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

Optimal. Leaf size=166 \[ -\frac{\left (11 a^2+18 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 a^3 f}+\frac{x \left (30 a^2 b+5 a^3+40 a b^2+16 b^3\right )}{16 a^4}-\frac{\sqrt{b} (a+b)^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a^4 f}+\frac{(3 a+2 b) \sin (e+f x) \cos ^3(e+f x)}{8 a^2 f}+\frac{\sin ^3(e+f x) \cos ^3(e+f x)}{6 a f} \]

[Out]

((5*a^3 + 30*a^2*b + 40*a*b^2 + 16*b^3)*x)/(16*a^4) - (Sqrt[b]*(a + b)^(5/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqr
t[a + b]])/(a^4*f) - ((11*a^2 + 18*a*b + 8*b^2)*Cos[e + f*x]*Sin[e + f*x])/(16*a^3*f) + ((3*a + 2*b)*Cos[e + f
*x]^3*Sin[e + f*x])/(8*a^2*f) + (Cos[e + f*x]^3*Sin[e + f*x]^3)/(6*a*f)

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Rubi [A]  time = 0.335182, antiderivative size = 166, 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 = {4132, 470, 578, 527, 522, 203, 205} \[ -\frac{\left (11 a^2+18 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 a^3 f}+\frac{x \left (30 a^2 b+5 a^3+40 a b^2+16 b^3\right )}{16 a^4}-\frac{\sqrt{b} (a+b)^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a^4 f}+\frac{(3 a+2 b) \sin (e+f x) \cos ^3(e+f x)}{8 a^2 f}+\frac{\sin ^3(e+f x) \cos ^3(e+f x)}{6 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a^3 + 30*a^2*b + 40*a*b^2 + 16*b^3)*x)/(16*a^4) - (Sqrt[b]*(a + b)^(5/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqr
t[a + b]])/(a^4*f) - ((11*a^2 + 18*a*b + 8*b^2)*Cos[e + f*x]*Sin[e + f*x])/(16*a^3*f) + ((3*a + 2*b)*Cos[e + f
*x]^3*Sin[e + f*x])/(8*a^2*f) + (Cos[e + f*x]^3*Sin[e + f*x]^3)/(6*a*f)

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/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 \sec ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^4 \left (a+b+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 f}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 (a+b)-3 (2 a+b) x^2\right )}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{6 a f}\\ &=\frac{(3 a+2 b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f}-\frac{\operatorname{Subst}\left (\int \frac{3 (a+b) (3 a+2 b)-3 \left (8 a^2+13 a b+6 b^2\right ) x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{24 a^2 f}\\ &=-\frac{\left (11 a^2+18 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 a^3 f}+\frac{(3 a+2 b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f}+\frac{\operatorname{Subst}\left (\int \frac{3 (a+b) (a+2 b) (5 a+4 b)-3 b \left (11 a^2+18 a b+8 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{48 a^3 f}\\ &=-\frac{\left (11 a^2+18 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 a^3 f}+\frac{(3 a+2 b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f}-\frac{\left (b (a+b)^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{a^4 f}+\frac{\left (5 a^3+30 a^2 b+40 a b^2+16 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 a^4 f}\\ &=\frac{\left (5 a^3+30 a^2 b+40 a b^2+16 b^3\right ) x}{16 a^4}-\frac{\sqrt{b} (a+b)^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a^4 f}-\frac{\left (11 a^2+18 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 a^3 f}+\frac{(3 a+2 b) \cos ^3(e+f x) \sin (e+f x)}{8 a^2 f}+\frac{\cos ^3(e+f x) \sin ^3(e+f x)}{6 a f}\\ \end{align*}

Mathematica [C]  time = 4.34908, size = 357, normalized size = 2.15 \[ \frac{\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\sqrt{b (\cos (e)-i \sin (e))^4} \left (2 \sqrt{b} \sqrt{a+b} \left (-3 a \left (15 a^2+32 a b+16 b^2\right ) \sin (2 (e+f x))+3 a^2 (3 a+2 b) \sin (4 (e+f x))+360 a^2 b f x-a^3 \sin (6 (e+f x))-12 a^3 e+60 a^3 f x+480 a b^2 f x+192 b^3 f x\right )+3 a^3 (9 a+8 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )\right )+3 \sqrt{b} \left (384 a^2 b^2+136 a^3 b+9 a^4+384 a b^3+128 b^4\right ) (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )\right )}{768 a^4 \sqrt{b} f \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4} \left (a+b \sec ^2(e+f x)\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(3*Sqrt[b]*(9*a^4 + 136*a^3*b + 384*a^2*b^2 + 384*a*b^3 + 128*b
^4)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b
*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]) + Sqrt[b*(Cos[e] - I*Sin[e])^4]*(3*a^3*(9*a + 8*b)*ArcTan[(S
qrt[b]*Tan[e + f*x])/Sqrt[a + b]] + 2*Sqrt[b]*Sqrt[a + b]*(-12*a^3*e + 60*a^3*f*x + 360*a^2*b*f*x + 480*a*b^2*
f*x + 192*b^3*f*x - 3*a*(15*a^2 + 32*a*b + 16*b^2)*Sin[2*(e + f*x)] + 3*a^2*(3*a + 2*b)*Sin[4*(e + f*x)] - a^3
*Sin[6*(e + f*x)]))))/(768*a^4*Sqrt[b]*Sqrt[a + b]*f*(a + b*Sec[e + f*x]^2)*Sqrt[b*(Cos[e] - I*Sin[e])^4])

