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

Optimal. Leaf size=154 \[ \frac{b^2 (a+b) \cos (e+f x)}{4 a^4 f \left (a \cos ^2(e+f x)+b\right )^2}-\frac{b (9 a+13 b) \cos (e+f x)}{8 a^4 f \left (a \cos ^2(e+f x)+b\right )}-\frac{(a+3 b) \cos (e+f x)}{a^4 f}+\frac{5 \sqrt{b} (3 a+7 b) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{8 a^{9/2} f}+\frac{\cos ^3(e+f x)}{3 a^3 f} \]

[Out]

(5*Sqrt[b]*(3*a + 7*b)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(8*a^(9/2)*f) - ((a + 3*b)*Cos[e + f*x])/(a^4*f
) + Cos[e + f*x]^3/(3*a^3*f) + (b^2*(a + b)*Cos[e + f*x])/(4*a^4*f*(b + a*Cos[e + f*x]^2)^2) - (b*(9*a + 13*b)
*Cos[e + f*x])/(8*a^4*f*(b + a*Cos[e + f*x]^2))

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Rubi [A]  time = 0.188327, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4133, 455, 1814, 1153, 205} \[ \frac{b^2 (a+b) \cos (e+f x)}{4 a^4 f \left (a \cos ^2(e+f x)+b\right )^2}-\frac{b (9 a+13 b) \cos (e+f x)}{8 a^4 f \left (a \cos ^2(e+f x)+b\right )}-\frac{(a+3 b) \cos (e+f x)}{a^4 f}+\frac{5 \sqrt{b} (3 a+7 b) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{8 a^{9/2} f}+\frac{\cos ^3(e+f x)}{3 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*Sqrt[b]*(3*a + 7*b)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(8*a^(9/2)*f) - ((a + 3*b)*Cos[e + f*x])/(a^4*f
) + Cos[e + f*x]^3/(3*a^3*f) + (b^2*(a + b)*Cos[e + f*x])/(4*a^4*f*(b + a*Cos[e + f*x]^2)^2) - (b*(9*a + 13*b)
*Cos[e + f*x])/(8*a^4*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 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^6 \left (1-x^2\right )}{\left (b+a x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{f}\\ &=\frac{b^2 (a+b) \cos (e+f x)}{4 a^4 f \left (b+a \cos ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{-b^2 (a+b)+4 a b (a+b) x^2-4 a^2 (a+b) x^4+4 a^3 x^6}{\left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{4 a^4 f}\\ &=\frac{b^2 (a+b) \cos (e+f x)}{4 a^4 f \left (b+a \cos ^2(e+f x)\right )^2}-\frac{b (9 a+13 b) \cos (e+f x)}{8 a^4 f \left (b+a \cos ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-b^2 (7 a+11 b)+8 a b (a+2 b) x^2-8 a^2 b x^4}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{8 a^4 b f}\\ &=\frac{b^2 (a+b) \cos (e+f x)}{4 a^4 f \left (b+a \cos ^2(e+f x)\right )^2}-\frac{b (9 a+13 b) \cos (e+f x)}{8 a^4 f \left (b+a \cos ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \left (8 b (a+3 b)-8 a b x^2-\frac{5 \left (3 a b^2+7 b^3\right )}{b+a x^2}\right ) \, dx,x,\cos (e+f x)\right )}{8 a^4 b f}\\ &=-\frac{(a+3 b) \cos (e+f x)}{a^4 f}+\frac{\cos ^3(e+f x)}{3 a^3 f}+\frac{b^2 (a+b) \cos (e+f x)}{4 a^4 f \left (b+a \cos ^2(e+f x)\right )^2}-\frac{b (9 a+13 b) \cos (e+f x)}{8 a^4 f \left (b+a \cos ^2(e+f x)\right )}+\frac{(5 b (3 a+7 b)) \operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{8 a^4 f}\\ &=\frac{5 \sqrt{b} (3 a+7 b) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{8 a^{9/2} f}-\frac{(a+3 b) \cos (e+f x)}{a^4 f}+\frac{\cos ^3(e+f x)}{3 a^3 f}+\frac{b^2 (a+b) \cos (e+f x)}{4 a^4 f \left (b+a \cos ^2(e+f x)\right )^2}-\frac{b (9 a+13 b) \cos (e+f x)}{8 a^4 f \left (b+a \cos ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 11.6896, size = 1392, normalized size = 9.04 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(3*((-3*ArcTan[(Sqrt[a] - Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]])/Sqrt[a] - (3*ArcTan[(Sqrt[a] + Sqrt[a + b]*T
an[(e + f*x)/2])/Sqrt[b]])/Sqrt[a] - (2*Sqrt[b]*Cos[e + f*x]*(3*a + 10*b + 3*a*Cos[2*(e + f*x)]))/(a + 2*b + a
*Cos[2*(e + f*x)])^2)*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6)/(8192*b^(5/2)*f*(a + b*Sec[e + f*x]^2)^
3) + (((3*a - 4*b)*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 - 4*b)*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]] + (2*Sqrt[a]*Sqrt[b]*Cos[e + f*x]*(3*a^2 + 6*a*b + 8*b^2 + a*(3*a - 4*b)*
Cos[2*(e + f*x)]))/(a + 2*b + a*Cos[2*(e + f*x)])^2)*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6)/(2048*a^
(3/2)*b^(5/2)*f*(a + b*Sec[e + f*x]^2)^3) - ((-3*(3*a^4 - 40*a^3*b + 720*a^2*b^2 + 6720*a*b^3 + 8960*b^4)*ArcT
an[((-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*(3*a^4 - 40*a^3*b + 720*a^2*b^2 + 6720*a*b^3 + 8960*b
^4)*ArcTan[((-Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] + Sqr
t[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]] - (2*Sqrt[a]*Sqrt[b]*Cos[e + f*x]*(9*a^5 - 90*a^4
*b - 10144*a^3*b^2 - 48672*a^2*b^3 - 85120*a*b^4 - 53760*b^5 + a*(9*a^4 - 120*a^3*b - 12432*a^2*b^2 - 47936*a*
b^3 - 44800*b^4)*Cos[2*(e + f*x)] - 128*a^2*b^2*(15*a + 28*b)*Cos[4*(e + f*x)] + 128*a^3*b^2*Cos[6*(e + f*x)])
)/(a + 2*b + a*Cos[2*(e + f*x)])^2)*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6)/(49152*a^(9/2)*b^(5/2)*f*
(a + b*Sec[e + f*x]^2)^3) - (3*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*((3*(a^3 - 8*a^2*b + 80*a*b^2 +
320*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]])/b^(5/2) + (3*(a^3 - 8*a^2*b + 80*a*b^2 + 3
20*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]])/b^(5/2) - 512*Sqrt[a]*Cos[e]*Cos[f*x] + (8*S
qrt[a]*(a^3 + 24*a^2*b + 80*a*b^2 + 64*b^3)*Cos[e + f*x])/(b*(a + 2*b + a*Cos[2*(e + f*x)])^2) + (2*Sqrt[a]*(3
*a^3 - 24*a^2*b - 400*a*b^2 - 576*b^3)*Cos[e + f*x])/(b^2*(a + 2*b + a*Cos[2*(e + f*x)])) + 512*Sqrt[a]*Sin[e]
*Sin[f*x]))/(16384*a^(7/2)*f*(a + b*Sec[e + f*x]^2)^3)

