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

Optimal. Leaf size=116 \[ \frac{5 \cos ^3(e+f x)}{8 a^2 f \left (a \cos ^2(e+f x)+b\right )}+\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{8 a^{7/2} f}-\frac{15 \cos (e+f x)}{8 a^3 f}+\frac{\cos ^5(e+f x)}{4 a f \left (a \cos ^2(e+f x)+b\right )^2} \]

[Out]

(15*Sqrt[b]*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(8*a^(7/2)*f) - (15*Cos[e + f*x])/(8*a^3*f) + Cos[e + f*x]
^5/(4*a*f*(b + a*Cos[e + f*x]^2)^2) + (5*Cos[e + f*x]^3)/(8*a^2*f*(b + a*Cos[e + f*x]^2))

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Rubi [A]  time = 0.0681461, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {4133, 288, 321, 205} \[ \frac{5 \cos ^3(e+f x)}{8 a^2 f \left (a \cos ^2(e+f x)+b\right )}+\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{8 a^{7/2} f}-\frac{15 \cos (e+f x)}{8 a^3 f}+\frac{\cos ^5(e+f x)}{4 a f \left (a \cos ^2(e+f x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*Sqrt[b]*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(8*a^(7/2)*f) - (15*Cos[e + f*x])/(8*a^3*f) + Cos[e + f*x]
^5/(4*a*f*(b + a*Cos[e + f*x]^2)^2) + (5*Cos[e + f*x]^3)/(8*a^2*f*(b + a*Cos[e + f*x]^2))

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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (b+a x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{f}\\ &=\frac{\cos ^5(e+f x)}{4 a f \left (b+a \cos ^2(e+f x)\right )^2}-\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{\left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{4 a f}\\ &=\frac{\cos ^5(e+f x)}{4 a f \left (b+a \cos ^2(e+f x)\right )^2}+\frac{5 \cos ^3(e+f x)}{8 a^2 f \left (b+a \cos ^2(e+f x)\right )}-\frac{15 \operatorname{Subst}\left (\int \frac{x^2}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{8 a^2 f}\\ &=-\frac{15 \cos (e+f x)}{8 a^3 f}+\frac{\cos ^5(e+f x)}{4 a f \left (b+a \cos ^2(e+f x)\right )^2}+\frac{5 \cos ^3(e+f x)}{8 a^2 f \left (b+a \cos ^2(e+f x)\right )}+\frac{(15 b) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{8 a^3 f}\\ &=\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{8 a^{7/2} f}-\frac{15 \cos (e+f x)}{8 a^3 f}+\frac{\cos ^5(e+f x)}{4 a f \left (b+a \cos ^2(e+f x)\right )^2}+\frac{5 \cos ^3(e+f x)}{8 a^2 f \left (b+a \cos ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 7.14912, size = 656, normalized size = 5.66 \[ \frac{\sec ^6(e+f x) (a \cos (2 (e+f x))+a+2 b)^3 \left (15 \left (a^3+64 b^3\right ) \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}-i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}-\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )+15 \left (a^3+64 b^3\right ) \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}+i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}+\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )+\frac{\sqrt{a} \left (-144 a^2 b^{5/2} \cos (e+f x)-24 a^3 b^{3/2} \cos (e+f x)-72 a^3 b^{3/2} \cos (e+f x) \cos (2 (e+f x))+72 a^2 b^{3/2} \cos (e+f x) (a \cos (2 (e+f x))+a+2 b)+24 a^4 \sqrt{b} \cos (e+f x)-24 a^3 \sqrt{b} \cos (e+f x) (a \cos (2 (e+f x))+a+2 b)-15 a^{5/2} (a \cos (2 (e+f x))+a+2 b)^2 \tan ^{-1}\left (\frac{\sqrt{a}-\sqrt{a+b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )-15 a^{5/2} (a \cos (2 (e+f x))+a+2 b)^2 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan \left (\frac{1}{2} (e+f x)\right )+\sqrt{a}}{\sqrt{b}}\right )+6 a^4 \sqrt{b} \sin (4 (e+f x)) \csc (e+f x)-1152 b^{7/2} \cos (e+f x) (a \cos (2 (e+f x))+a+2 b)-512 b^{5/2} \cos (e) \cos (f x) (a \cos (2 (e+f x))+a+2 b)^2+512 b^{5/2} \sin (e) \sin (f x) (a \cos (2 (e+f x))+a+2 b)^2+512 b^{9/2} \cos (e+f x)\right )}{(a \cos (2 (e+f x))+a+2 b)^2}\right )}{4096 a^{7/2} b^{5/2} f \left (a+b \sec ^2(e+f x)\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])^3*Sec[e + f*x]^6*(15*(a^3 + 64*b^3)*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Co
s[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]] + 15*(a^3 + 64*b^3)*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]] + (Sqrt[a]*(24*a^4
*Sqrt[b]*Cos[e + f*x] - 24*a^3*b^(3/2)*Cos[e + f*x] - 144*a^2*b^(5/2)*Cos[e + f*x] + 512*b^(9/2)*Cos[e + f*x]
- 72*a^3*b^(3/2)*Cos[e + f*x]*Cos[2*(e + f*x)] - 24*a^3*Sqrt[b]*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) +
72*a^2*b^(3/2)*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) - 1152*b^(7/2)*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e +
f*x)]) - 15*a^(5/2)*ArcTan[(Sqrt[a] - Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2
- 15*a^(5/2)*ArcTan[(Sqrt[a] + Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 - 512*
b^(5/2)*Cos[e]*Cos[f*x]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 512*b^(5/2)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Sin[e]
*Sin[f*x] + 6*a^4*Sqrt[b]*Csc[e + f*x]*Sin[4*(e + f*x)]))/(a + 2*b + a*Cos[2*(e + f*x)])^2))/(4096*a^(7/2)*b^(
5/2)*f*(a + b*Sec[e + f*x]^2)^3)

