3.121 \(\int \frac{\sin (e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=97 \[ -\frac{8 b \sec (e+f x)}{3 a^3 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{4 b \sec (e+f x)}{3 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cos (e+f x)}{a f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0684935, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4134, 271, 192, 191} \[ -\frac{8 b \sec (e+f x)}{3 a^3 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{4 b \sec (e+f x)}{3 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\cos (e+f x)}{a f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

-(Cos[e + f*x]/(a*f*(a + b*Sec[e + f*x]^2)^(3/2))) - (4*b*Sec[e + f*x])/(3*a^2*f*(a + b*Sec[e + f*x]^2)^(3/2))
- (8*b*Sec[e + f*x])/(3*a^3*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 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -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 )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x)}{a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{a f}\\ &=-\frac{\cos (e+f x)}{a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{4 b \sec (e+f x)}{3 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{(8 b) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{3 a^2 f}\\ &=-\frac{\cos (e+f x)}{a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{4 b \sec (e+f x)}{3 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{8 b \sec (e+f x)}{3 a^3 f \sqrt{a+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.45492, size = 88, normalized size = 0.91 \[ -\frac{\sec ^5(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (3 a^2 \cos (4 (e+f x))+12 a (a+4 b) \cos (2 (e+f x))+(3 a+8 b)^2\right )}{48 a^3 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

-((a + 2*b + a*Cos[2*(e + f*x)])*((3*a + 8*b)^2 + 12*a*(a + 4*b)*Cos[2*(e + f*x)] + 3*a^2*Cos[4*(e + f*x)])*Se
c[e + f*x]^5)/(48*a^3*f*(a + b*Sec[e + f*x]^2)^(5/2))

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Maple [A]  time = 0.052, size = 90, normalized size = 0.9 \begin{align*}{\frac{1}{f} \left ( -{\frac{1}{a\sec \left ( fx+e \right ) } \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}-4\,{\frac{b}{a} \left ( 1/3\,{\frac{\sec \left ( fx+e \right ) }{a \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{3/2}}}+2/3\,{\frac{\sec \left ( fx+e \right ) }{{a}^{2}\sqrt{a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2}}}} \right ) } \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)^(5/2),x)

[Out]

1/f*(-1/a/sec(f*x+e)/(a+b*sec(f*x+e)^2)^(3/2)-4*b/a*(1/3*sec(f*x+e)/a/(a+b*sec(f*x+e)^2)^(3/2)+2/3/a^2*sec(f*x
+e)/(a+b*sec(f*x+e)^2)^(1/2)))

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Maxima [A]  time = 1.00032, size = 116, normalized size = 1.2 \begin{align*} -\frac{\frac{3 \, \sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{3}} + \frac{6 \,{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )} b \cos \left (f x + e\right )^{2} - b^{2}}{{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} a^{3} \cos \left (f x + e\right )^{3}}}{3 \, 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)^(5/2),x, algorithm="maxima")

[Out]

-1/3*(3*sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e)/a^3 + (6*(a + b/cos(f*x + e)^2)*b*cos(f*x + e)^2 - b^2)/((a +
b/cos(f*x + e)^2)^(3/2)*a^3*cos(f*x + e)^3))/f

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Fricas [A]  time = 0.812738, size = 243, normalized size = 2.51 \begin{align*} -\frac{{\left (3 \, a^{2} \cos \left (f x + e\right )^{5} + 12 \, a b \cos \left (f x + e\right )^{3} + 8 \, b^{2} \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}}}}{3 \,{\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}} \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)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(3*a^2*cos(f*x + e)^5 + 12*a*b*cos(f*x + e)^3 + 8*b^2*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x +
e)^2)/(a^5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b^2*f)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(5/2),x)

[Out]

Timed out

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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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)/(b*sec(f*x + e)^2 + a)^(5/2), x)