\(\int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx\) [807]

Optimal result

Integrand size = 27, antiderivative size = 60 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2 a^2 \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (a+a \sin (c+d x))^2}{3 d}-\frac {2 a^2 \tan (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2934, 2748, 3852, 8} \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2 a^2 \tan (c+d x)}{3 d}-\frac {2 a^2 \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d} \]

[In]

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

[Out]

(-2*a^2*Sec[c + d*x])/(3*d) + (Sec[c + d*x]^3*(a + a*Sin[c + d*x])^2)/(3*d) - (2*a^2*Tan[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 {\sec ^3(c+d x) (a+a \sin (c+d x))^2}{3 d}-\frac {1}{3} (2 a) \int \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx \\ & = -\frac {2 a^2 \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (a+a \sin (c+d x))^2}{3 d}-\frac {1}{3} \left (2 a^2\right ) \int \sec ^2(c+d x) \, dx \\ & = -\frac {2 a^2 \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (a+a \sin (c+d x))^2}{3 d}+\frac {\left (2 a^2\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d} \\ & = -\frac {2 a^2 \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (a+a \sin (c+d x))^2}{3 d}-\frac {2 a^2 \tan (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.20 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=\frac {a^2 \left (3 \cos \left (\frac {1}{2} (c+d x)\right )-2 \cos \left (\frac {3}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.58

method	result	size
parallelrisch	\(\frac {2 a^{2} \left (1-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(35\)
risch	\(-\frac {2 a^{2} \left (-3 i {\mathrm e}^{i \left (d x +c \right )}+3 \,{\mathrm e}^{2 i \left (d x +c \right )}-2\right )}{3 d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3}}\)	\(48\)
derivativedivides	\(\frac {a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {2 a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+\frac {a^{2}}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(99\)
default	\(\frac {a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {2 a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+\frac {a^{2}}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(99\)
norman	\(\frac {\frac {2 a^{2}}{3 d}-\frac {16 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {32 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {16 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {28 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(174\)

[In]

int(sec(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.63 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=\frac {2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right ) - a^{2} - {\left (2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

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

[Out]

1/3*(2*a^2*cos(d*x + c)^2 + a^2*cos(d*x + c) - a^2 - (2*a^2*cos(d*x + c) + 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]

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

[In]

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

[Out]

a**2*(Integral(sin(c + d*x)*sec(c + d*x)**4, x) + Integral(2*sin(c + d*x)**2*sec(c + d*x)**4, x) + Integral(si
n(c + d*x)**3*sec(c + d*x)**4, x))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=\frac {2 \, a^{2} \tan \left (d x + c\right )^{3} - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2}}{\cos \left (d x + c\right )^{3}} + \frac {a^{2}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.63 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}\right )}}{3 \, d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.57 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2\,a^2\,\left (3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3} \]

[In]

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

[Out]

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