3.203 \(\int \frac{\sin ^6(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=230 \[ \frac{\sin ^3(c+d x) \left (8 b \left (a^2-b^2\right )-a \left (5 a^2-6 b^2\right ) \cos (c+d x)\right )}{24 a^4 d}+\frac{\sin (c+d x) \left (16 b \left (a^2-b^2\right )^2-a \left (-14 a^2 b^2+5 a^4+8 b^4\right ) \cos (c+d x)\right )}{16 a^6 d}+\frac{x \left (-30 a^4 b^2+40 a^2 b^4+5 a^6-16 b^6\right )}{16 a^7}+\frac{\sin ^5(c+d x) (6 b-5 a \cos (c+d x))}{30 a^2 d}-\frac{2 b (a-b)^{5/2} (a+b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^7 d} \]

[Out]

((5*a^6 - 30*a^4*b^2 + 40*a^2*b^4 - 16*b^6)*x)/(16*a^7) - (2*(a - b)^(5/2)*b*(a + b)^(5/2)*ArcTanh[(Sqrt[a - b
]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^7*d) + ((16*b*(a^2 - b^2)^2 - a*(5*a^4 - 14*a^2*b^2 + 8*b^4)*Cos[c + d*x]
)*Sin[c + d*x])/(16*a^6*d) + ((8*b*(a^2 - b^2) - a*(5*a^2 - 6*b^2)*Cos[c + d*x])*Sin[c + d*x]^3)/(24*a^4*d) +
((6*b - 5*a*Cos[c + d*x])*Sin[c + d*x]^5)/(30*a^2*d)

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Rubi [A]  time = 0.607589, antiderivative size = 230, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3872, 2865, 2735, 2659, 208} \[ \frac{\sin ^3(c+d x) \left (8 b \left (a^2-b^2\right )-a \left (5 a^2-6 b^2\right ) \cos (c+d x)\right )}{24 a^4 d}+\frac{\sin (c+d x) \left (16 b \left (a^2-b^2\right )^2-a \left (-14 a^2 b^2+5 a^4+8 b^4\right ) \cos (c+d x)\right )}{16 a^6 d}+\frac{x \left (-30 a^4 b^2+40 a^2 b^4+5 a^6-16 b^6\right )}{16 a^7}+\frac{\sin ^5(c+d x) (6 b-5 a \cos (c+d x))}{30 a^2 d}-\frac{2 b (a-b)^{5/2} (a+b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^6/(a + b*Sec[c + d*x]),x]

[Out]

((5*a^6 - 30*a^4*b^2 + 40*a^2*b^4 - 16*b^6)*x)/(16*a^7) - (2*(a - b)^(5/2)*b*(a + b)^(5/2)*ArcTanh[(Sqrt[a - b
]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^7*d) + ((16*b*(a^2 - b^2)^2 - a*(5*a^4 - 14*a^2*b^2 + 8*b^4)*Cos[c + d*x]
)*Sin[c + d*x])/(16*a^6*d) + ((8*b*(a^2 - b^2) - a*(5*a^2 - 6*b^2)*Cos[c + d*x])*Sin[c + d*x]^3)/(24*a^4*d) +
((6*b - 5*a*Cos[c + d*x])*Sin[c + d*x]^5)/(30*a^2*d)

