\(\int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) [960]

Optimal result

Integrand size = 29, antiderivative size = 157 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a (5 A-B) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {a^4 (A+B)}{24 d (a-a \sin (c+d x))^3}+\frac {a^3 (3 A+B)}{32 d (a-a \sin (c+d x))^2}+\frac {3 a^2 A}{16 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{32 d (a+a \sin (c+d x))^2}-\frac {a^2 (2 A-B)}{16 d (a+a \sin (c+d x))} \]

[Out]

1/16*a*(5*A-B)*arctanh(sin(d*x+c))/d+1/24*a^4*(A+B)/d/(a-a*sin(d*x+c))^3+1/32*a^3*(3*A+B)/d/(a-a*sin(d*x+c))^2
+3/16*a^2*A/d/(a-a*sin(d*x+c))-1/32*a^3*(A-B)/d/(a+a*sin(d*x+c))^2-1/16*a^2*(2*A-B)/d/(a+a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 78, 212} \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a^4 (A+B)}{24 d (a-a \sin (c+d x))^3}+\frac {a^3 (3 A+B)}{32 d (a-a \sin (c+d x))^2}-\frac {a^3 (A-B)}{32 d (a \sin (c+d x)+a)^2}-\frac {a^2 (2 A-B)}{16 d (a \sin (c+d x)+a)}+\frac {3 a^2 A}{16 d (a-a \sin (c+d x))}+\frac {a (5 A-B) \text {arctanh}(\sin (c+d x))}{16 d} \]

[In]

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

[Out]

(a*(5*A - B)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^4*(A + B))/(24*d*(a - a*Sin[c + d*x])^3) + (a^3*(3*A + B))/(32
*d*(a - a*Sin[c + d*x])^2) + (3*a^2*A)/(16*d*(a - a*Sin[c + d*x])) - (a^3*(A - B))/(32*d*(a + a*Sin[c + d*x])^
2) - (a^2*(2*A - B))/(16*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)^3} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^7 \text {Subst}\left (\int \left (\frac {A+B}{8 a^3 (a-x)^4}+\frac {3 A+B}{16 a^4 (a-x)^3}+\frac {3 A}{16 a^5 (a-x)^2}+\frac {A-B}{16 a^4 (a+x)^3}+\frac {2 A-B}{16 a^5 (a+x)^2}+\frac {5 A-B}{16 a^5 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^4 (A+B)}{24 d (a-a \sin (c+d x))^3}+\frac {a^3 (3 A+B)}{32 d (a-a \sin (c+d x))^2}+\frac {3 a^2 A}{16 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{32 d (a+a \sin (c+d x))^2}-\frac {a^2 (2 A-B)}{16 d (a+a \sin (c+d x))}+\frac {\left (a^2 (5 A-B)\right ) \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{16 d} \\ & = \frac {a (5 A-B) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {a^4 (A+B)}{24 d (a-a \sin (c+d x))^3}+\frac {a^3 (3 A+B)}{32 d (a-a \sin (c+d x))^2}+\frac {3 a^2 A}{16 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{32 d (a+a \sin (c+d x))^2}-\frac {a^2 (2 A-B)}{16 d (a+a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.66 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a \left (6 (5 A-B) \text {arctanh}(\sin (c+d x))-\frac {4 (A+B)}{(-1+\sin (c+d x))^3}+\frac {3 (3 A+B)}{(-1+\sin (c+d x))^2}-\frac {18 A}{-1+\sin (c+d x)}-\frac {3 (A-B)}{(1+\sin (c+d x))^2}+\frac {6 (-2 A+B)}{1+\sin (c+d x)}\right )}{96 d} \]

[In]

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

[Out]

(a*(6*(5*A - B)*ArcTanh[Sin[c + d*x]] - (4*(A + B))/(-1 + Sin[c + d*x])^3 + (3*(3*A + B))/(-1 + Sin[c + d*x])^
2 - (18*A)/(-1 + Sin[c + d*x]) - (3*(A - B))/(1 + Sin[c + d*x])^2 + (6*(-2*A + B))/(1 + Sin[c + d*x])))/(96*d)

