\(\int \sec ^2(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) [964]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2934, 8} \[ \int \sec ^2(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {(A+B) \sec (c+d x) (a \sin (c+d x)+a)}{d}-a B x \]

[In]

Int[Sec[c + d*x]^2*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2934

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c + a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p +
1))), x] + Dist[b*((a*d*m + b*c*(m + p + 1))/(a*g^2*(p + 1))), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*
x])^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {(A+B) \sec (c+d x) (a+a \sin (c+d x))}{d}-(a B) \int 1 \, dx \\ & = -a B x+\frac {(A+B) \sec (c+d x) (a+a \sin (c+d x))}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a (-B \arctan (\tan (c+d x))+(A+B) (\sec (c+d x)+\tan (c+d x)))}{d} \]

[In]

Integrate[Sec[c + d*x]^2*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

(a*(-(B*ArcTan[Tan[c + d*x]]) + (A + B)*(Sec[c + d*x] + Tan[c + d*x])))/d

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52

method	result	size
parallelrisch	\(\frac {a \left (-B \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d x +d x B -2 A -2 B \right )}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(44\)
risch	\(-a B x +\frac {2 a A}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}+\frac {2 a B}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\)	\(49\)
derivativedivides	\(\frac {\frac {a A}{\cos \left (d x +c \right )}+B a \left (\tan \left (d x +c \right )-d x -c \right )+a A \tan \left (d x +c \right )+\frac {B a}{\cos \left (d x +c \right )}}{d}\)	\(54\)
default	\(\frac {\frac {a A}{\cos \left (d x +c \right )}+B a \left (\tan \left (d x +c \right )-d x -c \right )+a A \tan \left (d x +c \right )+\frac {B a}{\cos \left (d x +c \right )}}{d}\)	\(54\)
norman	\(\frac {a B x +a B x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 a A +2 B a}{d}-\frac {\left (2 a A +2 B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (4 a A +4 B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-a B x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a B x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 \left (A +B \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 \left (A +B \right ) a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (A +B \right ) a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(206\)

[In]

int(sec(d*x+c)^2*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (29) = 58\).

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.52 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {B a d x - {\left (A + B\right )} a + {\left (B a d x - {\left (A + B\right )} a\right )} \cos \left (d x + c\right ) - {\left (B a d x + {\left (A + B\right )} a\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d} \]

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.93 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {{\left (d x + c - \tan \left (d x + c\right )\right )} B a - A a \tan \left (d x + c\right ) - \frac {A a}{\cos \left (d x + c\right )} - \frac {B a}{\cos \left (d x + c\right )}}{d} \]

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

-((d*x + c - tan(d*x + c))*B*a - A*a*tan(d*x + c) - A*a/cos(d*x + c) - B*a/cos(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {{\left (d x + c\right )} B a + \frac {2 \, {\left (A a + B a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}}{d} \]

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.92 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {2\,A\,a+2\,B\,a}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}-B\,a\,x \]

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x)))/cos(c + d*x)^2,x)

[Out]

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