3.49 \(\int (a+a \sec (c+d x))^3 \sin ^6(c+d x) \, dx\)

Optimal. Leaf size=182 \[ -\frac{3 a^3 \sin ^5(c+d x)}{5 d}-\frac{2 a^3 \sin ^3(c+d x)}{3 d}-\frac{a^3 \sin (c+d x)}{d}+\frac{3 a^3 \tan (c+d x)}{d}+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{a^3 \sin (c+d x) \cos ^5(c+d x)}{6 d}-\frac{5 a^3 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{43 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{a^3 \tan (c+d x) \sec (c+d x)}{2 d}-\frac{85 a^3 x}{16} \]

[Out]

(-85*a^3*x)/16 + (a^3*ArcTanh[Sin[c + d*x]])/(2*d) - (a^3*Sin[c + d*x])/d + (43*a^3*Cos[c + d*x]*Sin[c + d*x])
/(16*d) - (5*a^3*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) - (a^3*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (2*a^3*Sin[c
+ d*x]^3)/(3*d) - (3*a^3*Sin[c + d*x]^5)/(5*d) + (3*a^3*Tan[c + d*x])/d + (a^3*Sec[c + d*x]*Tan[c + d*x])/(2*d
)

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Rubi [A]  time = 0.273681, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3872, 2872, 2637, 2635, 8, 2633, 3767, 3768, 3770} \[ -\frac{3 a^3 \sin ^5(c+d x)}{5 d}-\frac{2 a^3 \sin ^3(c+d x)}{3 d}-\frac{a^3 \sin (c+d x)}{d}+\frac{3 a^3 \tan (c+d x)}{d}+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{a^3 \sin (c+d x) \cos ^5(c+d x)}{6 d}-\frac{5 a^3 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{43 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{a^3 \tan (c+d x) \sec (c+d x)}{2 d}-\frac{85 a^3 x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-85*a^3*x)/16 + (a^3*ArcTanh[Sin[c + d*x]])/(2*d) - (a^3*Sin[c + d*x])/d + (43*a^3*Cos[c + d*x]*Sin[c + d*x])
/(16*d) - (5*a^3*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) - (a^3*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (2*a^3*Sin[c
+ d*x]^3)/(3*d) - (3*a^3*Sin[c + d*x]^5)/(5*d) + (3*a^3*Tan[c + d*x])/d + (a^3*Sec[c + d*x]*Tan[c + d*x])/(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 2872

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, 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 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 3767

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

Rule 3768

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int (a+a \sec (c+d x))^3 \sin ^6(c+d x) \, dx &=-\int (-a-a \cos (c+d x))^3 \sin ^3(c+d x) \tan ^3(c+d x) \, dx\\ &=-\frac{\int \left (8 a^9+6 a^9 \cos (c+d x)-6 a^9 \cos ^2(c+d x)-8 a^9 \cos ^3(c+d x)+3 a^9 \cos ^5(c+d x)+a^9 \cos ^6(c+d x)-3 a^9 \sec ^2(c+d x)-a^9 \sec ^3(c+d x)\right ) \, dx}{a^6}\\ &=-8 a^3 x-a^3 \int \cos ^6(c+d x) \, dx+a^3 \int \sec ^3(c+d x) \, dx-\left (3 a^3\right ) \int \cos ^5(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^2(c+d x) \, dx-\left (6 a^3\right ) \int \cos (c+d x) \, dx+\left (6 a^3\right ) \int \cos ^2(c+d x) \, dx+\left (8 a^3\right ) \int \cos ^3(c+d x) \, dx\\ &=-8 a^3 x-\frac{6 a^3 \sin (c+d x)}{d}+\frac{3 a^3 \cos (c+d x) \sin (c+d x)}{d}-\frac{a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} a^3 \int \sec (c+d x) \, dx-\frac{1}{6} \left (5 a^3\right ) \int \cos ^4(c+d x) \, dx+\left (3 a^3\right ) \int 1 \, dx-\frac{\left (3 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}+\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac{\left (8 a^3\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=-5 a^3 x+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{a^3 \sin (c+d x)}{d}+\frac{3 a^3 \cos (c+d x) \sin (c+d x)}{d}-\frac{5 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 a^3 \sin ^3(c+d x)}{3 d}-\frac{3 a^3 \sin ^5(c+d x)}{5 d}+\frac{3 a^3 \tan (c+d x)}{d}+\frac{a^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{8} \left (5 a^3\right ) \int \cos ^2(c+d x) \, dx\\ &=-5 a^3 x+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{a^3 \sin (c+d x)}{d}+\frac{43 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac{5 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 a^3 \sin ^3(c+d x)}{3 d}-\frac{3 a^3 \sin ^5(c+d x)}{5 d}+\frac{3 a^3 \tan (c+d x)}{d}+\frac{a^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{16} \left (5 a^3\right ) \int 1 \, dx\\ &=-\frac{85 a^3 x}{16}+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{a^3 \sin (c+d x)}{d}+\frac{43 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac{5 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 a^3 \sin ^3(c+d x)}{3 d}-\frac{3 a^3 \sin ^5(c+d x)}{5 d}+\frac{3 a^3 \tan (c+d x)}{d}+\frac{a^3 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.86938, size = 136, normalized size = 0.75 \[ -\frac{a^3 \sec ^2(c+d x) \left (-460 \sin (c+d x)-8145 \sin (2 (c+d x))+1156 \sin (3 (c+d x))-1120 \sin (4 (c+d x))-268 \sin (5 (c+d x))+55 \sin (6 (c+d x))+36 \sin (7 (c+d x))+5 \sin (8 (c+d x))+10200 (c+d x) \cos (2 (c+d x))-1920 \cos ^2(c+d x) \tanh ^{-1}(\sin (c+d x))+10200 c+10200 d x\right )}{3840 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Sec[c + d*x]^2*(10200*c + 10200*d*x - 1920*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + 10200*(c + d*x)*Cos[2*
(c + d*x)] - 460*Sin[c + d*x] - 8145*Sin[2*(c + d*x)] + 1156*Sin[3*(c + d*x)] - 1120*Sin[4*(c + d*x)] - 268*Si
n[5*(c + d*x)] + 55*Sin[6*(c + d*x)] + 36*Sin[7*(c + d*x)] + 5*Sin[8*(c + d*x)]))/(3840*d)

