3.3.27 \(\int \frac {\sin (a+b x)}{\sqrt {d \sec (a+b x)}} \, dx\) [227]

Optimal. Leaf size=20 \[ -\frac {2 d}{3 b (d \sec (a+b x))^{3/2}} \]

[Out]

-2/3*d/b/(d*sec(b*x+a))^(3/2)

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Rubi [A]
time = 0.02, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2702, 30} \begin {gather*} -\frac {2 d}{3 b (d \sec (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]/Sqrt[d*Sec[a + b*x]],x]

[Out]

(-2*d)/(3*b*(d*Sec[a + b*x])^(3/2))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2702

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rubi steps

\begin {align*} \int \frac {\sin (a+b x)}{\sqrt {d \sec (a+b x)}} \, dx &=\frac {d \text {Subst}\left (\int \frac {1}{x^{5/2}} \, dx,x,d \sec (a+b x)\right )}{b}\\ &=-\frac {2 d}{3 b (d \sec (a+b x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 20, normalized size = 1.00 \begin {gather*} -\frac {2 d}{3 b (d \sec (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]/Sqrt[d*Sec[a + b*x]],x]

[Out]

(-2*d)/(3*b*(d*Sec[a + b*x])^(3/2))

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Maple [A]
time = 0.44, size = 17, normalized size = 0.85

method	result	size
derivativedivides	\(-\frac {2 d}{3 b \left (d \sec \left (b x +a \right )\right )^{\frac {3}{2}}}\)	\(17\)
default	\(-\frac {2 d}{3 b \left (d \sec \left (b x +a \right )\right )^{\frac {3}{2}}}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*d/b/(d*sec(b*x+a))^(3/2)

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Maxima [A]
time = 0.29, size = 23, normalized size = 1.15 \begin {gather*} -\frac {2 \, \cos \left (b x + a\right )}{3 \, b \sqrt {\frac {d}{\cos \left (b x + a\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*cos(b*x + a)/(b*sqrt(d/cos(b*x + a)))

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Fricas [A]
time = 2.40, size = 28, normalized size = 1.40 \begin {gather*} -\frac {2 \, \sqrt {\frac {d}{\cos \left (b x + a\right )}} \cos \left (b x + a\right )^{2}}{3 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*sqrt(d/cos(b*x + a))*cos(b*x + a)^2/(b*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin {\left (a + b x \right )}}{\sqrt {d \sec {\left (a + b x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b*x)/sqrt(d*sec(a + b*x)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (16) = 32\).
time = 0.44, size = 33, normalized size = 1.65 \begin {gather*} -\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \cos \left (b x + a\right )}{3 \, b d \mathrm {sgn}\left (\cos \left (b x + a\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)/(d*sec(b*x+a))^(1/2),x, algorithm="giac")

[Out]

-2/3*sqrt(d*cos(b*x + a))*cos(b*x + a)/(b*d*sgn(cos(b*x + a)))

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Mupad [B]
time = 0.24, size = 28, normalized size = 1.40 \begin {gather*} -\frac {2\,{\cos \left (a+b\,x\right )}^2\,\sqrt {\frac {d}{\cos \left (a+b\,x\right )}}}{3\,b\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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