\(\int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\) [998]

Optimal result

Integrand size = 31, antiderivative size = 91 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {3}{2} a^3 (2 A+3 B) x+\frac {2 a^3 (2 A+3 B) \cos (c+d x)}{d}+\frac {a^3 (2 A+3 B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {(A+B) \sec (c+d x) (a+a \sin (c+d x))^3}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2934, 2723} \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {2 a^3 (2 A+3 B) \cos (c+d x)}{d}+\frac {a^3 (2 A+3 B) \sin (c+d x) \cos (c+d x)}{2 d}-\frac {3}{2} a^3 x (2 A+3 B)+\frac {(A+B) \sec (c+d x) (a \sin (c+d x)+a)^3}{d} \]

[In]

Int[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

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

Rule 2723

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(2*a^2 + b^2)*(x/2), x] + (-Simp[2*a*b*(Cos[c
+ d*x]/d), x] - Simp[b^2*Cos[c + d*x]*(Sin[c + d*x]/(2*d)), x]) /; FreeQ[{a, b, 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))^3}{d}-(a (2 A+3 B)) \int (a+a \sin (c+d x))^2 \, dx \\ & = -\frac {3}{2} a^3 (2 A+3 B) x+\frac {2 a^3 (2 A+3 B) \cos (c+d x)}{d}+\frac {a^3 (2 A+3 B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {(A+B) \sec (c+d x) (a+a \sin (c+d x))^3}{d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.22 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.90 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {\sec (c+d x) \left (4 \sqrt {2} a^3 (2 A+3 B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2},\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt {1+\sin (c+d x)}-B (a+a \sin (c+d x))^3\right )}{2 d} \]

[In]

Integrate[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(Sec[c + d*x]*(4*Sqrt[2]*a^3*(2*A + 3*B)*Hypergeometric2F1[-3/2, -1/2, 1/2, (1 - Sin[c + d*x])/2]*Sqrt[1 + Sin
[c + d*x]] - B*(a + a*Sin[c + d*x])^3))/(2*d)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.29

method	result	size
parallelrisch	\(\frac {a^{3} \left (-24 d x A \cos \left (d x +c \right )-36 d x B \cos \left (d x +c \right )+4 A \cos \left (2 d x +2 c \right )+32 A \sin \left (d x +c \right )+40 A \cos \left (d x +c \right )+B \sin \left (3 d x +3 c \right )+12 B \cos \left (2 d x +2 c \right )+33 B \sin \left (d x +c \right )+56 B \cos \left (d x +c \right )+36 A +44 B \right )}{8 d \cos \left (d x +c \right )}\)	\(117\)
risch	\(-3 a^{3} x A -\frac {9 a^{3} B x}{2}+\frac {a^{3} {\mathrm e}^{i \left (d x +c \right )} A}{2 d}+\frac {3 a^{3} {\mathrm e}^{i \left (d x +c \right )} B}{2 d}+\frac {a^{3} {\mathrm e}^{-i \left (d x +c \right )} A}{2 d}+\frac {3 a^{3} {\mathrm e}^{-i \left (d x +c \right )} B}{2 d}+\frac {8 a^{3} A}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}+\frac {8 a^{3} B}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{3}}{4 d}\)	\(152\)
derivativedivides	\(\frac {A \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+B \,a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 A \,a^{3} \left (\tan \left (d x +c \right )-d x -c \right )+3 B \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+\frac {3 A \,a^{3}}{\cos \left (d x +c \right )}+3 B \,a^{3} \left (\tan \left (d x +c \right )-d x -c \right )+A \,a^{3} \tan \left (d x +c \right )+\frac {B \,a^{3}}{\cos \left (d x +c \right )}}{d}\)	\(219\)
default	\(\frac {A \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+B \,a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 A \,a^{3} \left (\tan \left (d x +c \right )-d x -c \right )+3 B \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+\frac {3 A \,a^{3}}{\cos \left (d x +c \right )}+3 B \,a^{3} \left (\tan \left (d x +c \right )-d x -c \right )+A \,a^{3} \tan \left (d x +c \right )+\frac {B \,a^{3}}{\cos \left (d x +c \right )}}{d}\)	\(219\)
norman	\(\frac {\left (3 A \,a^{3}+\frac {9}{2} B \,a^{3}\right ) x +\left (-9 A \,a^{3}-\frac {27}{2} B \,a^{3}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-3 A \,a^{3}-\frac {9}{2} B \,a^{3}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (9 A \,a^{3}+\frac {27}{2} B \,a^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-6 A \,a^{3}-9 B \,a^{3}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 A \,a^{3}+9 B \,a^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {10 A \,a^{3}+14 B \,a^{3}}{d}-\frac {\left (6 A \,a^{3}+2 B \,a^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (14 A \,a^{3}+10 B \,a^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (24 A \,a^{3}+24 B \,a^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (36 A \,a^{3}+44 B \,a^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {32 a^{3} \left (A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {32 a^{3} \left (A +B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3} \left (8 A +9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a^{3} \left (8 A +9 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{3} \left (24 A +23 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(432\)

