3.2.48 \(\int \frac {(b \sec (c+d x))^{3/2}}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [148]

Optimal. Leaf size=33 \[ \frac {b \sqrt {b \sec (c+d x)} \sin (c+d x)}{d \sqrt {\sec (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 33, 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, 2717} \begin {gather*} \frac {b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d \sqrt {\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[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 2717

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

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Mathematica [A]
time = 0.05, size = 32, normalized size = 0.97 \begin {gather*} \frac {(b \sec (c+d x))^{3/2} \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 34.12, size = 41, normalized size = 1.24

method	result	size
default	\(\frac {\sin \left (d x +c \right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}{d \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \cos \left (d x +c \right )}\)	\(41\)
risch	\(-\frac {i b \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{i \left (d x +c \right )}}{2 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {i b \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{-i \left (d x +c \right )}}{2 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(136\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.64, size = 13, normalized size = 0.39 \begin {gather*} \frac {b^{\frac {3}{2}} \sin \left (d x + c\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^(3/2)*sin(d*x + c)/d

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Fricas [A]
time = 3.17, size = 31, normalized size = 0.94 \begin {gather*} \frac {b \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 29.16, size = 46, normalized size = 1.39 \begin {gather*} \begin {cases} \frac {\left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan {\left (c + d x \right )}}{d \sec ^{\frac {5}{2}}{\left (c + d x \right )}} & \text {for}\: d \neq 0 \\\frac {x \left (b \sec {\left (c \right )}\right )^{\frac {3}{2}}}{\sec ^{\frac {5}{2}}{\left (c \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((b*sec(c + d*x))**(3/2)*tan(c + d*x)/(d*sec(c + d*x)**(5/2)), Ne(d, 0)), (x*(b*sec(c))**(3/2)/sec(c
)**(5/2), True))

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

[Out]

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

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Mupad [B]
time = 0.25, size = 33, normalized size = 1.00 \begin {gather*} \frac {b\,\sin \left (c+d\,x\right )\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}}{d\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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