3.190 \(\int \frac {\sin (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=20 \[ \frac {2}{b d \sqrt {d \cos (a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.03, 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 = {2565, 30} \[ \frac {2}{b d \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]/(d*Cos[a + b*x])^(3/2),x]

[Out]

2/(b*d*Sqrt[d*Cos[a + b*x]])

Rule 30

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

Rule 2565

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 20, normalized size = 1.00 \[ \frac {2}{b d \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]/(d*Cos[a + b*x])^(3/2),x]

[Out]

2/(b*d*Sqrt[d*Cos[a + b*x]])

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fricas [A]  time = 0.42, size = 26, normalized size = 1.30 \[ \frac {2 \, \sqrt {d \cos \left (b x + a\right )}}{b d^{2} \cos \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)/(d*cos(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.84, size = 18, normalized size = 0.90 \[ \frac {2}{\sqrt {d \cos \left (b x + a\right )} b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)/(d*cos(b*x+a))^(3/2),x, algorithm="giac")

[Out]

2/(sqrt(d*cos(b*x + a))*b*d)

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maple [A]  time = 0.01, size = 19, normalized size = 0.95 \[ \frac {2}{b d \sqrt {d \cos \left (b x +a \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)/(d*cos(b*x+a))^(3/2),x)

[Out]

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

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maxima [A]  time = 0.53, size = 18, normalized size = 0.90 \[ \frac {2}{\sqrt {d \cos \left (b x + a\right )} b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)/(d*cos(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

2/(sqrt(d*cos(b*x + a))*b*d)

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mupad [B]  time = 0.18, size = 37, normalized size = 1.85 \[ \frac {4\,\cos \left (a+b\,x\right )\,\sqrt {d\,\cos \left (a+b\,x\right )}}{b\,d^2\,\left (\cos \left (2\,a+2\,b\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)/(d*cos(a + b*x))^(3/2),x)

[Out]

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

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sympy [A]  time = 5.96, size = 31, normalized size = 1.55 \[ \begin {cases} \frac {2}{b d^{\frac {3}{2}} \sqrt {\cos {\left (a + b x \right )}}} & \text {for}\: b \neq 0 \\\frac {x \sin {\relax (a )}}{\left (d \cos {\relax (a )}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)/(d*cos(b*x+a))**(3/2),x)

[Out]

Piecewise((2/(b*d**(3/2)*sqrt(cos(a + b*x))), Ne(b, 0)), (x*sin(a)/(d*cos(a))**(3/2), True))

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