\(\int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) [1534]

Optimal result

Integrand size = 29, antiderivative size = 88 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {(3 a A-b B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\sec ^4(c+d x) (A b+a B+(a A+b B) \sin (c+d x))}{4 d}+\frac {(3 a A-b B) \sec (c+d x) \tan (c+d x)}{8 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2916, 792, 205, 212} \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {(3 a A-b B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\sec ^4(c+d x) ((a A+b B) \sin (c+d x)+a B+A b)}{4 d}+\frac {(3 a A-b B) \tan (c+d x) \sec (c+d x)}{8 d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.93 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\sec ^4(c+d x) \left (2 (A b+a B)+(3 a A-b B) \text {arctanh}(\sin (c+d x)) \cos ^4(c+d x)+(5 a A+b B) \sin (c+d x)+(-3 a A+b B) \sin ^3(c+d x)\right )}{8 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.60

method	result	size
derivativedivides	\(\frac {a A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B a}{4 \cos \left (d x +c \right )^{4}}+\frac {A b}{4 \cos \left (d x +c \right )^{4}}+B b \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(141\)
default	\(\frac {a A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B a}{4 \cos \left (d x +c \right )^{4}}+\frac {A b}{4 \cos \left (d x +c \right )^{4}}+B b \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(141\)
parallelrisch	\(\frac {-12 \left (a A -\frac {B b}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 \left (a A -\frac {B b}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-8 A b -8 B a \right ) \cos \left (2 d x +2 c \right )+\left (-2 A b -2 B a \right ) \cos \left (4 d x +4 c \right )+\left (6 a A -2 B b \right ) \sin \left (3 d x +3 c \right )+\left (22 a A +14 B b \right ) \sin \left (d x +c \right )+10 A b +10 B a}{8 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(200\)
risch	\(\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (-3 A a \,{\mathrm e}^{6 i \left (d x +c \right )}+B b \,{\mathrm e}^{6 i \left (d x +c \right )}-11 A a \,{\mathrm e}^{4 i \left (d x +c \right )}-7 B b \,{\mathrm e}^{4 i \left (d x +c \right )}+11 A a \,{\mathrm e}^{2 i \left (d x +c \right )}-16 i A b \,{\mathrm e}^{3 i \left (d x +c \right )}+7 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-16 i B a \,{\mathrm e}^{3 i \left (d x +c \right )}+3 a A -B b \right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a A}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a A}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{8 d}\)	\(242\)
norman	\(\frac {\frac {\left (4 A b +4 B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 A b +4 B a \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (5 a A +B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (5 a A +B b \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (7 a A +11 B b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (7 a A +11 B b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (13 a A +9 B b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (13 a A +9 B b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {2 \left (A b +B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (A b +B a \right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (A b +B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\left (3 a A -B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 a A -B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(349\)

[In]

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

[Out]

1/d*(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+1
/4*A*b/cos(d*x+c)^4+B*b*(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))))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.30 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A a - B b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A a - B b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, B a + 4 \, A b + 2 \, {\left ({\left (3 \, A a - B b\right )} \cos \left (d x + c\right )^{2} + 2 \, A a + 2 \, B b\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

1/16*((3*A*a - B*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (3*A*a - B*b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1)
+ 4*B*a + 4*A*b + 2*((3*A*a - B*b)*cos(d*x + c)^2 + 2*A*a + 2*B*b)*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.27 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A a - B b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A a - B b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (3 \, A a - B b\right )} \sin \left (d x + c\right )^{3} - 2 \, B a - 2 \, A b - {\left (5 \, A a + B b\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.30 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (3 \, A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a \sin \left (d x + c\right )^{3} - B b \sin \left (d x + c\right )^{3} - 5 \, A a \sin \left (d x + c\right ) - B b \sin \left (d x + c\right ) - 2 \, B a - 2 \, A b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

[In]

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

[Out]

1/16*((3*A*a - B*b)*log(abs(sin(d*x + c) + 1)) - (3*A*a - B*b)*log(abs(sin(d*x + c) - 1)) - 2*(3*A*a*sin(d*x +
c)^3 - B*b*sin(d*x + c)^3 - 5*A*a*sin(d*x + c) - B*b*sin(d*x + c) - 2*B*a - 2*A*b)/(sin(d*x + c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.03 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\left (\frac {B\,b}{8}-\frac {3\,A\,a}{8}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {5\,A\,a}{8}+\frac {B\,b}{8}\right )\,\sin \left (c+d\,x\right )+\frac {A\,b}{4}+\frac {B\,a}{4}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {3\,A\,a}{8}-\frac {B\,b}{8}\right )}{d} \]

[In]

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

[Out]

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