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {\frac {a A}{6 \cos \left (d x +c \right )^{6}}+B a \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+a A \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B a}{6 \cos \left (d x +c \right )^{6}}}{d}\)	\(169\)
default	\(\frac {\frac {a A}{6 \cos \left (d x +c \right )^{6}}+B a \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+a A \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B a}{6 \cos \left (d x +c \right )^{6}}}{d}\)	\(169\)
parallelrisch	\(-\frac {15 \left (\left (A -\frac {B}{5}\right ) \left (\frac {2 \sin \left (d x +c \right )}{3}+\sin \left (3 d x +3 c \right )-\frac {8 \cos \left (2 d x +2 c \right )}{3}+\frac {\sin \left (5 d x +5 c \right )}{3}-\frac {2 \cos \left (4 d x +4 c \right )}{3}-2\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (A -\frac {B}{5}\right ) \left (\frac {2 \sin \left (d x +c \right )}{3}+\sin \left (3 d x +3 c \right )-\frac {8 \cos \left (2 d x +2 c \right )}{3}+\frac {\sin \left (5 d x +5 c \right )}{3}-\frac {2 \cos \left (4 d x +4 c \right )}{3}-2\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {2 \left (A -\frac {B}{5}\right ) \cos \left (4 d x +4 c \right )}{9}+2 \left (A -\frac {B}{5}\right ) \sin \left (3 d x +3 c \right )+\frac {2 \left (A -\frac {B}{5}\right ) \sin \left (5 d x +5 c \right )}{9}+\frac {16 \left (A -\frac {B}{5}\right ) \sin \left (d x +c \right )}{3}-\frac {14 A}{15}+\frac {18 B}{5}\right ) a}{16 d \left (3 \sin \left (3 d x +3 c \right )+2 \sin \left (d x +c \right )+\sin \left (5 d x +5 c \right )-8 \cos \left (2 d x +2 c \right )-2 \cos \left (4 d x +4 c \right )-6\right )}\)	\(270\)
risch	\(-\frac {i a \,{\mathrm e}^{i \left (d x +c \right )} \left (6 i B \,{\mathrm e}^{7 i \left (d x +c \right )}+15 A \,{\mathrm e}^{8 i \left (d x +c \right )}-110 i A \,{\mathrm e}^{5 i \left (d x +c \right )}-3 B \,{\mathrm e}^{8 i \left (d x +c \right )}+110 i A \,{\mathrm e}^{3 i \left (d x +c \right )}+40 A \,{\mathrm e}^{6 i \left (d x +c \right )}-22 i B \,{\mathrm e}^{3 i \left (d x +c \right )}-8 B \,{\mathrm e}^{6 i \left (d x +c \right )}+22 i B \,{\mathrm e}^{5 i \left (d x +c \right )}+18 A \,{\mathrm e}^{4 i \left (d x +c \right )}-6 i B \,{\mathrm e}^{i \left (d x +c \right )}+150 B \,{\mathrm e}^{4 i \left (d x +c \right )}+30 i A \,{\mathrm e}^{i \left (d x +c \right )}+40 A \,{\mathrm e}^{2 i \left (d x +c \right )}-30 i A \,{\mathrm e}^{7 i \left (d x +c \right )}-8 B \,{\mathrm e}^{2 i \left (d x +c \right )}+15 A -3 B \right )}{24 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d}-\frac {5 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{16 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{16 d}+\frac {5 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{16 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{16 d}\)	\(361\)
norman	\(\frac {\frac {2 \left (a A +B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (a A +B a \right ) \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (a A +B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (a A +B a \right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {10 \left (4 a A +4 B a \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (13 a A +13 B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (13 a A +13 B a \right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {a \left (11 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {a \left (11 A +B \right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {7 a \left (19 A +25 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {7 a \left (19 A +25 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {a \left (71 A +53 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {a \left (71 A +53 B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {a \left (275 A +281 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {a \left (275 A +281 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{6} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {a \left (5 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 d}+\frac {a \left (5 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d}\)	\(437\)

[In]

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

[Out]

1/d*(1/6*a*A/cos(d*x+c)^6+B*a*(1/6*sin(d*x+c)^3/cos(d*x+c)^6+1/8*sin(d*x+c)^3/cos(d*x+c)^4+1/16*sin(d*x+c)^3/c
os(d*x+c)^2+1/16*sin(d*x+c)-1/16*ln(sec(d*x+c)+tan(d*x+c)))+a*A*(-(-1/6*sec(d*x+c)^5-5/24*sec(d*x+c)^3-5/16*se
c(d*x+c))*tan(d*x+c)+5/16*ln(sec(d*x+c)+tan(d*x+c)))+1/6*B*a/cos(d*x+c)^6)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.41 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {6 \, {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4} - 2 \, {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{2} - 4 \, {\left (A - 5 \, B\right )} a - 3 \, {\left ({\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A - B\right )} a\right )} \sin \left (d x + c\right )}{96 \, {\left (d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{4}\right )}} \]

[In]

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

[Out]

