3.101 \(\int \frac {\sin ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=129 \[ \frac {3 \sin ^5(c+d x)}{5 a^3 d}-\frac {7 \sin ^3(c+d x)}{3 a^3 d}+\frac {4 \sin (c+d x)}{a^3 d}-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 a^3 d}-\frac {23 \sin (c+d x) \cos ^3(c+d x)}{24 a^3 d}-\frac {23 \sin (c+d x) \cos (c+d x)}{16 a^3 d}-\frac {23 x}{16 a^3} \]

[Out]

-23/16*x/a^3+4*sin(d*x+c)/a^3/d-23/16*cos(d*x+c)*sin(d*x+c)/a^3/d-23/24*cos(d*x+c)^3*sin(d*x+c)/a^3/d-1/6*cos(
d*x+c)^5*sin(d*x+c)/a^3/d-7/3*sin(d*x+c)^3/a^3/d+3/5*sin(d*x+c)^5/a^3/d

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Rubi [A]  time = 0.29, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3872, 2869, 2757, 2633, 2635, 8} \[ \frac {3 \sin ^5(c+d x)}{5 a^3 d}-\frac {7 \sin ^3(c+d x)}{3 a^3 d}+\frac {4 \sin (c+d x)}{a^3 d}-\frac {\sin (c+d x) \cos ^5(c+d x)}{6 a^3 d}-\frac {23 \sin (c+d x) \cos ^3(c+d x)}{24 a^3 d}-\frac {23 \sin (c+d x) \cos (c+d x)}{16 a^3 d}-\frac {23 x}{16 a^3} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^6/(a + a*Sec[c + d*x])^3,x]

[Out]

(-23*x)/(16*a^3) + (4*Sin[c + d*x])/(a^3*d) - (23*Cos[c + d*x]*Sin[c + d*x])/(16*a^3*d) - (23*Cos[c + d*x]^3*S
in[c + d*x])/(24*a^3*d) - (Cos[c + d*x]^5*Sin[c + d*x])/(6*a^3*d) - (7*Sin[c + d*x]^3)/(3*a^3*d) + (3*Sin[c +
d*x]^5)/(5*a^3*d)

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

Rule 2869

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[a^(2*m), Int[(d*Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f,
n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 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))^3} \, dx &=-\int \frac {\cos ^3(c+d x) \sin ^6(c+d x)}{(-a-a \cos (c+d x))^3} \, dx\\ &=-\frac {\int \cos ^3(c+d x) (-a+a \cos (c+d x))^3 \, dx}{a^6}\\ &=-\frac {\int \left (-a^3 \cos ^3(c+d x)+3 a^3 \cos ^4(c+d x)-3 a^3 \cos ^5(c+d x)+a^3 \cos ^6(c+d x)\right ) \, dx}{a^6}\\ &=\frac {\int \cos ^3(c+d x) \, dx}{a^3}-\frac {\int \cos ^6(c+d x) \, dx}{a^3}-\frac {3 \int \cos ^4(c+d x) \, dx}{a^3}+\frac {3 \int \cos ^5(c+d x) \, dx}{a^3}\\ &=-\frac {3 \cos ^3(c+d x) \sin (c+d x)}{4 a^3 d}-\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a^3 d}-\frac {5 \int \cos ^4(c+d x) \, dx}{6 a^3}-\frac {9 \int \cos ^2(c+d x) \, dx}{4 a^3}-\frac {\operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^3 d}-\frac {3 \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{a^3 d}\\ &=\frac {4 \sin (c+d x)}{a^3 d}-\frac {9 \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac {23 \cos ^3(c+d x) \sin (c+d x)}{24 a^3 d}-\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a^3 d}-\frac {7 \sin ^3(c+d x)}{3 a^3 d}+\frac {3 \sin ^5(c+d x)}{5 a^3 d}-\frac {5 \int \cos ^2(c+d x) \, dx}{8 a^3}-\frac {9 \int 1 \, dx}{8 a^3}\\ &=-\frac {9 x}{8 a^3}+\frac {4 \sin (c+d x)}{a^3 d}-\frac {23 \cos (c+d x) \sin (c+d x)}{16 a^3 d}-\frac {23 \cos ^3(c+d x) \sin (c+d x)}{24 a^3 d}-\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a^3 d}-\frac {7 \sin ^3(c+d x)}{3 a^3 d}+\frac {3 \sin ^5(c+d x)}{5 a^3 d}-\frac {5 \int 1 \, dx}{16 a^3}\\ &=-\frac {23 x}{16 a^3}+\frac {4 \sin (c+d x)}{a^3 d}-\frac {23 \cos (c+d x) \sin (c+d x)}{16 a^3 d}-\frac {23 \cos ^3(c+d x) \sin (c+d x)}{24 a^3 d}-\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a^3 d}-\frac {7 \sin ^3(c+d x)}{3 a^3 d}+\frac {3 \sin ^5(c+d x)}{5 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 1.96, size = 111, normalized size = 0.86 \[ \frac {\cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (5040 \sin (c+d x)-1890 \sin (2 (c+d x))+760 \sin (3 (c+d x))-270 \sin (4 (c+d x))+72 \sin (5 (c+d x))-10 \sin (6 (c+d x))+9 \tan \left (\frac {c}{2}\right )-2760 d x\right )}{240 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^6/(a + a*Sec[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]^6*Sec[c + d*x]^3*(-2760*d*x + 5040*Sin[c + d*x] - 1890*Sin[2*(c + d*x)] + 760*Sin[3*(c + d*x
)] - 270*Sin[4*(c + d*x)] + 72*Sin[5*(c + d*x)] - 10*Sin[6*(c + d*x)] + 9*Tan[c/2]))/(240*a^3*d*(1 + Sec[c + d
*x])^3)

