\(\int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) [1535]

Optimal result

Integrand size = 29, antiderivative size = 118 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {(5 a A-b B) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {\sec ^6(c+d x) (A b+a B+(a A+b B) \sin (c+d x))}{6 d}+\frac {(5 a A-b B) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {(5 a A-b B) \sec ^3(c+d x) \tan (c+d x)}{24 d} \]

[Out]

1/16*(5*A*a-B*b)*arctanh(sin(d*x+c))/d+1/6*sec(d*x+c)^6*(A*b+B*a+(A*a+B*b)*sin(d*x+c))/d+1/16*(5*A*a-B*b)*sec(
d*x+c)*tan(d*x+c)/d+1/24*(5*A*a-B*b)*sec(d*x+c)^3*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2916, 792, 205, 212} \[ \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {(5 a A-b B) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {\sec ^6(c+d x) ((a A+b B) \sin (c+d x)+a B+A b)}{6 d}+\frac {(5 a A-b B) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac {(5 a A-b B) \tan (c+d x) \sec (c+d x)}{16 d} \]

[In]

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

[Out]

((5*a*A - b*B)*ArcTanh[Sin[c + d*x]])/(16*d) + (Sec[c + d*x]^6*(A*b + a*B + (a*A + b*B)*Sin[c + d*x]))/(6*d) +
((5*a*A - b*B)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + ((5*a*A - b*B)*Sec[c + d*x]^3*Tan[c + d*x])/(24*d)

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 792

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a*(e*f + d*g) - (
c*d*f - a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 2916

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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b^7 \text {Subst}\left (\int \frac {(a+x) \left (A+\frac {B x}{b}\right )}{\left (b^2-x^2\right )^4} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^6(c+d x) (A b+a B+(a A+b B) \sin (c+d x))}{6 d}+\frac {\left (b^5 (5 a A-b B)\right ) \text {Subst}\left (\int \frac {1}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{6 d} \\ & = \frac {\sec ^6(c+d x) (A b+a B+(a A+b B) \sin (c+d x))}{6 d}+\frac {(5 a A-b B) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {\left (b^3 (5 a A-b B)\right ) \text {Subst}\left (\int \frac {1}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = \frac {\sec ^6(c+d x) (A b+a B+(a A+b B) \sin (c+d x))}{6 d}+\frac {(5 a A-b B) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {(5 a A-b B) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {(b (5 a A-b B)) \text {Subst}\left (\int \frac {1}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{16 d} \\ & = \frac {(5 a A-b B) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {\sec ^6(c+d x) (A b+a B+(a A+b B) \sin (c+d x))}{6 d}+\frac {(5 a A-b B) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {(5 a A-b B) \sec ^3(c+d x) \tan (c+d x)}{24 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.88 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {\sec ^6(c+d x) \left (-8 (A b+a B)-3 (5 a A-b B) \text {arctanh}(\sin (c+d x)) \cos ^6(c+d x)-3 (11 a A+b B) \sin (c+d x)+8 (5 a A-b B) \sin ^3(c+d x)+(-15 a A+3 b B) \sin ^5(c+d x)\right )}{48 d} \]

[In]

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

[Out]

-1/48*(Sec[c + d*x]^6*(-8*(A*b + a*B) - 3*(5*a*A - b*B)*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^6 - 3*(11*a*A + b*B
)*Sin[c + d*x] + 8*(5*a*A - b*B)*Sin[c + d*x]^3 + (-15*a*A + 3*b*B)*Sin[c + d*x]^5))/d

Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.43

method	result	size
derivativedivides	\(\frac {a A \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B a}{6 \cos \left (d x +c \right )^{6}}+\frac {A b}{6 \cos \left (d x +c \right )^{6}}+B b \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\)	\(169\)
default	\(\frac {a A \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B a}{6 \cos \left (d x +c \right )^{6}}+\frac {A b}{6 \cos \left (d x +c \right )^{6}}+B b \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\)	\(169\)
parallelrisch	\(\frac {-15 \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \left (a A -\frac {B b}{5}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \left (a A -\frac {B b}{5}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-120 A b -120 B a \right ) \cos \left (2 d x +2 c \right )+\left (-48 A b -48 B a \right ) \cos \left (4 d x +4 c \right )+\left (-8 A b -8 B a \right ) \cos \left (6 d x +6 c \right )+\left (170 a A -34 B b \right ) \sin \left (3 d x +3 c \right )+\left (30 a A -6 B b \right ) \sin \left (5 d x +5 c \right )+\left (396 a A +228 B b \right ) \sin \left (d x +c \right )+176 A b +176 B a}{48 d \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right )}\)	\(271\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (15 A a \,{\mathrm e}^{10 i \left (d x +c \right )}-3 B b \,{\mathrm e}^{10 i \left (d x +c \right )}+85 A a \,{\mathrm e}^{8 i \left (d x +c \right )}-17 B b \,{\mathrm e}^{8 i \left (d x +c \right )}+198 A a \,{\mathrm e}^{6 i \left (d x +c \right )}+114 B b \,{\mathrm e}^{6 i \left (d x +c \right )}-198 A a \,{\mathrm e}^{4 i \left (d x +c \right )}+256 i A b \,{\mathrm e}^{5 i \left (d x +c \right )}-114 B b \,{\mathrm e}^{4 i \left (d x +c \right )}+256 i B a \,{\mathrm e}^{5 i \left (d x +c \right )}-85 A a \,{\mathrm e}^{2 i \left (d x +c \right )}+17 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-15 a A +3 B b \right )}{24 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a A}{16 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{16 d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a A}{16 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{16 d}\)	\(303\)
norman	\(\frac {\frac {\left (11 a A +B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {\left (11 a A +B b \right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {7 \left (19 a A +25 B b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {7 \left (19 a A +25 B b \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (71 a A +53 B b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (71 a A +53 B b \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (275 a A +281 B b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {\left (275 a A +281 B b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {2 \left (A b +B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (A b +B a \right ) \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (A b +B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (A b +B a \right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {10 \left (4 A b +4 B a \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (13 A b +13 B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (13 A b +13 B a \right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 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 {\left (5 a A -B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 d}+\frac {\left (5 a A -B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d}\)	\(449\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.14 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A a - B b\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A a - B b\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 16 \, B a + 16 \, A b + 2 \, {\left (3 \, {\left (5 \, A a - B b\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (5 \, A a - B b\right )} \cos \left (d x + c\right )^{2} + 8 \, A a + 8 \, B b\right )} \sin \left (d x + c\right )}{96 \, d \cos \left (d x + c\right )^{6}} \]

