3.154 \(\int \sqrt {\sec (c+d x)} (b \sec (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=78 \[ \frac {b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {b \sec (c+d x)}}{2 d}+\frac {b^2 \sqrt {b \sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{2 d \sqrt {\sec (c+d x)}} \]

[Out]

1/2*b^2*sec(d*x+c)^(3/2)*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/d+1/2*b^2*arctanh(sin(d*x+c))*(b*sec(d*x+c))^(1/2)/d/
sec(d*x+c)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {17, 3768, 3770} \[ \frac {b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {b \sec (c+d x)}}{2 d}+\frac {b^2 \sqrt {b \sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{2 d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.08, size = 50, normalized size = 0.64 \[ \frac {(b \sec (c+d x))^{5/2} \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )}{2 d \sec ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.60, size = 208, normalized size = 2.67 \[ \left [\frac {b^{\frac {5}{2}} \cos \left (d x + c\right ) \log \left (-\frac {b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b}{\cos \left (d x + c\right )^{2}}\right ) + \frac {2 \, b^{2} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, d \cos \left (d x + c\right )}, -\frac {\sqrt {-b} b^{2} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b}\right ) \cos \left (d x + c\right ) - \frac {b^{2} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{2 \, d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)*(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/4*(b^(5/2)*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*b^2*sqrt(b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)), -1
/2*(sqrt(-b)*b^2*arctan(sqrt(-b)*sqrt(b/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/b)*cos(d*x + c) - b^2*sq
rt(b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)*(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.96, size = 114, normalized size = 1.46 \[ -\frac {\left (\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\frac {-\sin \left (d x +c \right )-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.94, size = 747, normalized size = 9.58 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(1/2)*(b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-1/4*(4*(b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) -
4*(b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*cos(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))*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) + (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))*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*(b^2*cos(4*d*x + 4*c) + 2*b^2*cos(2*d*x + 2*c) + b^2)*sin(3/2*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(b^2*cos(4*d*x + 4*c) + 2*b^2*cos(2*d*x + 2*c) + b^2)*sin(1/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sqrt(b)/((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)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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