3.118 \(\int \frac {\sin (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\)

Optimal. Leaf size=37 \[ -\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{f (a-b)} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 37, 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 = {3664, 264} \[ -\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{f (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]/Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

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

Rule 264

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

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.60, size = 52, normalized size = 1.41 \[ \frac {\cos (e+f x) \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}{\sqrt {2} f (b-a)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]/Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

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

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fricas [A]  time = 0.62, size = 45, normalized size = 1.22 \[ -\frac {\sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a - b\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

-sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/((a - 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*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.37, size = 78, normalized size = 2.11 \[ -\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{f \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}}\, \cos \left (f x +e \right ) \left (a -b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2),x)

[Out]

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

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maxima [A]  time = 0.62, size = 35, normalized size = 0.95 \[ -\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a - b\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\sin \left (e+f\,x\right )}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)/(a + b*tan(e + f*x)^2)^(1/2),x)

[Out]

int(sin(e + f*x)/(a + b*tan(e + f*x)^2)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)/(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

Integral(sin(e + f*x)/sqrt(a + b*tan(e + f*x)**2), x)

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