3.84 \(\int \frac {\sin ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=104 \[ -\frac {\sin ^3(c+d x) (a-a \cos (c+d x))^3}{6 a^5 d}-\frac {\sin ^5(c+d x)}{10 a^2 d}-\frac {\sin ^3(c+d x) \cos (c+d x)}{8 a^2 d}-\frac {3 \sin (c+d x) \cos (c+d x)}{16 a^2 d}+\frac {3 x}{16 a^2} \]

[Out]

3/16*x/a^2-3/16*cos(d*x+c)*sin(d*x+c)/a^2/d-1/8*cos(d*x+c)*sin(d*x+c)^3/a^2/d-1/6*(a-a*cos(d*x+c))^3*sin(d*x+c
)^3/a^5/d-1/10*sin(d*x+c)^5/a^2/d

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Rubi [A]  time = 0.31, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3872, 2875, 2870, 2669, 2635, 8} \[ -\frac {\sin ^5(c+d x)}{10 a^2 d}-\frac {\sin ^3(c+d x) (a-a \cos (c+d x))^3}{6 a^5 d}-\frac {\sin ^3(c+d x) \cos (c+d x)}{8 a^2 d}-\frac {3 \sin (c+d x) \cos (c+d x)}{16 a^2 d}+\frac {3 x}{16 a^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^6/(a + a*Sec[c + d*x])^2,x]

[Out]

(3*x)/(16*a^2) - (3*Cos[c + d*x]*Sin[c + d*x])/(16*a^2*d) - (Cos[c + d*x]*Sin[c + d*x]^3)/(8*a^2*d) - ((a - a*
Cos[c + d*x])^3*Sin[c + d*x]^3)/(6*a^5*d) - Sin[c + d*x]^5/(10*a^2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2669

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

Rule 2870

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

Rule 2875

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

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]

Rubi steps

\begin {align*} \int \frac {\sin ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=\int \frac {\cos ^2(c+d x) \sin ^6(c+d x)}{(-a-a \cos (c+d x))^2} \, dx\\ &=\frac {\int \cos ^2(c+d x) (-a+a \cos (c+d x))^2 \sin ^2(c+d x) \, dx}{a^4}\\ &=-\frac {(a-a \cos (c+d x))^3 \sin ^3(c+d x)}{6 a^5 d}-\frac {\int (-a+a \cos (c+d x)) \sin ^4(c+d x) \, dx}{2 a^3}\\ &=-\frac {(a-a \cos (c+d x))^3 \sin ^3(c+d x)}{6 a^5 d}-\frac {\sin ^5(c+d x)}{10 a^2 d}+\frac {\int \sin ^4(c+d x) \, dx}{2 a^2}\\ &=-\frac {\cos (c+d x) \sin ^3(c+d x)}{8 a^2 d}-\frac {(a-a \cos (c+d x))^3 \sin ^3(c+d x)}{6 a^5 d}-\frac {\sin ^5(c+d x)}{10 a^2 d}+\frac {3 \int \sin ^2(c+d x) \, dx}{8 a^2}\\ &=-\frac {3 \cos (c+d x) \sin (c+d x)}{16 a^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{8 a^2 d}-\frac {(a-a \cos (c+d x))^3 \sin ^3(c+d x)}{6 a^5 d}-\frac {\sin ^5(c+d x)}{10 a^2 d}+\frac {3 \int 1 \, dx}{16 a^2}\\ &=\frac {3 x}{16 a^2}-\frac {3 \cos (c+d x) \sin (c+d x)}{16 a^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{8 a^2 d}-\frac {(a-a \cos (c+d x))^3 \sin ^3(c+d x)}{6 a^5 d}-\frac {\sin ^5(c+d x)}{10 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.95, size = 111, normalized size = 1.07 \[ \frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (-480 \sin (c+d x)+30 \sin (2 (c+d x))+80 \sin (3 (c+d x))-90 \sin (4 (c+d x))+48 \sin (5 (c+d x))-10 \sin (6 (c+d x))+25 \tan \left (\frac {c}{2}\right )+360 d x\right )}{480 a^2 d (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^6/(a + a*Sec[c + d*x])^2,x]

