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

Optimal. Leaf size=114 \[ \frac {\sqrt {b} (3 a+5 b) \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{2 a^{7/2} f}-\frac {b (a+b) \cos (e+f x)}{2 a^3 f \left (a \cos ^2(e+f x)+b\right )}-\frac {(a+2 b) \cos (e+f x)}{a^3 f}+\frac {\cos ^3(e+f x)}{3 a^2 f} \]

[Out]

-(a+2*b)*cos(f*x+e)/a^3/f+1/3*cos(f*x+e)^3/a^2/f-1/2*b*(a+b)*cos(f*x+e)/a^3/f/(b+a*cos(f*x+e)^2)+1/2*(3*a+5*b)
*arctan(cos(f*x+e)*a^(1/2)/b^(1/2))*b^(1/2)/a^(7/2)/f

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Rubi [A]  time = 0.11, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4133, 455, 1153, 205} \[ -\frac {b (a+b) \cos (e+f x)}{2 a^3 f \left (a \cos ^2(e+f x)+b\right )}-\frac {(a+2 b) \cos (e+f x)}{a^3 f}+\frac {\sqrt {b} (3 a+5 b) \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{2 a^{7/2} f}+\frac {\cos ^3(e+f x)}{3 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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
]

Rubi steps

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

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Mathematica [C]  time = 3.47, size = 403, normalized size = 3.54 \[ \frac {-\frac {9 a^3 \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{b^{3/2}}-\frac {9 a^3 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )+\sqrt {a}}{\sqrt {b}}\right )}{b^{3/2}}+\frac {3 \left (3 a^3+192 a b^2+320 b^3\right ) \tan ^{-1}\left (\frac {\sin (e) \tan \left (\frac {f x}{2}\right ) \left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )}{b^{3/2}}+\frac {3 \left (3 a^3+192 a b^2+320 b^3\right ) \tan ^{-1}\left (\frac {\sin (e) \tan \left (\frac {f x}{2}\right ) \left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )}{b^{3/2}}-\frac {32 \sqrt {a} \cos (e+f x) \left (a^2 (-\cos (4 (e+f x)))+9 a^2+4 a (2 a+5 b) \cos (2 (e+f x))+56 a b+60 b^2\right )}{a \cos (2 (e+f x))+a+2 b}}{384 a^{7/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*(3*a^3 + 192*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^(3/2) + (3*(3*a^
3 + 192*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^(3/2) - (9*a^3*ArcTan[(Sq
rt[a] - Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]])/b^(3/2) - (9*a^3*ArcTan[(Sqrt[a] + Sqrt[a + b]*Tan[(e + f*x)/2
])/Sqrt[b]])/b^(3/2) - (32*Sqrt[a]*Cos[e + f*x]*(9*a^2 + 56*a*b + 60*b^2 + 4*a*(2*a + 5*b)*Cos[2*(e + f*x)] -
a^2*Cos[4*(e + f*x)]))/(a + 2*b + a*Cos[2*(e + f*x)]))/(384*a^(7/2)*f)

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fricas [A]  time = 0.58, size = 297, normalized size = 2.61 \[ \left [\frac {4 \, a^{2} \cos \left (f x + e\right )^{5} - 4 \, {\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left ({\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 5 \, b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) - 6 \, {\left (3 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )}{12 \, {\left (a^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}}, \frac {2 \, a^{2} \cos \left (f x + e\right )^{5} - 2 \, {\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left ({\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 5 \, b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right ) - 3 \, {\left (3 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )}{6 \, {\left (a^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(4*a^2*cos(f*x + e)^5 - 4*(3*a^2 + 5*a*b)*cos(f*x + e)^3 + 3*((3*a^2 + 5*a*b)*cos(f*x + e)^2 + 3*a*b + 5
*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)) - 6*(3*a*b
+ 5*b^2)*cos(f*x + e))/(a^4*f*cos(f*x + e)^2 + a^3*b*f), 1/6*(2*a^2*cos(f*x + e)^5 - 2*(3*a^2 + 5*a*b)*cos(f*x
+ e)^3 + 3*((3*a^2 + 5*a*b)*cos(f*x + e)^2 + 3*a*b + 5*b^2)*sqrt(b/a)*arctan(a*sqrt(b/a)*cos(f*x + e)/b) - 3*
(3*a*b + 5*b^2)*cos(f*x + e))/(a^4*f*cos(f*x + e)^2 + a^3*b*f)]

