∫ cos5 _ 2 (c + dx)∘ ________ b cos(c + dx) (A + B cos(c + dx)) dx

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

Optimal. Leaf size=172 \[ -\frac {A \sin ^3(c+d x) \sqrt {b \cos (c+d x)}}{3 d \sqrt {\cos (c+d x)}}+\frac {A \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {3 B x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{4 d}+\frac {3 B \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{8 d} \]

[Out]

1/4*B*cos(d*x+c)^(5/2)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+3/8*B*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+A*sin(d

*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)-1/3*A*sin(d*x+c)^3*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)+3/8*B

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

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Rubi [A] time = 0.07, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {17, 2748, 2633, 2635, 8} \[ -\frac {A \sin ^3(c+d x) \sqrt {b \cos (c+d x)}}{3 d \sqrt {\cos (c+d x)}}+\frac {A \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {3 B x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)}}{4 d}+\frac {3 B \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}{8 d} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(3*B*x*Sqrt[b*Cos[c + d*x]])/(8*Sqrt[Cos[c + d*x]]) + (A*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[Cos[c + d*

x]]) + (3*B*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(8*d) + (B*Cos[c + d*x]^(5/2)*Sqrt[b*Cos[c +

d*x]]*Sin[c + d*x])/(4*d) - (A*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x]^3)/(3*d*Sqrt[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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x

, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

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] && Integer

Q[2*n]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S

in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rubi steps

\begin {align*} \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx &=\frac {\sqrt {b \cos (c+d x)} \int \cos ^3(c+d x) (A+B \cos (c+d x)) \, dx}{\sqrt {\cos (c+d x)}}\\ &=\frac {\left (A \sqrt {b \cos (c+d x)}\right ) \int \cos ^3(c+d x) \, dx}{\sqrt {\cos (c+d x)}}+\frac {\left (B \sqrt {b \cos (c+d x)}\right ) \int \cos ^4(c+d x) \, dx}{\sqrt {\cos (c+d x)}}\\ &=\frac {B \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {\left (3 B \sqrt {b \cos (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{4 \sqrt {\cos (c+d x)}}-\frac {\left (A \sqrt {b \cos (c+d x)}\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d \sqrt {\cos (c+d x)}}\\ &=\frac {A \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {3 B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {B \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}-\frac {A \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {\left (3 B \sqrt {b \cos (c+d x)}\right ) \int 1 \, dx}{8 \sqrt {\cos (c+d x)}}\\ &=\frac {3 B x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {A \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {3 B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {B \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}-\frac {A \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [A] time = 0.17, size = 81, normalized size = 0.47 \[ \frac {\sqrt {b \cos (c+d x)} (72 A \sin (c+d x)+8 A \sin (3 (c+d x))+24 B \sin (2 (c+d x))+3 B \sin (4 (c+d x))+36 B c+36 B d x)}{96 d \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[b*Cos[c + d*x]]*(36*B*c + 36*B*d*x + 72*A*Sin[c + d*x] + 24*B*Sin[2*(c + d*x)] + 8*A*Sin[3*(c + d*x)] +

3*B*Sin[4*(c + d*x)]))/(96*d*Sqrt[Cos[c + d*x]])

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fricas [A] time = 0.84, size = 252, normalized size = 1.47 \[ \left [\frac {9 \, B \sqrt {-b} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (6 \, B \cos \left (d x + c\right )^{3} + 8 \, A \cos \left (d x + c\right )^{2} + 9 \, B \cos \left (d x + c\right ) + 16 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )}, \frac {9 \, B \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (6 \, B \cos \left (d x + c\right )^{3} + 8 \, A \cos \left (d x + c\right )^{2} + 9 \, B \cos \left (d x + c\right ) + 16 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )}\right ] \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

[1/48*(9*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))*s

in(d*x + c) - b) + 2*(6*B*cos(d*x + c)^3 + 8*A*cos(d*x + c)^2 + 9*B*cos(d*x + c) + 16*A)*sqrt(b*cos(d*x + c))*

sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), 1/24*(9*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) + (6*B*cos(d*x + c)^3 + 8*A*cos(d*x + c)^2 + 9*B*cos(d*x + c) + 16*

A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/