3.2.70 \(\int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx\) [170]

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

[Out]

arctanh(sin(d*x+c))*sec(d*x+c)^(1/2)/b/d/(b*sec(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, 3855} \begin {gather*} \frac {\sqrt {\sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{b d \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(5/2)/(b*Sec[c + d*x])^(3/2),x]

[Out]

(ArcTanh[Sin[c + d*x]]*Sqrt[Sec[c + d*x]])/(b*d*Sqrt[b*Sec[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 3855

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

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Mathematica [A]
time = 0.03, size = 33, normalized size = 0.92 \begin {gather*} \frac {\tanh ^{-1}(\sin (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{d (b \sec (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^(5/2)/(b*Sec[c + d*x])^(3/2),x]

[Out]

(ArcTanh[Sin[c + d*x]]*Sec[c + d*x]^(3/2))/(d*(b*Sec[c + d*x])^(3/2))

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Maple [A]
time = 37.47, size = 52, normalized size = 1.44

method	result	size
default	\(-\frac {2 \cos \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right )}{d \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}\)	\(52\)
risch	\(\frac {\sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}-\frac {\sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(5/2)/(b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (32) = 64\).
time = 0.62, size = 65, normalized size = 1.81 \begin {gather*} \frac {\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 ) - \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 )}{2 \, b^{\frac {3}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)/(b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.32, size = 114, normalized size = 3.17 \begin {gather*} \left [\frac {\log \left (-\frac {b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b}{\cos \left (d x + c\right )^{2}}\right )}{2 \, b^{\frac {3}{2}} d}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b}\right )}{b^{2} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)/(b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**(5/2)/(b*sec(d*x+c))**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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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(sec(d*x+c)^(5/2)/(b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^(5/2)/(b*sec(d*x + c))^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(c + d*x))^(5/2)/(b/cos(c + d*x))^(3/2),x)

[Out]

int((1/cos(c + d*x))^(5/2)/(b/cos(c + d*x))^(3/2), x)

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