\(\int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [976]

Optimal result

Integrand size = 31, antiderivative size = 132 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 (2 A-B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^5 (A+B)}{12 d (a-a \sin (c+d x))^3}+\frac {a^4 A}{8 d (a-a \sin (c+d x))^2}+\frac {a^3 (3 A-B)}{16 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{16 d (a+a \sin (c+d x))} \]

[Out]

1/8*a^2*(2*A-B)*arctanh(sin(d*x+c))/d+1/12*a^5*(A+B)/d/(a-a*sin(d*x+c))^3+1/8*a^4*A/d/(a-a*sin(d*x+c))^2+1/16*
a^3*(3*A-B)/d/(a-a*sin(d*x+c))-1/16*a^3*(A-B)/d/(a+a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 132, 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} \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^5 (A+B)}{12 d (a-a \sin (c+d x))^3}+\frac {a^4 A}{8 d (a-a \sin (c+d x))^2}+\frac {a^3 (3 A-B)}{16 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{16 d (a \sin (c+d x)+a)}+\frac {a^2 (2 A-B) \text {arctanh}(\sin (c+d x))}{8 d} \]

[In]

Int[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*(2*A - B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^5*(A + B))/(12*d*(a - a*Sin[c + d*x])^3) + (a^4*A)/(8*d*(a -
a*Sin[c + d*x])^2) + (a^3*(3*A - B))/(16*d*(a - a*Sin[c + d*x])) - (a^3*(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)^2} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^7 \text {Subst}\left (\int \left (\frac {A+B}{4 a^2 (a-x)^4}+\frac {A}{4 a^3 (a-x)^3}+\frac {3 A-B}{16 a^4 (a-x)^2}+\frac {A-B}{16 a^4 (a+x)^2}+\frac {2 A-B}{8 a^4 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^5 (A+B)}{12 d (a-a \sin (c+d x))^3}+\frac {a^4 A}{8 d (a-a \sin (c+d x))^2}+\frac {a^3 (3 A-B)}{16 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{16 d (a+a \sin (c+d x))}+\frac {\left (a^3 (2 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^2 (2 A-B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^5 (A+B)}{12 d (a-a \sin (c+d x))^3}+\frac {a^4 A}{8 d (a-a \sin (c+d x))^2}+\frac {a^3 (3 A-B)}{16 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{16 d (a+a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.68 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 \left (6 (2 A-B) \text {arctanh}(\sin (c+d x))-\frac {4 (A+B)}{(-1+\sin (c+d x))^3}+\frac {6 A}{(-1+\sin (c+d x))^2}+\frac {-9 A+3 B}{-1+\sin (c+d x)}-\frac {3 (A-B)}{1+\sin (c+d x)}\right )}{48 d} \]

[In]

Integrate[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*(6*(2*A - B)*ArcTanh[Sin[c + d*x]] - (4*(A + B))/(-1 + Sin[c + d*x])^3 + (6*A)/(-1 + Sin[c + d*x])^2 + (-
9*A + 3*B)/(-1 + Sin[c + d*x]) - (3*(A - B))/(1 + Sin[c + d*x])))/(48*d)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.81

method	result	size
parallelrisch	\(-\frac {\left (\left (-\frac {5}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\sin \left (3 d x +3 c \right )+\sin \left (d x +c \right )-\cos \left (2 d x +2 c \right )\right ) \left (A -\frac {B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (-\frac {5}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\sin \left (3 d x +3 c \right )+\sin \left (d x +c \right )-\cos \left (2 d x +2 c \right )\right ) \left (A -\frac {B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {2 \left (A -2 B \right ) \cos \left (2 d x +2 c \right )}{3}+\frac {\left (A +\frac {B}{4}\right ) \cos \left (4 d x +4 c \right )}{3}+\frac {\left (5 A +\frac {7 B}{2}\right ) \sin \left (3 d x +3 c \right )}{6}+\frac {\left (7 A -\frac {3 B}{2}\right ) \sin \left (d x +c \right )}{2}-A +\frac {5 B}{4}\right ) a^{2}}{d \left (\cos \left (4 d x +4 c \right )-5-4 \cos \left (2 d x +2 c \right )+4 \sin \left (3 d x +3 c \right )+4 \sin \left (d x +c \right )\right )}\)	\(239\)
derivativedivides	\(\frac {A \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+B \,a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )+\frac {A \,a^{2}}{3 \cos \left (d x +c \right )^{6}}+2 B \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+A \,a^{2} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B \,a^{2}}{6 \cos \left (d x +c \right )^{6}}}{d}\)	\(304\)
default	\(\frac {A \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+B \,a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )+\frac {A \,a^{2}}{3 \cos \left (d x +c \right )^{6}}+2 B \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+A \,a^{2} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B \,a^{2}}{6 \cos \left (d x +c \right )^{6}}}{d}\)	\(304\)
risch	\(-\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )} \left (-24 i A \,{\mathrm e}^{5 i \left (d x +c \right )}+6 A \,{\mathrm e}^{6 i \left (d x +c \right )}+12 i B \,{\mathrm e}^{5 i \left (d x +c \right )}-3 B \,{\mathrm e}^{6 i \left (d x +c \right )}-16 i A \,{\mathrm e}^{3 i \left (d x +c \right )}-26 A \,{\mathrm e}^{4 i \left (d x +c \right )}-40 i B \,{\mathrm e}^{3 i \left (d x +c \right )}+13 B \,{\mathrm e}^{4 i \left (d x +c \right )}-24 i A \,{\mathrm e}^{i \left (d x +c \right )}+26 A \,{\mathrm e}^{2 i \left (d x +c \right )}+12 i B \,{\mathrm e}^{i \left (d x +c \right )}-13 B \,{\mathrm e}^{2 i \left (d x +c \right )}-6 A +3 B \right )}{12 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{4 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}\)	\(313\)

