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

Optimal. Leaf size=76 \[ -\frac{2 b \sec (e+f x)}{f (a-b)^2 \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{\cos (e+f x)}{f (a-b) \sqrt{a+b \sec ^2(e+f x)-b}} \]

[Out]

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

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Rubi [A]  time = 0.0638157, antiderivative size = 76, 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 = {3664, 271, 191} \[ -\frac{2 b \sec (e+f x)}{f (a-b)^2 \sqrt{a+b \sec ^2(e+f x)-b}}-\frac{\cos (e+f x)}{f (a-b) \sqrt{a+b \sec ^2(e+f x)-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

Mathematica [A]  time = 1.55434, size = 72, normalized size = 0.95 \[ -\frac{\sec (e+f x) ((a-b) \cos (2 (e+f x))+a+3 b)}{\sqrt{2} f (a-b)^2 \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 103, normalized size = 1.4 \begin{align*} -{\frac{ \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+2\,b \right ) }{f \left ( \cos \left ( fx+e \right ) \right ) ^{3} \left ( a-b \right ) ^{2}} \left ({\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02001, size = 112, normalized size = 1.47 \begin{align*} -\frac{\frac{\sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{2} - 2 \, a b + b^{2}} + \frac{b}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{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*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.45921, size = 236, normalized size = 3.11 \begin{align*} -\frac{{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Integral(sin(e + f*x)/(a + b*tan(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 \tan \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*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

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