3.2.12 \(\int \frac {(b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) [112]

Optimal. Leaf size=65 \[ \frac {b^2 C x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {17, 3091, 8} \begin {gather*} \frac {A b^2 \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \cos ^{\frac {3}{2}}(c+d x)}+\frac {b^2 C x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

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

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 3091

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A*Cos[e +
f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx &=\frac {\left (b^2 \sqrt {b \cos (c+d x)}\right ) \int \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx}{\sqrt {\cos (c+d x)}}\\ &=\frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (b^2 C \sqrt {b \cos (c+d x)}\right ) \int 1 \, dx}{\sqrt {\cos (c+d x)}}\\ &=\frac {b^2 C x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 45, normalized size = 0.69 \begin {gather*} \frac {(b \cos (c+d x))^{5/2} (C d x \cos (c+d x)+A \sin (c+d x))}{d \cos ^{\frac {7}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

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

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Maple [A]
time = 0.24, size = 45, normalized size = 0.69

method	result	size
default	\(\frac {\left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}} \left (C \cos \left (d x +c \right ) \left (d x +c \right )+A \sin \left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{\frac {7}{2}}}\)	\(45\)
risch	\(\frac {b^{2} C x \sqrt {b \cos \left (d x +c \right )}}{\sqrt {\cos \left (d x +c \right )}}+\frac {2 i b^{2} \sqrt {b \cos \left (d x +c \right )}\, A}{\sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.58, size = 80, normalized size = 1.23 \begin {gather*} \frac {2 \, {\left (C b^{\frac {5}{2}} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + \frac {A b^{\frac {5}{2}} \sin \left (2 \, d x + 2 \, c\right )}{\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1}\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.44, size = 194, normalized size = 2.98 \begin {gather*} \left [\frac {C \sqrt {-b} b^{2} \cos \left (d x + c\right )^{2} \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 )} A b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )^{2}}, \frac {C b^{\frac {5}{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 )^{2} + \sqrt {b \cos \left (d x + c\right )} A b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

[1/2*(C*sqrt(-b)*b^2*cos(d*x + c)^2*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*sqrt(b*cos(d*x + c))*A*b^2*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2), (C*b^(
5/2)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c)^2 + sqrt(b*cos(d*x +
c))*A*b^2*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)]

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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((b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(9/2),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((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(5/2)/cos(d*x + c)^(9/2), x)

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Mupad [B]
time = 1.19, size = 84, normalized size = 1.29 \begin {gather*} \frac {b^2\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (A\,\sin \left (2\,c+2\,d\,x\right )+C\,d\,x+C\,d\,x\,\cos \left (2\,c+2\,d\,x\right )+A\,1{}\mathrm {i}+A\,\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\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(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2))/cos(c + d*x)^(9/2),x)

[Out]

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

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