[In]

int(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-((-5/4+1/4*cos(4*d*x+4*c)+sin(3*d*x+3*c)+sin(d*x+c)-cos(2*d*x+2*c))*(A-1/2*B)*ln(tan(1/2*d*x+1/2*c)-1)-(-5/4+
1/4*cos(4*d*x+4*c)+sin(3*d*x+3*c)+sin(d*x+c)-cos(2*d*x+2*c))*(A-1/2*B)*ln(tan(1/2*d*x+1/2*c)+1)+2/3*(A-2*B)*co
s(2*d*x+2*c)+1/3*(A+1/4*B)*cos(4*d*x+4*c)+1/6*(5*A+7/2*B)*sin(3*d*x+3*c)+1/2*(7*A-3/2*B)*sin(d*x+c)-A+5/4*B)*a
^2/d/(cos(4*d*x+4*c)-5-4*cos(2*d*x+2*c)+4*sin(3*d*x+3*c)+4*sin(d*x+c))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (125) = 250\).

Time = 0.29 (sec) , antiderivative size = 271, normalized size of antiderivative = 2.05 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {12 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} - 8 \, {\left (A - 2 \, B\right )} a^{2} - 3 \, {\left ({\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} - 4 \, {\left (2 \, A - B\right )} a^{2}\right )} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} + 2 \, d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/48*(12*(2*A - B)*a^2*cos(d*x + c)^2 - 8*(A - 2*B)*a^2 - 3*((2*A - B)*a^2*cos(d*x + c)^4 + 2*(2*A - B)*a^2*c
os(d*x + c)^2*sin(d*x + c) - 2*(2*A - B)*a^2*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + 3*((2*A - B)*a^2*cos(d*x
+ c)^4 + 2*(2*A - B)*a^2*cos(d*x + c)^2*sin(d*x + c) - 2*(2*A - B)*a^2*cos(d*x + c)^2)*log(-sin(d*x + c) + 1)
- 2*(3*(2*A - B)*a^2*cos(d*x + c)^2 - 4*(2*A - B)*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4 + 2*d*cos(d*x + c)^2*si
n(d*x + c) - 2*d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**7*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.12 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (2 \, A - B\right )} a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A - B\right )} a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right )^{3} - 6 \, {\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right )^{2} + {\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right ) + 2 \, {\left (4 \, A + B\right )} a^{2}\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{3} + 2 \, \sin \left (d x + c\right ) - 1}}{48 \, d} \]

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.58 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {6 \, {\left (2 \, A a^{2} - B a^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 6 \, {\left (2 \, A a^{2} - B a^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (2 \, A a^{2} \sin \left (d x + c\right ) - B a^{2} \sin \left (d x + c\right ) + 3 \, A a^{2} - 2 \, B a^{2}\right )}}{\sin \left (d x + c\right ) + 1} + \frac {22 \, A a^{2} \sin \left (d x + c\right )^{3} - 11 \, B a^{2} \sin \left (d x + c\right )^{3} - 84 \, A a^{2} \sin \left (d x + c\right )^{2} + 39 \, B a^{2} \sin \left (d x + c\right )^{2} + 114 \, A a^{2} \sin \left (d x + c\right ) - 45 \, B a^{2} \sin \left (d x + c\right ) - 60 \, A a^{2} + 9 \, B a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{3}}}{96 \, d} \]

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/96*(6*(2*A*a^2 - B*a^2)*log(abs(sin(d*x + c) + 1)) - 6*(2*A*a^2 - B*a^2)*log(abs(sin(d*x + c) - 1)) - 6*(2*A
*a^2*sin(d*x + c) - B*a^2*sin(d*x + c) + 3*A*a^2 - 2*B*a^2)/(sin(d*x + c) + 1) + (22*A*a^2*sin(d*x + c)^3 - 11
*B*a^2*sin(d*x + c)^3 - 84*A*a^2*sin(d*x + c)^2 + 39*B*a^2*sin(d*x + c)^2 + 114*A*a^2*sin(d*x + c) - 45*B*a^2*
sin(d*x + c) - 60*A*a^2 + 9*B*a^2)/(sin(d*x + c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 9.64 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.03 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (2\,A-B\right )}{8\,d}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {A\,a^2}{4}-\frac {B\,a^2}{8}\right )-{\sin \left (c+d\,x\right )}^2\,\left (\frac {A\,a^2}{2}-\frac {B\,a^2}{4}\right )+\frac {A\,a^2}{3}+\frac {B\,a^2}{12}+\sin \left (c+d\,x\right )\,\left (\frac {A\,a^2}{12}-\frac {B\,a^2}{24}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^3+2\,\sin \left (c+d\,x\right )-1\right )} \]

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2)/cos(c + d*x)^7,x)

[Out]

(a^2*atanh(sin(c + d*x))*(2*A - B))/(8*d) - (sin(c + d*x)^3*((A*a^2)/4 - (B*a^2)/8) - sin(c + d*x)^2*((A*a^2)/
2 - (B*a^2)/4) + (A*a^2)/3 + (B*a^2)/12 + sin(c + d*x)*((A*a^2)/12 - (B*a^2)/24))/(d*(2*sin(c + d*x) - 2*sin(c
+ d*x)^3 + sin(c + d*x)^4 - 1))