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

Optimal. Leaf size=204 \[ -\frac{\left (5 a^2+10 a b-b^2\right ) \cos (e+f x)}{5 f (a-b)^4}-\frac{b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{10 f (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )}+\frac{(10 a-3 b) \cos ^3(e+f x)}{15 f (a-b)^3}-\frac{\cos ^5(e+f x)}{5 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )}-\frac{a \sqrt{b} (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{2 f (a-b)^{9/2}} \]

[Out]

-(a*Sqrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(2*(a - b)^(9/2)*f) - ((5*a^2 + 10*a*b - b
^2)*Cos[e + f*x])/(5*(a - b)^4*f) + ((10*a - 3*b)*Cos[e + f*x]^3)/(15*(a - b)^3*f) - Cos[e + f*x]^5/(5*(a - b)
*f*(a - b + b*Sec[e + f*x]^2)) - (b*(5*a^2 + 2*b^2)*Sec[e + f*x])/(10*(a - b)^4*f*(a - b + b*Sec[e + f*x]^2))

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Rubi [A]  time = 0.309352, antiderivative size = 204, 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 = {3664, 462, 456, 1261, 205} \[ -\frac{\left (5 a^2+10 a b-b^2\right ) \cos (e+f x)}{5 f (a-b)^4}-\frac{b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{10 f (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )}+\frac{(10 a-3 b) \cos ^3(e+f x)}{15 f (a-b)^3}-\frac{\cos ^5(e+f x)}{5 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )}-\frac{a \sqrt{b} (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{2 f (a-b)^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Sqrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(2*(a - b)^(9/2)*f) - ((5*a^2 + 10*a*b - b
^2)*Cos[e + f*x])/(5*(a - b)^4*f) + ((10*a - 3*b)*Cos[e + f*x]^3)/(15*(a - b)^3*f) - Cos[e + f*x]^5/(5*(a - b)
*f*(a - b + b*Sec[e + f*x]^2)) - (b*(5*a^2 + 2*b^2)*Sec[e + f*x])/(10*(a - b)^4*f*(a - b + b*Sec[e + f*x]^2))

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 462

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

Rule 456

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[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 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 ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2}{x^6 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{-10 a+3 b+5 (a-b) x^2}{x^4 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f}\\ &=-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{10 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{b \operatorname{Subst}\left (\int \frac{\frac{2 (10 a-3 b)}{(a-b) b}-\frac{2 \left (5 a^2+2 b^2\right ) x^2}{(a-b)^2 b}+\frac{\left (5 a^2+2 b^2\right ) x^4}{(a-b)^3}}{x^4 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{10 (a-b) f}\\ &=-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{10 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{b \operatorname{Subst}\left (\int \left (\frac{2 (10 a-3 b)}{(a-b)^2 b x^4}-\frac{2 \left (5 a^2+10 a b-b^2\right )}{(a-b)^3 b x^2}+\frac{5 a (3 a+4 b)}{(a-b)^3 \left (a-b+b x^2\right )}\right ) \, dx,x,\sec (e+f x)\right )}{10 (a-b) f}\\ &=-\frac{\left (5 a^2+10 a b-b^2\right ) \cos (e+f x)}{5 (a-b)^4 f}+\frac{(10 a-3 b) \cos ^3(e+f x)}{15 (a-b)^3 f}-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{10 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{(a b (3 a+4 b)) \operatorname{Subst}\left (\int \frac{1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{2 (a-b)^4 f}\\ &=-\frac{a \sqrt{b} (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{2 (a-b)^{9/2} f}-\frac{\left (5 a^2+10 a b-b^2\right ) \cos (e+f x)}{5 (a-b)^4 f}+\frac{(10 a-3 b) \cos ^3(e+f x)}{15 (a-b)^3 f}-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{b \left (5 a^2+2 b^2\right ) \sec (e+f x)}{10 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 3.71694, size = 215, normalized size = 1.05 \[ \frac{\frac{(a-b) (5 (5 a+3 b) \cos (3 (e+f x))+3 (b-a) \cos (5 (e+f x)))-30 \cos (e+f x) \left (a^2 \left (\frac{8 b}{(a-b) \cos (2 (e+f x))+a+b}+5\right )+18 a b+b^2\right )}{(a-b)^4}+\frac{120 a \sqrt{b} (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{a-b}-\sqrt{a} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )}{(a-b)^{9/2}}+\frac{120 a \sqrt{b} (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{a-b}+\sqrt{a} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )}{(a-b)^{9/2}}}{240 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((120*a*Sqrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(9/2) + (120*a*S
qrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(9/2) + (-30*Cos[e + f*x]
*(18*a*b + b^2 + a^2*(5 + (8*b)/(a + b + (a - b)*Cos[2*(e + f*x)]))) + (a - b)*(5*(5*a + 3*b)*Cos[3*(e + f*x)]
+ 3*(-a + b)*Cos[5*(e + f*x)]))/(a - b)^4)/(240*f)

