3.201 \(\int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=116 \[ \frac {3 \sin (c+d x)}{8 b^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {\sin (c+d x)}{4 b^2 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {3 \sqrt {\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{8 b^2 d \sqrt {b \cos (c+d x)}} \]
[Out]
1/4*sin(d*x+c)/b^2/d/cos(d*x+c)^(7/2)/(b*cos(d*x+c))^(1/2)+3/8*sin(d*x+c)/b^2/d/cos(d*x+c)^(3/2)/(b*cos(d*x+c)
)^(1/2)+3/8*arctanh(sin(d*x+c))*cos(d*x+c)^(1/2)/b^2/d/(b*cos(d*x+c))^(1/2)
________________________________________________________________________________________
Rubi [A] time = 0.04, antiderivative size = 116, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used =
{18, 3768, 3770} \[ \frac {3 \sin (c+d x)}{8 b^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {\sin (c+d x)}{4 b^2 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {3 \sqrt {\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{8 b^2 d \sqrt {b \cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[1/(Cos[c + d*x]^(5/2)*(b*Cos[c + d*x])^(5/2)),x]
[Out]
(3*ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(8*b^2*d*Sqrt[b*Cos[c + d*x]]) + Sin[c + d*x]/(4*b^2*d*Cos[c + d*
x]^(7/2)*Sqrt[b*Cos[c + d*x]]) + (3*Sin[c + d*x])/(8*b^2*d*Cos[c + d*x]^(3/2)*Sqrt[b*Cos[c + d*x]])
Rule 18
Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m - 1/2)*b^(n + 1/2)*Sqrt[a*v])/Sqrt[b*v]
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] && !IntegerQ[m] && ILtQ[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 {1}{\cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^{5/2}} \, dx &=\frac {\sqrt {\cos (c+d x)} \int \sec ^5(c+d x) \, dx}{b^2 \sqrt {b \cos (c+d x)}}\\ &=\frac {\sin (c+d x)}{4 b^2 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {\left (3 \sqrt {\cos (c+d x)}\right ) \int \sec ^3(c+d x) \, dx}{4 b^2 \sqrt {b \cos (c+d x)}}\\ &=\frac {\sin (c+d x)}{4 b^2 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {3 \sin (c+d x)}{8 b^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {\left (3 \sqrt {\cos (c+d x)}\right ) \int \sec (c+d x) \, dx}{8 b^2 \sqrt {b \cos (c+d x)}}\\ &=\frac {3 \tanh ^{-1}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{8 b^2 d \sqrt {b \cos (c+d x)}}+\frac {\sin (c+d x)}{4 b^2 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {b \cos (c+d x)}}+\frac {3 \sin (c+d x)}{8 b^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.04, size = 66, normalized size = 0.57 \[ \frac {\sin (c+d x) \left (3 \cos ^2(c+d x)+2\right )+3 \cos ^4(c+d x) \tanh ^{-1}(\sin (c+d x))}{8 d \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(Cos[c + d*x]^(5/2)*(b*Cos[c + d*x])^(5/2)),x]
[Out]
(3*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^4 + (2 + 3*Cos[c + d*x]^2)*Sin[c + d*x])/(8*d*Cos[c + d*x]^(3/2)*(b*Cos[
c + d*x])^(5/2))
________________________________________________________________________________________
fricas [A] time = 0.83, size = 233, normalized size = 2.01 \[ \left [\frac {3 \, \sqrt {b} \cos \left (d x + c\right )^{5} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} {\left (3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{16 \, b^{3} d \cos \left (d x + c\right )^{5}}, -\frac {3 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{5} - \sqrt {b \cos \left (d x + c\right )} {\left (3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, b^{3} d \cos \left (d x + c\right )^{5}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
[1/16*(3*sqrt(b)*cos(d*x + c)^5*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d*x + c))*sqrt(b)*sqrt(cos(d*x + c))*sin
(d*x + c) - 2*b*cos(d*x + c))/cos(d*x + c)^3) + 2*sqrt(b*cos(d*x + c))*(3*cos(d*x + c)^2 + 2)*sqrt(cos(d*x + c
))*sin(d*x + c))/(b^3*d*cos(d*x + c)^5), -1/8*(3*sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b
*sqrt(cos(d*x + c))))*cos(d*x + c)^5 - sqrt(b*cos(d*x + c))*(3*cos(d*x + c)^2 + 2)*sqrt(cos(d*x + c))*sin(d*x
+ c))/(b^3*d*cos(d*x + c)^5)]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate(1/((b*cos(d*x + c))^(5/2)*cos(d*x + c)^(5/2)), x)