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Maple [B]  time = 0.102, size = 460, normalized size = 2.8 \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*sec(f*x+e)^2),x)

[Out]

-9/8/f/a^2/(tan(f*x+e)^2+1)^3*tan(f*x+e)^5*b-1/2/f/a^3/(tan(f*x+e)^2+1)^3*tan(f*x+e)^5*b^2-11/16/f/a/(tan(f*x+
e)^2+1)^3*tan(f*x+e)^5-2/f/a^2/(tan(f*x+e)^2+1)^3*tan(f*x+e)^3*b-1/f/a^3/(tan(f*x+e)^2+1)^3*tan(f*x+e)^3*b^2-5
/6/f/a/(tan(f*x+e)^2+1)^3*tan(f*x+e)^3-5/16/f/a/(tan(f*x+e)^2+1)^3*tan(f*x+e)-7/8/f/a^2/(tan(f*x+e)^2+1)^3*tan
(f*x+e)*b-1/2/f/a^3/(tan(f*x+e)^2+1)^3*tan(f*x+e)*b^2+15/8/f/a^2*arctan(tan(f*x+e))*b+5/2/f/a^3*arctan(tan(f*x
+e))*b^2+1/f/a^4*arctan(tan(f*x+e))*b^3+5/16/f/a*arctan(tan(f*x+e))-1/f*b/a/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*
b/((a+b)*b)^(1/2))-3/f*b^2/a^2/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-3/f*b^3/a^3/((a+b)*b)^(1/2
)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-1/f*b^4/a^4/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*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)^6/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.624802, size = 1019, normalized size = 6.14 \begin{align*} \left [\frac{3 \,{\left (5 \, a^{3} + 30 \, a^{2} b + 40 \, a b^{2} + 16 \, b^{3}\right )} f x + 12 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{-a b - b^{2}} \log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt{-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) -{\left (8 \, a^{3} \cos \left (f x + e\right )^{5} - 2 \,{\left (13 \, a^{3} + 6 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (11 \, a^{3} + 18 \, a^{2} b + 8 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, a^{4} f}, \frac{3 \,{\left (5 \, a^{3} + 30 \, a^{2} b + 40 \, a b^{2} + 16 \, b^{3}\right )} f x + 24 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{a b + b^{2}} \arctan \left (\frac{{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt{a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) -{\left (8 \, a^{3} \cos \left (f x + e\right )^{5} - 2 \,{\left (13 \, a^{3} + 6 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (11 \, a^{3} + 18 \, a^{2} b + 8 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, a^{4} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(3*(5*a^3 + 30*a^2*b + 40*a*b^2 + 16*b^3)*f*x + 12*(a^2 + 2*a*b + b^2)*sqrt(-a*b - b^2)*log(((a^2 + 8*a*
b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 + 4*((a + 2*b)*cos(f*x + e)^3 - b*cos(f*x + e))*s
qrt(-a*b - b^2)*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) - (8*a^3*cos(f*x + e)^5
- 2*(13*a^3 + 6*a^2*b)*cos(f*x + e)^3 + 3*(11*a^3 + 18*a^2*b + 8*a*b^2)*cos(f*x + e))*sin(f*x + e))/(a^4*f),
1/48*(3*(5*a^3 + 30*a^2*b + 40*a*b^2 + 16*b^3)*f*x + 24*(a^2 + 2*a*b + b^2)*sqrt(a*b + b^2)*arctan(1/2*((a + 2
*b)*cos(f*x + e)^2 - b)/(sqrt(a*b + b^2)*cos(f*x + e)*sin(f*x + e))) - (8*a^3*cos(f*x + e)^5 - 2*(13*a^3 + 6*a
^2*b)*cos(f*x + e)^3 + 3*(11*a^3 + 18*a^2*b + 8*a*b^2)*cos(f*x + e))*sin(f*x + e))/(a^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*sec(f*x+e)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.27881, size = 338, normalized size = 2.04 \begin{align*} \frac{\frac{3 \,{\left (5 \, a^{3} + 30 \, a^{2} b + 40 \, a b^{2} + 16 \, b^{3}\right )}{\left (f x + e\right )}}{a^{4}} - \frac{48 \,{\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )}{\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 + b^{2}}}\right )\right )}}{\sqrt{a b + b^{2}} a^{4}} - \frac{33 \, a^{2} \tan \left (f x + e\right )^{5} + 54 \, a b \tan \left (f x + e\right )^{5} + 24 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} + 96 \, a b \tan \left (f x + e\right )^{3} + 48 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right ) + 42 \, a b \tan \left (f x + e\right ) + 24 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3} a^{3}}}{48 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*(5*a^3 + 30*a^2*b + 40*a*b^2 + 16*b^3)*(f*x + e)/a^4 - 48*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*(pi*floo
r((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/(sqrt(a*b + b^2)*a^4) - (33*a^2*tan(f*x
+ e)^5 + 54*a*b*tan(f*x + e)^5 + 24*b^2*tan(f*x + e)^5 + 40*a^2*tan(f*x + e)^3 + 96*a*b*tan(f*x + e)^3 + 48*b
^2*tan(f*x + e)^3 + 15*a^2*tan(f*x + e) + 42*a*b*tan(f*x + e) + 24*b^2*tan(f*x + e))/((tan(f*x + e)^2 + 1)^3*a
^3))/f