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Maple [B]  time = 0.082, size = 388, normalized size = 1.9 \begin{align*} -{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{5}{a}^{2}}{5\,f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) ^{2}}}+{\frac{2\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}ab}{5\,f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) ^{2}}}-{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{5}{b}^{2}}{5\,f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) ^{2}}}+{\frac{2\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}{a}^{2}}{3\,f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) ^{2}}}-{\frac{2\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}ab}{3\,f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) ^{2}}}-{\frac{{a}^{2}\cos \left ( fx+e \right ) }{f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) ^{2}}}-2\,{\frac{\cos \left ( fx+e \right ) ab}{f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) ^{2}}}-{\frac{{a}^{2}b\cos \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{4} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) }}+{\frac{3\,{a}^{2}b}{2\,f \left ( a-b \right ) ^{4}}\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 ) }}}}+2\,{\frac{a{b}^{2}}{f \left ( a-b \right ) ^{4}\sqrt{b \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \cos \left ( fx+e \right ) }{\sqrt{b \left ( a-b \right ) }}} \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)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.59003, size = 1330, normalized size = 6.52 \begin{align*} \left [-\frac{12 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 4 \,{\left (10 \, a^{3} - 23 \, a^{2} b + 16 \, a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + 20 \,{\left (3 \, a^{3} + a^{2} b - 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \,{\left (3 \, a^{2} b + 4 \, a b^{2} +{\left (3 \, a^{3} + a^{2} b - 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \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 \,{\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )}{60 \,{\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f\right )}}, -\frac{6 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 2 \,{\left (10 \, a^{3} - 23 \, a^{2} b + 16 \, a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + 10 \,{\left (3 \, a^{3} + a^{2} b - 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \,{\left (3 \, a^{2} b + 4 \, a b^{2} +{\left (3 \, a^{3} + a^{2} b - 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \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 \,{\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )}{30 \,{\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f\right )}}\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)^2,x, algorithm="fricas")

[Out]

[-1/60*(12*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 4*(10*a^3 - 23*a^2*b + 16*a*b^2 - 3*b^3)*cos(f*x +
e)^5 + 20*(3*a^3 + a^2*b - 4*a*b^2)*cos(f*x + e)^3 - 15*(3*a^2*b + 4*a*b^2 + (3*a^3 + a^2*b - 4*a*b^2)*cos(f*
x + e)^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*(3*a^2*b + 4*a*b^2)*cos(f*x + e))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b
^4 - b^5)*f*cos(f*x + e)^2 + (a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f), -1/30*(6*(a^3 - 3*a^2*b + 3*a
*b^2 - b^3)*cos(f*x + e)^7 - 2*(10*a^3 - 23*a^2*b + 16*a*b^2 - 3*b^3)*cos(f*x + e)^5 + 10*(3*a^3 + a^2*b - 4*a
*b^2)*cos(f*x + e)^3 + 15*(3*a^2*b + 4*a*b^2 + (3*a^3 + a^2*b - 4*a*b^2)*cos(f*x + e)^2)*sqrt(b/(a - b))*arcta
n(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) + 15*(3*a^2*b + 4*a*b^2)*cos(f*x + e))/((a^5 - 5*a^4*b + 10*a^3*b^2
- 10*a^2*b^3 + 5*a*b^4 - b^5)*f*cos(f*x + e)^2 + (a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*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)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.43139, size = 757, normalized size = 3.71 \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*tan(f*x+e)^2)^2,x, algorithm="giac")

[Out]

-1/30*(15*(3*a^2*b + 4*a*b^2)*arctan(-(a*cos(f*x + e) - b*cos(f*x + e) - b)/(sqrt(a*b - b^2)*cos(f*x + e) + sq
rt(a*b - b^2)))/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*sqrt(a*b - b^2)) + 30*(a^2*b + a^2*b*(cos(f*x + e
) - 1)/(cos(f*x + e) + 1) - 2*a*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b
^3 + b^4)*(a + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos(
f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)) - 4*(8*a^2 + 34*a*b + 3*b^2 - 40*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) +
1) - 140*a*b*(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 + 30*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 + 30*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^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*((cos(f*x + e) - 1)/(cos(f*x + e)
+ 1) - 1)^5))/f