[In]

int(sec(d*x+c)^2*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/8/d*a^3*(-24*d*x*A*cos(d*x+c)-36*d*x*B*cos(d*x+c)+4*A*cos(2*d*x+2*c)+32*A*sin(d*x+c)+40*A*cos(d*x+c)+B*sin(3
*d*x+3*c)+12*B*cos(2*d*x+2*c)+33*B*sin(d*x+c)+56*B*cos(d*x+c)+36*A+44*B)/cos(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.90 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {B a^{3} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, A + 3 \, B\right )} a^{3} d x + 2 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, {\left (A + B\right )} a^{3} - {\left (3 \, {\left (2 \, A + 3 \, B\right )} a^{3} d x - {\left (10 \, A + 13 \, B\right )} a^{3}\right )} \cos \left (d x + c\right ) + {\left (3 \, {\left (2 \, A + 3 \, B\right )} a^{3} d x + B a^{3} \cos \left (d x + c\right )^{2} - {\left (2 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right ) + 8 \, {\left (A + B\right )} a^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \]

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(B*a^3*cos(d*x + c)^3 - 3*(2*A + 3*B)*a^3*d*x + 2*(A + 3*B)*a^3*cos(d*x + c)^2 + 8*(A + B)*a^3 - (3*(2*A +
3*B)*a^3*d*x - (10*A + 13*B)*a^3)*cos(d*x + c) + (3*(2*A + 3*B)*a^3*d*x + B*a^3*cos(d*x + c)^2 - (2*A + 5*B)*
a^3*cos(d*x + c) + 8*(A + B)*a^3)*sin(d*x + c))/(d*cos(d*x + c) - d*sin(d*x + c) + d)

Sympy [F]

\[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=a^{3} \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sin ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \sin ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sin ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a**3*(Integral(A*sec(c + d*x)**2, x) + Integral(3*A*sin(c + d*x)*sec(c + d*x)**2, x) + Integral(3*A*sin(c + d*
x)**2*sec(c + d*x)**2, x) + Integral(A*sin(c + d*x)**3*sec(c + d*x)**2, x) + Integral(B*sin(c + d*x)*sec(c + d
*x)**2, x) + Integral(3*B*sin(c + d*x)**2*sec(c + d*x)**2, x) + Integral(3*B*sin(c + d*x)**3*sec(c + d*x)**2,
x) + Integral(B*sin(c + d*x)**4*sec(c + d*x)**2, x))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.84 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} A a^{3} + {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} B a^{3} + 6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} B a^{3} - 2 \, A a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - 6 \, B a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - 2 \, A a^{3} \tan \left (d x + c\right ) - \frac {6 \, A a^{3}}{\cos \left (d x + c\right )} - \frac {2 \, B a^{3}}{\cos \left (d x + c\right )}}{2 \, d} \]

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.66 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.62 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {3 \, {\left (2 \, A a^{3} + 3 \, B a^{3}\right )} {\left (d x + c\right )} + \frac {16 \, {\left (A a^{3} + B a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} + \frac {2 \, {\left (B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A a^{3} - 6 \, B a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/2*(3*(2*A*a^3 + 3*B*a^3)*(d*x + c) + 16*(A*a^3 + B*a^3)/(tan(1/2*d*x + 1/2*c) - 1) + 2*(B*a^3*tan(1/2*d*x +
1/2*c)^3 - 2*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 6*B*a^3*tan(1/2*d*x + 1/2*c)^2 - B*a^3*tan(1/2*d*x + 1/2*c) - 2*A
*a^3 - 6*B*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d

Mupad [B] (verification not implemented)

Time = 11.81 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.57 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {10\,A\,a^3-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,a^3+5\,B\,a^3\right )+14\,B\,a^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A\,a^3+7\,B\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (8\,A\,a^3+9\,B\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (18\,A\,a^3+21\,B\,a^3\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}-\frac {3\,a^3\,\mathrm {atan}\left (\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A+3\,B\right )}{6\,A\,a^3+9\,B\,a^3}\right )\,\left (2\,A+3\,B\right )}{d} \]

[In]

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

[Out]

- (10*A*a^3 - tan(c/2 + (d*x)/2)*(2*A*a^3 + 5*B*a^3) + 14*B*a^3 - tan(c/2 + (d*x)/2)^3*(2*A*a^3 + 7*B*a^3) + t
an(c/2 + (d*x)/2)^4*(8*A*a^3 + 9*B*a^3) + tan(c/2 + (d*x)/2)^2*(18*A*a^3 + 21*B*a^3))/(d*(tan(c/2 + (d*x)/2) -
2*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^3 - tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^5 - 1)) - (3*a^3*
atan((3*a^3*tan(c/2 + (d*x)/2)*(2*A + 3*B))/(6*A*a^3 + 9*B*a^3))*(2*A + 3*B))/d