\(\int \sec ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [980]

Optimal result

Integrand size = 31, antiderivative size = 55 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-a^2 (A+2 B) x+\frac {a^2 (A+2 B) \cos (c+d x)}{d}+\frac {(A+B) \sec (c+d x) (a+a \sin (c+d x))^2}{d} \]

[Out]

-a^2*(A+2*B)*x+a^2*(A+2*B)*cos(d*x+c)/d+(A+B)*sec(d*x+c)*(a+a*sin(d*x+c))^2/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2934, 2718} \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 (A+2 B) \cos (c+d x)}{d}-\left (a^2 x (A+2 B)\right )+\frac {(A+B) \sec (c+d x) (a \sin (c+d x)+a)^2}{d} \]

[In]

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

[Out]

-(a^2*(A + 2*B)*x) + (a^2*(A + 2*B)*Cos[c + d*x])/d + ((A + B)*Sec[c + d*x]*(a + a*Sin[c + d*x])^2)/d

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, 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))^2}{d}-(a (A+2 B)) \int (a+a \sin (c+d x)) \, dx \\ & = -a^2 (A+2 B) x+\frac {(A+B) \sec (c+d x) (a+a \sin (c+d x))^2}{d}-\left (a^2 (A+2 B)\right ) \int \sin (c+d x) \, dx \\ & = -a^2 (A+2 B) x+\frac {a^2 (A+2 B) \cos (c+d x)}{d}+\frac {(A+B) \sec (c+d x) (a+a \sin (c+d x))^2}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.65 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 \sec (c+d x) \left (4 A+5 B+4 (A+2 B) \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {\cos ^2(c+d x)}+B \cos (2 (c+d x))+4 A \sin (c+d x)+4 B \sin (c+d x)\right )}{2 d} \]

[In]

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

[Out]

(a^2*Sec[c + d*x]*(4*A + 5*B + 4*(A + 2*B)*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[Cos[c + d*x]^2] + B*Cos
[2*(c + d*x)] + 4*A*Sin[c + d*x] + 4*B*Sin[c + d*x]))/(2*d)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.33

method	result	size
parallelrisch	\(-\frac {\left (-\frac {B \cos \left (2 d x +2 c \right )}{2}+\left (d x A +2 d x B -2 A -3 B \right ) \cos \left (d x +c \right )+\left (-2 B -2 A \right ) \sin \left (d x +c \right )-2 A -\frac {5 B}{2}\right ) a^{2}}{d \cos \left (d x +c \right )}\)	\(73\)
risch	\(-a^{2} x A -2 a^{2} x B +\frac {B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {B \,a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {4 a^{2} A}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}+\frac {4 a^{2} B}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\)	\(98\)
derivativedivides	\(\frac {A \,a^{2} \left (\tan \left (d x +c \right )-d x -c \right )+B \,a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+\frac {2 A \,a^{2}}{\cos \left (d x +c \right )}+2 B \,a^{2} \left (\tan \left (d x +c \right )-d x -c \right )+A \,a^{2} \tan \left (d x +c \right )+\frac {B \,a^{2}}{\cos \left (d x +c \right )}}{d}\)	\(123\)
default	\(\frac {A \,a^{2} \left (\tan \left (d x +c \right )-d x -c \right )+B \,a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+\frac {2 A \,a^{2}}{\cos \left (d x +c \right )}+2 B \,a^{2} \left (\tan \left (d x +c \right )-d x -c \right )+A \,a^{2} \tan \left (d x +c \right )+\frac {B \,a^{2}}{\cos \left (d x +c \right )}}{d}\)	\(123\)
norman	\(\frac {\left (A \,a^{2}+2 B \,a^{2}\right ) x +\left (-2 A \,a^{2}-4 B \,a^{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A \,a^{2}-2 B \,a^{2}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 A \,a^{2}+4 B \,a^{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4 A \,a^{2}+6 B \,a^{2}}{d}-\frac {2 \left (2 A \,a^{2}+B \,a^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (6 A \,a^{2}+7 B \,a^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {12 a^{2} \left (A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {12 a^{2} \left (A +B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (A +B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \left (6 A +5 B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(312\)

[In]

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

[Out]

-(-1/2*B*cos(2*d*x+2*c)+(A*d*x+2*B*d*x-2*A-3*B)*cos(d*x+c)+(-2*B-2*A)*sin(d*x+c)-2*A-5/2*B)*a^2/d/cos(d*x+c)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (55) = 110\).

Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.33 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {{\left (A + 2 \, B\right )} a^{2} d x - B a^{2} \cos \left (d x + c\right )^{2} - 2 \, {\left (A + B\right )} a^{2} + {\left ({\left (A + 2 \, B\right )} a^{2} d x - {\left (2 \, A + 3 \, B\right )} a^{2}\right )} \cos \left (d x + c\right ) - {\left ({\left (A + 2 \, B\right )} a^{2} d x - B a^{2} \cos \left (d x + c\right ) + 2 \, {\left (A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d} \]

[In]

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

[Out]

-((A + 2*B)*a^2*d*x - B*a^2*cos(d*x + c)^2 - 2*(A + B)*a^2 + ((A + 2*B)*a^2*d*x - (2*A + 3*B)*a^2)*cos(d*x + c
) - ((A + 2*B)*a^2*d*x - B*a^2*cos(d*x + c) + 2*(A + B)*a^2)*sin(d*x + c))/(d*cos(d*x + c) - d*sin(d*x + c) +
d)

Sympy [F]

\[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=a^{2} \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 A \sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 B \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sin ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a**2*(Integral(A*sec(c + d*x)**2, x) + Integral(2*A*sin(c + d*x)*sec(c + d*x)**2, x) + Integral(A*sin(c + d*x)
**2*sec(c + d*x)**2, x) + Integral(B*sin(c + d*x)*sec(c + d*x)**2, x) + Integral(2*B*sin(c + d*x)**2*sec(c + d
*x)**2, x) + Integral(B*sin(c + d*x)**3*sec(c + d*x)**2, x))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.89 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {{\left (d x + c - \tan \left (d x + c\right )\right )} A a^{2} + 2 \, {\left (d x + c - \tan \left (d x + c\right )\right )} B a^{2} - B a^{2} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - A a^{2} \tan \left (d x + c\right ) - \frac {2 \, A a^{2}}{\cos \left (d x + c\right )} - \frac {B a^{2}}{\cos \left (d x + c\right )}}{d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (55) = 110\).

Time = 0.31 (sec) , antiderivative size = 125, normalized size of antiderivative = 2.27 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {{\left (A a^{2} + 2 \, B a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (2 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A a^{2} + 3 \, B a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}}{d} \]

[In]

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

[Out]

-((A*a^2 + 2*B*a^2)*(d*x + c) + 2*(2*A*a^2*tan(1/2*d*x + 1/2*c)^2 + 2*B*a^2*tan(1/2*d*x + 1/2*c)^2 - B*a^2*tan
(1/2*d*x + 1/2*c) + 2*A*a^2 + 3*B*a^2)/(tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c)
- 1))/d

Mupad [B] (verification not implemented)

Time = 10.12 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.00 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {4\,A\,a^2+6\,B\,a^2+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (4\,A\,a^2+4\,B\,a^2\right )-2\,B\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}-A\,a^2\,x-2\,B\,a^2\,x \]

[In]

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

[Out]

- (4*A*a^2 + 6*B*a^2 + tan(c/2 + (d*x)/2)^2*(4*A*a^2 + 4*B*a^2) - 2*B*a^2*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d
*x)/2) - tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^3 - 1)) - A*a^2*x - 2*B*a^2*x