3.108 \(\int \frac{\sin (e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=62 \[ -\frac{2 b \sec (e+f x)}{a^2 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{\cos (e+f x)}{a f \sqrt{a+b \sec ^2(e+f x)}} \]

[Out]

-(Cos[e + f*x]/(a*f*Sqrt[a + b*Sec[e + f*x]^2])) - (2*b*Sec[e + f*x])/(a^2*f*Sqrt[a + b*Sec[e + f*x]^2])

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Rubi [A]  time = 0.0561524, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4134, 271, 191} \[ -\frac{2 b \sec (e+f x)}{a^2 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{\cos (e+f x)}{a f \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]/(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

-(Cos[e + f*x]/(a*f*Sqrt[a + b*Sec[e + f*x]^2])) - (2*b*Sec[e + f*x])/(a^2*f*Sqrt[a + b*Sec[e + f*x]^2])

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [A]  time = 1.42073, size = 64, normalized size = 1.03 \[ -\frac{\sec ^3(e+f x) (a \cos (2 (e+f x))+a+2 b) (a \cos (2 (e+f x))+a+4 b)}{4 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]/(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

-((a + 2*b + a*Cos[2*(e + f*x)])*(a + 4*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^3)/(4*a^2*f*(a + b*Sec[e + f*x]^2
)^(3/2))

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Maple [A]  time = 0.052, size = 59, normalized size = 1. \begin{align*}{\frac{1}{f} \left ( -{\frac{1}{a\sec \left ( fx+e \right ) }{\frac{1}{\sqrt{a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2}}}}}-2\,{\frac{b\sec \left ( fx+e \right ) }{{a}^{2}\sqrt{a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)/(a+b*sec(f*x+e)^2)^(3/2),x)

[Out]

1/f*(-1/a/sec(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2)-2*b/a^2*sec(f*x+e)/(a+b*sec(f*x+e)^2)^(1/2))

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Maxima [A]  time = 1.00267, size = 77, normalized size = 1.24 \begin{align*} -\frac{\frac{\sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{2}} + \frac{b}{\sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} a^{2} \cos \left (f x + e\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

-(sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e)/a^2 + b/(sqrt(a + b/cos(f*x + e)^2)*a^2*cos(f*x + e)))/f

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Fricas [A]  time = 0.607297, size = 158, normalized size = 2.55 \begin{align*} -\frac{{\left (a \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{a^{3} f \cos \left (f x + e\right )^{2} + a^{2} b f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)/(a+b*sec(f*x+e)**2)**(3/2),x)

[Out]

Integral(sin(e + f*x)/(a + b*sec(e + f*x)**2)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)/(b*sec(f*x + e)^2 + a)^(3/2), x)