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

Optimal. Leaf size=71 \[ -\frac{(a+b) \cos (e+f x)}{a^2 f}+\frac{\sqrt{b} (a+b) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{a^{5/2} f}+\frac{\cos ^3(e+f x)}{3 a f} \]

[Out]

(Sqrt[b]*(a + b)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(a^(5/2)*f) - ((a + b)*Cos[e + f*x])/(a^2*f) + Cos[e
+ f*x]^3/(3*a*f)

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Rubi [A]  time = 0.0840075, antiderivative size = 71, 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 = {4133, 459, 321, 205} \[ -\frac{(a+b) \cos (e+f x)}{a^2 f}+\frac{\sqrt{b} (a+b) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{a^{5/2} f}+\frac{\cos ^3(e+f x)}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*(a + b)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(a^(5/2)*f) - ((a + b)*Cos[e + f*x])/(a^2*f) + Cos[e
+ f*x]^3/(3*a*f)

Rule 4133

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

Rule 459

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

Rule 321

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sin ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (1-x^2\right )}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x)}{3 a f}-\frac{(a+b) \operatorname{Subst}\left (\int \frac{x^2}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{a f}\\ &=-\frac{(a+b) \cos (e+f x)}{a^2 f}+\frac{\cos ^3(e+f x)}{3 a f}+\frac{(b (a+b)) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{a^2 f}\\ &=\frac{\sqrt{b} (a+b) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{a^{5/2} f}-\frac{(a+b) \cos (e+f x)}{a^2 f}+\frac{\cos ^3(e+f x)}{3 a f}\\ \end{align*}

Mathematica [C]  time = 1.62599, size = 376, normalized size = 5.3 \[ \frac{\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (3 \left (a^2+8 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}-i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}-\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )+3 \left (a^2+8 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}+i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}+\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )-3 a^2 \tan ^{-1}\left (\frac{\sqrt{a}-\sqrt{a+b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )-3 a^2 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan \left (\frac{1}{2} (e+f x)\right )+\sqrt{a}}{\sqrt{b}}\right )+4 \sqrt{a} \sqrt{b} \cos (e+f x) (a \cos (2 (e+f x))-5 a-6 b)\right )}{48 a^{5/2} \sqrt{b} f \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*(3*(a^2 + 8*a*b + 8*b^2)*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Si
n[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[
b]] + 3*(a^2 + 8*a*b + 8*b^2)*ArcTan[((-Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)/
2] + Cos[e]*(Sqrt[a] + Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]] - 3*a^2*ArcTan[(Sqrt[a]
- Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]] - 3*a^2*ArcTan[(Sqrt[a] + Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]] + 4
*Sqrt[a]*Sqrt[b]*Cos[e + f*x]*(-5*a - 6*b + a*Cos[2*(e + f*x)]))*Sec[e + f*x]^2)/(48*a^(5/2)*Sqrt[b]*f*(a + b*
Sec[e + f*x]^2))

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Maple [A]  time = 0.064, size = 103, normalized size = 1.5 \begin{align*}{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{3\,af}}-{\frac{\cos \left ( fx+e \right ) }{af}}-{\frac{b\cos \left ( fx+e \right ) }{f{a}^{2}}}+{\frac{b}{af}\arctan \left ({a\cos \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{{b}^{2}}{f{a}^{2}}\arctan \left ({a\cos \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \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),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.551744, size = 381, normalized size = 5.37 \begin{align*} \left [\frac{2 \, a \cos \left (f x + e\right )^{3} + 3 \,{\left (a + b\right )} \sqrt{-\frac{b}{a}} \log \left (-\frac{a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt{-\frac{b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) - 6 \,{\left (a + b\right )} \cos \left (f x + e\right )}{6 \, a^{2} f}, \frac{a \cos \left (f x + e\right )^{3} + 3 \,{\left (a + b\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}} \cos \left (f x + e\right )}{b}\right ) - 3 \,{\left (a + b\right )} \cos \left (f x + e\right )}{3 \, a^{2} 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),x, algorithm="fricas")

[Out]

[1/6*(2*a*cos(f*x + e)^3 + 3*(a + b)*sqrt(-b/a)*log(-(a*cos(f*x + e)^2 + 2*a*sqrt(-b/a)*cos(f*x + e) - b)/(a*c
os(f*x + e)^2 + b)) - 6*(a + b)*cos(f*x + e))/(a^2*f), 1/3*(a*cos(f*x + e)^3 + 3*(a + b)*sqrt(b/a)*arctan(a*sq
rt(b/a)*cos(f*x + e)/b) - 3*(a + b)*cos(f*x + e))/(a^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)**3/(a+b*sec(f*x+e)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.18679, size = 120, normalized size = 1.69 \begin{align*} \frac{{\left (a b + b^{2}\right )} \arctan \left (\frac{a \cos \left (f x + e\right )}{\sqrt{a b}}\right )}{\sqrt{a b} a^{2} f} + \frac{a^{2} f^{5} \cos \left (f x + e\right )^{3} - 3 \, a^{2} f^{5} \cos \left (f x + e\right ) - 3 \, a b f^{5} \cos \left (f x + e\right )}{3 \, a^{3} f^{6}} \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),x, algorithm="giac")

[Out]

(a*b + b^2)*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^2*f) + 1/3*(a^2*f^5*cos(f*x + e)^3 - 3*a^2*f^5*cos(f
*x + e) - 3*a*b*f^5*cos(f*x + e))/(a^3*f^6)