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

Optimal. Leaf size=35 \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{b^2 d \sqrt {b \sec (c+d x)}} \]

[Out]

sin(d*x+c)*sec(d*x+c)^(1/2)/b^2/d/(b*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {17, 2637} \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{b^2 d \sqrt {b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 35, normalized size = 1.00 \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{b^2 d \sqrt {b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.76, size = 33, normalized size = 0.94 \[ \frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(b/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(b^3*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{\frac {3}{2}}}{\left (b \sec \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.84, size = 41, normalized size = 1.17 \[ \frac {\sin \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}{d \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.90, size = 13, normalized size = 0.37 \[ \frac {\sin \left (d x + c\right )}{b^{\frac {5}{2}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(d*x + c)/(b^(5/2)*d)

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mupad [B]  time = 0.39, size = 39, normalized size = 1.11 \[ \frac {\sin \left (2\,c+2\,d\,x\right )\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{2\,b^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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