\(\int \sec ^4(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx\) [874]

Optimal result

Integrand size = 27, antiderivative size = 31 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=\frac {a^5 \sin ^2(c+d x)}{2 d (a-a \sin (c+d x))^2} \]

[Out]

1/2*a^5*sin(d*x+c)^2/d/(a-a*sin(d*x+c))^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2915, 12, 37} \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=\frac {a^5 \sin ^2(c+d x)}{2 d (a-a \sin (c+d x))^2} \]

[In]

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^3*Tan[c + d*x],x]

[Out]

(a^5*Sin[c + d*x]^2)/(2*d*(a - a*Sin[c + d*x])^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 37

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

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=\frac {a^3 \sin ^2(c+d x)}{2 d (1-\sin (c+d x))^2} \]

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^3*Tan[c + d*x],x]

[Out]

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

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06

method	result	size
parallelrisch	\(\frac {2 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}\)	\(33\)
derivativedivides	\(-\frac {a^{3} \left (-\frac {1}{\sin \left (d x +c \right )-1}-\frac {1}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}\right )}{d}\)	\(34\)
default	\(-\frac {a^{3} \left (-\frac {1}{\sin \left (d x +c \right )-1}-\frac {1}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}\right )}{d}\)	\(34\)
risch	\(\frac {2 i \left (-a^{3} {\mathrm e}^{i \left (d x +c \right )}+a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}\)	\(64\)
norman	\(\frac {\frac {2 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{3} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {44 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {44 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {32 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {48 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {32 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {18 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {18 a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(242\)

[In]

int(sec(d*x+c)^5*sin(d*x+c)*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=-\frac {2 \, a^{3} \sin \left (d x + c\right ) - a^{3}}{2 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*sin(d*x+c)*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=\frac {2 \, a^{3} \sin \left (d x + c\right ) - a^{3}}{2 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )} d} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=\frac {2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{4}} \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.83 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=\frac {a^3\,{\sin \left (c+d\,x\right )}^2}{8\,d\,{\cos \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d\,x}{2}\right )}^4} \]

[In]

int((sin(c + d*x)*(a + a*sin(c + d*x))^3)/cos(c + d*x)^5,x)

[Out]

(a^3*sin(c + d*x)^2)/(8*d*cos(c/2 + pi/4 + (d*x)/2)^4)