3.2.65 \(\int \frac {1}{\sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}} \, dx\) [165]

Optimal. Leaf size=32 \[ \frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {b \sec (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 32, 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 = {18, 2717} \begin {gather*} \frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Sec[c + d*x]]*Sqrt[b*Sec[c + d*x]]),x]

[Out]

(Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*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 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 {1}{\sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}} \, dx &=\frac {\sqrt {\sec (c+d x)} \int \cos (c+d x) \, dx}{\sqrt {b \sec (c+d x)}}\\ &=\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {b \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 32, normalized size = 1.00 \begin {gather*} \frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Sec[c + d*x]]*Sqrt[b*Sec[c + d*x]]),x]

[Out]

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

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Maple [A]
time = 35.23, size = 41, normalized size = 1.28

method	result	size
default	\(\frac {\sin \left (d x +c \right )}{d \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sqrt {\frac {b}{\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )}\)	\(41\)
risch	\(-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{2 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {i}{2 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(151\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.59, size = 13, normalized size = 0.41 \begin {gather*} \frac {\sin \left (d x + c\right )}{\sqrt {b} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(d*x + c)/(sqrt(b)*d)

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Fricas [A]
time = 3.35, size = 33, normalized size = 1.03 \begin {gather*} \frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 20.37, size = 46, normalized size = 1.44 \begin {gather*} \begin {cases} \frac {\tan {\left (c + d x \right )}}{d \sqrt {b \sec {\left (c + d x \right )}} \sqrt {\sec {\left (c + d x \right )}}} & \text {for}\: d \neq 0 \\\frac {x}{\sqrt {b \sec {\left (c \right )}} \sqrt {\sec {\left (c \right )}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((tan(c + d*x)/(d*sqrt(b*sec(c + d*x))*sqrt(sec(c + d*x))), Ne(d, 0)), (x/(sqrt(b*sec(c))*sqrt(sec(c)
)), 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(1/sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(b*sec(d*x + c))*sqrt(sec(d*x + c))), x)

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Mupad [B]
time = 0.31, size = 35, normalized size = 1.09 \begin {gather*} \frac {\sin \left (c+d\,x\right )\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}}{b\,d\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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