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Maple [A]  time = 0.093, size = 231, normalized size = 1.5 \begin{align*}{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{3\,{a}^{3}f}}-{\frac{\cos \left ( fx+e \right ) }{{a}^{3}f}}-3\,{\frac{b\cos \left ( fx+e \right ) }{f{a}^{4}}}-{\frac{9\,b \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{8\,f{a}^{2} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{13\,{b}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{8\,{a}^{3}f \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{7\,{b}^{2}\cos \left ( fx+e \right ) }{8\,{a}^{3}f \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{11\,{b}^{3}\cos \left ( fx+e \right ) }{8\,f{a}^{4} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{15\,b}{8\,{a}^{3}f}\arctan \left ({a\cos \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{35\,{b}^{2}}{8\,f{a}^{4}}\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)^3,x)

[Out]

1/3*cos(f*x+e)^3/a^3/f-cos(f*x+e)/a^3/f-3/f/a^4*b*cos(f*x+e)-9/8/f*b/a^2/(b+a*cos(f*x+e)^2)^2*cos(f*x+e)^3-13/
8/f*b^2/a^3/(b+a*cos(f*x+e)^2)^2*cos(f*x+e)^3-7/8/f*b^2/a^3/(b+a*cos(f*x+e)^2)^2*cos(f*x+e)-11/8/f*b^3/a^4/(b+
a*cos(f*x+e)^2)^2*cos(f*x+e)+15/8/f*b/a^3/(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))+35/8/f*b^2/a^4/(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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.67883, size = 1007, normalized size = 6.54 \begin{align*} \left [\frac{16 \, a^{3} \cos \left (f x + e\right )^{7} - 16 \,{\left (3 \, a^{3} + 7 \, a^{2} b\right )} \cos \left (f x + e\right )^{5} - 50 \,{\left (3 \, a^{2} b + 7 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \,{\left ({\left (3 \, a^{3} + 7 \, a^{2} b\right )} \cos \left (f x + e\right )^{4} + 3 \, a b^{2} + 7 \, b^{3} + 2 \,{\left (3 \, a^{2} b + 7 \, a b^{2}\right )} \cos \left (f x + e\right )^{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 ) - 30 \,{\left (3 \, a b^{2} + 7 \, b^{3}\right )} \cos \left (f x + e\right )}{48 \,{\left (a^{6} f \cos \left (f x + e\right )^{4} + 2 \, a^{5} b f \cos \left (f x + e\right )^{2} + a^{4} b^{2} f\right )}}, \frac{8 \, a^{3} \cos \left (f x + e\right )^{7} - 8 \,{\left (3 \, a^{3} + 7 \, a^{2} b\right )} \cos \left (f x + e\right )^{5} - 25 \,{\left (3 \, a^{2} b + 7 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \,{\left ({\left (3 \, a^{3} + 7 \, a^{2} b\right )} \cos \left (f x + e\right )^{4} + 3 \, a b^{2} + 7 \, b^{3} + 2 \,{\left (3 \, a^{2} b + 7 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}} \cos \left (f x + e\right )}{b}\right ) - 15 \,{\left (3 \, a b^{2} + 7 \, b^{3}\right )} \cos \left (f x + e\right )}{24 \,{\left (a^{6} f \cos \left (f x + e\right )^{4} + 2 \, a^{5} b f \cos \left (f x + e\right )^{2} + a^{4} b^{2} f\right )}}\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,x, algorithm="fricas")

[Out]

[1/48*(16*a^3*cos(f*x + e)^7 - 16*(3*a^3 + 7*a^2*b)*cos(f*x + e)^5 - 50*(3*a^2*b + 7*a*b^2)*cos(f*x + e)^3 + 1
5*((3*a^3 + 7*a^2*b)*cos(f*x + e)^4 + 3*a*b^2 + 7*b^3 + 2*(3*a^2*b + 7*a*b^2)*cos(f*x + e)^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)) - 30*(3*a*b^2 + 7*b^3)*cos(f*x +
e))/(a^6*f*cos(f*x + e)^4 + 2*a^5*b*f*cos(f*x + e)^2 + a^4*b^2*f), 1/24*(8*a^3*cos(f*x + e)^7 - 8*(3*a^3 + 7*a
^2*b)*cos(f*x + e)^5 - 25*(3*a^2*b + 7*a*b^2)*cos(f*x + e)^3 + 15*((3*a^3 + 7*a^2*b)*cos(f*x + e)^4 + 3*a*b^2
+ 7*b^3 + 2*(3*a^2*b + 7*a*b^2)*cos(f*x + e)^2)*sqrt(b/a)*arctan(a*sqrt(b/a)*cos(f*x + e)/b) - 15*(3*a*b^2 + 7
*b^3)*cos(f*x + e))/(a^6*f*cos(f*x + e)^4 + 2*a^5*b*f*cos(f*x + e)^2 + a^4*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)**3/(a+b*sec(f*x+e)**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.21004, size = 247, normalized size = 1.6 \begin{align*} \frac{5 \,{\left (3 \, a b + 7 \, b^{2}\right )} \arctan \left (\frac{a \cos \left (f x + e\right )}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a^{4} f} - \frac{\frac{9 \, a^{2} b \cos \left (f x + e\right )^{3}}{f} + \frac{13 \, a b^{2} \cos \left (f x + e\right )^{3}}{f} + \frac{7 \, a b^{2} \cos \left (f x + e\right )}{f} + \frac{11 \, b^{3} \cos \left (f x + e\right )}{f}}{8 \,{\left (a \cos \left (f x + e\right )^{2} + b\right )}^{2} a^{4}} + \frac{a^{6} f^{17} \cos \left (f x + e\right )^{3} - 3 \, a^{6} f^{17} \cos \left (f x + e\right ) - 9 \, a^{5} b f^{17} \cos \left (f x + e\right )}{3 \, a^{9} f^{18}} \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,x, algorithm="giac")

[Out]

5/8*(3*a*b + 7*b^2)*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^4*f) - 1/8*(9*a^2*b*cos(f*x + e)^3/f + 13*a*
b^2*cos(f*x + e)^3/f + 7*a*b^2*cos(f*x + e)/f + 11*b^3*cos(f*x + e)/f)/((a*cos(f*x + e)^2 + b)^2*a^4) + 1/3*(a
^6*f^17*cos(f*x + e)^3 - 3*a^6*f^17*cos(f*x + e) - 9*a^5*b*f^17*cos(f*x + e))/(a^9*f^18)