3.28 \(\int \frac {\sin ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=98 \[ \frac {\sqrt {b} (a+b)^2 \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{a^{7/2} f}-\frac {(a+b)^2 \cos (e+f x)}{a^3 f}+\frac {(2 a+b) \cos ^3(e+f x)}{3 a^2 f}-\frac {\cos ^5(e+f x)}{5 a f} \]

[Out]

-(a+b)^2*cos(f*x+e)/a^3/f+1/3*(2*a+b)*cos(f*x+e)^3/a^2/f-1/5*cos(f*x+e)^5/a/f+(a+b)^2*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 = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4133, 461, 205} \[ \frac {(2 a+b) \cos ^3(e+f x)}{3 a^2 f}-\frac {(a+b)^2 \cos (e+f x)}{a^3 f}+\frac {\sqrt {b} (a+b)^2 \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{a^{7/2} f}-\frac {\cos ^5(e+f x)}{5 a f} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^5/(a + b*Sec[e + f*x]^2),x]

[Out]

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

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 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

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

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Mathematica [C]  time = 3.26, size = 425, normalized size = 4.34 \[ \frac {\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-75 a^3 \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )-75 a^3 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )+\sqrt {a}}{\sqrt {b}}\right )-8 \sqrt {a} \sqrt {b} \cos (e+f x) \left (3 a^2 \cos (4 (e+f x))+89 a^2-4 a (7 a+5 b) \cos (2 (e+f x))+220 a b+120 b^2\right )+15 \left (5 a^3+64 a^2 b+128 a b^2+64 b^3\right ) \tan ^{-1}\left (\frac {\sin (e) \tan \left (\frac {f x}{2}\right ) \left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )+15 \left (5 a^3+64 a^2 b+128 a b^2+64 b^3\right ) \tan ^{-1}\left (\frac {\sin (e) \tan \left (\frac {f x}{2}\right ) \left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )\right )}{1920 a^{7/2} \sqrt {b} f \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^5/(a + b*Sec[e + f*x]^2),x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*(15*(5*a^3 + 64*a^2*b + 128*a*b^2 + 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]*T
an[(f*x)/2]))/Sqrt[b]] + 15*(5*a^3 + 64*a^2*b + 128*a*b^2 + 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]] - 75*a^3*ArcTan[(Sqrt[a] - Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]] - 75*a^3*ArcTan[(Sqrt[a] + Sq
rt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]] - 8*Sqrt[a]*Sqrt[b]*Cos[e + f*x]*(89*a^2 + 220*a*b + 120*b^2 - 4*a*(7*a +
5*b)*Cos[2*(e + f*x)] + 3*a^2*Cos[4*(e + f*x)]))*Sec[e + f*x]^2)/(1920*a^(7/2)*Sqrt[b]*f*(a + b*Sec[e + f*x]^
2))

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fricas [A]  time = 0.72, size = 229, normalized size = 2.34 \[ \left [-\frac {6 \, a^{2} \cos \left (f x + e\right )^{5} - 10 \, {\left (2 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 30 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{30 \, a^{3} f}, -\frac {3 \, a^{2} \cos \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right ) + 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{15 \, a^{3} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

[-1/30*(6*a^2*cos(f*x + e)^5 - 10*(2*a^2 + a*b)*cos(f*x + e)^3 - 15*(a^2 + 2*a*b + 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)) + 30*(a^2 + 2*a*b + b^2)*cos(f*x + e))/
(a^3*f), -1/15*(3*a^2*cos(f*x + e)^5 - 5*(2*a^2 + a*b)*cos(f*x + e)^3 - 15*(a^2 + 2*a*b + b^2)*sqrt(b/a)*arcta
n(a*sqrt(b/a)*cos(f*x + e)/b) + 15*(a^2 + 2*a*b + b^2)*cos(f*x + e))/(a^3*f)]

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giac [B]  time = 0.27, size = 373, normalized size = 3.81 \[ -\frac {\frac {15 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {a \cos \left (f x + e\right ) - b}{\sqrt {a b} \cos \left (f x + e\right ) + \sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (8 \, a^{2} + 25 \, a b + 15 \, b^{2} - \frac {40 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {110 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {60 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {80 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {160 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {90 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {90 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {60 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {15 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{a^{3} {\left (\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 1\right )}^{5}}}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

-1/15*(15*(a^2*b + 2*a*b^2 + b^3)*arctan(-(a*cos(f*x + e) - b)/(sqrt(a*b)*cos(f*x + e) + sqrt(a*b)))/(sqrt(a*b
)*a^3) - 2*(8*a^2 + 25*a*b + 15*b^2 - 40*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 110*a*b*(cos(f*x + e) - 1
)/(cos(f*x + e) + 1) - 60*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 80*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e
) + 1)^2 + 160*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 90*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^
2 - 90*a*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 60*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 15*a
*b*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 15*b^2*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4)/(a^3*((cos(f*
x + e) - 1)/(cos(f*x + e) + 1) - 1)^5))/f

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maple [B]  time = 0.90, size = 183, normalized size = 1.87 \[ -\frac {\cos ^{5}\left (f x +e \right )}{5 a f}+\frac {2 \left (\cos ^{3}\left (f x +e \right )\right )}{3 a f}+\frac {\left (\cos ^{3}\left (f x +e \right )\right ) b}{3 f \,a^{2}}-\frac {\cos \left (f x +e \right )}{a f}-\frac {2 b \cos \left (f x +e \right )}{f \,a^{2}}-\frac {\cos \left (f x +e \right ) b^{2}}{f \,a^{3}}+\frac {b \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{f a \sqrt {a b}}+\frac {2 b^{2} \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{f \,a^{2} \sqrt {a b}}+\frac {b^{3} \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{f \,a^{3} \sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^5/(a+b*sec(f*x+e)^2),x)

[Out]

-1/5*cos(f*x+e)^5/a/f+2/3*cos(f*x+e)^3/a/f+1/3/f/a^2*cos(f*x+e)^3*b-cos(f*x+e)/a/f-2/f/a^2*b*cos(f*x+e)-1/f/a^
3*cos(f*x+e)*b^2+1/f*b/a/(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))+2/f*b^2/a^2/(a*b)^(1/2)*arctan(a*cos(f*x
+e)/(a*b)^(1/2))+1/f*b^3/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 = 102, normalized size = 1.04 \[ \frac {\frac {15 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {3 \, a^{2} \cos \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{a^{3}}}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

1/15*(15*(a^2*b + 2*a*b^2 + b^3)*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^3) - (3*a^2*cos(f*x + e)^5 - 5*
(2*a^2 + a*b)*cos(f*x + e)^3 + 15*(a^2 + 2*a*b + b^2)*cos(f*x + e))/a^3)/f

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mupad [B]  time = 4.31, size = 123, normalized size = 1.26 \[ \frac {{\cos \left (e+f\,x\right )}^3\,\left (\frac {b}{3\,a^2}+\frac {2}{3\,a}\right )}{f}-\frac {{\cos \left (e+f\,x\right )}^5}{5\,a\,f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {1}{a}+\frac {b\,\left (\frac {b}{a^2}+\frac {2}{a}\right )}{a}\right )}{f}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\cos \left (e+f\,x\right )\,{\left (a+b\right )}^2}{a^2\,b+2\,a\,b^2+b^3}\right )\,{\left (a+b\right )}^2}{a^{7/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^5/(a + b/cos(e + f*x)^2),x)

[Out]

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

[Out]

Timed out

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