\(\int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) [1532]

Optimal result

Integrand size = 27, antiderivative size = 64 \[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {(a+b) (A+B) \log (1-\sin (c+d x))}{2 d}+\frac {(a-b) (A-B) \log (1+\sin (c+d x))}{2 d}-\frac {b B \sin (c+d x)}{d} \]

[Out]

-1/2*(a+b)*(A+B)*ln(1-sin(d*x+c))/d+1/2*(a-b)*(A-B)*ln(1+sin(d*x+c))/d-b*B*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2916, 788, 647, 31} \[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {(a+b) (A+B) \log (1-\sin (c+d x))}{2 d}+\frac {(a-b) (A-B) \log (\sin (c+d x)+1)}{2 d}-\frac {b B \sin (c+d x)}{d} \]

[In]

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

[Out]

-1/2*((a + b)*(A + B)*Log[1 - Sin[c + d*x]])/d + ((a - b)*(A - B)*Log[1 + Sin[c + d*x]])/(2*d) - (b*B*Sin[c +
d*x])/d

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 788

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

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 \text {Subst}\left (\int \frac {(a+x) \left (A+\frac {B x}{b}\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = -\frac {b B \sin (c+d x)}{d}-\frac {b \text {Subst}\left (\int \frac {-a A-b B-\left (A+\frac {a B}{b}\right ) x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = -\frac {b B \sin (c+d x)}{d}-\frac {((a-b) (A-B)) \text {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{2 d}+\frac {((a+b) (A+B)) \text {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{2 d} \\ & = -\frac {(a+b) (A+B) \log (1-\sin (c+d x))}{2 d}+\frac {(a-b) (A-B) \log (1+\sin (c+d x))}{2 d}-\frac {b B \sin (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.06 \[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a A \text {arctanh}(\sin (c+d x))}{d}+\frac {b B \text {arctanh}(\sin (c+d x))}{d}-\frac {A b \log (\cos (c+d x))}{d}-\frac {a B \log (\cos (c+d x))}{d}-\frac {b B \sin (c+d x)}{d} \]

[In]

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

[Out]

(a*A*ArcTanh[Sin[c + d*x]])/d + (b*B*ArcTanh[Sin[c + d*x]])/d - (A*b*Log[Cos[c + d*x]])/d - (a*B*Log[Cos[c + d
*x]])/d - (b*B*Sin[c + d*x])/d

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-B a \ln \left (\cos \left (d x +c \right )\right )-A b \ln \left (\cos \left (d x +c \right )\right )+B b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\)	\(71\)
default	\(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-B a \ln \left (\cos \left (d x +c \right )\right )-A b \ln \left (\cos \left (d x +c \right )\right )+B b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\)	\(71\)
parallelrisch	\(\frac {\left (A b +B a \right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (a +b \right ) \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a -b \right ) \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-B b \sin \left (d x +c \right )}{d}\)	\(79\)
norman	\(\frac {-\frac {2 B b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 B b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (A b +B a \right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (a A -A b -B a +B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {\left (a A +A b +B a +B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(139\)
risch	\(\frac {2 i B a c}{d}+\frac {2 i A b c}{d}+i B a x +\frac {i B b \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i B b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+i A b x -\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a A}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a A}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{d}\)	\(224\)

[In]

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

[Out]

1/d*(a*A*ln(sec(d*x+c)+tan(d*x+c))-B*a*ln(cos(d*x+c))-A*b*ln(cos(d*x+c))+B*b*(-sin(d*x+c)+ln(sec(d*x+c)+tan(d*
x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03 \[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {2 \, B b \sin \left (d x + c\right ) - {\left ({\left (A - B\right )} a - {\left (A - B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A + B\right )} a + {\left (A + B\right )} b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(sec(d*x+c)*(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), x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00 \[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {2 \, B b \sin \left (d x + c\right ) - {\left ({\left (A - B\right )} a - {\left (A - B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A + B\right )} a + {\left (A + B\right )} b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05 \[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {2 \, B b \sin \left (d x + c\right ) - {\left (A a - B a - A b + B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + {\left (A a + B a + A b + B b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, d} \]

[In]

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

[Out]

-1/2*(2*B*b*sin(d*x + c) - (A*a - B*a - A*b + B*b)*log(abs(sin(d*x + c) + 1)) + (A*a + B*a + A*b + B*b)*log(ab
s(sin(d*x + c) - 1)))/d

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.83 \[ \int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {B\,b\,\sin \left (c+d\,x\right )-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (A-B\right )\,\left (a-b\right )}{2}+\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (a+b\right )\,\left (A+B\right )}{2}}{d} \]

[In]

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

[Out]

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