[In]

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

[Out]

1/96*(3*(5*A*a - B*b)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 3*(5*A*a - B*b)*cos(d*x + c)^6*log(-sin(d*x + c)
+ 1) + 16*B*a + 16*A*b + 2*(3*(5*A*a - B*b)*cos(d*x + c)^4 + 2*(5*A*a - B*b)*cos(d*x + c)^2 + 8*A*a + 8*B*b)*s
in(d*x + c))/(d*cos(d*x + c)^6)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.21 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A a - B b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A a - B b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (5 \, A a - B b\right )} \sin \left (d x + c\right )^{5} - 8 \, {\left (5 \, A a - B b\right )} \sin \left (d x + c\right )^{3} + 8 \, B a + 8 \, A b + 3 \, {\left (11 \, A a + B b\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1}}{96 \, d} \]

[In]

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

[Out]

1/96*(3*(5*A*a - B*b)*log(sin(d*x + c) + 1) - 3*(5*A*a - B*b)*log(sin(d*x + c) - 1) - 2*(3*(5*A*a - B*b)*sin(d
*x + c)^5 - 8*(5*A*a - B*b)*sin(d*x + c)^3 + 8*B*a + 8*A*b + 3*(11*A*a + B*b)*sin(d*x + c))/(sin(d*x + c)^6 -
3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1))/d

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.18 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (5 \, A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, A a \sin \left (d x + c\right )^{5} - 3 \, B b \sin \left (d x + c\right )^{5} - 40 \, A a \sin \left (d x + c\right )^{3} + 8 \, B b \sin \left (d x + c\right )^{3} + 33 \, A a \sin \left (d x + c\right ) + 3 \, B b \sin \left (d x + c\right ) + 8 \, B a + 8 \, A b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{96 \, d} \]

[In]

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

[Out]

1/96*(3*(5*A*a - B*b)*log(abs(sin(d*x + c) + 1)) - 3*(5*A*a - B*b)*log(abs(sin(d*x + c) - 1)) - 2*(15*A*a*sin(
d*x + c)^5 - 3*B*b*sin(d*x + c)^5 - 40*A*a*sin(d*x + c)^3 + 8*B*b*sin(d*x + c)^3 + 33*A*a*sin(d*x + c) + 3*B*b
*sin(d*x + c) + 8*B*a + 8*A*b)/(sin(d*x + c)^2 - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 12.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.02 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {5\,A\,a}{16}-\frac {B\,b}{16}\right )}{d}-\frac {\left (\frac {5\,A\,a}{16}-\frac {B\,b}{16}\right )\,{\sin \left (c+d\,x\right )}^5+\left (\frac {B\,b}{6}-\frac {5\,A\,a}{6}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {11\,A\,a}{16}+\frac {B\,b}{16}\right )\,\sin \left (c+d\,x\right )+\frac {A\,b}{6}+\frac {B\,a}{6}}{d\,\left ({\sin \left (c+d\,x\right )}^6-3\,{\sin \left (c+d\,x\right )}^4+3\,{\sin \left (c+d\,x\right )}^2-1\right )} \]

[In]

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

[Out]

(atanh(sin(c + d*x))*((5*A*a)/16 - (B*b)/16))/d - ((A*b)/6 + (B*a)/6 + sin(c + d*x)*((11*A*a)/16 + (B*b)/16) -
sin(c + d*x)^3*((5*A*a)/6 - (B*b)/6) + sin(c + d*x)^5*((5*A*a)/16 - (B*b)/16))/(d*(3*sin(c + d*x)^2 - 3*sin(c
+ d*x)^4 + sin(c + d*x)^6 - 1))