\(\int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\) [993]

Optimal result

Integrand size = 31, antiderivative size = 105 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3 (A-B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^6 (A+B)}{6 d (a-a \sin (c+d x))^3}+\frac {a^5 (A-B)}{8 d (a-a \sin (c+d x))^2}+\frac {a^4 (A-B)}{8 d (a-a \sin (c+d x))} \]

[Out]

1/8*a^3*(A-B)*arctanh(sin(d*x+c))/d+1/6*a^6*(A+B)/d/(a-a*sin(d*x+c))^3+1/8*a^5*(A-B)/d/(a-a*sin(d*x+c))^2+1/8*
a^4*(A-B)/d/(a-a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2915, 78, 212} \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^6 (A+B)}{6 d (a-a \sin (c+d x))^3}+\frac {a^5 (A-B)}{8 d (a-a \sin (c+d x))^2}+\frac {a^4 (A-B)}{8 d (a-a \sin (c+d x))}+\frac {a^3 (A-B) \text {arctanh}(\sin (c+d x))}{8 d} \]

[In]

Int[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(a^3*(A - B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^6*(A + B))/(6*d*(a - a*Sin[c + d*x])^3) + (a^5*(A - B))/(8*d*(a
- a*Sin[c + d*x])^2) + (a^4*(A - B))/(8*d*(a - a*Sin[c + d*x]))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 212

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

Rule 2915

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

Rubi steps \begin{align*} \text {integral}& = \frac {a^7 \text {Subst}\left (\int \frac {A+\frac {B x}{a}}{(a-x)^4 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^7 \text {Subst}\left (\int \left (\frac {A+B}{2 a (a-x)^4}+\frac {A-B}{4 a^2 (a-x)^3}+\frac {A-B}{8 a^3 (a-x)^2}+\frac {A-B}{8 a^3 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^6 (A+B)}{6 d (a-a \sin (c+d x))^3}+\frac {a^5 (A-B)}{8 d (a-a \sin (c+d x))^2}+\frac {a^4 (A-B)}{8 d (a-a \sin (c+d x))}+\frac {\left (a^4 (A-B)\right ) \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 d} \\ & = \frac {a^3 (A-B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^6 (A+B)}{6 d (a-a \sin (c+d x))^3}+\frac {a^5 (A-B)}{8 d (a-a \sin (c+d x))^2}+\frac {a^4 (A-B)}{8 d (a-a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3 \left (2 (-5 A+B)-3 (A-B) \text {arctanh}(\sin (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^6+9 (A-B) \sin (c+d x)-3 (A-B) \sin ^2(c+d x)\right )}{24 d (-1+\sin (c+d x))^3} \]

[In]

Integrate[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(a^3*(2*(-5*A + B) - 3*(A - B)*ArcTanh[Sin[c + d*x]]*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^6 + 9*(A - B)*Sin[c
+ d*x] - 3*(A - B)*Sin[c + d*x]^2))/(24*d*(-1 + Sin[c + d*x])^3)

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.83

method	result	size
parallelrisch	\(-\frac {3 \left (\left (A -B \right ) \left (\cos \left (2 d x +2 c \right )+\frac {5 \sin \left (d x +c \right )}{2}-\frac {\sin \left (3 d x +3 c \right )}{6}-\frac {5}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (A -B \right ) \left (\cos \left (2 d x +2 c \right )+\frac {5 \sin \left (d x +c \right )}{2}-\frac {\sin \left (3 d x +3 c \right )}{6}-\frac {5}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {2 \cos \left (2 d x +2 c \right ) \left (A -B \right )}{3}+\frac {\left (-A +B \right ) \sin \left (3 d x +3 c \right )}{6}+\frac {\sin \left (d x +c \right ) \left (A -B \right )}{2}+\frac {8 A}{9}+\frac {8 B}{9}\right ) a^{3}}{4 d \left (-10+15 \sin \left (d x +c \right )-\sin \left (3 d x +3 c \right )+6 \cos \left (2 d x +2 c \right )\right )}\)	\(192\)
risch	\(-\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )} \left (-18 i A \,{\mathrm e}^{3 i \left (d x +c \right )}+3 A \,{\mathrm e}^{4 i \left (d x +c \right )}+18 i B \,{\mathrm e}^{3 i \left (d x +c \right )}-3 B \,{\mathrm e}^{4 i \left (d x +c \right )}+18 i A \,{\mathrm e}^{i \left (d x +c \right )}-46 A \,{\mathrm e}^{2 i \left (d x +c \right )}-18 i B \,{\mathrm e}^{i \left (d x +c \right )}+14 B \,{\mathrm e}^{2 i \left (d x +c \right )}+3 A -3 B \right )}{12 d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}\)	\(241\)
derivativedivides	\(\frac {A \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )+B \,a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{5}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{48 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{48}-\frac {\sin \left (d x +c \right )}{16}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+3 A \,a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+3 B \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )+\frac {A \,a^{3}}{2 \cos \left (d x +c \right )^{6}}+3 B \,a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+A \,a^{3} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B \,a^{3}}{6 \cos \left (d x +c \right )^{6}}}{d}\)	\(442\)
default	\(\frac {A \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )+B \,a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{5}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{48 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{48}-\frac {\sin \left (d x +c \right )}{16}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+3 A \,a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+3 B \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )+\frac {A \,a^{3}}{2 \cos \left (d x +c \right )^{6}}+3 B \,a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+A \,a^{3} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B \,a^{3}}{6 \cos \left (d x +c \right )^{6}}}{d}\)	\(442\)

