3.132 \(\int \sec ^{\frac{9}{2}}(c+d x) \sqrt{b \sec (c+d x)} \, dx\)
Optimal. Leaf size=107 \[ \frac{\sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) \sqrt{b \sec (c+d x)}}{4 d}+\frac{3 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \sqrt{b \sec (c+d x)}}{8 d}+\frac{3 \sqrt{b \sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{8 d \sqrt{\sec (c+d x)}} \]
[Out]
(3*ArcTanh[Sin[c + d*x]]*Sqrt[b*Sec[c + d*x]])/(8*d*Sqrt[Sec[c + d*x]]) + (3*Sec[c + d*x]^(3/2)*Sqrt[b*Sec[c +
d*x]]*Sin[c + d*x])/(8*d) + (Sec[c + d*x]^(7/2)*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/(4*d)
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Rubi [A] time = 0.0332845, 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 =
{17, 3768, 3770} \[ \frac{\sin (c+d x) \sec ^{\frac{7}{2}}(c+d x) \sqrt{b \sec (c+d x)}}{4 d}+\frac{3 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \sqrt{b \sec (c+d x)}}{8 d}+\frac{3 \sqrt{b \sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{8 d \sqrt{\sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[Sec[c + d*x]^(9/2)*Sqrt[b*Sec[c + d*x]],x]
[Out]
(3*ArcTanh[Sin[c + d*x]]*Sqrt[b*Sec[c + d*x]])/(8*d*Sqrt[Sec[c + d*x]]) + (3*Sec[c + d*x]^(3/2)*Sqrt[b*Sec[c +
d*x]]*Sin[c + d*x])/(8*d) + (Sec[c + d*x]^(7/2)*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/(4*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 \sec ^{\frac{9}{2}}(c+d x) \sqrt{b \sec (c+d x)} \, dx &=\frac{\sqrt{b \sec (c+d x)} \int \sec ^5(c+d x) \, dx}{\sqrt{\sec (c+d x)}}\\ &=\frac{\sec ^{\frac{7}{2}}(c+d x) \sqrt{b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{\left (3 \sqrt{b \sec (c+d x)}\right ) \int \sec ^3(c+d x) \, dx}{4 \sqrt{\sec (c+d x)}}\\ &=\frac{3 \sec ^{\frac{3}{2}}(c+d x) \sqrt{b \sec (c+d x)} \sin (c+d x)}{8 d}+\frac{\sec ^{\frac{7}{2}}(c+d x) \sqrt{b \sec (c+d x)} \sin (c+d x)}{4 d}+\frac{\left (3 \sqrt{b \sec (c+d x)}\right ) \int \sec (c+d x) \, dx}{8 \sqrt{\sec (c+d x)}}\\ &=\frac{3 \tanh ^{-1}(\sin (c+d x)) \sqrt{b \sec (c+d x)}}{8 d \sqrt{\sec (c+d x)}}+\frac{3 \sec ^{\frac{3}{2}}(c+d x) \sqrt{b \sec (c+d x)} \sin (c+d x)}{8 d}+\frac{\sec ^{\frac{7}{2}}(c+d x) \sqrt{b \sec (c+d x)} \sin (c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 0.141481, size = 64, normalized size = 0.6 \[ \frac{\sqrt{b \sec (c+d x)} \left (3 \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x) \left (2 \sec ^2(c+d x)+3\right )\right )}{8 d \sqrt{\sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sec[c + d*x]^(9/2)*Sqrt[b*Sec[c + d*x]],x]
[Out]
(Sqrt[b*Sec[c + d*x]]*(3*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(3 + 2*Sec[c + d*x]^2)*Tan[c + d*x]))/(8*d*Sqrt[
Sec[c + d*x]])
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Maple [A] time = 0.178, size = 131, normalized size = 1.2 \begin{align*}{\frac{\cos \left ( dx+c \right ) }{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 ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{9}{2}}}\sqrt{{\frac{b}{\cos \left ( dx+c \right ) }}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)^(9/2)*(b*sec(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)*(1/cos(d*x+c))^(9/2)*(b/cos(d*x+c))^(1/2)
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Maxima [B] time = 2.48916, 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(sec(d*x+c)^(9/2)*(b*sec(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))))*sqrt(b)/((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)*d)
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Fricas [A] time = 1.60803, size = 608, normalized size = 5.68 \begin{align*} \left [\frac{3 \, \sqrt{b} \cos \left (d x + c\right )^{3} \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 \,{\left (3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sqrt{\frac{b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{16 \, d \cos \left (d x + c\right )^{3}}, -\frac{3 \, \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 )^{3} - \frac{{\left (3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sqrt{\frac{b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{8 \, d \cos \left (d x + c\right )^{3}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^(9/2)*(b*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
[1/16*(3*sqrt(b)*cos(d*x + c)^3*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*(3*cos(d*x + c)^2 + 2)*sqrt(b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c
)))/(d*cos(d*x + c)^3), -1/8*(3*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)^3 - (3*cos(d*x + c)^2 + 2)*sqrt(b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x +
c)^3)]
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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)**(9/2)*(b*sec(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 \sqrt{b \sec \left (d x + c\right )} \sec \left (d x + c\right )^{\frac{9}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^(9/2)*(b*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate(sqrt(b*sec(d*x + c))*sec(d*x + c)^(9/2), x)