\(\int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx\) [814]

Optimal result

Integrand size = 27, antiderivative size = 64 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=a^3 x+\frac {\sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-\frac {2 a^5 \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 64, 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, 2749, 2759, 8} \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=a^3 x-\frac {2 a^5 \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )}+\frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d} \]

[In]

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

[Out]

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

Rule 8

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

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 2759

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[2*g*(g
*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(2*m +
p + 1))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

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

Mathematica [A] (verified)

Time = 5.74 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.67 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=-\frac {a^3 \left (-9 (2+c+d x) \cos \left (\frac {1}{2} (c+d x)\right )+(14+3 c+3 d x) \cos \left (\frac {3}{2} (c+d x)\right )+6 (2 (2+c+d x)+(c+d x) \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

[In]

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

[Out]

-1/6*(a^3*(-9*(2 + c + d*x)*Cos[(c + d*x)/2] + (14 + 3*c + 3*d*x)*Cos[(3*(c + d*x))/2] + 6*(2*(2 + c + d*x) +
(c + d*x)*Cos[c + d*x])*Sin[(c + d*x)/2]))/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84

method	result	size
risch	\(a^{3} x -\frac {2 a^{3} \left (-12 i {\mathrm e}^{i \left (d x +c \right )}+9 \,{\mathrm e}^{2 i \left (d x +c \right )}-7\right )}{3 d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3}}\)	\(54\)
parallelrisch	\(\frac {a^{3} \left (3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) d x -9 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) x d +9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d x +6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 d x -24 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+10\right )}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(95\)
derivativedivides	\(\frac {a^{3} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+3 a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+\frac {a^{3}}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(126\)
default	\(\frac {a^{3} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+3 a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+\frac {a^{3}}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(126\)
norman	\(\frac {a^{3} x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a^{3} x +\frac {10 a^{3}}{3 d}+\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {26 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {36 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {116 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {36 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {26 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+3 a^{3} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{3} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {18 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {28 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(304\)

[In]

int(sec(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

a^3*x-2/3*a^3*(-12*I*exp(I*(d*x+c))+9*exp(2*I*(d*x+c))-7)/d/(exp(I*(d*x+c))-I)^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (63) = 126\).

Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.23 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=-\frac {6 \, a^{3} d x + 2 \, a^{3} - {\left (3 \, a^{3} d x + 7 \, a^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, a^{3} d x - 5 \, a^{3}\right )} \cos \left (d x + c\right ) - {\left (6 \, a^{3} d x - 2 \, a^{3} + {\left (3 \, a^{3} d x - 7 \, a^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/3*(6*a^3*d*x + 2*a^3 - (3*a^3*d*x + 7*a^3)*cos(d*x + c)^2 + (3*a^3*d*x - 5*a^3)*cos(d*x + c) - (6*a^3*d*x -
2*a^3 + (3*a^3*d*x - 7*a^3)*cos(d*x + c))*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(-1)]

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

[In]

integrate(sec(d*x+c)**4*sin(d*x+c)*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.31 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=\frac {3 \, a^{3} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{3} - \frac {3 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{3}}{\cos \left (d x + c\right )^{3}} + \frac {a^{3}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=\frac {3 \, {\left (d x + c\right )} a^{3} + \frac {2 \, {\left (3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/3*(3*(d*x + c)*a^3 + 2*(3*a^3*tan(1/2*d*x + 1/2*c)^2 - 12*a^3*tan(1/2*d*x + 1/2*c) + 5*a^3)/(tan(1/2*d*x + 1
/2*c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 10.70 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.59 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=a^3\,x+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^3\,\left (9\,d\,x-24\right )}{3}-3\,a^3\,d\,x\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^3\,\left (9\,d\,x-6\right )}{3}-3\,a^3\,d\,x\right )-\frac {a^3\,\left (3\,d\,x-10\right )}{3}+a^3\,d\,x}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3} \]

[In]

int((sin(c + d*x)*(a + a*sin(c + d*x))^3)/cos(c + d*x)^4,x)

[Out]

a^3*x + (tan(c/2 + (d*x)/2)*((a^3*(9*d*x - 24))/3 - 3*a^3*d*x) - tan(c/2 + (d*x)/2)^2*((a^3*(9*d*x - 6))/3 - 3
*a^3*d*x) - (a^3*(3*d*x - 10))/3 + a^3*d*x)/(d*(tan(c/2 + (d*x)/2) - 1)^3)