3.175 \(\int \frac{1}{\sec ^{\frac{5}{2}}(c+d x) (b \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=107 \[ \frac{3 x \sqrt{\sec (c+d x)}}{8 b \sqrt{b \sec (c+d x)}}+\frac{\sin (c+d x)}{4 b d \sec ^{\frac{5}{2}}(c+d x) \sqrt{b \sec (c+d x)}}+\frac{3 \sin (c+d x)}{8 b d \sqrt{\sec (c+d x)} \sqrt{b \sec (c+d x)}} \]

[Out]

(3*x*Sqrt[Sec[c + d*x]])/(8*b*Sqrt[b*Sec[c + d*x]]) + Sin[c + d*x]/(4*b*d*Sec[c + d*x]^(5/2)*Sqrt[b*Sec[c + d*
x]]) + (3*Sin[c + d*x])/(8*b*d*Sqrt[Sec[c + d*x]]*Sqrt[b*Sec[c + d*x]])

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Rubi [A]  time = 0.027397, 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, 2635, 8} \[ \frac{3 x \sqrt{\sec (c+d x)}}{8 b \sqrt{b \sec (c+d x)}}+\frac{\sin (c+d x)}{4 b d \sec ^{\frac{5}{2}}(c+d x) \sqrt{b \sec (c+d x)}}+\frac{3 \sin (c+d x)}{8 b d \sqrt{\sec (c+d x)} \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sec[c + d*x]^(5/2)*(b*Sec[c + d*x])^(3/2)),x]

[Out]

(3*x*Sqrt[Sec[c + d*x]])/(8*b*Sqrt[b*Sec[c + d*x]]) + Sin[c + d*x]/(4*b*d*Sec[c + d*x]^(5/2)*Sqrt[b*Sec[c + d*
x]]) + (3*Sin[c + d*x])/(8*b*d*Sqrt[Sec[c + d*x]]*Sqrt[b*Sec[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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x) (b \sec (c+d x))^{3/2}} \, dx &=\frac{\sqrt{\sec (c+d x)} \int \cos ^4(c+d x) \, dx}{b \sqrt{b \sec (c+d x)}}\\ &=\frac{\sin (c+d x)}{4 b d \sec ^{\frac{5}{2}}(c+d x) \sqrt{b \sec (c+d x)}}+\frac{\left (3 \sqrt{\sec (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{4 b \sqrt{b \sec (c+d x)}}\\ &=\frac{\sin (c+d x)}{4 b d \sec ^{\frac{5}{2}}(c+d x) \sqrt{b \sec (c+d x)}}+\frac{3 \sin (c+d x)}{8 b d \sqrt{\sec (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{\left (3 \sqrt{\sec (c+d x)}\right ) \int 1 \, dx}{8 b \sqrt{b \sec (c+d x)}}\\ &=\frac{3 x \sqrt{\sec (c+d x)}}{8 b \sqrt{b \sec (c+d x)}}+\frac{\sin (c+d x)}{4 b d \sec ^{\frac{5}{2}}(c+d x) \sqrt{b \sec (c+d x)}}+\frac{3 \sin (c+d x)}{8 b d \sqrt{\sec (c+d x)} \sqrt{b \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.121046, size = 55, normalized size = 0.51 \[ \frac{(12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x))) \sec ^{\frac{3}{2}}(c+d x)}{32 d (b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sec[c + d*x]^(5/2)*(b*Sec[c + d*x])^(3/2)),x]

[Out]

(Sec[c + d*x]^(3/2)*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(32*d*(b*Sec[c + d*x])^(3/2))

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Maple [A]  time = 0.139, size = 74, normalized size = 0.7 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +3\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +3\,dx+3\,c}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{-{\frac{5}{2}}} \left ({\frac{b}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sec(d*x+c)^(5/2)/(b*sec(d*x+c))^(3/2),x)

[Out]

1/8/d*(2*cos(d*x+c)^3*sin(d*x+c)+3*cos(d*x+c)*sin(d*x+c)+3*d*x+3*c)/(1/cos(d*x+c))^(5/2)/(b/cos(d*x+c))^(3/2)/
cos(d*x+c)^4

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Maxima [A]  time = 1.87057, size = 66, normalized size = 0.62 \begin{align*} \frac{12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )}{32 \, b^{\frac{3}{2}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(5/2)/(b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/32*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))))/(b^(3/2)*d)

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Fricas [A]  time = 2.15391, size = 552, normalized size = 5.16 \begin{align*} \left [\frac{\frac{2 \,{\left (2 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2}\right )} \sqrt{\frac{b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}} - 3 \, \sqrt{-b} \log \left (2 \, \sqrt{-b} \sqrt{\frac{b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac{3}{2}} \sin \left (d x + c\right ) + 2 \, b \cos \left (d x + c\right )^{2} - b\right )}{16 \, b^{2} d}, \frac{\frac{{\left (2 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2}\right )} \sqrt{\frac{b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}} + 3 \, \sqrt{b} \arctan \left (\frac{\sqrt{\frac{b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt{b} \sqrt{\cos \left (d x + c\right )}}\right )}{8 \, b^{2} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(5/2)/(b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/16*(2*(2*cos(d*x + c)^4 + 3*cos(d*x + c)^2)*sqrt(b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) - 3*sqrt(-
b)*log(2*sqrt(-b)*sqrt(b/cos(d*x + c))*cos(d*x + c)^(3/2)*sin(d*x + c) + 2*b*cos(d*x + c)^2 - b))/(b^2*d), 1/8
*((2*cos(d*x + c)^4 + 3*cos(d*x + c)^2)*sqrt(b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 3*sqrt(b)*arcta
n(sqrt(b/cos(d*x + c))*sin(d*x + c)/(sqrt(b)*sqrt(cos(d*x + c)))))/(b^2*d)]

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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/sec(d*x+c)**(5/2)/(b*sec(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \sec \left (d x + c\right )\right )^{\frac{3}{2}} \sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(5/2)/(b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*sec(d*x + c))^(3/2)*sec(d*x + c)^(5/2)), x)