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

Optimal. Leaf size=161 \[ -\frac{(a+b)^2 \cos ^5(e+f x)}{2 a^2 b f \left (a \cos ^2(e+f x)+b\right )}+\frac{(a+b) (3 a+7 b) \cos ^3(e+f x)}{6 a^3 b f}-\frac{(a+b) (3 a+7 b) \cos (e+f x)}{2 a^4 f}+\frac{\sqrt{b} (a+b) (3 a+7 b) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{2 a^{9/2} f}-\frac{\cos ^5(e+f x)}{5 a^2 f} \]

[Out]

(Sqrt[b]*(a + b)*(3*a + 7*b)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(2*a^(9/2)*f) - ((a + b)*(3*a + 7*b)*Cos[
e + f*x])/(2*a^4*f) + ((a + b)*(3*a + 7*b)*Cos[e + f*x]^3)/(6*a^3*b*f) - Cos[e + f*x]^5/(5*a^2*f) - ((a + b)^2
*Cos[e + f*x]^5)/(2*a^2*b*f*(b + a*Cos[e + f*x]^2))

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Rubi [A]  time = 0.177968, antiderivative size = 161, 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, 463, 459, 302, 205} \[ -\frac{(a+b)^2 \cos ^5(e+f x)}{2 a^2 b f \left (a \cos ^2(e+f x)+b\right )}+\frac{(a+b) (3 a+7 b) \cos ^3(e+f x)}{6 a^3 b f}-\frac{(a+b) (3 a+7 b) \cos (e+f x)}{2 a^4 f}+\frac{\sqrt{b} (a+b) (3 a+7 b) \tan ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{b}}\right )}{2 a^{9/2} f}-\frac{\cos ^5(e+f x)}{5 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*(a + b)*(3*a + 7*b)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(2*a^(9/2)*f) - ((a + b)*(3*a + 7*b)*Cos[
e + f*x])/(2*a^4*f) + ((a + b)*(3*a + 7*b)*Cos[e + f*x]^3)/(6*a^3*b*f) - Cos[e + f*x]^5/(5*a^2*f) - ((a + b)^2
*Cos[e + f*x]^5)/(2*a^2*b*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 463

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

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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

Mathematica [C]  time = 6.82331, size = 454, normalized size = 2.82 \[ \frac{-\frac{16 \sqrt{a} \cos (e+f x) \left (a \left (125 a^2+688 a b+560 b^2\right ) \cos (2 (e+f x))-2 a^2 (11 a+14 b) \cos (4 (e+f x))+1436 a^2 b+3 a^3 \cos (6 (e+f x))+150 a^3+2960 a b^2+1680 b^3\right )}{a \cos (2 (e+f x))+a+2 b}-\frac{45 a^4 \tan ^{-1}\left (\frac{\sqrt{a}-\sqrt{a+b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )}{b^{3/2}}-\frac{45 a^4 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan \left (\frac{1}{2} (e+f x)\right )+\sqrt{a}}{\sqrt{b}}\right )}{b^{3/2}}+\frac{15 \left (384 a^2 b^2+3 a^4+1280 a b^3+896 b^4\right ) \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}-i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}-\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )}{b^{3/2}}+\frac{15 \left (384 a^2 b^2+3 a^4+1280 a b^3+896 b^4\right ) \tan ^{-1}\left (\frac{\sin (e) \tan \left (\frac{f x}{2}\right ) \left (-\sqrt{a}+i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt{a}+\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan \left (\frac{f x}{2}\right )\right )}{\sqrt{b}}\right )}{b^{3/2}}}{3840 a^{9/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((15*(3*a^4 + 384*a^2*b^2 + 1280*a*b^3 + 896*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] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]])/b^
(3/2) + (15*(3*a^4 + 384*a^2*b^2 + 1280*a*b^3 + 896*b^4)*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]])/b^(3/2) - (45*a^4*ArcTan[(Sqrt[a] - Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]])/b^(3/2) - (45*a^4*ArcTan[(Sqr
t[a] + Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]])/b^(3/2) - (16*Sqrt[a]*Cos[e + f*x]*(150*a^3 + 1436*a^2*b + 2960
*a*b^2 + 1680*b^3 + a*(125*a^2 + 688*a*b + 560*b^2)*Cos[2*(e + f*x)] - 2*a^2*(11*a + 14*b)*Cos[4*(e + f*x)] +
3*a^3*Cos[6*(e + f*x)]))/(a + 2*b + a*Cos[2*(e + f*x)]))/(3840*a^(9/2)*f)

