\(\int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) [959]

Optimal result

Integrand size = 29, antiderivative size = 100 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a (3 A-B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 (A+B)}{8 d (a-a \sin (c+d x))^2}+\frac {a^2 A}{4 d (a-a \sin (c+d x))}-\frac {a^2 (A-B)}{8 d (a+a \sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 100, 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} \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a^3 (A+B)}{8 d (a-a \sin (c+d x))^2}-\frac {a^2 (A-B)}{8 d (a \sin (c+d x)+a)}+\frac {a^2 A}{4 d (a-a \sin (c+d x))}+\frac {a (3 A-B) \text {arctanh}(\sin (c+d x))}{8 d} \]

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.67 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a \left ((3 A-B) \text {arctanh}(\sin (c+d x))+\frac {A+B}{(-1+\sin (c+d x))^2}-\frac {2 A}{-1+\sin (c+d x)}+\frac {-A+B}{1+\sin (c+d x)}\right )}{8 d} \]

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

(a*((3*A - B)*ArcTanh[Sin[c + d*x]] + (A + B)/(-1 + Sin[c + d*x])^2 - (2*A)/(-1 + Sin[c + d*x]) + (-A + B)/(1
+ Sin[c + d*x])))/(8*d)

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.41

method	result	size
derivativedivides	\(\frac {\frac {a A}{4 \cos \left (d x +c \right )^{4}}+B a \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B a}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(141\)
default	\(\frac {\frac {a A}{4 \cos \left (d x +c \right )^{4}}+B a \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B a}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(141\)
parallelrisch	\(-\frac {3 \left (\left (-\frac {\sin \left (3 d x +3 c \right )}{2}-\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \left (A -\frac {B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (-\frac {\sin \left (3 d x +3 c \right )}{2}-\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \left (A -\frac {B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-\frac {A}{3}+B \right ) \cos \left (2 d x +2 c \right )+\frac {\left (-A -B \right ) \sin \left (3 d x +3 c \right )}{3}+\frac {\left (-7 A +B \right ) \sin \left (d x +c \right )}{3}+\frac {A}{3}-B \right ) a}{4 d \left (2-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )\right )}\)	\(191\)
risch	\(-\frac {i a \,{\mathrm e}^{i \left (d x +c \right )} \left (-6 i A \,{\mathrm e}^{3 i \left (d x +c \right )}+3 A \,{\mathrm e}^{4 i \left (d x +c \right )}+2 i B \,{\mathrm e}^{3 i \left (d x +c \right )}-B \,{\mathrm e}^{4 i \left (d x +c \right )}+6 i A \,{\mathrm e}^{i \left (d x +c \right )}+2 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2 i B \,{\mathrm e}^{i \left (d x +c \right )}+10 B \,{\mathrm e}^{2 i \left (d x +c \right )}+3 A -B \right )}{4 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}\)	\(245\)
norman	\(\frac {\frac {\left (4 a A +4 B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 a A +4 B a \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (a A +B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (a A +B a \right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (a A +B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a \left (5 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (5 A +B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a \left (7 A +11 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {a \left (7 A +11 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {a \left (13 A +9 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a \left (13 A +9 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {a \left (3 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a \left (3 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(339\)

[In]

int(sec(d*x+c)^5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/4*a*A/cos(d*x+c)^4+B*a*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/cos(d*x+c)^2+1/8*sin(d*x+c)-1/8*
ln(sec(d*x+c)+tan(d*x+c)))+a*A*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+
1/4*B*a/cos(d*x+c)^4)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.82 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {2 \, {\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A - B\right )} a \sin \left (d x + c\right ) - 2 \, {\left (A - 3 \, B\right )} a - {\left ({\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - {\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - {\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/16*(2*(3*A - B)*a*cos(d*x + c)^2 + 2*(3*A - B)*a*sin(d*x + c) - 2*(A - 3*B)*a - ((3*A - B)*a*cos(d*x + c)^2
*sin(d*x + c) - (3*A - B)*a*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((3*A - B)*a*cos(d*x + c)^2*sin(d*x + c) -
(3*A - B)*a*cos(d*x + c)^2)*log(-sin(d*x + c) + 1))/(d*cos(d*x + c)^2*sin(d*x + c) - d*cos(d*x + c)^2)

Sympy [F]

\[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=a \left (\int A \sec ^{5}{\left (c + d x \right )}\, dx + \int A \sin {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int B \sin ^{2}{\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate(sec(d*x+c)**5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

a*(Integral(A*sec(c + d*x)**5, x) + Integral(A*sin(c + d*x)*sec(c + d*x)**5, x) + Integral(B*sin(c + d*x)*sec(
c + d*x)**5, x) + Integral(B*sin(c + d*x)**2*sec(c + d*x)**5, x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.15 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A - B\right )} a \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A - B\right )} a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (3 \, A - B\right )} a \sin \left (d x + c\right )^{2} - {\left (3 \, A - B\right )} a \sin \left (d x + c\right ) - 2 \, {\left (A + B\right )} a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/16*((3*A - B)*a*log(sin(d*x + c) + 1) - (3*A - B)*a*log(sin(d*x + c) - 1) - 2*((3*A - B)*a*sin(d*x + c)^2 -
(3*A - B)*a*sin(d*x + c) - 2*(A + B)*a)/(sin(d*x + c)^3 - sin(d*x + c)^2 - sin(d*x + c) + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.52 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {2 \, {\left (3 \, A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 2 \, {\left (3 \, A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a \sin \left (d x + c\right ) - B a \sin \left (d x + c\right ) + 5 \, A a - 3 \, B a\right )}}{\sin \left (d x + c\right ) + 1} + \frac {9 \, A a \sin \left (d x + c\right )^{2} - 3 \, B a \sin \left (d x + c\right )^{2} - 26 \, A a \sin \left (d x + c\right ) + 6 \, B a \sin \left (d x + c\right ) + 21 \, A a + B a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{32 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/32*(2*(3*A*a - B*a)*log(abs(sin(d*x + c) + 1)) - 2*(3*A*a - B*a)*log(abs(sin(d*x + c) - 1)) - 2*(3*A*a*sin(d
*x + c) - B*a*sin(d*x + c) + 5*A*a - 3*B*a)/(sin(d*x + c) + 1) + (9*A*a*sin(d*x + c)^2 - 3*B*a*sin(d*x + c)^2
- 26*A*a*sin(d*x + c) + 6*B*a*sin(d*x + c) + 21*A*a + B*a)/(sin(d*x + c) - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 9.84 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.98 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (3\,A-B\right )}{8\,d}-\frac {\left (\frac {B\,a}{8}-\frac {3\,A\,a}{8}\right )\,{\sin \left (c+d\,x\right )}^2+\left (\frac {3\,A\,a}{8}-\frac {B\,a}{8}\right )\,\sin \left (c+d\,x\right )+\frac {A\,a}{4}+\frac {B\,a}{4}}{d\,\left (-{\sin \left (c+d\,x\right )}^3+{\sin \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )-1\right )} \]

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x)))/cos(c + d*x)^5,x)

[Out]

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