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giac [A]  time = 0.64, size = 143, normalized size = 1.25 \[ \frac {{\left (3 \, a b + 5 \, b^{2}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3} f} - \frac {\frac {a b \cos \left (f x + e\right )}{f} + \frac {b^{2} \cos \left (f x + e\right )}{f}}{2 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )} a^{3}} + \frac {a^{4} f^{11} \cos \left (f x + e\right )^{3} - 3 \, a^{4} f^{11} \cos \left (f x + e\right ) - 6 \, a^{3} b f^{11} \cos \left (f x + e\right )}{3 \, a^{6} f^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*a*b + 5*b^2)*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^3*f) - 1/2*(a*b*cos(f*x + e)/f + b^2*cos(f*x
+ e)/f)/((a*cos(f*x + e)^2 + b)*a^3) + 1/3*(a^4*f^11*cos(f*x + e)^3 - 3*a^4*f^11*cos(f*x + e) - 6*a^3*b*f^11*
cos(f*x + e))/(a^6*f^12)

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maple [A]  time = 0.91, size = 165, normalized size = 1.45 \[ \frac {\cos ^{3}\left (f x +e \right )}{3 a^{2} f}-\frac {\cos \left (f x +e \right )}{a^{2} f}-\frac {2 \cos \left (f x +e \right ) b}{f \,a^{3}}-\frac {b \cos \left (f x +e \right )}{2 f \,a^{2} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}-\frac {b^{2} \cos \left (f x +e \right )}{2 f \,a^{3} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}+\frac {3 b \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 f \,a^{2} \sqrt {a b}}+\frac {5 b^{2} \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 f \,a^{3} \sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 104, normalized size = 0.91 \[ -\frac {\frac {3 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )}{a^{4} \cos \left (f x + e\right )^{2} + a^{3} b} - \frac {3 \, {\left (3 \, a b + 5 \, b^{2}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (a \cos \left (f x + e\right )^{3} - 3 \, {\left (a + 2 \, b\right )} \cos \left (f x + e\right )\right )}}{a^{3}}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(a*b + b^2)*cos(f*x + e)/(a^4*cos(f*x + e)^2 + a^3*b) - 3*(3*a*b + 5*b^2)*arctan(a*cos(f*x + e)/sqrt(a
*b))/(sqrt(a*b)*a^3) - 2*(a*cos(f*x + e)^3 - 3*(a + 2*b)*cos(f*x + e))/a^3)/f

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mupad [B]  time = 0.14, size = 130, normalized size = 1.14 \[ \frac {{\cos \left (e+f\,x\right )}^3}{3\,a^2\,f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {2\,b}{a^3}+\frac {1}{a^2}\right )}{f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {b^2}{2}+\frac {a\,b}{2}\right )}{f\,\left (a^4\,{\cos \left (e+f\,x\right )}^2+b\,a^3\right )}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\cos \left (e+f\,x\right )\,\left (3\,a+5\,b\right )}{5\,b^2+3\,a\,b}\right )\,\left (3\,a+5\,b\right )}{2\,a^{7/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos(e + f*x)^3/(3*a^2*f) - (cos(e + f*x)*((2*b)/a^3 + 1/a^2))/f - (cos(e + f*x)*((a*b)/2 + b^2/2))/(f*(a^3*b +
a^4*cos(e + f*x)^2)) + (b^(1/2)*atan((a^(1/2)*b^(1/2)*cos(e + f*x)*(3*a + 5*b))/(3*a*b + 5*b^2))*(3*a + 5*b))
/(2*a^(7/2)*f)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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