[Out]

(Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*(360*d*x - 480*Sin[c + d*x] + 30*Sin[2*(c + d*x)] + 80*Sin[3*(c + d*x)] - 9
0*Sin[4*(c + d*x)] + 48*Sin[5*(c + d*x)] - 10*Sin[6*(c + d*x)] + 25*Tan[c/2]))/(480*a^2*d*(1 + Sec[c + d*x])^2
)

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fricas [A]  time = 1.43, size = 71, normalized size = 0.68 \[ \frac {45 \, d x - {\left (40 \, \cos \left (d x + c\right )^{5} - 96 \, \cos \left (d x + c\right )^{4} + 50 \, \cos \left (d x + c\right )^{3} + 32 \, \cos \left (d x + c\right )^{2} - 45 \, \cos \left (d x + c\right ) + 64\right )} \sin \left (d x + c\right )}{240 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/240*(45*d*x - (40*cos(d*x + c)^5 - 96*cos(d*x + c)^4 + 50*cos(d*x + c)^3 + 32*cos(d*x + c)^2 - 45*cos(d*x +
c) + 64)*sin(d*x + c))/(a^2*d)

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giac [A]  time = 0.26, size = 113, normalized size = 1.09 \[ \frac {\frac {45 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 1025 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 174 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 594 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 255 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6} a^{2}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/240*(45*(d*x + c)/a^2 + 2*(45*tan(1/2*d*x + 1/2*c)^11 - 1025*tan(1/2*d*x + 1/2*c)^9 - 174*tan(1/2*d*x + 1/2*
c)^7 - 594*tan(1/2*d*x + 1/2*c)^5 - 255*tan(1/2*d*x + 1/2*c)^3 - 45*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*
c)^2 + 1)^6*a^2))/d

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maple [B]  time = 0.63, size = 222, normalized size = 2.13 \[ \frac {3 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {205 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {29 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {99 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {17 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^6/(a+a*sec(d*x+c))^2,x)

[Out]

3/8/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^11-205/24/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x
+1/2*c)^9-29/20/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^7-99/20/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*t
an(1/2*d*x+1/2*c)^5-17/8/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^3-3/8/d/a^2/(1+tan(1/2*d*x+1/2*c)
^2)^6*tan(1/2*d*x+1/2*c)+3/8/d/a^2*arctan(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.55, size = 292, normalized size = 2.81 \[ -\frac {\frac {\frac {45 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {255 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {594 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {174 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {1025 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {45 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}}{a^{2} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {20 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {a^{2} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}} - \frac {45 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/120*((45*sin(d*x + c)/(cos(d*x + c) + 1) + 255*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 594*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 + 174*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 1025*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 45*si
n(d*x + c)^11/(cos(d*x + c) + 1)^11)/(a^2 + 6*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a^2*sin(d*x + c)^4/
(cos(d*x + c) + 1)^4 + 20*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8
+ 6*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) - 45*arctan(sin(d*
x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 3.75, size = 107, normalized size = 1.03 \[ \frac {3\,x}{16\,a^2}-\frac {-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {205\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\frac {29\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{20}+\frac {99\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{a^2\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^6/(a + a/cos(c + d*x))^2,x)

[Out]

(3*x)/(16*a^2) - ((3*tan(c/2 + (d*x)/2))/8 + (17*tan(c/2 + (d*x)/2)^3)/8 + (99*tan(c/2 + (d*x)/2)^5)/20 + (29*
tan(c/2 + (d*x)/2)^7)/20 + (205*tan(c/2 + (d*x)/2)^9)/24 - (3*tan(c/2 + (d*x)/2)^11)/8)/(a^2*d*(tan(c/2 + (d*x
)/2)^2 + 1)^6)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sin ^{6}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**6/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(sin(c + d*x)**6/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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