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

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Rubi [A]  time = 0.06, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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]

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 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])

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*}

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Mathematica [A]  time = 1.27, 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]

-1/4*((a + 2*b + a*Cos[2*(e + f*x)])*(a + 4*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^3)/(a^2*f*(a + b*Sec[e + f*x]
^2)^(3/2))

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fricas [A]  time = 0.51, size = 67, normalized size = 1.08 \[ -\frac {{\left (a \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{a^{3} f \cos \left (f x + e\right )^{2} + a^{2} b f} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*2*(2*(1/8*b^2/
a^2/sign(tan((f*x+exp(1))/2)^2-1)/b+1/8*b^2*tan((f*x+exp(1))/2)^2/a^2/sign(tan((f*x+exp(1))/2)^2-1)/b)/sqrt(a*
tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b)+(-tan((
f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))
/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))/a/(2*sqrt(a+b)*(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))
/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))-(-tan((f*x+exp(1))/2)^
2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1)
)/2)^2+a+b))^2+3*a-b)/sign(tan((f*x+exp(1))/2)^2-1))

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maple [A]  time = 0.29, size = 59, normalized size = 0.95 \[ \frac {-\frac {1}{a \sec \left (f x +e \right ) \sqrt {a +b \left (\sec ^{2}\left (f x +e \right )\right )}}-\frac {2 b \sec \left (f x +e \right )}{a^{2} \sqrt {a +b \left (\sec ^{2}\left (f x +e \right )\right )}}}{f} \]

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 = 0.35, size = 57, normalized size = 0.92 \[ -\frac {\frac {\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{2}} + \frac {b}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a^{2} \cos \left (f x + e\right )}}{f} \]

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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mupad [B]  time = 10.87, size = 155, normalized size = 2.50 \[ -\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {a+\frac {b}{{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}\right )}^2}}\,\left (a+2\,a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}+8\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\right )}{2\,a^2\,f\,\left (a+2\,a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}+4\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(exp(- e*1i - f*x*1i)*(exp(e*2i + f*x*2i) + 1)*(a + b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2)^2)^(1/2
)*(a + 2*a*exp(e*2i + f*x*2i) + a*exp(e*4i + f*x*4i) + 8*b*exp(e*2i + f*x*2i)))/(2*a^2*f*(a + 2*a*exp(e*2i + f
*x*2i) + a*exp(e*4i + f*x*4i) + 4*b*exp(e*2i + f*x*2i)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin {\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

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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