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fricas [A]  time = 0.72, size = 70, normalized size = 0.54 \[ -\frac {345 \, d x + {\left (40 \, \cos \left (d x + c\right )^{5} - 144 \, \cos \left (d x + c\right )^{4} + 230 \, \cos \left (d x + c\right )^{3} - 272 \, \cos \left (d x + c\right )^{2} + 345 \, \cos \left (d x + c\right ) - 544\right )} \sin \left (d x + c\right )}{240 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/240*(345*d*x + (40*cos(d*x + c)^5 - 144*cos(d*x + c)^4 + 230*cos(d*x + c)^3 - 272*cos(d*x + c)^2 + 345*cos(
d*x + c) - 544)*sin(d*x + c))/(a^3*d)

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giac [A]  time = 0.35, size = 113, normalized size = 0.88 \[ -\frac {\frac {345 \, {\left (d x + c\right )}}{a^{3}} - \frac {2 \, {\left (1575 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 3165 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 5814 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4554 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1955 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 345 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6} a^{3}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-1/240*(345*(d*x + c)/a^3 - 2*(1575*tan(1/2*d*x + 1/2*c)^11 + 3165*tan(1/2*d*x + 1/2*c)^9 + 5814*tan(1/2*d*x +
1/2*c)^7 + 4554*tan(1/2*d*x + 1/2*c)^5 + 1955*tan(1/2*d*x + 1/2*c)^3 + 345*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*
x + 1/2*c)^2 + 1)^6*a^3))/d

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maple [A]  time = 0.73, size = 222, normalized size = 1.72 \[ \frac {105 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {211 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {969 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {759 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {391 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {23 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^6/(a+a*sec(d*x+c))^3,x)

[Out]

105/8/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^11+211/8/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*
x+1/2*c)^9+969/20/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^7+759/20/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^
6*tan(1/2*d*x+1/2*c)^5+391/24/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^6*tan(1/2*d*x+1/2*c)^3+23/8/d/a^3/(1+tan(1/2*d*x+
1/2*c)^2)^6*tan(1/2*d*x+1/2*c)-23/8/d/a^3*arctan(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.50, size = 292, normalized size = 2.26 \[ \frac {\frac {\frac {345 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {1955 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {4554 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5814 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {3165 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {1575 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}}{a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {20 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}} - \frac {345 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^6/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/120*((345*sin(d*x + c)/(cos(d*x + c) + 1) + 1955*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 4554*sin(d*x + c)^5/(
cos(d*x + c) + 1)^5 + 5814*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 3165*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 15
75*sin(d*x + c)^11/(cos(d*x + c) + 1)^11)/(a^3 + 6*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a^3*sin(d*x +
c)^4/(cos(d*x + c) + 1)^4 + 20*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a^3*sin(d*x + c)^8/(cos(d*x + c) +
1)^8 + 6*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + a^3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) - 345*arctan(
sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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mupad [B]  time = 3.67, size = 106, normalized size = 0.82 \[ \frac {\frac {105\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {211\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{8}+\frac {969\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{20}+\frac {759\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {391\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {23\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{a^3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6}-\frac {23\,x}{16\,a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^6/(a + a/cos(c + d*x))^3,x)

[Out]

((23*tan(c/2 + (d*x)/2))/8 + (391*tan(c/2 + (d*x)/2)^3)/24 + (759*tan(c/2 + (d*x)/2)^5)/20 + (969*tan(c/2 + (d
*x)/2)^7)/20 + (211*tan(c/2 + (d*x)/2)^9)/8 + (105*tan(c/2 + (d*x)/2)^11)/8)/(a^3*d*(tan(c/2 + (d*x)/2)^2 + 1)
^6) - (23*x)/(16*a^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**6/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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