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*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)

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Rubi [A]  time = 0.127833, antiderivative size = 127, normalized size of antiderivative = 1., 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 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 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 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 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 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 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 8

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

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*}

Mathematica [A]  time = 0.184909, 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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Maple [A]  time = 0.085, size = 130, normalized size = 1. \begin{align*} -{\frac{a\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{6\,d}}-{\frac{5\,a\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{24\,d}}-{\frac{5\,a\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{16\,d}}+{\frac{5\,ax}{16}}+{\frac{5\,ac}{16\,d}}-{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5\,d}}-{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{a\sin \left ( dx+c \right ) }{d}}+{\frac{a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

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*a*c-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 = 0.998566, size = 143, normalized size = 1.13 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 1.87778, size = 290, normalized size = 2.28 \begin{align*} \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} \end{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.48382, size = 197, normalized size = 1.55 \begin{align*} \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} \end{align*}

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