\(\int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) [957]

Optimal result

Integrand size = 27, antiderivative size = 34 \[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {a (A+B) \log (1-\sin (c+d x))}{d}-\frac {a B \sin (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2915, 45} \[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {a (A+B) \log (1-\sin (c+d x))}{d}-\frac {a B \sin (c+d x)}{d} \]

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 \text {Subst}\left (\int \frac {A+\frac {B x}{a}}{a-x} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a \text {Subst}\left (\int \left (-\frac {B}{a}+\frac {A+B}{a-x}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {a (A+B) \log (1-\sin (c+d x))}{d}-\frac {a B \sin (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.00 \[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a A \text {arctanh}(\sin (c+d x))}{d}+\frac {a B \text {arctanh}(\sin (c+d x))}{d}-\frac {a A \log (\cos (c+d x))}{d}-\frac {a B \log (\cos (c+d x))}{d}-\frac {a B \sin (c+d x)}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(-\frac {a \left (B \sin \left (d x +c \right )+\left (A +B \right ) \ln \left (\sin \left (d x +c \right )-1\right )\right )}{d}\)	\(29\)
default	\(-\frac {a \left (B \sin \left (d x +c \right )+\left (A +B \right ) \ln \left (\sin \left (d x +c \right )-1\right )\right )}{d}\)	\(29\)
parallelrisch	\(\frac {\left (\left (A +B \right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 B -2 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-B \sin \left (d x +c \right )\right ) a}{d}\)	\(52\)
norman	\(\frac {-\frac {2 B a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 B a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (A +B \right ) a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (A +B \right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(96\)
risch	\(i x a A +i x a B +\frac {i B a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i B a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i a A c}{d}+\frac {2 i a B c}{d}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\)	\(106\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {{\left (A + B\right )} a \log \left (-\sin \left (d x + c\right ) + 1\right ) + B a \sin \left (d x + c\right )}{d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {{\left (A + B\right )} a \log \left (\sin \left (d x + c\right ) - 1\right ) + B a \sin \left (d x + c\right )}{d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (34) = 68\).

Time = 0.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 3.35 \[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (A a + B a\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 2 \, {\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a + B a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}}{d} \]

[In]

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

[Out]

((A*a + B*a)*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 2*(A*a + B*a)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - (A*a*tan(1/2
*d*x + 1/2*c)^2 + B*a*tan(1/2*d*x + 1/2*c)^2 + 2*B*a*tan(1/2*d*x + 1/2*c) + A*a + B*a)/(tan(1/2*d*x + 1/2*c)^2
+ 1))/d

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (A\,a+B\,a\right )}{d}-\frac {B\,a\,\sin \left (c+d\,x\right )}{d} \]

[In]

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

[Out]

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