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Maple [A]  time = 0.097, size = 276, normalized size = 1.7 \begin{align*} -{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{5}}{5\,{a}^{2}f}}+{\frac{2\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{3\,{a}^{2}f}}+{\frac{2\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}b}{3\,f{a}^{3}}}-{\frac{\cos \left ( fx+e \right ) }{{a}^{2}f}}-4\,{\frac{b\cos \left ( fx+e \right ) }{f{a}^{3}}}-3\,{\frac{{b}^{2}\cos \left ( fx+e \right ) }{f{a}^{4}}}-{\frac{b\cos \left ( fx+e \right ) }{2\,{a}^{2}f \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{{b}^{2}\cos \left ( fx+e \right ) }{f{a}^{3} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{{b}^{3}\cos \left ( fx+e \right ) }{2\,f{a}^{4} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{3\,b}{2\,{a}^{2}f}\arctan \left ({a\cos \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+5\,{\frac{{b}^{2}}{f{a}^{3}\sqrt{ab}}\arctan \left ({\frac{a\cos \left ( fx+e \right ) }{\sqrt{ab}}} \right ) }+{\frac{7\,{b}^{3}}{2\,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)^5/(a+b*sec(f*x+e)^2)^2,x)

[Out]

-1/5*cos(f*x+e)^5/a^2/f+2/3*cos(f*x+e)^3/a^2/f+2/3/f/a^3*cos(f*x+e)^3*b-cos(f*x+e)/a^2/f-4/f/a^3*b*cos(f*x+e)-
3/f/a^4*b^2*cos(f*x+e)-1/2/f*b/a^2*cos(f*x+e)/(b+a*cos(f*x+e)^2)-1/f*b^2/a^3*cos(f*x+e)/(b+a*cos(f*x+e)^2)-1/2
/f*b^3/a^4*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/f*b^2/a^3/
(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))+7/2/f*b^3/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)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.650045, size = 941, normalized size = 5.84 \begin{align*} \left [-\frac{12 \, a^{3} \cos \left (f x + e\right )^{7} - 4 \,{\left (10 \, a^{3} + 7 \, a^{2} b\right )} \cos \left (f x + e\right )^{5} + 20 \,{\left (3 \, a^{3} + 10 \, a^{2} b + 7 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \,{\left (3 \, a^{2} b + 10 \, a b^{2} + 7 \, b^{3} +{\left (3 \, a^{3} + 10 \, 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^{2} b + 10 \, a b^{2} + 7 \, b^{3}\right )} \cos \left (f x + e\right )}{60 \,{\left (a^{5} f \cos \left (f x + e\right )^{2} + a^{4} b f\right )}}, -\frac{6 \, a^{3} \cos \left (f x + e\right )^{7} - 2 \,{\left (10 \, a^{3} + 7 \, a^{2} b\right )} \cos \left (f x + e\right )^{5} + 10 \,{\left (3 \, a^{3} + 10 \, a^{2} b + 7 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \,{\left (3 \, a^{2} b + 10 \, a b^{2} + 7 \, b^{3} +{\left (3 \, a^{3} + 10 \, 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^{2} b + 10 \, a b^{2} + 7 \, b^{3}\right )} \cos \left (f x + e\right )}{30 \,{\left (a^{5} f \cos \left (f x + e\right )^{2} + a^{4} b f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/60*(12*a^3*cos(f*x + e)^7 - 4*(10*a^3 + 7*a^2*b)*cos(f*x + e)^5 + 20*(3*a^3 + 10*a^2*b + 7*a*b^2)*cos(f*x
+ e)^3 - 15*(3*a^2*b + 10*a*b^2 + 7*b^3 + (3*a^3 + 10*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^2*b + 10*a*b^2 + 7*b^3)*cos(f*
x + e))/(a^5*f*cos(f*x + e)^2 + a^4*b*f), -1/30*(6*a^3*cos(f*x + e)^7 - 2*(10*a^3 + 7*a^2*b)*cos(f*x + e)^5 +
10*(3*a^3 + 10*a^2*b + 7*a*b^2)*cos(f*x + e)^3 - 15*(3*a^2*b + 10*a*b^2 + 7*b^3 + (3*a^3 + 10*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^2*b + 10*a*b^2 + 7*b^3)*cos(f*x + e))/
(a^5*f*cos(f*x + e)^2 + a^4*b*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)**5/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.16877, size = 736, normalized size = 4.57 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(15*(3*a^2*b + 10*a*b^2 + 7*b^3)*arctan(-(a*cos(f*x + e) - b)/(sqrt(a*b)*cos(f*x + e) + sqrt(a*b)))/(sqr
t(a*b)*a^4) + 30*(a^2*b + 2*a*b^2 + b^3 + a^2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - b^3*(cos(f*x + e) - 1)
/(cos(f*x + e) + 1))/((a + b + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*b*(cos(f*x + e) - 1)/(cos(f*x + e
) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*a^4) - 4*(
8*a^2 + 50*a*b + 45*b^2 - 40*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 220*a*b*(cos(f*x + e) - 1)/(cos(f*x +
e) + 1) - 180*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 +
320*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 270*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 180*a*
b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 180*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 30*a*b*(cos(
f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 45*b^2*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4)/(a^4*((cos(f*x + e) -
1)/(cos(f*x + e) + 1) - 1)^5))/f