3.2.54 \(\int \sqrt {\sec (c+d x)} (b \sec (c+d x))^{5/2} \, dx\) [154]

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

[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)

________________________________________________________________________________________

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, 3853, 3855} \begin {gather*} \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)}} \end {gather*}

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 3853

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 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 \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*}

________________________________________________________________________________________

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

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))

________________________________________________________________________________________

Maple [A]
time = 36.99, size = 112, normalized size = 1.44

method	result	size
default	\(\frac {\left (\ln \left (-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-\ln \left (-\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{2 d}\)	\(112\)
risch	\(-\frac {i b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \cos \left (d x +c \right )}{d}+\frac {b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \cos \left (d x +c \right )}{d}\)	\(241\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 747 vs. \(2 (66) = 132\).
time = 0.63, size = 747, normalized size = 9.58 \begin {gather*} -\frac {{\left (4 \, {\left (b^{2} \sin \left (4 \, d x + 4 \, c\right ) + 2 \, b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) - 4 \, {\left (b^{2} \sin \left (4 \, d x + 4 \, c\right ) + 2 \, b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) - {\left (b^{2} \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, b^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{2} \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, b^{2} \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, b^{2} \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2} + 2 \, {\left (2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 1\right ) + {\left (b^{2} \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, b^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{2} \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, b^{2} \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, b^{2} \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2} + 2 \, {\left (2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 1\right ) - 4 \, {\left (b^{2} \cos \left (4 \, d x + 4 \, c\right ) + 2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) + 4 \, {\left (b^{2} \cos \left (4 \, d x + 4 \, c\right ) + 2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )\right )} \sqrt {b}}{4 \, {\left (2 \, {\left (2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + \cos \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (4 \, d x + 4 \, c\right )^{2} + 4 \, \sin \left (4 \, d x + 4 \, c\right ) \sin \left (2 \, d x + 2 \, c\right ) + 4 \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} d} \end {gather*}

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)

________________________________________________________________________________________

Fricas [A]
time = 3.68, size = 208, normalized size = 2.67 \begin {gather*} \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 ] \end {gather*}

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))]

________________________________________________________________________________________

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

________________________________________________________________________________________

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)^(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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________