\(\int (a+a \sec (c+d x))^3 \sin (c+d x) \, dx\) [42]

Optimal result

Integrand size = 19, antiderivative size = 62 \[ \int (a+a \sec (c+d x))^3 \sin (c+d x) \, dx=-\frac {a^3 \cos (c+d x)}{d}-\frac {3 a^3 \log (\cos (c+d x))}{d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3957, 2912, 12, 45} \[ \int (a+a \sec (c+d x))^3 \sin (c+d x) \, dx=-\frac {a^3 \cos (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d}+\frac {3 a^3 \sec (c+d x)}{d}-\frac {3 a^3 \log (\cos (c+d x))}{d} \]

[In]

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

[Out]

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

Rule 12

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

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

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.05 \[ \int (a+a \sec (c+d x))^3 \sin (c+d x) \, dx=-\frac {a^3 (-4-9 \cos (c+d x)+\cos (3 (c+d x))+6 \log (\cos (c+d x))+\cos (2 (c+d x)) (-2+6 \log (\cos (c+d x)))) \sec ^2(c+d x)}{4 d} \]

[In]

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

[Out]

-1/4*(a^3*(-4 - 9*Cos[c + d*x] + Cos[3*(c + d*x)] + 6*Log[Cos[c + d*x]] + Cos[2*(c + d*x)]*(-2 + 6*Log[Cos[c +
d*x]]))*Sec[c + d*x]^2)/d

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {a^{3} \left (\frac {\sec \left (d x +c \right )^{2}}{2}+3 \sec \left (d x +c \right )+3 \ln \left (\sec \left (d x +c \right )\right )-\frac {1}{\sec \left (d x +c \right )}\right )}{d}\)	\(46\)
default	\(\frac {a^{3} \left (\frac {\sec \left (d x +c \right )^{2}}{2}+3 \sec \left (d x +c \right )+3 \ln \left (\sec \left (d x +c \right )\right )-\frac {1}{\sec \left (d x +c \right )}\right )}{d}\)	\(46\)
parts	\(-\frac {a^{3} \cos \left (d x +c \right )}{d}+\frac {a^{3} \sec \left (d x +c \right )^{2}}{2 d}+\frac {3 a^{3} \ln \left (\sec \left (d x +c \right )\right )}{d}+\frac {3 a^{3} \sec \left (d x +c \right )}{d}\)	\(61\)
risch	\(3 i a^{3} x -\frac {a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {6 i a^{3} c}{d}+\frac {2 a^{3} \left (3 \,{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(126\)
norman	\(\frac {\frac {4 a^{3}}{d}+\frac {6 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {6 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}-\frac {3 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {3 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}+\frac {3 a^{3} \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}\)	\(142\)
parallelrisch	\(\frac {a^{3} \left (-6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (2 d x +2 c \right )-6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (2 d x +2 c \right )+6 \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \cos \left (2 d x +2 c \right )+9 \cos \left (d x +c \right )-\cos \left (3 d x +3 c \right )+3 \cos \left (2 d x +2 c \right )-6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+6 \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+5\right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(165\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.05 \[ \int (a+a \sec (c+d x))^3 \sin (c+d x) \, dx=-\frac {2 \, a^{3} \cos \left (d x + c\right )^{3} + 6 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) - 6 \, a^{3} \cos \left (d x + c\right ) - a^{3}}{2 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.89 \[ \int (a+a \sec (c+d x))^3 \sin (c+d x) \, dx=-\frac {2 \, a^{3} \cos \left (d x + c\right ) + 6 \, a^{3} \log \left (\cos \left (d x + c\right )\right ) - \frac {6 \, a^{3}}{\cos \left (d x + c\right )} - \frac {a^{3}}{\cos \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.03 \[ \int (a+a \sec (c+d x))^3 \sin (c+d x) \, dx=-\frac {a^{3} \cos \left (d x + c\right )}{d} - \frac {3 \, a^{3} \log \left (\frac {{\left | \cos \left (d x + c\right ) \right |}}{{\left | d \right |}}\right )}{d} + \frac {6 \, a^{3} \cos \left (d x + c\right ) + a^{3}}{2 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

-a^3*cos(d*x + c)/d - 3*a^3*log(abs(cos(d*x + c))/abs(d))/d + 1/2*(6*a^3*cos(d*x + c) + a^3)/(d*cos(d*x + c)^2
)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.84 \[ \int (a+a \sec (c+d x))^3 \sin (c+d x) \, dx=\frac {a^3\,\left (3\,\cos \left (c+d\,x\right )-{\cos \left (c+d\,x\right )}^3-3\,{\cos \left (c+d\,x\right )}^2\,\ln \left (\cos \left (c+d\,x\right )\right )+\frac {1}{2}\right )}{d\,{\cos \left (c+d\,x\right )}^2} \]

[In]

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

[Out]

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