\(\int \sec ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\) [1000]

Optimal result

Integrand size = 31, antiderivative size = 107 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^5 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac {(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^3}{5 d}+\frac {a^5 (2 A-3 B) \cos (c+d x)}{15 d \left (a^2-a^2 \sin (c+d x)\right )} \]

[Out]

1/15*a^5*(2*A-3*B)*cos(d*x+c)/d/(a-a*sin(d*x+c))^2+1/5*(A+B)*sec(d*x+c)^5*(a+a*sin(d*x+c))^3/d+1/15*a^5*(2*A-3
*B)*cos(d*x+c)/d/(a^2-a^2*sin(d*x+c))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2934, 2749, 2729, 2727} \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^5 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac {a^5 (2 A-3 B) \cos (c+d x)}{15 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac {(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^3}{5 d} \]

[In]

Int[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(a^5*(2*A - 3*B)*Cos[c + d*x])/(15*d*(a - a*Sin[c + d*x])^2) + ((A + B)*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3)
/(5*d) + (a^5*(2*A - 3*B)*Cos[c + d*x])/(15*d*(a^2 - a^2*Sin[c + d*x]))

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2749

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

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 ^5(c+d x) (a+a \sin (c+d x))^3}{5 d}+\frac {1}{5} (a (2 A-3 B)) \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx \\ & = \frac {(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^3}{5 d}+\frac {1}{5} \left (a^5 (2 A-3 B)\right ) \int \frac {1}{(a-a \sin (c+d x))^2} \, dx \\ & = \frac {a^5 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac {(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^3}{5 d}+\frac {1}{15} \left (a^4 (2 A-3 B)\right ) \int \frac {1}{a-a \sin (c+d x)} \, dx \\ & = \frac {a^5 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac {a^4 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))}+\frac {(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^3}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.88 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {a^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (-16 A+9 B+(2 A-3 B) \cos (2 (c+d x))+6 (2 A-3 B) \sin (c+d x))}{30 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5} \]

[In]

Integrate[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

-1/30*(a^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-16*A + 9*B + (2*A - 3*B)*Cos[2*(c + d*x)] + 6*(2*A - 3*B)*S
in[c + d*x]))/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5)

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.87

method	result	size
parallelrisch	\(-\frac {2 a^{3} \left (A \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (B -2 A \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {8 A}{3}-B \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {4 A}{3}+B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {7 A}{15}-\frac {B}{5}\right )}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}\)	\(93\)
risch	\(\frac {2 i a^{3} \left (20 i A \,{\mathrm e}^{2 i \left (d x +c \right )}-15 i B \,{\mathrm e}^{2 i \left (d x +c \right )}+15 B \,{\mathrm e}^{3 i \left (d x +c \right )}-2 i A +10 A \,{\mathrm e}^{i \left (d x +c \right )}+3 i B -15 B \,{\mathrm e}^{i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{5}}\)	\(95\)
derivativedivides	\(\frac {A \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )+\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{5 \cos \left (d x +c \right )^{5}}+3 A \,a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+3 B \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )+\frac {3 A \,a^{3}}{5 \cos \left (d x +c \right )^{5}}+3 B \,a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )-A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+\frac {B \,a^{3}}{5 \cos \left (d x +c \right )^{5}}}{d}\)	\(333\)
default	\(\frac {A \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )+\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{5 \cos \left (d x +c \right )^{5}}+3 A \,a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+3 B \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )+\frac {3 A \,a^{3}}{5 \cos \left (d x +c \right )^{5}}+3 B \,a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )-A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+\frac {B \,a^{3}}{5 \cos \left (d x +c \right )^{5}}}{d}\)	\(333\)

[In]

int(sec(d*x+c)^6*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-2*a^3*(A*tan(1/2*d*x+1/2*c)^4+(B-2*A)*tan(1/2*d*x+1/2*c)^3+(8/3*A-B)*tan(1/2*d*x+1/2*c)^2+(-4/3*A+B)*tan(1/2*
d*x+1/2*c)+7/15*A-1/5*B)/d/(tan(1/2*d*x+1/2*c)-1)^5

