3.2.33 \(\int \frac {\cos ^{\frac {7}{2}}(c+d x) (A+C \cos ^2(c+d x))}{(b \cos (c+d x))^{5/2}} \, dx\) [133]

Optimal. Leaf size=80 \[ \frac {(A+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{b^2 d \sqrt {b \cos (c+d x)}}-\frac {C \sqrt {\cos (c+d x)} \sin ^3(c+d x)}{3 b^2 d \sqrt {b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {17, 3092} \begin {gather*} \frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{b^2 d \sqrt {b \cos (c+d x)}}-\frac {C \sin ^3(c+d x) \sqrt {\cos (c+d x)}}{3 b^2 d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^(7/2)*(A + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(5/2),x]

[Out]

((A + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(b^2*d*Sqrt[b*Cos[c + d*x]]) - (C*Sqrt[Cos[c + d*x]]*Sin[c + d*x]^3)
/(3*b^2*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 3092

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[-f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx &=\frac {\sqrt {\cos (c+d x)} \int \cos (c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx}{b^2 \sqrt {b \cos (c+d x)}}\\ &=-\frac {\sqrt {\cos (c+d x)} \text {Subst}\left (\int \left (A+C-C x^2\right ) \, dx,x,-\sin (c+d x)\right )}{b^2 d \sqrt {b \cos (c+d x)}}\\ &=\frac {(A+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{b^2 d \sqrt {b \cos (c+d x)}}-\frac {C \sqrt {\cos (c+d x)} \sin ^3(c+d x)}{3 b^2 d \sqrt {b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 55, normalized size = 0.69 \begin {gather*} \frac {\sqrt {\cos (c+d x)} (6 A+5 C+C \cos (2 (c+d x))) \sin (c+d x)}{6 b^2 d \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^(7/2)*(A + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(5/2),x]

[Out]

(Sqrt[Cos[c + d*x]]*(6*A + 5*C + C*Cos[2*(c + d*x)])*Sin[c + d*x])/(6*b^2*d*Sqrt[b*Cos[c + d*x]])

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Maple [A]
time = 0.28, size = 47, normalized size = 0.59

method	result	size
default	\(\frac {\left (C \left (\cos ^{2}\left (d x +c \right )\right )+3 A +2 C \right ) \sin \left (d x +c \right ) \left (\cos ^{\frac {5}{2}}\left (d x +c \right )\right )}{3 d \left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}}}\)	\(47\)
risch	\(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (4 A +3 C \right ) \sin \left (d x +c \right )}{4 b^{2} \sqrt {b \cos \left (d x +c \right )}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) C \sin \left (3 d x +3 c \right )}{12 b^{2} \sqrt {b \cos \left (d x +c \right )}\, d}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(7/2)*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.62, size = 57, normalized size = 0.71 \begin {gather*} \frac {\frac {C {\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 {5}{2}}} + \frac {12 \, A \sin \left (d x + c\right )}{b^{\frac {5}{2}}}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/12*(C*(sin(3*d*x + 3*c) + 9*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))/b^(5/2) + 12*A*sin(d*x + c
)/b^(5/2))/d

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Fricas [A]
time = 0.39, size = 49, normalized size = 0.61 \begin {gather*} \frac {{\left (C \cos \left (d x + c\right )^{2} + 3 \, A + 2 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, b^{3} d \sqrt {\cos \left (d x + c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

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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+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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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)^(7/2)*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^(7/2)/(b*cos(d*x + c))^(5/2), x)

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Mupad [B]
time = 0.84, size = 75, normalized size = 0.94 \begin {gather*} \frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (12\,A\,\sin \left (2\,c+2\,d\,x\right )+10\,C\,\sin \left (2\,c+2\,d\,x\right )+C\,\sin \left (4\,c+4\,d\,x\right )\right )}{12\,b^3\,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 + C*cos(c + d*x)^2))/(b*cos(c + d*x))^(5/2),x)

[Out]

(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^(1/2)*(12*A*sin(2*c + 2*d*x) + 10*C*sin(2*c + 2*d*x) + C*sin(4*c + 4*d*x)
))/(12*b^3*d*(cos(2*c + 2*d*x) + 1))

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