3.166 \(\int \frac{(b \cos (c+d x))^{5/2}}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=36 \[ \frac{b^2 \sqrt{b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt{\cos (c+d x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0094418, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {17, 3770} \[ \frac{b^2 \sqrt{b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*ArcTanh[Sin[c + d*x]]*Sqrt[b*Cos[c + d*x]])/(d*Sqrt[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 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{7}{2}}(c+d x)} \, dx &=\frac{\left (b^2 \sqrt{b \cos (c+d x)}\right ) \int \sec (c+d x) \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{b^2 \tanh ^{-1}(\sin (c+d x)) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0221454, size = 33, normalized size = 0.92 \[ \frac{(b \cos (c+d x))^{5/2} \tanh ^{-1}(\sin (c+d x))}{d \cos ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.145, size = 42, normalized size = 1.2 \begin{align*} -2\,{\frac{ \left ( b\cos \left ( dx+c \right ) \right ) ^{5/2}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{5/2}}{\it Artanh} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) } \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)^(7/2),x)

[Out]

-2/d*arctanh((-1+cos(d*x+c))/sin(d*x+c))*(b*cos(d*x+c))^(5/2)/cos(d*x+c)^(5/2)

________________________________________________________________________________________

Maxima [B]  time = 1.8455, size = 97, normalized size = 2.69 \begin{align*} \frac{{\left (b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} \sqrt{b}}{2 \, d} \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)^(7/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.12115, size = 316, normalized size = 8.78 \begin{align*} \left [\frac{b^{\frac{5}{2}} \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 \, d}, -\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 )}{d}\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)^(7/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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)**(7/2),x)

[Out]

Timed out

________________________________________________________________________________________

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{7}{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)^(7/2),x, algorithm="giac")

[Out]

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