[In]

int(sec(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-3/4*((A-B)*(cos(2*d*x+2*c)+5/2*sin(d*x+c)-1/6*sin(3*d*x+3*c)-5/3)*ln(tan(1/2*d*x+1/2*c)-1)-(A-B)*(cos(2*d*x+2
*c)+5/2*sin(d*x+c)-1/6*sin(3*d*x+3*c)-5/3)*ln(tan(1/2*d*x+1/2*c)+1)+2/3*cos(2*d*x+2*c)*(A-B)+1/6*(-A+B)*sin(3*
d*x+3*c)+1/2*sin(d*x+c)*(A-B)+8/9*A+8/9*B)*a^3/d/(-10+15*sin(d*x+c)-sin(3*d*x+3*c)+6*cos(2*d*x+2*c))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 242 vs. \(2 (100) = 200\).

Time = 0.29 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.30 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {6 \, {\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} + 18 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right ) - 2 \, {\left (13 \, A - 5 \, B\right )} a^{3} + 3 \, {\left (3 \, {\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} - 4 \, {\left (A - B\right )} a^{3} - {\left ({\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} - 4 \, {\left (A - B\right )} a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, {\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} - 4 \, {\left (A - B\right )} a^{3} - {\left ({\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} - 4 \, {\left (A - B\right )} a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{48 \, {\left (3 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{2} - 4 \, d\right )} \sin \left (d x + c\right ) - 4 \, d\right )}} \]

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/48*(6*(A - B)*a^3*cos(d*x + c)^2 + 18*(A - B)*a^3*sin(d*x + c) - 2*(13*A - 5*B)*a^3 + 3*(3*(A - B)*a^3*cos(d
*x + c)^2 - 4*(A - B)*a^3 - ((A - B)*a^3*cos(d*x + c)^2 - 4*(A - B)*a^3)*sin(d*x + c))*log(sin(d*x + c) + 1) -
3*(3*(A - B)*a^3*cos(d*x + c)^2 - 4*(A - B)*a^3 - ((A - B)*a^3*cos(d*x + c)^2 - 4*(A - B)*a^3)*sin(d*x + c))*
log(-sin(d*x + c) + 1))/(3*d*cos(d*x + c)^2 - (d*cos(d*x + c)^2 - 4*d)*sin(d*x + c) - 4*d)

Sympy [F(-1)]

Timed out. \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**7*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.17 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (A - B\right )} a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A - B\right )} a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right )^{2} - 9 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right ) + 2 \, {\left (5 \, A - B\right )} a^{3}\right )}}{\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 1}}{48 \, d} \]

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/48*(3*(A - B)*a^3*log(sin(d*x + c) + 1) - 3*(A - B)*a^3*log(sin(d*x + c) - 1) - 2*(3*(A - B)*a^3*sin(d*x + c
)^2 - 9*(A - B)*a^3*sin(d*x + c) + 2*(5*A - B)*a^3)/(sin(d*x + c)^3 - 3*sin(d*x + c)^2 + 3*sin(d*x + c) - 1))/
d

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.50 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {6 \, {\left (A a^{3} - B a^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 6 \, {\left (A a^{3} - B a^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {11 \, A a^{3} \sin \left (d x + c\right )^{3} - 11 \, B a^{3} \sin \left (d x + c\right )^{3} - 45 \, A a^{3} \sin \left (d x + c\right )^{2} + 45 \, B a^{3} \sin \left (d x + c\right )^{2} + 69 \, A a^{3} \sin \left (d x + c\right ) - 69 \, B a^{3} \sin \left (d x + c\right ) - 51 \, A a^{3} + 19 \, B a^{3}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{3}}}{96 \, d} \]

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/96*(6*(A*a^3 - B*a^3)*log(abs(sin(d*x + c) + 1)) - 6*(A*a^3 - B*a^3)*log(abs(sin(d*x + c) - 1)) + (11*A*a^3*
sin(d*x + c)^3 - 11*B*a^3*sin(d*x + c)^3 - 45*A*a^3*sin(d*x + c)^2 + 45*B*a^3*sin(d*x + c)^2 + 69*A*a^3*sin(d*
x + c) - 69*B*a^3*sin(d*x + c) - 51*A*a^3 + 19*B*a^3)/(sin(d*x + c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 9.61 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.07 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (A-B\right )}{8\,d}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {A\,a^3}{8}-\frac {B\,a^3}{8}\right )+\frac {5\,A\,a^3}{12}-\frac {B\,a^3}{12}-\sin \left (c+d\,x\right )\,\left (\frac {3\,A\,a^3}{8}-\frac {3\,B\,a^3}{8}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^3-3\,{\sin \left (c+d\,x\right )}^2+3\,\sin \left (c+d\,x\right )-1\right )} \]

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3)/cos(c + d*x)^7,x)

[Out]

(a^3*atanh(sin(c + d*x))*(A - B))/(8*d) - (sin(c + d*x)^2*((A*a^3)/8 - (B*a^3)/8) + (5*A*a^3)/12 - (B*a^3)/12
- sin(c + d*x)*((3*A*a^3)/8 - (3*B*a^3)/8))/(d*(3*sin(c + d*x) - 3*sin(c + d*x)^2 + sin(c + d*x)^3 - 1))