\(\int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [982]

Optimal result

Integrand size = 31, antiderivative size = 104 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 (3 A-2 B) \sec ^3(c+d x)}{15 d}+\frac {(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^2}{5 d}+\frac {a^2 (3 A-2 B) \tan (c+d x)}{5 d}+\frac {a^2 (3 A-2 B) \tan ^3(c+d x)}{15 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2934, 2748, 3852} \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 (3 A-2 B) \tan ^3(c+d x)}{15 d}+\frac {a^2 (3 A-2 B) \tan (c+d x)}{5 d}+\frac {a^2 (3 A-2 B) \sec ^3(c+d x)}{15 d}+\frac {(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^2}{5 d} \]

[In]

Int[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

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

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 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]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.99 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 \left ((6 A+B) \sec ^5(c+d x)+15 A \sec ^4(c+d x) \tan (c+d x)+5 B \sec ^3(c+d x) \tan ^2(c+d x)-5 (3 A-2 B) \sec ^2(c+d x) \tan ^3(c+d x)+2 (3 A-2 B) \tan ^5(c+d x)\right )}{15 d} \]

[In]

Integrate[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*((6*A + B)*Sec[c + d*x]^5 + 15*A*Sec[c + d*x]^4*Tan[c + d*x] + 5*B*Sec[c + d*x]^3*Tan[c + d*x]^2 - 5*(3*A
- 2*B)*Sec[c + d*x]^2*Tan[c + d*x]^3 + 2*(3*A - 2*B)*Tan[c + d*x]^5))/(15*d)

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.17

method	result	size
parallelrisch	\(-\frac {2 \left (A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (B -2 A \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (A -\frac {2 B}{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4 B \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {\left (-3 A -\frac {4 B}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+\frac {2 A}{5}+\frac {B}{15}\right ) a^{2}}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}\)	\(122\)
risch	\(-\frac {4 \left (15 i A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+12 A \,a^{2} {\mathrm e}^{i \left (d x +c \right )}-3 i A \,a^{2}+10 B \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-10 i B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-8 B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}+2 i B \,a^{2}\right )}{15 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{5} d}\)	\(126\)
derivativedivides	\(\frac {A \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+B \,a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )+\frac {2 A \,a^{2}}{5 \cos \left (d x +c \right )^{5}}+2 B \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )-A \,a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+\frac {B \,a^{2}}{5 \cos \left (d x +c \right )^{5}}}{d}\)	\(231\)
default	\(\frac {A \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+B \,a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )+\frac {2 A \,a^{2}}{5 \cos \left (d x +c \right )^{5}}+2 B \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )-A \,a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+\frac {B \,a^{2}}{5 \cos \left (d x +c \right )^{5}}}{d}\)	\(231\)
norman	\(\frac {-\frac {12 A \,a^{2}+2 B \,a^{2}}{15 d}-\frac {2 \left (2 A \,a^{2}+B \,a^{2}\right ) \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (6 A \,a^{2}+5 B \,a^{2}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (18 A \,a^{2}+13 B \,a^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 \left (30 A \,a^{2}+35 B \,a^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 \left (78 A \,a^{2}+73 B \,a^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 \left (186 A \,a^{2}+181 B \,a^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 A \,a^{2} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \left (9 A +8 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{2} \left (9 A +8 B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{2} \left (42 A +47 B \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{2} \left (111 A +136 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 a^{2} \left (111 A +136 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 a^{2} \left (213 A +208 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 a^{2} \left (213 A +208 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(444\)

[In]

int(sec(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.09 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {4 \, {\left (3 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 3 \, {\left (2 \, A - 3 \, B\right )} a^{2} - {\left (2 \, {\left (3 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 3 \, {\left (3 \, A - 2 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )\right )}} \]

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**6*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.41 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{2} + {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} A a^{2} + 2 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} B a^{2} - \frac {{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} B a^{2}}{\cos \left (d x + c\right )^{5}} + \frac {6 \, A a^{2}}{\cos \left (d x + c\right )^{5}} + \frac {3 \, B a^{2}}{\cos \left (d x + c\right )^{5}}}{15 \, d} \]

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/15*((3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^2 + (3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*A
*a^2 + 2*(3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*B*a^2 - (5*cos(d*x + c)^2 - 3)*B*a^2/cos(d*x + c)^5 + 6*A*a^2/c
os(d*x + c)^5 + 3*B*a^2/cos(d*x + c)^5)/d

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.85 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {\frac {15 \, {\left (A a^{2} - B a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} + \frac {105 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 15 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 270 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 360 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 40 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 210 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 50 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 63 \, A a^{2} - 7 \, B a^{2}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{5}}}{60 \, d} \]

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/60*(15*(A*a^2 - B*a^2)/(tan(1/2*d*x + 1/2*c) + 1) + (105*A*a^2*tan(1/2*d*x + 1/2*c)^4 + 15*B*a^2*tan(1/2*d*
x + 1/2*c)^4 - 270*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 30*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 360*A*a^2*tan(1/2*d*x + 1/
2*c)^2 - 40*B*a^2*tan(1/2*d*x + 1/2*c)^2 - 210*A*a^2*tan(1/2*d*x + 1/2*c) + 50*B*a^2*tan(1/2*d*x + 1/2*c) + 63
*A*a^2 - 7*B*a^2)/(tan(1/2*d*x + 1/2*c) - 1)^5)/d

Mupad [B] (verification not implemented)

Time = 10.31 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.68 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {2\,a^2\,\left (\frac {5\,B\,\sin \left (c+d\,x\right )}{2}-\frac {15\,A\,\cos \left (c+d\,x\right )}{4}-\frac {5\,B\,\cos \left (c+d\,x\right )}{8}-\frac {15\,A\,\sin \left (c+d\,x\right )}{4}-\frac {5\,B}{2}-3\,A\,\cos \left (2\,c+2\,d\,x\right )+\frac {3\,A\,\cos \left (3\,c+3\,d\,x\right )}{4}+2\,B\,\cos \left (2\,c+2\,d\,x\right )+\frac {B\,\cos \left (3\,c+3\,d\,x\right )}{8}+3\,A\,\sin \left (2\,c+2\,d\,x\right )+\frac {3\,A\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {B\,\sin \left (2\,c+2\,d\,x\right )}{2}-\frac {B\,\sin \left (3\,c+3\,d\,x\right )}{2}\right )}{15\,d\,\left (\frac {\cos \left (3\,c+3\,d\,x\right )}{4}-\frac {5\,\cos \left (c+d\,x\right )}{4}+\sin \left (2\,c+2\,d\,x\right )\right )} \]

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2)/cos(c + d*x)^6,x)

[Out]

(2*a^2*((5*B*sin(c + d*x))/2 - (15*A*cos(c + d*x))/4 - (5*B*cos(c + d*x))/8 - (15*A*sin(c + d*x))/4 - (5*B)/2
- 3*A*cos(2*c + 2*d*x) + (3*A*cos(3*c + 3*d*x))/4 + 2*B*cos(2*c + 2*d*x) + (B*cos(3*c + 3*d*x))/8 + 3*A*sin(2*
c + 2*d*x) + (3*A*sin(3*c + 3*d*x))/4 + (B*sin(2*c + 2*d*x))/2 - (B*sin(3*c + 3*d*x))/2))/(15*d*(cos(3*c + 3*d
*x)/4 - (5*cos(c + d*x))/4 + sin(2*c + 2*d*x)))