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

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

[Out]

arctanh(sin(d*x+c))*sec(d*x+c)^(1/2)/b^2/d/(b*sec(d*x+c))^(1/2)

________________________________________________________________________________________

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} \[ \frac {\sqrt {\sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{b^2 d \sqrt {b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTanh[Sin[c + d*x]]*Sqrt[Sec[c + d*x]])/(b^2*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 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 {\sec ^{\frac {7}{2}}(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx &=\frac {\sqrt {\sec (c+d x)} \int \sec (c+d x) \, dx}{b^2 \sqrt {b \sec (c+d x)}}\\ &=\frac {\tanh ^{-1}(\sin (c+d x)) \sqrt {\sec (c+d x)}}{b^2 d \sqrt {b \sec (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 33, normalized size = 0.92 \[ \frac {\sec ^{\frac {5}{2}}(c+d x) \tanh ^{-1}(\sin (c+d x))}{d (b \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.97, size = 114, normalized size = 3.17 \[ \left [\frac {\log \left (-\frac {b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b}{\cos \left (d x + c\right )^{2}}\right )}{2 \, b^{\frac {5}{2}} d}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b}\right )}{b^{3} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)/(b*sec(d*x+c))^(5/2),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^(5/2)*d), -sqrt(-b)*arctan(sqrt(-b)*sqrt(b/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/b)/(b^3*d)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)/(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.78, size = 52, normalized size = 1.44 \[ -\frac {2 \cos \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \arctanh \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )}{d \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.90, size = 65, normalized size = 1.81 \[ \frac {\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )}{2 \, b^{\frac {5}{2}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(c + d*x))^(7/2)/(b/cos(c + d*x))^(5/2),x)

[Out]

int((1/cos(c + d*x))^(7/2)/(b/cos(c + d*x))^(5/2), x)

________________________________________________________________________________________

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(sec(d*x+c)**(7/2)/(b*sec(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________