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Maple [A]  time = 0.049, size = 197, normalized size = 1.1 \begin{align*}{\frac{17\,{a}^{3}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{85\,{a}^{3}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{85\,{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{16\,d}}-{\frac{85\,{a}^{3}x}{16}}-{\frac{85\,{a}^{3}c}{16\,d}}-{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{10\,d}}-{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{6\,d}}-{\frac{{a}^{3}\sin \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{d\cos \left ( dx+c \right ) }}+{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

17/6*a^3*cos(d*x+c)*sin(d*x+c)^5/d+85/24*a^3*cos(d*x+c)*sin(d*x+c)^3/d+85/16*a^3*cos(d*x+c)*sin(d*x+c)/d-85/16
*a^3*x-85/16/d*a^3*c-1/10*a^3*sin(d*x+c)^5/d-1/6*a^3*sin(d*x+c)^3/d-1/2*a^3*sin(d*x+c)/d+1/2/d*a^3*ln(sec(d*x+
c)+tan(d*x+c))+3/d*a^3*sin(d*x+c)^7/cos(d*x+c)+1/2/d*a^3*sin(d*x+c)^7/cos(d*x+c)^2

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Maxima [A]  time = 1.53424, size = 324, normalized size = 1.78 \begin{align*} -\frac{96 \,{\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^{3} - 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^{3} - 80 \,{\left (4 \, \sin \left (d x + c\right )^{3} - \frac{6 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 24 \, \sin \left (d x + c\right )\right )} a^{3} + 360 \,{\left (15 \, d x + 15 \, c - \frac{9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} a^{3}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(96*(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^3 - 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^3 - 80
*(4*sin(d*x + c)^3 - 6*sin(d*x + c)/(sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)
+ 24*sin(d*x + c))*a^3 + 360*(15*d*x + 15*c - (9*tan(d*x + c)^3 + 7*tan(d*x + c))/(tan(d*x + c)^4 + 2*tan(d*x
+ c)^2 + 1) - 8*tan(d*x + c))*a^3)/d

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Fricas [A]  time = 1.97831, size = 467, normalized size = 2.57 \begin{align*} -\frac{1275 \, a^{3} d x \cos \left (d x + c\right )^{2} - 60 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + 60 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (40 \, a^{3} \cos \left (d x + c\right )^{7} + 144 \, a^{3} \cos \left (d x + c\right )^{6} + 50 \, a^{3} \cos \left (d x + c\right )^{5} - 448 \, a^{3} \cos \left (d x + c\right )^{4} - 645 \, a^{3} \cos \left (d x + c\right )^{3} + 544 \, a^{3} \cos \left (d x + c\right )^{2} - 720 \, a^{3} \cos \left (d x + c\right ) - 120 \, a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(1275*a^3*d*x*cos(d*x + c)^2 - 60*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1) + 60*a^3*cos(d*x + c)^2*log(
-sin(d*x + c) + 1) + (40*a^3*cos(d*x + c)^7 + 144*a^3*cos(d*x + c)^6 + 50*a^3*cos(d*x + c)^5 - 448*a^3*cos(d*x
+ c)^4 - 645*a^3*cos(d*x + c)^3 + 544*a^3*cos(d*x + c)^2 - 720*a^3*cos(d*x + c) - 120*a^3)*sin(d*x + c))/(d*c
os(d*x + c)^2)

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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))**3*sin(d*x+c)**6,x)

[Out]

Timed out

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Giac [A]  time = 1.28381, size = 286, normalized size = 1.57 \begin{align*} -\frac{1275 \,{\left (d x + c\right )} a^{3} - 120 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 120 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{240 \,{\left (5 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, a^{3} \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 )}^{2}} + \frac{2 \,{\left (795 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 4025 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 7614 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 5634 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 345 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 315 \, a^{3} \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))^3*sin(d*x+c)^6,x, algorithm="giac")

[Out]

-1/240*(1275*(d*x + c)*a^3 - 120*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 120*a^3*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 240*(5*a^3*tan(1/2*d*x + 1/2*c)^3 - 7*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 + 2*(7
95*a^3*tan(1/2*d*x + 1/2*c)^11 + 4025*a^3*tan(1/2*d*x + 1/2*c)^9 + 7614*a^3*tan(1/2*d*x + 1/2*c)^7 + 5634*a^3*
tan(1/2*d*x + 1/2*c)^5 - 345*a^3*tan(1/2*d*x + 1/2*c)^3 - 315*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^
2 + 1)^6)/d