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

Optimal. Leaf size=117 \[ -\frac{a^2 \cos (e+f x)}{f (a-b)^3}-\frac{a^2 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{f (a-b)^{7/2}}-\frac{\cos ^5(e+f x)}{5 f (a-b)}+\frac{(2 a-b) \cos ^3(e+f x)}{3 f (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.184031, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3664, 461, 205} \[ -\frac{a^2 \cos (e+f x)}{f (a-b)^3}-\frac{a^2 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{f (a-b)^{7/2}}-\frac{\cos ^5(e+f x)}{5 f (a-b)}+\frac{(2 a-b) \cos ^3(e+f x)}{3 f (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

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

Mathematica [A]  time = 3.04949, size = 177, normalized size = 1.51 \[ \frac{\sqrt{a-b} \cos (e+f x) \left (4 \left (7 a^2-9 a b+2 b^2\right ) \cos (2 (e+f x))-89 a^2-3 (a-b)^2 \cos (4 (e+f x))-42 a b+11 b^2\right )+120 a^2 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a-b}-\sqrt{a} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )+120 a^2 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a-b}+\sqrt{a} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )}{120 f (a-b)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(120*a^2*Sqrt[b]*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]] + 120*a^2*Sqrt[b]*ArcTan[(Sqrt[a - b
] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]] + Sqrt[a - b]*Cos[e + f*x]*(-89*a^2 - 42*a*b + 11*b^2 + 4*(7*a^2 - 9*a*
b + 2*b^2)*Cos[2*(e + f*x)] - 3*(a - b)^2*Cos[4*(e + f*x)]))/(120*(a - b)^(7/2)*f)

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Maple [A]  time = 0.062, size = 205, normalized size = 1.8 \begin{align*} -{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{5}{a}^{2}}{5\,f \left ( a-b \right ) ^{3}}}+{\frac{2\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}ab}{5\,f \left ( a-b \right ) ^{3}}}-{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{5}{b}^{2}}{5\,f \left ( a-b \right ) ^{3}}}+{\frac{2\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}{a}^{2}}{3\,f \left ( a-b \right ) ^{3}}}-{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{3}ab}{f \left ( a-b \right ) ^{3}}}+{\frac{{b}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{3\,f \left ( a-b \right ) ^{3}}}-{\frac{{a}^{2}\cos \left ( fx+e \right ) }{f \left ( a-b \right ) ^{3}}}+{\frac{{a}^{2}b}{f \left ( a-b \right ) ^{3}}\arctan \left ({ \left ( a-b \right ) \cos \left ( fx+e \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/f/(a-b)^3*cos(f*x+e)^5*a^2+2/5/f/(a-b)^3*cos(f*x+e)^5*a*b-1/5/f/(a-b)^3*cos(f*x+e)^5*b^2+2/3/f/(a-b)^3*co
s(f*x+e)^3*a^2-1/f/(a-b)^3*cos(f*x+e)^3*a*b+1/3/f/(a-b)^3*b^2*cos(f*x+e)^3-a^2*cos(f*x+e)/(a-b)^3/f+1/f*a^2*b/
(a-b)^3/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(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*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.38719, size = 675, normalized size = 5.77 \begin{align*} \left [-\frac{6 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 10 \,{\left (2 \, a^{2} - 3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, a^{2} \sqrt{-\frac{b}{a - b}} \log \left (-\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (a - b\right )} \sqrt{-\frac{b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) + 30 \, a^{2} \cos \left (f x + e\right )}{30 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}, -\frac{3 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 5 \,{\left (2 \, a^{2} - 3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, a^{2} \sqrt{\frac{b}{a - b}} \arctan \left (-\frac{{\left (a - b\right )} \sqrt{\frac{b}{a - b}} \cos \left (f x + e\right )}{b}\right ) + 15 \, a^{2} \cos \left (f x + e\right )}{15 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.42267, size = 509, normalized size = 4.35 \begin{align*} -\frac{\frac{15 \, a^{2} b \arctan \left (-\frac{a \cos \left (f x + e\right ) - b \cos \left (f x + e\right ) - b}{\sqrt{a b - b^{2}} \cos \left (f x + e\right ) + \sqrt{a b - b^{2}}}\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt{a b - b^{2}}} - \frac{2 \,{\left (8 \, a^{2} + 9 \, a b - 2 \, b^{2} - \frac{40 \, a^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{30 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{10 \, 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{10 \, 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{30 \, 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}}\right )}}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}{\left (\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 1\right )}^{5}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*a^2*b*arctan(-(a*cos(f*x + e) - b*cos(f*x + e) - b)/(sqrt(a*b - b^2)*cos(f*x + e) + sqrt(a*b - b^2))
)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b - b^2)) - 2*(8*a^2 + 9*a*b - 2*b^2 - 40*a^2*(cos(f*x + e) - 1)/(co
s(f*x + e) + 1) - 30*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 10*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 + 10*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 + 30*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)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*((cos(f*x + e) - 1)/(cos(f*x + e) + 1) -
1)^5))/f