3.30 \(\int (a+a \sec (c+d x))^2 \sin ^6(c+d x) \, dx\)

Optimal. Leaf size=157 \[ -\frac {2 a^2 \sin ^5(c+d x)}{5 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \tan (c+d x)}{d}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {7 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}-\frac {25 a^2 x}{16} \]

[Out]

-25/16*a^2*x+2*a^2*arctanh(sin(d*x+c))/d-2*a^2*sin(d*x+c)/d+7/16*a^2*cos(d*x+c)*sin(d*x+c)/d+7/24*a^2*cos(d*x+
c)^3*sin(d*x+c)/d-1/6*a^2*cos(d*x+c)^5*sin(d*x+c)/d-2/3*a^2*sin(d*x+c)^3/d-2/5*a^2*sin(d*x+c)^5/d+a^2*tan(d*x+
c)/d

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Rubi [A]  time = 0.27, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3872, 2872, 2637, 2633, 2635, 8, 3770, 3767} \[ -\frac {2 a^2 \sin ^5(c+d x)}{5 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \tan (c+d x)}{d}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {7 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{16 d}-\frac {25 a^2 x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-25*a^2*x)/16 + (2*a^2*ArcTanh[Sin[c + d*x]])/d - (2*a^2*Sin[c + d*x])/d + (7*a^2*Cos[c + d*x]*Sin[c + d*x])/
(16*d) + (7*a^2*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) - (a^2*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (2*a^2*Sin[c +
d*x]^3)/(3*d) - (2*a^2*Sin[c + d*x]^5)/(5*d) + (a^2*Tan[c + d*x])/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 2637

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

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 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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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))^2 \sin ^6(c+d x) \, dx &=\int (-a-a \cos (c+d x))^2 \sin ^4(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac {\int \left (-2 a^8-6 a^8 \cos (c+d x)+6 a^8 \cos ^3(c+d x)+2 a^8 \cos ^4(c+d x)-2 a^8 \cos ^5(c+d x)-a^8 \cos ^6(c+d x)+2 a^8 \sec (c+d x)+a^8 \sec ^2(c+d x)\right ) \, dx}{a^6}\\ &=-2 a^2 x-a^2 \int \cos ^6(c+d x) \, dx+a^2 \int \sec ^2(c+d x) \, dx+\left (2 a^2\right ) \int \cos ^4(c+d x) \, dx-\left (2 a^2\right ) \int \cos ^5(c+d x) \, dx+\left (2 a^2\right ) \int \sec (c+d x) \, dx-\left (6 a^2\right ) \int \cos (c+d x) \, dx+\left (6 a^2\right ) \int \cos ^3(c+d x) \, dx\\ &=-2 a^2 x+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {6 a^2 \sin (c+d x)}{d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{2 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {1}{6} \left (5 a^2\right ) \int \cos ^4(c+d x) \, dx+\frac {1}{2} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx-\frac {a^2 \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac {\left (6 a^2\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=-2 a^2 x+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{4 d}+\frac {7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin ^5(c+d x)}{5 d}+\frac {a^2 \tan (c+d x)}{d}-\frac {1}{8} \left (5 a^2\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{4} \left (3 a^2\right ) \int 1 \, dx\\ &=-\frac {5 a^2 x}{4}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin ^5(c+d x)}{5 d}+\frac {a^2 \tan (c+d x)}{d}-\frac {1}{16} \left (5 a^2\right ) \int 1 \, dx\\ &=-\frac {25 a^2 x}{16}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin ^5(c+d x)}{5 d}+\frac {a^2 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.56, size = 124, normalized size = 0.79 \[ -\frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (384 \sin ^5(c+d x)+640 \sin ^3(c+d x)+1920 \sin (c+d x)-255 \sin (2 (c+d x))-15 \sin (4 (c+d x))+5 \sin (6 (c+d x))+420 \tan ^{-1}(\tan (c+d x))-960 \tan (c+d x)-1920 \tanh ^{-1}(\sin (c+d x))+1080 c+1080 d x\right )}{3840 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3840*(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(1080*c + 1080*d*x + 420*ArcTan[Tan[c + d*x]] - 1920*ArcT
anh[Sin[c + d*x]] + 1920*Sin[c + d*x] + 640*Sin[c + d*x]^3 + 384*Sin[c + d*x]^5 - 255*Sin[2*(c + d*x)] - 15*Si
n[4*(c + d*x)] + 5*Sin[6*(c + d*x)] - 960*Tan[c + d*x]))/d