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.76 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {{\left (2 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} - 2 \, {\left (2 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} - 3 \, {\left (3 \, A - 2 \, B\right )} a^{3} \cos \left (d x + c\right ) - 3 \, {\left (A + B\right )} a^{3} + {\left ({\left (2 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right ) - 3 \, {\left (A + B\right )} a^{3}\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + 3 \, d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) - {\left (d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) - 4 \, d\right )} \sin \left (d x + c\right ) - 4 \, d\right )}} \]

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/15*((2*A - 3*B)*a^3*cos(d*x + c)^3 - 2*(2*A - 3*B)*a^3*cos(d*x + c)^2 - 3*(3*A - 2*B)*a^3*cos(d*x + c) - 3*(
A + B)*a^3 + ((2*A - 3*B)*a^3*cos(d*x + c)^2 + 3*(2*A - 3*B)*a^3*cos(d*x + c) - 3*(A + B)*a^3)*sin(d*x + c))/(
d*cos(d*x + c)^3 + 3*d*cos(d*x + c)^2 - 2*d*cos(d*x + c) - (d*cos(d*x + c)^2 - 2*d*cos(d*x + c) - 4*d)*sin(d*x
+ c) - 4*d)

Sympy [F(-1)]

Timed out. \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**6*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.76 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {3 \, B a^{3} \tan \left (d x + c\right )^{5} + {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 3 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} A a^{3} + 3 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} B a^{3} - \frac {{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} A a^{3}}{\cos \left (d x + c\right )^{5}} - \frac {3 \, {\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} B a^{3}}{\cos \left (d x + c\right )^{5}} + \frac {9 \, A a^{3}}{\cos \left (d x + c\right )^{5}} + \frac {3 \, B a^{3}}{\cos \left (d x + c\right )^{5}}}{15 \, d} \]

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/15*(3*B*a^3*tan(d*x + c)^5 + (3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^3 + 3*(3*tan(d*x +
c)^5 + 5*tan(d*x + c)^3)*A*a^3 + 3*(3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*B*a^3 - (5*cos(d*x + c)^2 - 3)*A*a^3
/cos(d*x + c)^5 - 3*(5*cos(d*x + c)^2 - 3)*B*a^3/cos(d*x + c)^5 + 9*A*a^3/cos(d*x + c)^5 + 3*B*a^3/cos(d*x + c
)^5)/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.36 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 20 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, A a^{3} - 3 \, B a^{3}\right )}}{15 \, d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{5}} \]

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-2/15*(15*A*a^3*tan(1/2*d*x + 1/2*c)^4 - 30*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 15*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 4
0*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 15*B*a^3*tan(1/2*d*x + 1/2*c)^2 - 20*A*a^3*tan(1/2*d*x + 1/2*c) + 15*B*a^3*ta
n(1/2*d*x + 1/2*c) + 7*A*a^3 - 3*B*a^3)/(d*(tan(1/2*d*x + 1/2*c) - 1)^5)

Mupad [B] (verification not implemented)

Time = 11.83 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.06 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {\sqrt {2}\,a^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,B-\frac {53\,A}{4}+4\,A\,\cos \left (c+d\,x\right )+\frac {3\,B\,\cos \left (c+d\,x\right )}{2}+\frac {25\,A\,\sin \left (c+d\,x\right )}{2}-\frac {15\,B\,\sin \left (c+d\,x\right )}{2}+\frac {9\,A\,\cos \left (2\,c+2\,d\,x\right )}{4}-\frac {3\,B\,\cos \left (2\,c+2\,d\,x\right )}{2}-\frac {5\,A\,\sin \left (2\,c+2\,d\,x\right )}{4}\right )}{60\,d\,{\cos \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d\,x}{2}\right )}^5} \]

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3)/cos(c + d*x)^6,x)

[Out]

-(2^(1/2)*a^3*cos(c/2 + (d*x)/2)*(3*B - (53*A)/4 + 4*A*cos(c + d*x) + (3*B*cos(c + d*x))/2 + (25*A*sin(c + d*x
))/2 - (15*B*sin(c + d*x))/2 + (9*A*cos(2*c + 2*d*x))/4 - (3*B*cos(2*c + 2*d*x))/2 - (5*A*sin(2*c + 2*d*x))/4)
)/(60*d*cos(c/2 + pi/4 + (d*x)/2)^5)