\(\int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [981]

Optimal result

Integrand size = 31, antiderivative size = 73 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 (A-2 B) \sec (c+d x)}{3 d}+\frac {(A+B) \sec ^3(c+d x) (a+a \sin (c+d x))^2}{3 d}+\frac {a^2 (A-2 B) \tan (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 73, 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, 2748, 3852, 8} \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 (A-2 B) \tan (c+d x)}{3 d}+\frac {a^2 (A-2 B) \sec (c+d x)}{3 d}+\frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d} \]

[In]

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

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

Rule 8

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

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.66 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {a^2 \sec (c+d x) (-2 A+B+(A-2 B) \sin (c+d x)) (\sec (c+d x)+\tan (c+d x))^2}{3 d} \]

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.78

method	result	size
parallelrisch	\(-\frac {2 \left (A \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 A}{3}-\frac {B}{3}\right ) a^{2}}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(57\)
risch	\(-\frac {2 \left (-3 i B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}+3 B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+A \,a^{2}-2 B \,a^{2}+3 i A \,a^{2} {\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}\)	\(79\)
derivativedivides	\(\frac {\frac {A \,a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+B \,a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {2 A \,a^{2}}{3 \cos \left (d x +c \right )^{3}}+\frac {2 B \,a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}-A \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {B \,a^{2}}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(162\)
default	\(\frac {\frac {A \,a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+B \,a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {2 A \,a^{2}}{3 \cos \left (d x +c \right )^{3}}+\frac {2 B \,a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}-A \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {B \,a^{2}}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(162\)
norman	\(\frac {-\frac {4 A \,a^{2}-2 B \,a^{2}}{3 d}-\frac {\left (8 A \,a^{2}+12 B \,a^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (12 A \,a^{2}+10 B \,a^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 A \,a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \left (2 A +B \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \left (2 A +B \right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (3 A +4 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (3 A +4 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (10 A +13 B \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{2} \left (11 A +8 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{2} \left (11 A +8 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(325\)

[In]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.64 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {{\left (A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right ) + {\left (A + B\right )} a^{2} - {\left ({\left (A - 2 \, B\right )} a^{2} \cos \left (d x + c\right ) - {\left (A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.48 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {A a^{2} \tan \left (d x + c\right )^{3} + 2 \, B a^{2} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} B a^{2}}{\cos \left (d x + c\right )^{3}} + \frac {2 \, A a^{2}}{\cos \left (d x + c\right )^{3}} + \frac {B a^{2}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.07 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A a^{2} - B a^{2}\right )}}{3 \, d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} \]

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.95 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.05 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {\sqrt {2}\,a^2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B}{2}-\frac {5\,A}{2}+\frac {A\,\cos \left (c+d\,x\right )}{2}+\frac {B\,\cos \left (c+d\,x\right )}{2}+\frac {3\,A\,\sin \left (c+d\,x\right )}{2}-\frac {3\,B\,\sin \left (c+d\,x\right )}{2}\right )}{6\,d\,{\cos \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d\,x}{2}\right )}^3} \]

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2)/cos(c + d*x)^4,x)

[Out]

-(2^(1/2)*a^2*cos(c/2 + (d*x)/2)*(B/2 - (5*A)/2 + (A*cos(c + d*x))/2 + (B*cos(c + d*x))/2 + (3*A*sin(c + d*x))
/2 - (3*B*sin(c + d*x))/2))/(6*d*cos(c/2 + pi/4 + (d*x)/2)^3)