________________________________________________________________________________________
maple [A] time = 0.16, size = 121, normalized size = 1.04 \[ -\frac {3 \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-3 \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\frac {-\sin \left (d x +c \right )-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-3 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-2 \sin \left (d x +c \right )}{8 d \left (b \cos \left (d x +c \right )\right )^{\frac {5}{2}} \cos \left (d x +c \right )^{\frac {3}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(5/2),x)
[Out]
-1/8/d*(3*cos(d*x+c)^4*ln(-(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))-3*cos(d*x+c)^4*ln(-(-sin(d*x+c)-1+cos(d*x+c)
)/sin(d*x+c))-3*cos(d*x+c)^2*sin(d*x+c)-2*sin(d*x+c))/(b*cos(d*x+c))^(5/2)/cos(d*x+c)^(3/2)
________________________________________________________________________________________
maxima [B] time = 1.94, size = 1729, normalized size = 14.91 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cos(d*x+c)^(5/2)/(b*cos(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
-1/16*(12*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos(7/2*arctan2(si
n(2*d*x + 2*c), cos(2*d*x + 2*c))) + 44*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*
d*x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 44*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) +
6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 12*(sin(8*d*x
+ 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(
2*d*x + 2*c))) - 3*(2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(8*d*x + 8*c) + co
s(8*d*x + 8*c)^2 + 8*(6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + 16*cos(6*d*x + 6*c)^2 +
12*(4*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 36*cos(4*d*x + 4*c)^2 + 16*cos(2*d*x + 2*c)^2 + 4*(2*sin(6*d*x
+ 6*c) + 3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + sin(8*d*x + 8*c)^2 + 16*(3*sin(4*d*x + 4*
c) + 2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 16*sin(6*d*x + 6*c)^2 + 36*sin(4*d*x + 4*c)^2 + 48*sin(4*d*x + 4*c
)*sin(2*d*x + 2*c) + 16*sin(2*d*x + 2*c)^2 + 8*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) + 3*(2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(8*d*x
+ 8*c) + cos(8*d*x + 8*c)^2 + 8*(6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6*c) + 16*cos(6*d*x
+ 6*c)^2 + 12*(4*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 36*cos(4*d*x + 4*c)^2 + 16*cos(2*d*x + 2*c)^2 + 4*(
2*sin(6*d*x + 6*c) + 3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + sin(8*d*x + 8*c)^2 + 16*(3*si
n(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 16*sin(6*d*x + 6*c)^2 + 36*sin(4*d*x + 4*c)^2 + 48*sin
(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*sin(2*d*x + 2*c)^2 + 8*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) - 12*(cos(8*d*x + 8*c) + 4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*co
s(2*d*x + 2*c) + 1)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 44*(cos(8*d*x + 8*c) + 4*cos(6*d*x
+ 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4
4*(cos(8*d*x + 8*c) + 4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*sin(3/2*arctan2(sin(2*
d*x + 2*c), cos(2*d*x + 2*c))) + 12*(cos(8*d*x + 8*c) + 4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*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(8*d*x + 8*c)^2 + 16*b^2*cos(6*d*x
+ 6*c)^2 + 36*b^2*cos(4*d*x + 4*c)^2 + 16*b^2*cos(2*d*x + 2*c)^2 + b^2*sin(8*d*x + 8*c)^2 + 16*b^2*sin(6*d*x +
6*c)^2 + 36*b^2*sin(4*d*x + 4*c)^2 + 48*b^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*b^2*sin(2*d*x + 2*c)^2 + 8
*b^2*cos(2*d*x + 2*c) + b^2 + 2*(4*b^2*cos(6*d*x + 6*c) + 6*b^2*cos(4*d*x + 4*c) + 4*b^2*cos(2*d*x + 2*c) + b^
2)*cos(8*d*x + 8*c) + 8*(6*b^2*cos(4*d*x + 4*c) + 4*b^2*cos(2*d*x + 2*c) + b^2)*cos(6*d*x + 6*c) + 12*(4*b^2*c
os(2*d*x + 2*c) + b^2)*cos(4*d*x + 4*c) + 4*(2*b^2*sin(6*d*x + 6*c) + 3*b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x
+ 2*c))*sin(8*d*x + 8*c) + 16*(3*b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c))*sqrt(b)*d)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(cos(c + d*x)^(5/2)*(b*cos(c + d*x))^(5/2)),x)
[Out]
int(1/(cos(c + d*x)^(5/2)*(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(1/cos(d*x+c)**(5/2)/(b*cos(d*x+c))**(5/2),x)
[Out]
Timed out
________________________________________________________________________________________