3.9.73 \(\int \frac {\cos ^{\frac {7}{2}}(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{3/2}} \, dx\) [873]

Optimal. Leaf size=148 \[ \frac {A x \sqrt {\cos (c+d x)}}{2 b \sqrt {b \cos (c+d x)}}+\frac {B \sqrt {\cos (c+d x)} \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}+\frac {A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 b d \sqrt {b \cos (c+d x)}}-\frac {B \sqrt {\cos (c+d x)} \sin ^3(c+d x)}{3 b d \sqrt {b \cos (c+d x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {17, 2827, 2715, 8, 2713} \begin {gather*} \frac {A x \sqrt {\cos (c+d x)}}{2 b \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 b d \sqrt {b \cos (c+d x)}}-\frac {B \sin ^3(c+d x) \sqrt {\cos (c+d x)}}{3 b d \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]^(7/2)*(A + B*Cos[c + d*x]))/(b*Cos[c + d*x])^(3/2),x]

[Out]

(A*x*Sqrt[Cos[c + d*x]])/(2*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]]) + (A*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*b*d*Sqrt[b*Cos[c + d*x]]) - (B*Sqrt[Cos[c + d*x]]*Sin[c + d*x]
^3)/(3*b*d*Sqrt[b*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 2713

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 2715

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] && Integ
erQ[2*n]

Rule 2827

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 \frac {\cos ^{\frac {7}{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 ^2(c+d x) (A+B \cos (c+d x)) \, dx}{b \sqrt {b \cos (c+d x)}}\\ &=\frac {\left (A \sqrt {\cos (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{b \sqrt {b \cos (c+d x)}}+\frac {\left (B \sqrt {\cos (c+d x)}\right ) \int \cos ^3(c+d x) \, dx}{b \sqrt {b \cos (c+d x)}}\\ &=\frac {A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 b d \sqrt {b \cos (c+d x)}}+\frac {\left (A \sqrt {\cos (c+d x)}\right ) \int 1 \, dx}{2 b \sqrt {b \cos (c+d x)}}-\frac {\left (B \sqrt {\cos (c+d x)}\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{b d \sqrt {b \cos (c+d x)}}\\ &=\frac {A x \sqrt {\cos (c+d x)}}{2 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)}}+\frac {A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 b d \sqrt {b \cos (c+d x)}}-\frac {B \sqrt {\cos (c+d x)} \sin ^3(c+d x)}{3 b d \sqrt {b \cos (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.09, size = 69, normalized size = 0.47 \begin {gather*} \frac {\cos ^{\frac {3}{2}}(c+d x) (6 A c+6 A d x+9 B \sin (c+d x)+3 A \sin (2 (c+d x))+B \sin (3 (c+d x)))}{12 d (b \cos (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^(7/2)*(A + B*Cos[c + d*x]))/(b*Cos[c + d*x])^(3/2),x]

[Out]

(Cos[c + d*x]^(3/2)*(6*A*c + 6*A*d*x + 9*B*Sin[c + d*x] + 3*A*Sin[2*(c + d*x)] + B*Sin[3*(c + d*x)]))/(12*d*(b
*Cos[c + d*x])^(3/2))

________________________________________________________________________________________

Maple [A]
time = 0.27, size = 74, normalized size = 0.50

method	result	size
default	\(\frac {\left (\cos ^{\frac {3}{2}}\left (d x +c \right )\right ) \left (2 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+3 A \sin \left (d x +c \right ) \cos \left (d x +c \right )+3 A \left (d x +c \right )+4 B \sin \left (d x +c \right )\right )}{6 d \left (b \cos \left (d x +c \right )\right )^{\frac {3}{2}}}\)	\(74\)
risch	\(\frac {A x \left (\sqrt {\cos }\left (d x +c \right )\right )}{2 b \sqrt {b \cos \left (d x +c \right )}}+\frac {3 B \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{4 b d \sqrt {b \cos \left (d x +c \right )}}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) B \sin \left (3 d x +3 c \right )}{12 b \sqrt {b \cos \left (d x +c \right )}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) A \sin \left (2 d x +2 c \right )}{4 b \sqrt {b \cos \left (d x +c \right )}\, d}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(7/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/6/d*cos(d*x+c)^(3/2)*(2*B*sin(d*x+c)*cos(d*x+c)^2+3*A*sin(d*x+c)*cos(d*x+c)+3*A*(d*x+c)+4*B*sin(d*x+c))/(b*c
os(d*x+c))^(3/2)

________________________________________________________________________________________

Maxima [A]
time = 0.76, size = 68, normalized size = 0.46 \begin {gather*} \frac {\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A}{b^{\frac {3}{2}}} + \frac {B {\left (\sin \left (3 \, d x + 3 \, c\right ) + 9 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )}}{b^{\frac {3}{2}}}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A/b^(3/2) + B*(sin(3*d*x + 3*c) + 9*sin(1/3*arctan2(sin(3*d*x + 3*c),
cos(3*d*x + 3*c))))/b^(3/2))/d

________________________________________________________________________________________

Fricas [A]
time = 0.40, size = 236, normalized size = 1.59 \begin {gather*} \left [-\frac {3 \, 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 \, {\left (2 \, B \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + 4 \, B\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, b^{2} d \cos \left (d x + c\right )}, \frac {3 \, 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 ) + {\left (2 \, B \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + 4 \, B\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, 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)^(7/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[-1/12*(3*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*(2*B*cos(d*x + c)^2 + 3*A*cos(d*x + c) + 4*B)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*si
n(d*x + c))/(b^2*d*cos(d*x + c)), 1/6*(3*A*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) + (2*B*cos(d*x + c)^2 + 3*A*cos(d*x + c) + 4*B)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c
))*sin(d*x + c))/(b^2*d*cos(d*x + c))]

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(7/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

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)^(7/2)*(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)^(7/2)/(b*cos(d*x + c))^(3/2), x)

________________________________________________________________________________________

Mupad [B]
time = 1.57, size = 95, normalized size = 0.64 \begin {gather*} \frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (3\,A\,\sin \left (c+d\,x\right )+3\,A\,\sin \left (3\,c+3\,d\,x\right )+10\,B\,\sin \left (2\,c+2\,d\,x\right )+B\,\sin \left (4\,c+4\,d\,x\right )+12\,A\,d\,x\,\cos \left (c+d\,x\right )\right )}{12\,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)^(7/2)*(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)*(3*A*sin(c + d*x) + 3*A*sin(3*c + 3*d*x) + 10*B*sin(2*c + 2*d*x) +
B*sin(4*c + 4*d*x) + 12*A*d*x*cos(c + d*x)))/(12*b^2*d*(cos(2*c + 2*d*x) + 1))

________________________________________________________________________________________