3.61 \(\int \frac{\sin ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=178 \[ -\frac{\left (11 a^2-4 a b+b^2\right ) \sin (e+f x) \cos (e+f x)}{16 f (a-b)^3}+\frac{x \left (15 a^2 b+5 a^3-5 a b^2+b^3\right )}{16 (a-b)^4}-\frac{a^{5/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{f (a-b)^4}+\frac{\sin ^3(e+f x) \cos ^3(e+f x)}{6 f (a-b)}+\frac{(3 a-b) \sin (e+f x) \cos ^3(e+f x)}{8 f (a-b)^2} \]

[Out]

((5*a^3 + 15*a^2*b - 5*a*b^2 + b^3)*x)/(16*(a - b)^4) - (a^(5/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]
])/((a - b)^4*f) - ((11*a^2 - 4*a*b + b^2)*Cos[e + f*x]*Sin[e + f*x])/(16*(a - b)^3*f) + ((3*a - b)*Cos[e + f*
x]^3*Sin[e + f*x])/(8*(a - b)^2*f) + (Cos[e + f*x]^3*Sin[e + f*x]^3)/(6*(a - b)*f)

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Rubi [A]  time = 0.292728, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3663, 470, 578, 527, 522, 203, 205} \[ -\frac{\left (11 a^2-4 a b+b^2\right ) \sin (e+f x) \cos (e+f x)}{16 f (a-b)^3}+\frac{x \left (15 a^2 b+5 a^3-5 a b^2+b^3\right )}{16 (a-b)^4}-\frac{a^{5/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{f (a-b)^4}+\frac{\sin ^3(e+f x) \cos ^3(e+f x)}{6 f (a-b)}+\frac{(3 a-b) \sin (e+f x) \cos ^3(e+f x)}{8 f (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^6/(a + b*Tan[e + f*x]^2),x]

[Out]

((5*a^3 + 15*a^2*b - 5*a*b^2 + b^3)*x)/(16*(a - b)^4) - (a^(5/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]
])/((a - b)^4*f) - ((11*a^2 - 4*a*b + b^2)*Cos[e + f*x]*Sin[e + f*x])/(16*(a - b)^3*f) + ((3*a - b)*Cos[e + f*
x]^3*Sin[e + f*x])/(8*(a - b)^2*f) + (Cos[e + f*x]^3*Sin[e + f*x]^3)/(6*(a - b)*f)

Rule 3663

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

Rule 470

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

Rule 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

Rule 527

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

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^4 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 (a-b) f}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a-3 (2 a-b) x^2\right )}{\left (1+x^2\right )^3 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{6 (a-b) f}\\ &=\frac{(3 a-b) \cos ^3(e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 (a-b) f}-\frac{\operatorname{Subst}\left (\int \frac{3 a (3 a-b)-3 \left (8 a^2-3 a b+b^2\right ) x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{24 (a-b)^2 f}\\ &=-\frac{\left (11 a^2-4 a b+b^2\right ) \cos (e+f x) \sin (e+f x)}{16 (a-b)^3 f}+\frac{(3 a-b) \cos ^3(e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 (a-b) f}+\frac{\operatorname{Subst}\left (\int \frac{3 a (5 a-b) (a+b)-3 b \left (11 a^2-4 a b+b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{48 (a-b)^3 f}\\ &=-\frac{\left (11 a^2-4 a b+b^2\right ) \cos (e+f x) \sin (e+f x)}{16 (a-b)^3 f}+\frac{(3 a-b) \cos ^3(e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 (a-b) f}-\frac{\left (a^3 b\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{(a-b)^4 f}+\frac{\left (5 a^3+15 a^2 b-5 a b^2+b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 (a-b)^4 f}\\ &=\frac{\left (5 a^3+15 a^2 b-5 a b^2+b^3\right ) x}{16 (a-b)^4}-\frac{a^{5/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{(a-b)^4 f}-\frac{\left (11 a^2-4 a b+b^2\right ) \cos (e+f x) \sin (e+f x)}{16 (a-b)^3 f}+\frac{(3 a-b) \cos ^3(e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 (a-b) f}\\ \end{align*}

Mathematica [A]  time = 0.562333, size = 140, normalized size = 0.79 \[ -\frac{-12 \left (15 a^2 b+5 a^3-5 a b^2+b^3\right ) (e+f x)+192 a^{5/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )+(a-b)^3 \sin (6 (e+f x))-3 (3 a-b) (a-b)^2 \sin (4 (e+f x))+3 (5 a-b) (3 a+b) (a-b) \sin (2 (e+f x))}{192 f (a-b)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^6/(a + b*Tan[e + f*x]^2),x]

[Out]

-(-12*(5*a^3 + 15*a^2*b - 5*a*b^2 + b^3)*(e + f*x) + 192*a^(5/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]
] + 3*(a - b)*(5*a - b)*(3*a + b)*Sin[2*(e + f*x)] - 3*(a - b)^2*(3*a - b)*Sin[4*(e + f*x)] + (a - b)^3*Sin[6*
(e + f*x)])/(192*(a - b)^4*f)

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Maple [B]  time = 0.075, size = 545, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^6/(a+b*tan(f*x+e)^2),x)

[Out]