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2865

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*m]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\sin ^6(c+d x)}{a+b \sec (c+d x)} \, dx &=-\int \frac{\cos (c+d x) \sin ^6(c+d x)}{-b-a \cos (c+d x)} \, dx\\ &=\frac{(6 b-5 a \cos (c+d x)) \sin ^5(c+d x)}{30 a^2 d}-\frac{\int \frac{\left (-a b+\left (5 a^2-6 b^2\right ) \cos (c+d x)\right ) \sin ^4(c+d x)}{-b-a \cos (c+d x)} \, dx}{6 a^2}\\ &=\frac{\left (8 b \left (a^2-b^2\right )-a \left (5 a^2-6 b^2\right ) \cos (c+d x)\right ) \sin ^3(c+d x)}{24 a^4 d}+\frac{(6 b-5 a \cos (c+d x)) \sin ^5(c+d x)}{30 a^2 d}-\frac{\int \frac{\left (-3 a b \left (3 a^2-2 b^2\right )+3 \left (5 a^4-14 a^2 b^2+8 b^4\right ) \cos (c+d x)\right ) \sin ^2(c+d x)}{-b-a \cos (c+d x)} \, dx}{24 a^4}\\ &=\frac{\left (16 b \left (a^2-b^2\right )^2-a \left (5 a^4-14 a^2 b^2+8 b^4\right ) \cos (c+d x)\right ) \sin (c+d x)}{16 a^6 d}+\frac{\left (8 b \left (a^2-b^2\right )-a \left (5 a^2-6 b^2\right ) \cos (c+d x)\right ) \sin ^3(c+d x)}{24 a^4 d}+\frac{(6 b-5 a \cos (c+d x)) \sin ^5(c+d x)}{30 a^2 d}-\frac{\int \frac{-3 a b \left (11 a^4-18 a^2 b^2+8 b^4\right )+3 \left (5 a^6-30 a^4 b^2+40 a^2 b^4-16 b^6\right ) \cos (c+d x)}{-b-a \cos (c+d x)} \, dx}{48 a^6}\\ &=\frac{\left (5 a^6-30 a^4 b^2+40 a^2 b^4-16 b^6\right ) x}{16 a^7}+\frac{\left (16 b \left (a^2-b^2\right )^2-a \left (5 a^4-14 a^2 b^2+8 b^4\right ) \cos (c+d x)\right ) \sin (c+d x)}{16 a^6 d}+\frac{\left (8 b \left (a^2-b^2\right )-a \left (5 a^2-6 b^2\right ) \cos (c+d x)\right ) \sin ^3(c+d x)}{24 a^4 d}+\frac{(6 b-5 a \cos (c+d x)) \sin ^5(c+d x)}{30 a^2 d}+\frac{\left (b \left (a^2-b^2\right )^3\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{a^7}\\ &=\frac{\left (5 a^6-30 a^4 b^2+40 a^2 b^4-16 b^6\right ) x}{16 a^7}+\frac{\left (16 b \left (a^2-b^2\right )^2-a \left (5 a^4-14 a^2 b^2+8 b^4\right ) \cos (c+d x)\right ) \sin (c+d x)}{16 a^6 d}+\frac{\left (8 b \left (a^2-b^2\right )-a \left (5 a^2-6 b^2\right ) \cos (c+d x)\right ) \sin ^3(c+d x)}{24 a^4 d}+\frac{(6 b-5 a \cos (c+d x)) \sin ^5(c+d x)}{30 a^2 d}+\frac{\left (2 b \left (a^2-b^2\right )^3\right ) \operatorname{Subst}\left (\int \frac{1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^7 d}\\ &=\frac{\left (5 a^6-30 a^4 b^2+40 a^2 b^4-16 b^6\right ) x}{16 a^7}-\frac{2 (a-b)^{5/2} b (a+b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^7 d}+\frac{\left (16 b \left (a^2-b^2\right )^2-a \left (5 a^4-14 a^2 b^2+8 b^4\right ) \cos (c+d x)\right ) \sin (c+d x)}{16 a^6 d}+\frac{\left (8 b \left (a^2-b^2\right )-a \left (5 a^2-6 b^2\right ) \cos (c+d x)\right ) \sin ^3(c+d x)}{24 a^4 d}+\frac{(6 b-5 a \cos (c+d x)) \sin ^5(c+d x)}{30 a^2 d}\\ \end{align*}

Mathematica [A]  time = 2.38069, size = 268, normalized size = 1.17 \[ \frac{-30 a^4 b^2 \sin (4 (c+d x))+80 a^3 b^3 \sin (3 (c+d x))+120 a b \left (-18 a^2 b^2+11 a^4+8 b^4\right ) \sin (c+d x)-15 \left (-32 a^4 b^2+16 a^2 b^4+15 a^6\right ) \sin (2 (c+d x))+1920 b \left (a^2-b^2\right )^{5/2} \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )-1800 a^4 b^2 c+2400 a^2 b^4 c-1800 a^4 b^2 d x+2400 a^2 b^4 d x-140 a^5 b \sin (3 (c+d x))+12 a^5 b \sin (5 (c+d x))+45 a^6 \sin (4 (c+d x))-5 a^6 \sin (6 (c+d x))+300 a^6 c+300 a^6 d x-960 b^6 c-960 b^6 d x}{960 a^7 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^6/(a + b*Sec[c + d*x]),x]

[Out]

