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(d*x+c))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.02, 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))

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fricas [A]  time = 0.55, size = 116, normalized size = 3.22 \[ \left [\frac {b^{\frac {5}{2}} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right )}{2 \, d}, -\frac {\sqrt {-b} b^{2} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right )}{d}\right ] \]

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]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.08, size = 42, normalized size = 1.17 \[ -\frac {2 \arctanh \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}}}{d \cos \left (d x +c \right )^{\frac {5}{2}}} \]

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)

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maxima [B]  time = 1.36, size = 72, normalized size = 2.00 \[ \frac {{\left (b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} \sqrt {b}}{2 \, d} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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