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

Optimal. Leaf size=114 \[ -\frac{2 b (3 a+4 b) \sec (e+f x)}{3 a^3 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{(3 a+4 b) \cos (e+f x)}{3 a^2 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\cos ^3(e+f x)}{3 a f \sqrt{a+b \sec ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.116966, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {4134, 453, 271, 191} \[ -\frac{2 b (3 a+4 b) \sec (e+f x)}{3 a^3 f \sqrt{a+b \sec ^2(e+f x)}}-\frac{(3 a+4 b) \cos (e+f x)}{3 a^2 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\cos ^3(e+f x)}{3 a f \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a + 4*b)*Cos[e + f*x])/(3*a^2*f*Sqrt[a + b*Sec[e + f*x]^2]) + Cos[e + f*x]^3/(3*a*f*Sqrt[a + b*Sec[e + f*
x]^2]) - (2*b*(3*a + 4*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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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 ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{-1+x^2}{x^4 \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x)}{3 a f \sqrt{a+b \sec ^2(e+f x)}}+\frac{(3 a+4 b) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{3 a f}\\ &=-\frac{(3 a+4 b) \cos (e+f x)}{3 a^2 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\cos ^3(e+f x)}{3 a f \sqrt{a+b \sec ^2(e+f x)}}-\frac{(2 b (3 a+4 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{(3 a+4 b) \cos (e+f x)}{3 a^2 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\cos ^3(e+f x)}{3 a f \sqrt{a+b \sec ^2(e+f x)}}-\frac{2 b (3 a+4 b) \sec (e+f x)}{3 a^3 f \sqrt{a+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 3.68715, size = 93, normalized size = 0.82 \[ -\frac{\sec ^3(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (a^2 (-\cos (4 (e+f x)))+9 a^2+8 a (a+2 b) \cos (2 (e+f x))+64 a b+64 b^2\right )}{48 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + 2*b + a*Cos[2*(e + f*x)])*(9*a^2 + 64*a*b + 64*b^2 + 8*a*(a + 2*b)*Cos[2*(e + f*x)] - a^2*Cos[4*(e + f*
x)])*Sec[e + f*x]^3)/(48*a^3*f*(a + b*Sec[e + f*x]^2)^(3/2))

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Maple [B]  time = 0.89, size = 12782, normalized size = 112.1 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [A]  time = 1.01991, size = 190, normalized size = 1.67 \begin{align*} -\frac{\frac{3 \, \sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{2}} - \frac{{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3} - 6 \, \sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )}{a^{3}} + \frac{3 \, b}{\sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} a^{2} \cos \left (f x + e\right )} + \frac{3 \, b^{2}}{\sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} a^{3} \cos \left (f x + e\right )}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.722797, size = 228, normalized size = 2. \begin{align*} \frac{{\left (a^{2} \cos \left (f x + e\right )^{5} -{\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{3} - 2 \,{\left (3 \, a b + 4 \, b^{2}\right )} \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^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a^2*cos(f*x + e)^5 - (3*a^2 + 4*a*b)*cos(f*x + e)^3 - 2*(3*a*b + 4*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e
)^2 + b)/cos(f*x + e)^2)/(a^4*f*cos(f*x + e)^2 + a^3*b*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)**3/(a+b*sec(f*x+e)**2)**(3/2),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 )^{3}}{{\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)^3/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

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