3.11 \(\int (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx\)

Optimal. Leaf size=127 \[ -\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \sin (c+d x)}{d}+\frac {a \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac {5 a \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {5 a \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a x}{16} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3872, 2838, 2592, 302, 206, 2635, 8} \[ -\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \sin ^3(c+d x)}{3 d}-\frac {a \sin (c+d x)}{d}+\frac {a \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac {5 a \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {5 a \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*x)/16 + (a*ArcTanh[Sin[c + d*x]])/d - (a*Sin[c + d*x])/d - (5*a*Cos[c + d*x]*Sin[c + d*x])/(16*d) - (a*Si
n[c + d*x]^3)/(3*d) - (5*a*Cos[c + d*x]*Sin[c + d*x]^3)/(24*d) - (a*Sin[c + d*x]^5)/(5*d) - (a*Cos[c + d*x]*Si
n[c + d*x]^5)/(6*d)

Rule 8

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

Rule 206

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

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

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 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

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 (a+a \sec (c+d x)) \sin ^6(c+d x) \, dx &=-\int (-a-a \cos (c+d x)) \sin ^5(c+d x) \tan (c+d x) \, dx\\ &=a \int \sin ^6(c+d x) \, dx+a \int \sin ^5(c+d x) \tan (c+d x) \, dx\\ &=-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac {1}{6} (5 a) \int \sin ^4(c+d x) \, dx+\frac {a \operatorname {Subst}\left (\int \frac {x^6}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {5 a \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac {1}{8} (5 a) \int \sin ^2(c+d x) \, dx+\frac {a \operatorname {Subst}\left (\int \left (-1-x^2-x^4+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {a \sin (c+d x)}{d}-\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a \sin ^3(c+d x)}{3 d}-\frac {5 a \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac {1}{16} (5 a) \int 1 \, dx+\frac {a \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {5 a x}{16}+\frac {a \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a \sin (c+d x)}{d}-\frac {5 a \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a \sin ^3(c+d x)}{3 d}-\frac {5 a \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {a \sin ^5(c+d x)}{5 d}-\frac {a \cos (c+d x) \sin ^5(c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 86, normalized size = 0.68 \[ \frac {a \left (-192 \sin ^5(c+d x)-320 \sin ^3(c+d x)-960 \sin (c+d x)+5 (-45 \sin (2 (c+d x))+9 \sin (4 (c+d x))-\sin (6 (c+d x))+60 c+60 d x)+960 \tanh ^{-1}(\sin (c+d x))\right )}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(960*ArcTanh[Sin[c + d*x]] - 960*Sin[c + d*x] - 320*Sin[c + d*x]^3 - 192*Sin[c + d*x]^5 + 5*(60*c + 60*d*x
- 45*Sin[2*(c + d*x)] + 9*Sin[4*(c + d*x)] - Sin[6*(c + d*x)])))/(960*d)

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fricas [A]  time = 0.77, size = 102, normalized size = 0.80 \[ \frac {75 \, a d x + 120 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 120 \, a \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (40 \, a \cos \left (d x + c\right )^{5} + 48 \, a \cos \left (d x + c\right )^{4} - 130 \, a \cos \left (d x + c\right )^{3} - 176 \, a \cos \left (d x + c\right )^{2} + 165 \, a \cos \left (d x + c\right ) + 368 \, a\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(75*a*d*x + 120*a*log(sin(d*x + c) + 1) - 120*a*log(-sin(d*x + c) + 1) - (40*a*cos(d*x + c)^5 + 48*a*cos
(d*x + c)^4 - 130*a*cos(d*x + c)^3 - 176*a*cos(d*x + c)^2 + 165*a*cos(d*x + c) + 368*a)*sin(d*x + c))/d

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giac [A]  time = 0.84, size = 146, normalized size = 1.15 \[ \frac {75 \, {\left (d x + c\right )} a + 240 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 240 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (165 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 1095 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 3138 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 5118 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1945 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 315 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(75*(d*x + c)*a + 240*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 240*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) -
2*(165*a*tan(1/2*d*x + 1/2*c)^11 + 1095*a*tan(1/2*d*x + 1/2*c)^9 + 3138*a*tan(1/2*d*x + 1/2*c)^7 + 5118*a*tan(
1/2*d*x + 1/2*c)^5 + 1945*a*tan(1/2*d*x + 1/2*c)^3 + 315*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^
6)/d

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maple [A]  time = 0.68, size = 130, normalized size = 1.02 \[ -\frac {a \cos \left (d x +c \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{6 d}-\frac {5 a \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{24 d}-\frac {5 a \cos \left (d x +c \right ) \sin \left (d x +c \right )}{16 d}+\frac {5 a x}{16}+\frac {5 c a}{16 d}-\frac {a \left (\sin ^{5}\left (d x +c \right )\right )}{5 d}-\frac {a \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a \sin \left (d x +c \right )}{d}+\frac {a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a*cos(d*x+c)*sin(d*x+c)^5/d-5/24*a*cos(d*x+c)*sin(d*x+c)^3/d-5/16*a*cos(d*x+c)*sin(d*x+c)/d+5/16*a*x+5/16
/d*c*a-1/5*a*sin(d*x+c)^5/d-1/3*a*sin(d*x+c)^3/d-a*sin(d*x+c)/d+1/d*a*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.47, size = 106, normalized size = 0.83 \[ -\frac {32 \, {\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} a - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(32*(6*sin(d*x + c)^5 + 10*sin(d*x + c)^3 - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1) + 30*si
n(d*x + c))*a - 5*(4*sin(2*d*x + 2*c)^3 + 60*d*x + 60*c + 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a)/d

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mupad [B]  time = 1.05, size = 120, normalized size = 0.94 \[ \frac {5\,a\,x}{16}+\frac {2\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {15\,a\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {7\,a\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {3\,a\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}-\frac {a\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}-\frac {a\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}-\frac {11\,a\,\sin \left (c+d\,x\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a*x)/16 + (2*a*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (15*a*sin(2*c + 2*d*x))/(64*d) + (7*a*sin(
3*c + 3*d*x))/(48*d) + (3*a*sin(4*c + 4*d*x))/(64*d) - (a*sin(5*c + 5*d*x))/(80*d) - (a*sin(6*c + 6*d*x))/(192
*d) - (11*a*sin(c + d*x))/(8*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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