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

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

[Out]

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

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Rubi [A]  time = 0.0202177, 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} \[ \frac{\sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{2 b^2 d \sqrt{b \sec (c+d x)}}+\frac{\sqrt{\sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{2 b^2 d \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTanh[Sin[c + d*x]]*Sqrt[Sec[c + d*x]])/(2*b^2*d*Sqrt[b*Sec[c + d*x]]) + (Sec[c + d*x]^(5/2)*Sin[c + d*x])/
(2*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 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{\sec ^{\frac{11}{2}}(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx &=\frac{\sqrt{\sec (c+d x)} \int \sec ^3(c+d x) \, dx}{b^2 \sqrt{b \sec (c+d x)}}\\ &=\frac{\sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 b^2 d \sqrt{b \sec (c+d x)}}+\frac{\sqrt{\sec (c+d x)} \int \sec (c+d x) \, dx}{2 b^2 \sqrt{b \sec (c+d x)}}\\ &=\frac{\tanh ^{-1}(\sin (c+d x)) \sqrt{\sec (c+d x)}}{2 b^2 d \sqrt{b \sec (c+d x)}}+\frac{\sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 b^2 d \sqrt{b \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0506845, size = 53, normalized size = 0.68 \[ \frac{\sqrt{\sec (c+d x)} \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )}{2 b^2 d \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.117, size = 112, normalized size = 1.4 \begin{align*}{\frac{\cos \left ( dx+c \right ) }{2\,d} \left ( \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 ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+\sin \left ( dx+c \right ) \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{11}{2}}} \left ({\frac{b}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*(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))/sin(d*x+c))*cos(
d*x+c)^2+sin(d*x+c))*(1/cos(d*x+c))^(11/2)*cos(d*x+c)/(b/cos(d*x+c))^(5/2)

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Maxima [B]  time = 2.12037, size = 929, normalized size = 11.91 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 4*(sin(
4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (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)*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) + (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)*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*(cos(4*d*x + 4*c) + 2*cos(2*d*x +
2*c) + 1)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)
*sin(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))*sqrt(b)*d)

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Fricas [A]  time = 1.95703, size = 539, normalized size = 6.91 \begin{align*} \left [\frac{\sqrt{b} \cos \left (d x + c\right ) \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 ) + \frac{2 \, \sqrt{\frac{b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{4 \, b^{3} d \cos \left (d x + c\right )}, -\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 ) \cos \left (d x + c\right ) - \frac{\sqrt{\frac{b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{2 \, b^{3} d \cos \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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