-1/f*b/(a-b)^4*a^3/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))-11/16/f/(a-b)^4/(1+tan(f*x+e)^2)^3*tan(f*x+e)^
5*a^3+15/16/f/(a-b)^4/(1+tan(f*x+e)^2)^3*tan(f*x+e)^5*a^2*b-5/16/f/(a-b)^4/(1+tan(f*x+e)^2)^3*tan(f*x+e)^5*a*b
^2+1/16/f/(a-b)^4/(1+tan(f*x+e)^2)^3*tan(f*x+e)^5*b^3-5/6/f/(a-b)^4/(1+tan(f*x+e)^2)^3*tan(f*x+e)^3*a^3+1/2/f/
(a-b)^4/(1+tan(f*x+e)^2)^3*tan(f*x+e)^3*a^2*b+1/2/f/(a-b)^4/(1+tan(f*x+e)^2)^3*tan(f*x+e)^3*a*b^2-1/6/f/(a-b)^
4/(1+tan(f*x+e)^2)^3*tan(f*x+e)^3*b^3-5/16/f/(a-b)^4/(1+tan(f*x+e)^2)^3*tan(f*x+e)*a^3+1/16/f/(a-b)^4/(1+tan(f
*x+e)^2)^3*tan(f*x+e)*a^2*b+5/16/f/(a-b)^4/(1+tan(f*x+e)^2)^3*tan(f*x+e)*a*b^2-1/16/f/(a-b)^4/(1+tan(f*x+e)^2)
^3*tan(f*x+e)*b^3+5/16/f/(a-b)^4*arctan(tan(f*x+e))*a^3+15/16/f/(a-b)^4*arctan(tan(f*x+e))*a^2*b-5/16/f/(a-b)^
4*arctan(tan(f*x+e))*a*b^2+1/16/f/(a-b)^4*arctan(tan(f*x+e))*b^3

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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)^6/(a+b*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.27819, size = 1214, normalized size = 6.82 \begin{align*} \left [\frac{12 \, \sqrt{-a b} a^{2} \log \left (\frac{{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt{-a b} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) + 3 \,{\left (5 \, a^{3} + 15 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )} f x -{\left (8 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{5} - 2 \,{\left (13 \, a^{3} - 33 \, a^{2} b + 27 \, a b^{2} - 7 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (11 \, a^{3} - 15 \, a^{2} b + 5 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \,{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f}, \frac{24 \, \sqrt{a b} a^{2} \arctan \left (\frac{{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt{a b}}{2 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + 3 \,{\left (5 \, a^{3} + 15 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )} f x -{\left (8 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{5} - 2 \,{\left (13 \, a^{3} - 33 \, a^{2} b + 27 \, a b^{2} - 7 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (11 \, a^{3} - 15 \, a^{2} b + 5 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \,{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^6/(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

[1/48*(12*sqrt(-a*b)*a^2*log(((a^2 + 6*a*b + b^2)*cos(f*x + e)^4 - 2*(3*a*b + b^2)*cos(f*x + e)^2 + 4*((a + b)
*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt(-a*b)*sin(f*x + e) + b^2)/((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 + 2*(a*b
- b^2)*cos(f*x + e)^2 + b^2)) + 3*(5*a^3 + 15*a^2*b - 5*a*b^2 + b^3)*f*x - (8*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*
cos(f*x + e)^5 - 2*(13*a^3 - 33*a^2*b + 27*a*b^2 - 7*b^3)*cos(f*x + e)^3 + 3*(11*a^3 - 15*a^2*b + 5*a*b^2 - b^
3)*cos(f*x + e))*sin(f*x + e))/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*f), 1/48*(24*sqrt(a*b)*a^2*arctan(
1/2*((a + b)*cos(f*x + e)^2 - b)*sqrt(a*b)/(a*b*cos(f*x + e)*sin(f*x + e))) + 3*(5*a^3 + 15*a^2*b - 5*a*b^2 +
b^3)*f*x - (8*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^5 - 2*(13*a^3 - 33*a^2*b + 27*a*b^2 - 7*b^3)*cos(f*
x + e)^3 + 3*(11*a^3 - 15*a^2*b + 5*a*b^2 - b^3)*cos(f*x + e))*sin(f*x + e))/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a
*b^3 + b^4)*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)**6/(a+b*tan(f*x+e)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.39315, size = 394, normalized size = 2.21 \begin{align*} -\frac{\frac{48 \,{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b}}\right )\right )} a^{3} b}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt{a b}} - \frac{3 \,{\left (5 \, a^{3} + 15 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )}{\left (f x + e\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac{33 \, a^{2} \tan \left (f x + e\right )^{5} - 12 \, a b \tan \left (f x + e\right )^{5} + 3 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} + 16 \, a b \tan \left (f x + e\right )^{3} - 8 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right ) + 12 \, a b \tan \left (f x + e\right ) - 3 \, b^{2} \tan \left (f x + e\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3}}}{48 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^6/(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

-1/48*(48*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*a^3*b/((a^4 - 4*a^3*b + 6*a
^2*b^2 - 4*a*b^3 + b^4)*sqrt(a*b)) - 3*(5*a^3 + 15*a^2*b - 5*a*b^2 + b^3)*(f*x + e)/(a^4 - 4*a^3*b + 6*a^2*b^2
- 4*a*b^3 + b^4) + (33*a^2*tan(f*x + e)^5 - 12*a*b*tan(f*x + e)^5 + 3*b^2*tan(f*x + e)^5 + 40*a^2*tan(f*x + e
)^3 + 16*a*b*tan(f*x + e)^3 - 8*b^2*tan(f*x + e)^3 + 15*a^2*tan(f*x + e) + 12*a*b*tan(f*x + e) - 3*b^2*tan(f*x
+ e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(tan(f*x + e)^2 + 1)^3))/f