3.168 \(\int \frac{(b \cos (c+d x))^{5/2}}{\cos ^{\frac{11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=78 \[ \frac{b^2 \sin (c+d x) \sqrt{b \cos (c+d x)}}{2 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{b^2 \sqrt{b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{2 d \sqrt{\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0212637, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {17, 3768, 3770} \[ \frac{b^2 \sin (c+d x) \sqrt{b \cos (c+d x)}}{2 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{b^2 \sqrt{b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{2 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^(5/2)/Cos[c + d*x]^(11/2),x]

[Out]

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

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{(b \cos (c+d x))^{5/2}}{\cos ^{\frac{11}{2}}(c+d x)} \, dx &=\frac{\left (b^2 \sqrt{b \cos (c+d x)}\right ) \int \sec ^3(c+d x) \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{b^2 \sqrt{b \cos (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{\left (b^2 \sqrt{b \cos (c+d x)}\right ) \int \sec (c+d x) \, dx}{2 \sqrt{\cos (c+d x)}}\\ &=\frac{b^2 \tanh ^{-1}(\sin (c+d x)) \sqrt{b \cos (c+d x)}}{2 d \sqrt{\cos (c+d x)}}+\frac{b^2 \sqrt{b \cos (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 0.0597304, size = 52, normalized size = 0.67 \[ \frac{(b \cos (c+d x))^{5/2} \left (\sin (c+d x)+\cos ^2(c+d x) \tanh ^{-1}(\sin (c+d x))\right )}{2 d \cos ^{\frac{9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^(5/2)/Cos[c + d*x]^(11/2),x]

[Out]

((b*Cos[c + d*x])^(5/2)*(ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + Sin[c + d*x]))/(2*d*Cos[c + d*x]^(9/2))

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Maple [A]  time = 0.162, size = 104, normalized size = 1.3 \begin{align*} -{\frac{1}{2\,d} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}\ln \left ( -{\frac{-1+\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) - \left ( \cos \left ( dx+c \right ) \right ) ^{2}\ln \left ( -{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) -\sin \left ( dx+c \right ) \right ) \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}} \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^(5/2)/cos(d*x+c)^(11/2),x)

[Out]

-1/2/d*(cos(d*x+c)^2*ln(-(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))-cos(d*x+c)^2*ln(-(-1+cos(d*x+c)-sin(d*x+c))/si
n(d*x+c))-sin(d*x+c))*(b*cos(d*x+c))^(5/2)/cos(d*x+c)^(9/2)

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Maxima [B]  time = 1.85142, size = 1008, normalized size = 12.92 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)/cos(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

-1/4*(4*(b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) -
4*(b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (b^2
*cos(4*d*x + 4*c)^2 + 4*b^2*cos(2*d*x + 2*c)^2 + b^2*sin(4*d*x + 4*c)^2 + 4*b^2*sin(4*d*x + 4*c)*sin(2*d*x + 2
*c) + 4*b^2*sin(2*d*x + 2*c)^2 + 4*b^2*cos(2*d*x + 2*c) + b^2 + 2*(2*b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d*x + 4
*c))*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (b^2*cos(4*d*x + 4*c)^2 + 4*b^2*cos
(2*d*x + 2*c)^2 + b^2*sin(4*d*x + 4*c)^2 + 4*b^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*b^2*sin(2*d*x + 2*c)^2
+ 4*b^2*cos(2*d*x + 2*c) + b^2 + 2*(2*b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d*x + 4*c))*log(cos(1/2*arctan2(sin(2*
d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sin(1/2*arctan2(
sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 4*(b^2*cos(4*d*x + 4*c) + 2*b^2*cos(2*d*x + 2*c) + b^2)*sin(3/2*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(b^2*cos(4*d*x + 4*c) + 2*b^2*cos(2*d*x + 2*c) + b^2)*sin(1/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sqrt(b)/((2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x
+ 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*
c)^2 + 4*cos(2*d*x + 2*c) + 1)*d)

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Fricas [A]  time = 2.1825, size = 575, normalized size = 7.37 \begin{align*} \left [\frac{b^{\frac{5}{2}} \cos \left (d x + c\right )^{3} \log \left (-\frac{b \cos \left (d x + c\right )^{3} - 2 \, \sqrt{b \cos \left (d x + c\right )} \sqrt{b} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt{b \cos \left (d x + c\right )} b^{2} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{3}}, -\frac{\sqrt{-b} b^{2} \arctan \left (\frac{\sqrt{b \cos \left (d x + c\right )} \sqrt{-b} \sin \left (d x + c\right )}{b \sqrt{\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{3} - \sqrt{b \cos \left (d x + c\right )} b^{2} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)/cos(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

[1/4*(b^(5/2)*cos(d*x + c)^3*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d*x + c))*sqrt(b)*sqrt(cos(d*x + c))*sin(d*
x + c) - 2*b*cos(d*x + c))/cos(d*x + c)^3) + 2*sqrt(b*cos(d*x + c))*b^2*sqrt(cos(d*x + c))*sin(d*x + c))/(d*co
s(d*x + c)^3), -1/2*(sqrt(-b)*b^2*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*co
s(d*x + c)^3 - sqrt(b*cos(d*x + c))*b^2*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))**(5/2)/cos(d*x+c)**(11/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}}}{\cos \left (d x + c\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^(5/2)/cos(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c))^(5/2)/cos(d*x + c)^(11/2), x)