(300*a^6*c - 1800*a^4*b^2*c + 2400*a^2*b^4*c - 960*b^6*c + 300*a^6*d*x - 1800*a^4*b^2*d*x + 2400*a^2*b^4*d*x -
960*b^6*d*x + 1920*b*(a^2 - b^2)^(5/2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + 120*a*b*(11*a^4
- 18*a^2*b^2 + 8*b^4)*Sin[c + d*x] - 15*(15*a^6 - 32*a^4*b^2 + 16*a^2*b^4)*Sin[2*(c + d*x)] - 140*a^5*b*Sin[3
*(c + d*x)] + 80*a^3*b^3*Sin[3*(c + d*x)] + 45*a^6*Sin[4*(c + d*x)] - 30*a^4*b^2*Sin[4*(c + d*x)] + 12*a^5*b*S
in[5*(c + d*x)] - 5*a^6*Sin[6*(c + d*x)])/(960*a^7*d)

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Maple [B]  time = 0.074, size = 1566, normalized size = 6.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^6/(a+b*sec(d*x+c)),x)

[Out]

5/8/d/a*arctan(tan(1/2*d*x+1/2*c))-2/d*b/a/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^
(1/2))+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)*b-6/d*b^5/a^5/((a+b)*(a-b))^(1/2)*arctanh((a-b)*t
an(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))+2/d*b^7/a^7/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)
*(a-b))^(1/2))-4/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)*b^3+2/d/a^6/(1+tan(1/2*d*x+1/2*c)^2)^6*ta
n(1/2*d*x+1/2*c)*b^5+7/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)*b^2-48/d/a^4/(1+tan(1/2*d*x+1/2*c
)^2)^6*tan(1/2*d*x+1/2*c)^5*b^3-11/2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^7*b^2+1/d/a^5/(1+tan(
1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^11*b^4+10/d/a^6/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^9*b^5-68/
3/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^9*b^3+172/5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x
+1/2*c)^7*b+2/d/a^6/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^11*b^5+38/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6
*tan(1/2*d*x+1/2*c)^9*b-29/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^9*b^2+3/d/a^5/(1+tan(1/2*d*x+
1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^9*b^4+20/d/a^6/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^5*b^5+38/3/d/a^2/(
1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^3*b+29/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^3*b^
2-48/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^7*b^3+2/d/a^5/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+
1/2*c)^7*b^4+20/d/a^6/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^7*b^5+11/2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^
6*tan(1/2*d*x+1/2*c)^5*b^2-2/d/a^5/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^5*b^4+172/5/d/a^2/(1+tan(1/2*
d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^5*b+6/d*b^3/a^3/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)
*(a-b))^(1/2))-1/d/a^5/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)*b^4+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*ta
n(1/2*d*x+1/2*c)^11*b-7/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^11*b^2-4/d/a^4/(1+tan(1/2*d*x+1/
2*c)^2)^6*tan(1/2*d*x+1/2*c)^11*b^3-68/3/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^3*b^3-3/d/a^5/(1+
tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^3*b^4+10/d/a^6/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^3*b^5-
33/4/d/a/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^5+33/4/d/a/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c
)^7+5/8/d/a/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^11+85/24/d/a/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+
1/2*c)^9-85/24/d/a/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^3-5/8/d/a/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*
d*x+1/2*c)-15/4/d/a^3*arctan(tan(1/2*d*x+1/2*c))*b^2+5/d/a^5*arctan(tan(1/2*d*x+1/2*c))*b^4-2/d/a^7*arctan(tan
(1/2*d*x+1/2*c))*b^6