-1/96*(6*(5*A - B)*a*cos(d*x + c)^4 - 2*(5*A - B)*a*cos(d*x + c)^2 - 4*(A - 5*B)*a - 3*((5*A - B)*a*cos(d*x +
c)^4*sin(d*x + c) - (5*A - B)*a*cos(d*x + c)^4)*log(sin(d*x + c) + 1) + 3*((5*A - B)*a*cos(d*x + c)^4*sin(d*x
+ c) - (5*A - B)*a*cos(d*x + c)^4)*log(-sin(d*x + c) + 1) + 2*(3*(5*A - B)*a*cos(d*x + c)^2 + 2*(5*A - B)*a)*s
in(d*x + c))/(d*cos(d*x + c)^4*sin(d*x + c) - d*cos(d*x + c)^4)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.09 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A - B\right )} a \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A - B\right )} a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (5 \, A - B\right )} a \sin \left (d x + c\right )^{4} - 3 \, {\left (5 \, A - B\right )} a \sin \left (d x + c\right )^{3} - 5 \, {\left (5 \, A - B\right )} a \sin \left (d x + c\right )^{2} + 5 \, {\left (5 \, A - B\right )} a \sin \left (d x + c\right ) + 8 \, {\left (A + B\right )} a\right )}}{\sin \left (d x + c\right )^{5} - \sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{3} + 2 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right ) - 1}}{96 \, d} \]

[In]

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

[Out]

1/96*(3*(5*A - B)*a*log(sin(d*x + c) + 1) - 3*(5*A - B)*a*log(sin(d*x + c) - 1) - 2*(3*(5*A - B)*a*sin(d*x + c
)^4 - 3*(5*A - B)*a*sin(d*x + c)^3 - 5*(5*A - B)*a*sin(d*x + c)^2 + 5*(5*A - B)*a*sin(d*x + c) + 8*(A + B)*a)/
(sin(d*x + c)^5 - sin(d*x + c)^4 - 2*sin(d*x + c)^3 + 2*sin(d*x + c)^2 + sin(d*x + c) - 1))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.28 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {6 \, {\left (5 \, A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 6 \, {\left (5 \, A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {3 \, {\left (15 \, A a \sin \left (d x + c\right )^{2} - 3 \, B a \sin \left (d x + c\right )^{2} + 38 \, A a \sin \left (d x + c\right ) - 10 \, B a \sin \left (d x + c\right ) + 25 \, A a - 9 \, B a\right )}}{{\left (\sin \left (d x + c\right ) + 1\right )}^{2}} + \frac {55 \, A a \sin \left (d x + c\right )^{3} - 11 \, B a \sin \left (d x + c\right )^{3} - 201 \, A a \sin \left (d x + c\right )^{2} + 33 \, B a \sin \left (d x + c\right )^{2} + 255 \, A a \sin \left (d x + c\right ) - 27 \, B a \sin \left (d x + c\right ) - 117 \, A a - 3 \, B a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{3}}}{192 \, d} \]

[In]

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

[Out]

1/192*(6*(5*A*a - B*a)*log(abs(sin(d*x + c) + 1)) - 6*(5*A*a - B*a)*log(abs(sin(d*x + c) - 1)) - 3*(15*A*a*sin
(d*x + c)^2 - 3*B*a*sin(d*x + c)^2 + 38*A*a*sin(d*x + c) - 10*B*a*sin(d*x + c) + 25*A*a - 9*B*a)/(sin(d*x + c)
+ 1)^2 + (55*A*a*sin(d*x + c)^3 - 11*B*a*sin(d*x + c)^3 - 201*A*a*sin(d*x + c)^2 + 33*B*a*sin(d*x + c)^2 + 25
5*A*a*sin(d*x + c) - 27*B*a*sin(d*x + c) - 117*A*a - 3*B*a)/(sin(d*x + c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 10.03 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.99 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (5\,A-B\right )}{16\,d}-\frac {\left (\frac {5\,A\,a}{16}-\frac {B\,a}{16}\right )\,{\sin \left (c+d\,x\right )}^4+\left (\frac {B\,a}{16}-\frac {5\,A\,a}{16}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {5\,B\,a}{48}-\frac {25\,A\,a}{48}\right )\,{\sin \left (c+d\,x\right )}^2+\left (\frac {25\,A\,a}{48}-\frac {5\,B\,a}{48}\right )\,\sin \left (c+d\,x\right )+\frac {A\,a}{6}+\frac {B\,a}{6}}{d\,\left ({\sin \left (c+d\,x\right )}^5-{\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^3+2\,{\sin \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )-1\right )} \]

[In]

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

[Out]

(a*atanh(sin(c + d*x))*(5*A - B))/(16*d) - ((A*a)/6 + (B*a)/6 + sin(c + d*x)*((25*A*a)/48 - (5*B*a)/48) - sin(
c + d*x)^3*((5*A*a)/16 - (B*a)/16) + sin(c + d*x)^4*((5*A*a)/16 - (B*a)/16) - sin(c + d*x)^2*((25*A*a)/48 - (5
*B*a)/48))/(d*(sin(c + d*x) + 2*sin(c + d*x)^2 - 2*sin(c + d*x)^3 - sin(c + d*x)^4 + sin(c + d*x)^5 - 1))