3.2.71 \(\int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx\) [171]

Optimal. Leaf size=27 \[ \frac {x \sqrt {\sec (c+d x)}}{b \sqrt {b \sec (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 27, 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, 8} \begin {gather*} \frac {x \sqrt {\sec (c+d x)}}{b \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rubi steps

\begin {align*} \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx &=\frac {\sqrt {\sec (c+d x)} \int 1 \, dx}{b \sqrt {b \sec (c+d x)}}\\ &=\frac {x \sqrt {\sec (c+d x)}}{b \sqrt {b \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 24, normalized size = 0.89 \begin {gather*} \frac {x \sec ^{\frac {3}{2}}(c+d x)}{(b \sec (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 34.40, size = 32, normalized size = 1.19

method	result	size
default	\(\frac {\left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (d x +c \right )}{d \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}\)	\(32\)
risch	\(\frac {\sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, x}{b \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.53, size = 26, normalized size = 0.96 \begin {gather*} \frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{b^{\frac {3}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 4.26, size = 101, normalized size = 3.74 \begin {gather*} \left [-\frac {\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 )}{2 \, b^{2} d}, \frac {\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 )}{b^{\frac {3}{2}} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*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), arctan(sqrt(b/cos(d*x + c))*sin(d*x + c)/(sqrt(b)*sqrt(cos(d*x + c))))/(b^(3/2)*d)]

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Sympy [A]
time = 13.45, size = 22, normalized size = 0.81 \begin {gather*} \frac {x \sec ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.25, size = 27, normalized size = 1.00 \begin {gather*} \frac {x\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}}{b^2\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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