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fricas [A]  time = 0.68, size = 158, normalized size = 1.01 \[ -\frac {375 \, a^{2} d x \cos \left (d x + c\right ) - 240 \, a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + 240 \, a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (40 \, a^{2} \cos \left (d x + c\right )^{6} + 96 \, a^{2} \cos \left (d x + c\right )^{5} - 70 \, a^{2} \cos \left (d x + c\right )^{4} - 352 \, a^{2} \cos \left (d x + c\right )^{3} - 105 \, a^{2} \cos \left (d x + c\right )^{2} + 736 \, a^{2} \cos \left (d x + c\right ) - 240 \, a^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(375*a^2*d*x*cos(d*x + c) - 240*a^2*cos(d*x + c)*log(sin(d*x + c) + 1) + 240*a^2*cos(d*x + c)*log(-sin(
d*x + c) + 1) + (40*a^2*cos(d*x + c)^6 + 96*a^2*cos(d*x + c)^5 - 70*a^2*cos(d*x + c)^4 - 352*a^2*cos(d*x + c)^
3 - 105*a^2*cos(d*x + c)^2 + 736*a^2*cos(d*x + c) - 240*a^2)*sin(d*x + c))/(d*cos(d*x + c))

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giac [A]  time = 0.51, size = 193, normalized size = 1.23 \[ -\frac {375 \, {\left (d x + c\right )} a^{2} - 480 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 480 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {480 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (615 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 3485 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 7926 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 8586 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2595 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 345 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(375*(d*x + c)*a^2 - 480*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 480*a^2*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 480*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(615*a^2*tan(1/2*d*x + 1/2*c)^11 + 3485*
a^2*tan(1/2*d*x + 1/2*c)^9 + 7926*a^2*tan(1/2*d*x + 1/2*c)^7 + 8586*a^2*tan(1/2*d*x + 1/2*c)^5 + 2595*a^2*tan(
1/2*d*x + 1/2*c)^3 + 345*a^2*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 = 0.69, size = 172, normalized size = 1.10 \[ \frac {5 a^{2} \cos \left (d x +c \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{6 d}+\frac {25 a^{2} \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{24 d}+\frac {25 a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{16 d}-\frac {25 a^{2} x}{16}-\frac {25 a^{2} c}{16 d}-\frac {2 a^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{5 d}-\frac {2 a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}-\frac {2 a^{2} \sin \left (d x +c \right )}{d}+\frac {2 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/6/d*a^2*cos(d*x+c)*sin(d*x+c)^5+25/24/d*a^2*cos(d*x+c)*sin(d*x+c)^3+25/16*a^2*cos(d*x+c)*sin(d*x+c)/d-25/16*
a^2*x-25/16/d*a^2*c-2/5*a^2*sin(d*x+c)^5/d-2/3*a^2*sin(d*x+c)^3/d-2*a^2*sin(d*x+c)/d+2/d*a^2*ln(sec(d*x+c)+tan
(d*x+c))+1/d*a^2*sin(d*x+c)^7/cos(d*x+c)

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maxima [A]  time = 0.54, size = 174, normalized size = 1.11 \[ -\frac {64 \, {\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} a^{2} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} + 120 \, {\left (15 \, d x + 15 \, c - \frac {9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} a^{2}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(64*(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^2 - 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^2 + 12
0*(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^2)/d

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mupad [B]  time = 2.16, size = 235, normalized size = 1.50 \[ \frac {\frac {57\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{8}+\frac {431\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{12}+\frac {8041\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{120}+\frac {91\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}-\frac {797\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{40}-\frac {27\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}-\frac {7\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {25\,a^2\,x}{16}+\frac {4\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

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]

((91*a^2*tan(c/2 + (d*x)/2)^7)/2 - (797*a^2*tan(c/2 + (d*x)/2)^5)/40 - (27*a^2*tan(c/2 + (d*x)/2)^3)/4 + (8041
*a^2*tan(c/2 + (d*x)/2)^9)/120 + (431*a^2*tan(c/2 + (d*x)/2)^11)/12 + (57*a^2*tan(c/2 + (d*x)/2)^13)/8 - (7*a^
2*tan(c/2 + (d*x)/2))/8)/(d*(5*tan(c/2 + (d*x)/2)^2 + 9*tan(c/2 + (d*x)/2)^4 + 5*tan(c/2 + (d*x)/2)^6 - 5*tan(
c/2 + (d*x)/2)^8 - 9*tan(c/2 + (d*x)/2)^10 - 5*tan(c/2 + (d*x)/2)^12 - tan(c/2 + (d*x)/2)^14 + 1)) - (25*a^2*x
)/16 + (4*a^2*atanh(tan(c/2 + (d*x)/2)))/d

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

[Out]

Timed out

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