*Sqrt[b*Cos[c + d*x]]) + (A*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(b*d*Sqrt[b*Cos[c +
d*x]]) + (B*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*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 2734

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 {5}{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 \cos (c+d x) (A+B \cos (c+d x)) \, dx}{b \sqrt {b \cos (c+d x)}}\\ &=\frac {B x \sqrt {\cos (c+d x)}}{2 b \sqrt {b \cos (c+d x)}}+\frac {A \sqrt {\cos (c+d x)} \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}+\frac {B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 b d \sqrt {b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 57, normalized size = 0.53 \[ \frac {\cos ^{\frac {3}{2}}(c+d x) (4 A \sin (c+d x)+B (2 (c+d x)+\sin (2 (c+d x))))}{4 d (b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x]))/(b*Cos[c + d*x])^(3/2),x]

[Out]

(Cos[c + d*x]^(3/2)*(4*A*Sin[c + d*x] + B*(2*(c + d*x) + Sin[2*(c + d*x)])))/(4*d*(b*Cos[c + d*x])^(3/2))

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fricas [A]  time = 1.16, size = 210, normalized size = 1.96 \[ \left [-\frac {B \sqrt {-b} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) - 2 \, {\left (B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, b^{2} d \cos \left (d x + c\right )}, \frac {B \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (B \cos \left (d x + c\right ) + 2 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{2} d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(3/2),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^2*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^2*d*cos(d*x + c))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(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)^(5/2)/(b*cos(d*x + c))^(3/2), x)

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maple [A]  time = 0.17, size = 55, normalized size = 0.51 \[ \frac {\left (\cos ^{\frac {3}{2}}\left (d x +c \right )\right ) \left (B \cos \left (d x +c \right ) \sin \left (d x +c \right )+2 A \sin \left (d x +c \right )+B \left (d x +c \right )\right )}{2 d \left (b \cos \left (d x +c \right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(3/2),x)

[Out]

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

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maxima [A]  time = 0.68, size = 40, normalized size = 0.37 \[ \frac {\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B}{b^{\frac {3}{2}}} + \frac {4 \, A \sin \left (d x + c\right )}{b^{\frac {3}{2}}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*B/b^(3/2) + 4*A*sin(d*x + c)/b^(3/2))/d

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mupad [B]  time = 0.69, size = 82, normalized size = 0.77 \[ \frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (B\,\sin \left (c+d\,x\right )+4\,A\,\sin \left (2\,c+2\,d\,x\right )+B\,\sin \left (3\,c+3\,d\,x\right )+4\,B\,d\,x\,\cos \left (c+d\,x\right )\right )}{4\,b^2\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^(5/2)*(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(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^2*d*(cos(2*c + 2*d*x) + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(5/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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