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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(d*x+c)^6/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.40288, size = 1296, normalized size = 5.63 \begin{align*} \left [\frac{15 \,{\left (5 \, a^{6} - 30 \, a^{4} b^{2} + 40 \, a^{2} b^{4} - 16 \, b^{6}\right )} d x + 120 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) -{\left (40 \, a^{6} \cos \left (d x + c\right )^{5} - 48 \, a^{5} b \cos \left (d x + c\right )^{4} - 368 \, a^{5} b + 560 \, a^{3} b^{3} - 240 \, a b^{5} - 10 \,{\left (13 \, a^{6} - 6 \, a^{4} b^{2}\right )} \cos \left (d x + c\right )^{3} + 16 \,{\left (11 \, a^{5} b - 5 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (11 \, a^{6} - 18 \, a^{4} b^{2} + 8 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, a^{7} d}, \frac{15 \,{\left (5 \, a^{6} - 30 \, a^{4} b^{2} + 40 \, a^{2} b^{4} - 16 \, b^{6}\right )} d x - 240 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) -{\left (40 \, a^{6} \cos \left (d x + c\right )^{5} - 48 \, a^{5} b \cos \left (d x + c\right )^{4} - 368 \, a^{5} b + 560 \, a^{3} b^{3} - 240 \, a b^{5} - 10 \,{\left (13 \, a^{6} - 6 \, a^{4} b^{2}\right )} \cos \left (d x + c\right )^{3} + 16 \,{\left (11 \, a^{5} b - 5 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (11 \, a^{6} - 18 \, a^{4} b^{2} + 8 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, a^{7} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/240*(15*(5*a^6 - 30*a^4*b^2 + 40*a^2*b^4 - 16*b^6)*d*x + 120*(a^4*b - 2*a^2*b^3 + b^5)*sqrt(a^2 - b^2)*log(
(2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a
^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - (40*a^6*cos(d*x + c)^5 - 48*a^5*b*cos(d*x + c)^4
- 368*a^5*b + 560*a^3*b^3 - 240*a*b^5 - 10*(13*a^6 - 6*a^4*b^2)*cos(d*x + c)^3 + 16*(11*a^5*b - 5*a^3*b^3)*cos
(d*x + c)^2 + 15*(11*a^6 - 18*a^4*b^2 + 8*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/(a^7*d), 1/240*(15*(5*a^6 - 30*
a^4*b^2 + 40*a^2*b^4 - 16*b^6)*d*x - 240*(a^4*b - 2*a^2*b^3 + b^5)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(
b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (40*a^6*cos(d*x + c)^5 - 48*a^5*b*cos(d*x + c)^4 - 368*a^5*b
+ 560*a^3*b^3 - 240*a*b^5 - 10*(13*a^6 - 6*a^4*b^2)*cos(d*x + c)^3 + 16*(11*a^5*b - 5*a^3*b^3)*cos(d*x + c)^2
+ 15*(11*a^6 - 18*a^4*b^2 + 8*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/(a^7*d)]

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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(d*x+c)**6/(a+b*sec(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.25831, size = 1054, normalized size = 4.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

1/240*(15*(5*a^6 - 30*a^4*b^2 + 40*a^2*b^4 - 16*b^6)*(d*x + c)/a^7 - 480*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)
*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))
/sqrt(-a^2 + b^2)))/(sqrt(-a^2 + b^2)*a^7) + 2*(75*a^5*tan(1/2*d*x + 1/2*c)^11 + 240*a^4*b*tan(1/2*d*x + 1/2*c
)^11 - 210*a^3*b^2*tan(1/2*d*x + 1/2*c)^11 - 480*a^2*b^3*tan(1/2*d*x + 1/2*c)^11 + 120*a*b^4*tan(1/2*d*x + 1/2
*c)^11 + 240*b^5*tan(1/2*d*x + 1/2*c)^11 + 425*a^5*tan(1/2*d*x + 1/2*c)^9 + 1520*a^4*b*tan(1/2*d*x + 1/2*c)^9
- 870*a^3*b^2*tan(1/2*d*x + 1/2*c)^9 - 2720*a^2*b^3*tan(1/2*d*x + 1/2*c)^9 + 360*a*b^4*tan(1/2*d*x + 1/2*c)^9
+ 1200*b^5*tan(1/2*d*x + 1/2*c)^9 + 990*a^5*tan(1/2*d*x + 1/2*c)^7 + 4128*a^4*b*tan(1/2*d*x + 1/2*c)^7 - 660*a
^3*b^2*tan(1/2*d*x + 1/2*c)^7 - 5760*a^2*b^3*tan(1/2*d*x + 1/2*c)^7 + 240*a*b^4*tan(1/2*d*x + 1/2*c)^7 + 2400*
b^5*tan(1/2*d*x + 1/2*c)^7 - 990*a^5*tan(1/2*d*x + 1/2*c)^5 + 4128*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 660*a^3*b^2*
tan(1/2*d*x + 1/2*c)^5 - 5760*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 240*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 2400*b^5*tan
(1/2*d*x + 1/2*c)^5 - 425*a^5*tan(1/2*d*x + 1/2*c)^3 + 1520*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 870*a^3*b^2*tan(1/2
*d*x + 1/2*c)^3 - 2720*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 360*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 1200*b^5*tan(1/2*d*
x + 1/2*c)^3 - 75*a^5*tan(1/2*d*x + 1/2*c) + 240*a^4*b*tan(1/2*d*x + 1/2*c) + 210*a^3*b^2*tan(1/2*d*x + 1/2*c)
- 480*a^2*b^3*tan(1/2*d*x + 1/2*c) - 120*a*b^4*tan(1/2*d*x + 1/2*c) + 240*b^5*tan(1/2*d*x + 1/2*c))/((tan(1/2
*d*x + 1/2*c)^2 + 1)^6*a^6))/d