3.181 \(\int \frac{1}{\cos ^{\frac{9}{2}}(c+d x) \sqrt{b \cos (c+d x)}} \, dx\)
Optimal. Leaf size=107 \[ \frac{3 \sin (c+d x)}{8 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{\sin (c+d x)}{4 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{3 \sqrt{\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{8 d \sqrt{b \cos (c+d x)}} \]
[Out]
(3*ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(8*d*Sqrt[b*Cos[c + d*x]]) + Sin[c + d*x]/(4*d*Cos[c + d*x]^(7/2)
*Sqrt[b*Cos[c + d*x]]) + (3*Sin[c + d*x])/(8*d*Cos[c + d*x]^(3/2)*Sqrt[b*Cos[c + d*x]])
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Rubi [A] time = 0.0349813, antiderivative size = 107, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used =
{18, 3768, 3770} \[ \frac{3 \sin (c+d x)}{8 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{\sin (c+d x)}{4 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{3 \sqrt{\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{8 d \sqrt{b \cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[1/(Cos[c + d*x]^(9/2)*Sqrt[b*Cos[c + d*x]]),x]
[Out]
(3*ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(8*d*Sqrt[b*Cos[c + d*x]]) + Sin[c + d*x]/(4*d*Cos[c + d*x]^(7/2)
*Sqrt[b*Cos[c + d*x]]) + (3*Sin[c + d*x])/(8*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{9}{2}}(c+d x) \sqrt{b \cos (c+d x)}} \, dx &=\frac{\sqrt{\cos (c+d x)} \int \sec ^5(c+d x) \, dx}{\sqrt{b \cos (c+d x)}}\\ &=\frac{\sin (c+d x)}{4 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 \sqrt{b \cos (c+d x)}}\\ &=\frac{\sin (c+d x)}{4 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{3 \sin (c+d x)}{8 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 \sqrt{b \cos (c+d x)}}\\ &=\frac{3 \tanh ^{-1}(\sin (c+d x)) \sqrt{\cos (c+d x)}}{8 d \sqrt{b \cos (c+d x)}}+\frac{\sin (c+d x)}{4 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{3 \sin (c+d x)}{8 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{b \cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.0785357, size = 66, normalized size = 0.62 \[ \frac{\sin (c+d x) \left (3 \cos ^2(c+d x)+2\right )+3 \cos ^4(c+d x) \tanh ^{-1}(\sin (c+d x))}{8 d \cos ^{\frac{7}{2}}(c+d x) \sqrt{b \cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(Cos[c + d*x]^(9/2)*Sqrt[b*Cos[c + d*x]]),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]^(7/2)*Sqrt[b*
Cos[c + d*x]])
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Maple [A] time = 0.282, size = 121, normalized size = 1.1 \begin{align*}{\frac{1}{8\,d} \left ( 3\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}\ln \left ( -{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) -3\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}\ln \left ( -{\frac{-1+\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +2\,\sin \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{7}{2}}}{\frac{1}{\sqrt{b\cos \left ( dx+c \right ) }}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^(1/2),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(-(-1+cos(d*x+c)+sin(d*x+c))
/sin(d*x+c))+3*cos(d*x+c)^2*sin(d*x+c)+2*sin(d*x+c))/cos(d*x+c)^(7/2)/(b*cos(d*x+c))^(1/2)
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Maxima [B] time = 1.89037, size = 2236, normalized size = 20.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^(1/2),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))))/((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*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)*sqrt(b)*d)
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Fricas [A] time = 2.23464, size = 633, normalized size = 5.92 \begin{align*} \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 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 d \cos \left (d x + c\right )^{5}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^(1/2),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*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*s
qrt(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*d*cos(d*x + c)^5)]
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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(1/cos(d*x+c)**(9/2)/(b*cos(d*x+c))**(1/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \cos \left (d x + c\right )} \cos \left (d x + c\right )^{\frac{9}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cos(d*x+c)^(9/2)/(b*cos(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate(1/(sqrt(b*cos(d*x + c))*cos(d*x + c)^(9/2)), x)