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Maple [A]  time = 0.046, size = 108, normalized size = 0.9 \begin{align*} -{\frac{7\,{b}^{2} \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{8\,f{a}^{3} \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{9\,b\sec \left ( fx+e \right ) }{8\,f{a}^{2} \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{15\,b}{8\,f{a}^{3}}\arctan \left ({b\sec \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{1}{f{a}^{3}\sec \left ( fx+e \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)^3,x)

[Out]

-7/8/f/a^3*b^2/(a+b*sec(f*x+e)^2)^2*sec(f*x+e)^3-9/8/f/a^2*b/(a+b*sec(f*x+e)^2)^2*sec(f*x+e)-15/8/f/a^3*b/(a*b
)^(1/2)*arctan(sec(f*x+e)*b/(a*b)^(1/2))-1/f/a^3/sec(f*x+e)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.603093, size = 713, normalized size = 6.15 \begin{align*} \left [-\frac{16 \, a^{2} \cos \left (f x + e\right )^{5} + 50 \, a b \cos \left (f x + e\right )^{3} + 30 \, b^{2} \cos \left (f x + e\right ) - 15 \,{\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\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 )}{16 \,{\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 )}}, -\frac{8 \, a^{2} \cos \left (f x + e\right )^{5} + 25 \, a b \cos \left (f x + e\right )^{3} + 15 \, b^{2} \cos \left (f x + e\right ) - 15 \,{\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}} \cos \left (f x + e\right )}{b}\right )}{8 \,{\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 )}}\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)^3,x, algorithm="fricas")

[Out]

[-1/16*(16*a^2*cos(f*x + e)^5 + 50*a*b*cos(f*x + e)^3 + 30*b^2*cos(f*x + e) - 15*(a^2*cos(f*x + e)^4 + 2*a*b*c
os(f*x + e)^2 + b^2)*sqrt(-b/a)*log(-(a*cos(f*x + e)^2 + 2*a*sqrt(-b/a)*cos(f*x + e) - b)/(a*cos(f*x + e)^2 +
b)))/(a^5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b^2*f), -1/8*(8*a^2*cos(f*x + e)^5 + 25*a*b*cos(f*
x + e)^3 + 15*b^2*cos(f*x + e) - 15*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(b/a)*arctan(a*sqrt(
b/a)*cos(f*x + e)/b))/(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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.33127, size = 131, normalized size = 1.13 \begin{align*} \frac{15 \, b \arctan \left (\frac{a \cos \left (f x + e\right )}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a^{3} f} - \frac{\cos \left (f x + e\right )}{a^{3} f} - \frac{\frac{9 \, a b \cos \left (f x + e\right )^{3}}{f} + \frac{7 \, b^{2} \cos \left (f x + e\right )}{f}}{8 \,{\left (a \cos \left (f x + e\right )^{2} + b\right )}^{2} a^{3}} \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)^3,x, algorithm="giac")

[Out]

15/8*b*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^3*f) - cos(f*x + e)/(a^3*f) - 1/8*(9*a*b*cos(f*x + e)^3/f
+ 7*b^2*cos(f*x + e)/f)/((a*cos(f*x + e)^2 + b)^2*a^3)